Electrostatics is a branch of physics that studies electric charges at rest.
“An electric field, also called an electrical field or an electrostatic field, surrounds any object that has charge. The electric field strength at any given distance from an object is directly proportional to the amount of charge on the object”
Electrostatic involves the buildup of charge on the surface of objects due to contact with other surfaces. There are many examples of electrostatic phenomena, from those as simple as the attraction of the plastic wrap to one's hand after it is removed from a package to the apparently spontaneous explosion of grain silos, the damage of electronic components during manufacturing, and photocopier & laser printer operation.
The familiar phenomenon of a static 'shock' is caused by the neutralization of charge built up in the body from contact with insulated surfaces.
charge :
Electric charge is the property associated with a body or a particle due to which it is able to produce as well as experience the electric and magnetic effects.
It is a scalar. Its S.I. unit is coulomb. Its dimensional formula is M0L0T1A1
Charges are of two types
(a) Positive charge
(b) Negative charge
Positive Charge :
iv) When a body loses electrons, it is said to acquire positive charge.
v) The mass of a body some what decreases when it is positively charged.
vi) A glass rod gets positively charged on rubbing it with silk.
Negative Charge :
vii) When a body gains electrons, it is said to acquire negative charge.
viii) When a negative charge is given to a body then its mass increases.
ix) When an ebonite rod is rubbed with fur, it gets negatively charged.
x) The root cause of two types of charges is the law of conservation of charge.
xi) The net charge on a neutral body is zero and it is equal to the sum of positive and negative charges,
(for a neutral body)
\(\text{q = }{{\text{q}}^{\text{+}}}+{{q}^{-}}=0\) .for a neutral body).
Quantization of Charge:
Ø The electric charge is discrete. It has been verified by Millikan’s oil drop experiment.
Ø Charge is quantised. The charge on any body is an integral multiple of the minimum charge or electron charge, i.e \(q\) if is the charge then \(q=\pm ne\) where is an integer, and e is the charge of electron = \( .1.6\times {{10}^{-19}}C\)
Ø The minimum charge possible is. \( .1.6\times {{10}^{-19}}C\)
If a body possesses \({{n}_{1}}\) protonsand \({{n}_{2}}\) electrons, then net charge on it will be \(\left( {{n}_{1}}-{{n}_{2}} \right)e,\)
i.e \({{n}_{1}}\left( e \right)+{{n}_{2\left( -e \right)}}=\left( {{n}_{1}}-{{n}_{2}} \right)e\) 3
Charge And Properties:
Ø Study of characteristics of electric charges at rest is known as electrostatics.
Ø Electric charge is the property associated with a body or a particle due to which it is able to produce as well as experience the electric and magnetic effects.
Ø Charge is a fundamental property of matter and never found free.
Ø The excess or deficiency of electrons in a body gives the concept of charge.
Ø There are two types of charges namely positive and negative charges.
Ø The deficiency of electrons in a body is known as positive charge.
Ø The excess of electrons in a body is known as negative charge.
Ø If a body gets positive charge, its mass slightly decreases.
Ø If a body is given negative charge, its mass slightly increases.
Charge is relativistically invariant, i.e. it does not change with motion of charged particle and no change in it is possible, whatsover may be the circumstances. i.e.
Ø Charge is a scalar. S.I. unit of charge is coulomb(C).
One electrostatic unit of charge
(esu) = \(\frac{1}{3\times {{10}^{9}}}\) coulomb.
One electromagnetic unit of charge (emu) = 10coulomb
Ø Charge is a derived physical quantity with dimensions [AT].
Law of conservation of charge :
The total net charge of an isolated physical system always remains constant,
i.e. \(q={{q}_{+}}+{{q}_{-}}=\)constant.
In every chemical or nuclear reaction, the total charge before and after the reaction remains constant.
This law is applicable to all types of processes like nuclear, atomic, molecular and the like.
Ø Charge is conserved. It can neither be created nor destroyed. It can only be transferred from one object to the other.
Ø Like charges repel each other and unlike charges attract each other.
Ø Charge always resides on the outer surface of a charged body. It accumulates more at sharp points.
Ø The total charge on a body is algebric sum of the charges located at different points on the body.
Electrification :
Ø A body can be charged by friction, conduction and induction.
Ø By Friction: When two bodies are rubbed together, equal and opposite charges are produced on both the bodies.
Ø By Conduction: An uncharged body acquiring charge when kept in contact with a charged body is called conduction. Conduction preceeds repulsion.
Ø By Induction: If a charged body is brought near a neutral body, the charged body will attract opposite charge and repel similar charge present in the neutral body. Opposite charge is induced at the near end and similar charge at the farther end. Inducing body neither gains nor loses charge. Induction always preceeds attraction.
Ø Repulsion is the sure test of electrification.
Ø Induced charge \({{q}^{1}}=-q\left[ 1-\frac{1}{K} \right]\) where K is a Dielectric constant
Charging a body by means of induction is preferable since the same charged body can be used to charge any number of bodies without loss of charge.
CONCEPTUAL UNDERSTANDING:
Ans : does not change.
Ans : more charge accumulates at regions of large curvature.
SOLVED PROBLEMS
Solution :.
\({{\vec{E}}_{P(A{A}')}}=\).Electric field at P due to charges at A and A¢ =
Solution :. 2x\(\left( \frac{1}{4\pi {{\varepsilon }_{0}}} \right)\frac{{{q}_{A}}{{x}_{P}}}{{{(y_{A}^{2}+x_{P}^{2})}^{3/2}}}\).\((-\hat{i})\) similarly \({{\vec{E}}_{P(B{B}')}}\) = 2\(\left( \frac{1}{4\pi {{\varepsilon }_{0}}} \right)\frac{{{q}_{B}}{{x}_{P}}}{{{(y_{B}^{2}+x_{P}^{2})}^{3/2}}}.(\hat{i})\) |
Hence, v0 should be just enough to enable the particle to reach P.
\([ Alternately V(x) = \frac{2}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{B}}}{{{(y_{B}^{2}+{{x}^{2}})}^{1/2}}}+\frac{{{Q}_{A}}}{{{(y_{A}^{2}+{{x}^{2}})}^{1/2}}} \right)\)
\( \frac{dV(x)}{dx}$= 0 Þ x = 0 and x = \sqrt{\frac{5}{2}}\)
V(x)|x = ¥= 0,
at x = \(\sqrt{\frac{5}{2}}\) , V(x) is positive Þ V(x) is maximum at x = \(\sqrt{\frac{5}{2}}\)
So, if the particle is able to cross x = \(\sqrt{\frac{5}{2}}\) then it would be able to come to x = 0. ]
Þ \(\frac{1}{2}mv_{0}^{2}\) = \(\frac{2q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{q}_{A}}}{\sqrt{y_{A}^{2}+x_{P}^{2}}}+\frac{{{q}_{B}}}{\sqrt{y_{B}^{2}+x_{P}^{2}}} \right]\) ---------
Solving we get, v0 = 3 m/s
Kinetic energy at origin = loss of PE = (PE)P – (PE)origin
= \(\frac{1}{2}mv_{0}^{2}-\frac{2q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{q}_{A}}}{{{y}_{A}}}+\frac{{{q}_{B}}}{{{y}_{B}}} \right]\) = 2.5 ´ 10-4 J.
Problem 2 :
A uniform electric field, E exists between the plates of a capacitor. The plate length is l and the separation of the plates is d.
(a) An electron and a proton start from the negative plate and positive plate respectively and go to the opposite plates. Which of them wins this race?
(b) An electron and a proton start from the midpoint of the separation of plates at one end of the plates. Which of the two will have greater deviation when they start with the
(i) same initial velocity
(ii) same initial kinetic energy, and (iii) same initial momentum?
Solution :
(a) In the chosen reference frame there is no force along x axis. The accelerations of the electron and the proton along y axis are as follows
\(_{{{a}_{e}}=\frac{-eE}{{{m}_{e}}},\,\,\,{{a}_{p}}=\frac{eE}{{{m}_{p}}}}\)
\(_{\begin{align} & \text{Using}\,\,\text{S=ut+}\frac{\text{1}}{\text{2}}a{{t}^{2}} \\ & here\,\,\,\,S=d,\,u=0 \\ & \Rightarrow {{t}_{e}}=\sqrt{\frac{2S}{{{a}_{e}}}}\,\,\,\,and\,\,\,{{t}_{p}}=\sqrt{\frac{2S}{{{a}_{p}}}} \\ \end{align}} \)
\(or\,\,\,\,\,\,\,\,\,{{t}_{e}}=\sqrt{\frac{2d{{m}_{e}}}{eE}}\,\,\,\,and\,\,\,\,{{t}_{p}}=\sqrt{\frac{2d.{{m}_{p}}}{eE}}\)
As me < mp, te < tp Þ Electron takes less time to cross over than the proton.
(b) When proton moves parallel to the plates, it is deflected to the negative plate.
Time taken by proton to cross over, \(t=\frac{l}{{{V}_{x}}}\) During this time, deflection along vertical direction \(y=\frac{1}{2}a{{t}^{_{2}}}\,\,\,\,or\,\,\,y=\frac{1}{2}\frac{eE}{m}{{\left[ \frac{l}{{{V}_{o}}} \right]}^{_{2}}}\) . |
(i) Thus \({{Y}_{p}}=\frac{1}{2}\frac{eE}{{{m}_{p}}}\frac{{{1}^{2}}}{V_{x}^{2}}\,\,and\,\,{{Y}_{^{e}}}=\frac{1}{2}\frac{eE}{{{m}_{e}}}.\frac{{{l}^{2}}}{V_{x}^{2}}\)
As mp > me, yp < ye Of course, electron will be deflected in the opposite direction.
(ii) Also \(y=\frac{1}{2}\frac{eE{{l}^{_{2}}}}{2\left[ \frac{m}{2}{{v}_{x}}^{2} \right]}=\frac{1}{4}\frac{eE{{l}^{2}}}{K}\,\,\,\,\,where\,\,K=\frac{1}{2}m{{v}_{x}}^{2}\) ,or initial kinetic energy
Þ yp = ye
Also, as \(K=\frac{{{p}^{_{2}}}}{2m}\) ,where p = momentum,
\(y=\frac{eE{{l}^{_{2}}}}{4(\frac{{{p}^{2}}}{2m})}.=\frac{eE{{l}^{2}}.m}{2{{p}^{2}}}\Rightarrow {{y}_{p}}=\frac{eE{{l}^{_{2}}}{{m}_{p}}}{2{{p}^{2}}},{{y}_{e}}=\frac{eE{{l}^{_{2}}}{{m}_{e}}}{2{{p}^{2}}}\)
Thus for: mp > me, yp > ye
Problem 1 : Two point charges each of charge +q are fixed at (+a, 0) and (-a, 0). Another positive point charge q placed at the origin is free to move along x- axis.
The charge q at origin in equilibrium will have |
Solution :
The net force on q at origin is
\(\vec{F}={{\vec{F}}_{1}}+{{\vec{F}}_{2}}\)
= \(\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{{{r}^{2}}}\hat{i}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{{{r}^{2}}}(-\hat{i})=\hat{0}\)
The P.E. of the charge q in between the extreme charges at a distance x from the origin along +ve x axis is
U = \(\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{\left( a-x \right)}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{(a+x)}\) = \(\frac{1}{4\pi {{\varepsilon }_{0}}}.{{q}^{2}}\left[ \frac{1}{a-x}+\frac{1}{a+x} \right]\) .
\(\frac{dU}{dx}=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}}\left[ -\frac{1}{{{(a-x)}^{2}}}+\frac{1}{{{(a+x)}^{2}}} \right]\)
For U to be minimum
\(\frac{dU}{dx}=0,\) \(\frac{{{d}^{2}}U}{d{{x}^{2}}}>0,\)
Þ (a-x)2 = (a + x)2 Þ a + x = ± (a – x)
Þ x = 0, because other solution is relevant.
Thus the charged particle at the origin will have minimum force and minimum P.E.
\ (D).
Problem 2 :
Two conducting spheres having radii a and b are charged to q1 & q2 respectively. The potential difference between 1 & 2 will be
(A) \(\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}a}-\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{0}}b}\) (B)\(\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{a}-\frac{1}{b} \right)\) |
Solution :
The potential on the surface of the sphere 1 is given by
v1 =\(\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{a}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{2}}}{b}\) . . . . (a)
The potential on the surface of the sphere 2 is given by,
v = v1 – v2
V2 = \(\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{b}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{2}}}{b}\) Þ v = \(\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{a}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{b}\)
Þ v= \(\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{a}-\frac{1}{b} \right)\) \ (C)
Application ability:
1.One million electrons are added to a glass rod. The total charge on the rod is.
Ans: \(-1.6\times {{10}^{-13}}C\)
2.A body has a charge of \(9.6\times {{10}^{-20}}\) coulomb. It is
Ans: not possible.
COULOMB’S INVERSE SQUARE LAW:
The force of attraction or repulsion between two stationary point charges Q1 and Q2 which are separated by a distance ‘d’, in air or vacuum is given by
\({{F}_{air}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}\)
where is permittivity of free space or air and \({{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}/N-{{m}^{2}}\)
In vector form, \(\overrightarrow{F}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{3}}}\overrightarrow{r}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}\overset{\hat{\ }}{\mathop{r}}\,\)
\(\overset{\hat{\ }}{\mathop{r}}\,\) is the unit vector along the direction of force In a medium,
\({{F}_{medium}}=\frac{1}{4\pi \varepsilon }\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}=\frac{1}{4\pi {{\varepsilon }_{0}}K}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}\)
Where is the absolute permittivity of medium and K is dielectric constant or relative permittivity or specific inductive capacity (or K)
\(K={{\varepsilon }_{r}}=\frac{\varepsilon }{{{\varepsilon }_{0}}}\) and also \(K=\frac{{{F}_{air}}}{{{F}_{medium}}}\)
For air (or) vacuum, K = 1
For all other insulating materials, K > 1
For all conductors, \(K=\infty \)
From figure\(\overline{{{F}_{1}}}=-\overline{\,{{F}_{2}}}\) (action-reaction pair)
\(\left| \overline{{{F}_{1}}} \right|=\left| \overline{{{F}_{2}}} \right|=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{\,2}}}\)
i) Coulomb’s electrostatic force is a conservative force.
ii) Coulomb’s force is a central force.
iii) Force between two point charges is independent of presence of other charges.
iv) Principle of super position : the net force on a given charge is the vector sum of all individual forces exerted by various charges separately.
\(\bar{F}={{\bar{F}}_{12}}+{{\bar{F}}_{13}}+{{\bar{F}}_{14}}+\cdots +{{\bar{F}}_{1n}}\)
Coulomb’s Law And Force due to Multiple charges
Ø\(F=\frac{1}{4\pi {{\in }_{0}}{{\in }_{r}}}\) \(\frac{{{q}_{1}}{{q}_{2}}}{{{d}^{2}}}\)
\({{\in }_{0}}\) - permittivity of free space or vacuum or air.
\({{\in }_{r}}\) - Relative permittivity or dielectric constant of the medium in which the charges are situated.
Ø \({{\in }_{0}}=8.857\times {{10}^{-12}}\) \(\frac{{{C}^{2}}}{N{{m}^{2}}}\) or \(\frac{farad}{metre},\)
and \(\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}\)
Ø Suppose the position vector of two charges \({{q}_{1}}\) and \({{q}_{2}}\) are \(\overrightarrow{{{r}_{1}}}\) and \(\overrightarrow{{{r}_{2}}}\) , then electric force on charge \({{q}_{1}}\)due to \({{q}_{2}}\) is,
\(\overrightarrow{{{F}_{1}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{\left| \overrightarrow{{{r}_{2}}}-\overrightarrow{{{r}_{1}}} \right|}^{3}}}\left( \overrightarrow{{{r}_{1}}}-\overrightarrow{{{r}_{2}}} \right)\)
Similarly, electric force on \({{q}_{2}}\) due to charge \({{q}_{1}}\) is
\(\overrightarrow{{{F}_{2}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{\left| \overrightarrow{{{r}_{2}}}-\overrightarrow{{{r}_{1}}} \right|}^{3}}}\left( \overrightarrow{{{r}_{2}}}-\overrightarrow{{{r}_{1}}} \right)\)
Here \({{q}_{1}}\) and \({{q}_{2}}\) are to be substitued with sign
\(\overrightarrow{{{r}_{1}}}={{x}_{1}}i+{{y}_{1}}j+{{z}_{1}}k\) and \(\overrightarrow{{{r}_{2}}}={{x}_{2}}i+{{y}_{2}}j+{{z}_{2}}k\)
where \( \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) \) and \(\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\) co-ordinates of charges\({{q}_{1}}\) and\({{q}_{2}}\).
Relative permittivity(\({{\in }_{r}}\) ):
The relative permittivity is the ratio of permittivity of the medium to the permittivity of the absolute free space \({{\in }_{r}}=\frac{\in }{{{\in }_{0}}}\)
\({{\in }_{r}}\) has no units and dimensional formula
i.e \([{{M}^{\circ }}{{L}^{\circ }}{{T}^{\circ }}{{A}^{\circ }}]\)
ØAnd also \({{\in }_{r}}=\frac{\text{Force}\,\text{between}\,\text{two}\,\text{charges}\,\text{in}\,\text{air}}{\begin{align} & \text{Force}\,\text{between}\,\text{the}\,\text{same}\,\text{two}\,\text{charges}\,\text{in}\,\text{the}\, \\ & \text{medium}\,\text{at}\,\text{same}\,\text{distance} \\ \end{align}}\,\)
\(=\frac{{{F}_{air}}}{{{F}_{medium}}}\)
Ø For air \(k=1\)
\(k>1\) for any dielectric medium;
\(k=\infty \) for conducting medium like metals
Coulomb’s law in vector form:
Ø \(\overrightarrow{{{F}_{12}}}=\frac{1}{4\pi {{\in }_{0}}}\,\,\frac{{{q}_{1}}{{q}_{2}}}{r_{2}^{2}}{{\widehat{r}}_{12}}\) and \(\overrightarrow{{{F}_{21}}}=-\overrightarrow{{{F}_{12}}}\)
Here \({{F}_{12}}\) is force exerted by \({{q}_{1}}\) on \({{q}_{2}}\)and\({{q}_{1}}\) is force exerted by \({{q}_{2}}\) on \({{q}_{1}}\)
Ø Coulomb’s law holds for stationary charges only which are point sized.
This law is valid for all types of charge distributions.
This law is valid at distances greater than \({{10}^{-15}}m.\)
This law obeys Newton’s third law.
This law represents central forces.
This law is analogous to Newton law of gravitation in mechanics.
Ø The electric force is an action reaction pair, i.e the two charges exert equal and opposite forces on each other.
Ø The electric force is conservative in nature.
Ø Coulomb force is central.
Ø Coulomb force is much stronger than gravitational force\(\left( {{10}^{36}}{{F}_{g}}={{F}_{E}} \right)\) .
Forces between multiple charges :-
Ø Force on a charged particle due to a number of point charges is the resultant of forces due to individual point charges
\(\overrightarrow{F}=\overrightarrow{{{F}_{1}}}+{{\overrightarrow{F}}_{2}}+{{\overrightarrow{F}}_{3}}+.....\)
Ø If the force between two charges in different media is the same for different seperations
\(F=\frac{1}{k}\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{d}^{2}}}=\)constant.
\(k{{d}^{2}}=\)constant or \({{k}_{1}}d_{1}^{2}={{k}_{2}}d_{2}^{2}\)
Ø If two charged spheres of radii r1 and r2 respectively are kept in contact for some time, then charge exchanges takes place between them until their electric potentials are equal. Now the new charges on two spheres are
\({{q}_{1}}\,\,=\,\,\left( \frac{{{r}_{1}}}{{{r}_{1}}\,\,+\,\,{{r}_{2}}} \right)\,\,q\,\,\And \,\,\,{{q}_{2}}\,\,=\,\,\left( \frac{{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}} \right)\,\,q\)
Where \('q'\) is the total charge on the two spheres
Ø Two point sized identical spheres carrying charges \({{q}_{1}}\) and\({{q}_{2}}\) on them are seperated by a certain distance. The mutual force between them is F. These two are brought in contact and kept at the same
separation. Now, the force between them is\({{F}^{1}}\) . Then .\(\frac{{{F}^{1}}}{F}=\frac{{{\left( {{q}_{1}}+{{q}_{2}} \right)}^{2}}}{4{{q}_{1}}{{q}_{2}}}\)
Ø The effective distance in vacuum for a dielectric slab of thickness t and dielectric constant k is
\( {{t}_{eff}}\) =\(t\,\sqrt{k}\) to have same electrostatic force .
Ø If a dielectric slab of dielectric constant k and thickness t is placed between two charges, then the effective distance between the charges is
\({{r}^{1}}=r-t+t\sqrt{k}\)
\(\therefore F=\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{\left( r-t+t\,\sqrt{k} \right)}^{2}}}\)
Ø Two identical charged spheres are suspended by strings of equal lengths in gravitational field. The strings make an angle of \('\theta '\) with each other due to repulsion. When same system is kept immersed in a liquid of density\('{{d}_{\ell }}'\) , even then the angle between them remains same. Then the dielectric constant of the liquid, \(K\,\,=\,\,\frac{{{d}_{b}}}{{{d}_{b}}\,\,-\,\,{{d}_{\ell }}}\) . Where \({{d}_{b}}\) - density of the body.
Note : If gravitational field is absent, then the angle between two strings is \({{180}^{0}}\) .
Linear charge density\(\left( \lambda \right)\) is defined as the charge per unit length.
\(\lambda =\frac{dq}{dl}\)
where dq is the charge on an infinitesimal length dl.
Units of \(\lambda \) are Coulomb / meter (C/m)
Examples:-Charged straight wire, circular charged ring.
Test charge:
That small positive charge, which does not affect the other charges present and by the help of which we determine the effect of other charges, is defined as test charge.
CONCEPTUAL UNDERSTANDING:
Ans : two point charges at rest.
2 ) The law, governing the force between electric charges is known as
Ans : Coulomb’s law.
Ans: increase
Application ability:
Ans : 5N
Ans : 100 dyne.
Higher order thinking :
Illustration 1 :
Two particles A and B having charges 8 x10-6 C and –2 x10-6C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force ?
Solution :
As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B
Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC=x and the charge on C is Q
\({{\vec{F}}_{CA}}\ =\,\frac{1}{4\pi \ {{\in }_{0}}}\ \frac{(8.0\times {{10}^{-6}})\ Q}{{{(0.2+x)}^{2}}}\ \hat{i}\)
and \({{\vec{F}}_{CB}}\ =\ \frac{-1}{4\pi {{\in }_{0}}}\ \frac{(2.0\times {{10}^{-6}})Q}{{{x}^{2}}}\ \hat{i}\)
\({{\vec{F}}_{C}}\ =\,{{\vec{F}}_{CA}}+{{\vec{F}}_{CB}}\,\,\,\,=\,\,\,\,\,\frac{1}{4\pi {{\in }_{0}}}\ \left[ \frac{(8.0\times {{10}^{-6}})\ Q}{{{(0.2+x)}^{2}}}\ -\ \frac{(2.0\times {{10}^{-6}})\,Q}{{{x}^{2}}} \right]\ i\)
But \(|{{\vec{F}}_{C}}|=0\)
Hence \(\frac{1}{4\pi {{\in }_{0}}}\ \left[ \frac{(8.0\times {{10}^{-6}})\ Q}{{{(0.2+x)}^{2}}}-\ \frac{(2.0\times {{10}^{-6}})\ Q}{{{x}^{2}}} \right]\ =0\)
Which gives x = 0.2 m
Illustration 2:
It is required to hold equal charges, q in equilibrium at the corners of a square. What charge when placed at the centre of the square will do this?
Solution :
As ABCD is a square of side a,
\(r=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}\)
\({{\vec{F}}_{BA}}=\frac{k{{q}^{2}}}{{{a}^{2}}}\hat{i},{{\vec{F}}_{BC}}=\frac{-K{{q}^{2}}}{{{a}^{2}}}\hat{j}\)
\({{\vec{F}}_{BD}}=\frac{k{{q}^{2}}}{{{(a\sqrt{2})}^{2}}}\left( \cos {{45}^{\circ }}\hat{i}-\sin {{45}^{\circ }}\hat{j} \right)\)
\({{\vec{F}}_{BQ}}=\frac{kQq}{{{(a/\sqrt{2})}^{2}}}\left( \cos {{45}^{\circ }}\hat{i}-\sin {{45}^{\circ }}\hat{j} \right)\)
Here \(\widehat{i}and\widehat{J}\) have usual meaning.
Net force on the charge at B is
\(\overset{\to }{\mathop{{{F}_{R}}}}\,=\left( \frac{k{{q}^{2}}}{{{a}^{2}}}+\frac{k{{q}^{2}}}{{{\left( a\sqrt{2} \right)}^{2}}}\cos {{45}^{\circ }}+\frac{kQ{{q}^{{}}}}{{{\left( a/\sqrt{2} \right)}^{2}}}\cos {{45}^{\circ }} \right)\hat{i}\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\left( \frac{k{{q}^{2}}}{{{a}^{2}}}+\frac{k{{q}^{2}}}{{{\left( a\sqrt{2} \right)}^{2}}}\sin {{45}^{\circ }}+\frac{kQ{{q}^{{}}}}{{{\left( a/\sqrt{2} \right)}^{2}}}\sin {{45}^{\circ }} \right)\hat{j}\)
For charge, q to be in equilibrium at B, the net force on it must be zero.
\(\therefore {{F}_{Bx}}={{F}_{BA}}+{{F}_{BD}}\cos {{45}^{o}}+{{F}_{BQ}}\cos {{45}^{o}}=0\)
\(\Rightarrow k\left[ \frac{{{q}^{2}}}{{{a}^{2}}}+\frac{{{q}^{2}}}{{{\left( a\sqrt{2} \right)}^{2}}}.\frac{1}{\sqrt{2}}+\frac{Qq}{{{\left( a/\sqrt{2} \right)}^{2}}}.\frac{1}{\sqrt{2}} \right]=0\)
\(\therefore Q=-\frac{q}{4}(1+2\sqrt{2})\)
Similarly,\({{F}_{By}}=0,if\,Q=-\frac{q}{4}(1+2\sqrt{2})\)
Solved problems:
Problem 1 : What charges will flow after the shorting of the switch S in the circuit given in the figure, through section 1 and 2 ?
Solution : When S is open, C1 and C2 are in series and their equivalent capacitance is \(\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\) Charge on the plates 1 and 3 is Q = +\(\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\varepsilon \) . When S is closed P.D. across C1 is zero. Þ Charge on the plate 1 is 0. |
If Charge flown through 2 is q, then Q + q = 0
Þ q = -Q = -\(\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\varepsilon \)
Charge on the plate 3 is C2e, which is also equal to the charge flown through 1.
Electric Field :
The region of space around an electric charge, in which its effect can be experienced is known as “electric field”.
(or)
The space around electric charge upto which its influence is felt is known as” electric field”.
* Electric field is a conservative field. Two types of electric fields are there
i) Uniform electric field:
The electric field, at every point of which a unit positive test charge expriences the same electric force is known as a uniform electric field.
In a uniform electric field, the intensity is same at every point both in magnitude and direction.
Ex : Electric field between the plates of parallel plate condenser.
ii) Non - uniform electric field:
The electric field, at different points of which a unit positive test charge experiences different forces is known as non - uniform electric field.
In a non-uniform electric field, the intensity is different at different points.
Ex : Electric field due to a point charge.
Intensity of electric field :
The electric field intensity at any point in an electric field is defined as the force experienced by a unit positive charge placed at that point.
The force experienced by a test charge q0 in an electric field of intensity E is given by
\(\overline{F}=\overline{E}\,{{q}_{0}}\) \(\therefore \,\overline{E}=\frac{\overline{F}}{{{q}_{0}}}\)
If q0 = 1 then E = F (numerically)
a) Intensity is a vector quantity.
b) If q0 is positive charge then the force acting on it is in the direction of the field.
c)If q0 is negative then the direction of this force is opposite to the field direction.
d) S.I unit of intensity of electric field is newton/colomb (or) volt/meter.
e) Its dimensional formula is M1 L1 T–3A–1.
iii) Motion of a charged particle in uniform electric field :
a) A charged body of mass ‘m’ and charge ‘q’ is initially at rest in a uniform electric field of intensity E.
The force acting on it, F = Eq
Here the direction of F is in the direction of field if ‘q’ is + ve and opposite to the field if ‘q’ is –ve.
b) The body travels in straight line path with uniform acceleration, \(a=\frac{F}{m}=\frac{Eq}{m}\) initial velocity, u= 0.
At an instant of time t ,
c) Its final velocity , \(v=u+at=\left( \frac{Eq}{m} \right)t\)
d) Displacement \(s=ut+\frac{1}{2}a{{t}^{2}}=\frac{1}{2}\left( \frac{Eq}{m} \right){{t}^{2}}\)
e) Momentum, \(P=mv=(Eq)t\)
f) Kinetic energy, K.E =\(\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\left( \frac{{{E}^{2}}{{q}^{2}}}{m} \right){{t}^{2}}\)
g) When a charged particle enters perpendicularly into a uniform electric field of intensity E with a velocity u then it describes parabolic path as shown in figure
Along the horizontal direction, there is no acceleration and hence x = ut.
Along the vertical direction, acceleration, \(a=\frac{F}{m}=\frac{Eq}{m}\) (here gravitational force is not considered)
h) Hence vertical displacement , \\(y=\frac{1}{2}\left( \frac{Eq}{m} \right){{t}^{2}}\)
\(y=\frac{1}{2}\left( \frac{qE}{m} \right){{\left( \frac{x}{u} \right)}^{2}}\) \(=\left( \frac{qE}{2m{{u}^{2}}} \right){{x}^{2}}\)
i) At any instant of time t, horizontal component of velocity , vx = u.
j) vertical componet of velocity
\({{v}_{y}}=at=\left( \frac{Eq}{m} \right)t\)
\(v=\left| {\bar{v}} \right|=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{{{u}^{2}}+\frac{{{E}^{2}}{{q}^{2}}{{t}^{2}}}{{{m}^{2}}}}\)
iv) Null point (or) Neutral Point :
The point where the resultant electric field intensity becomes zero is called a null point.
a) If two like charges q1 and q2 are separated by a distance ‘d’ as shown in the figure then null point ‘N’ will be formed in between them and nearer to charge of smaller magnitude.
\(x=\frac{d}{\sqrt{\frac{{{q}_{2}}}{{{q}_{1}}}}+1}\) x is from .
In the above case, if q1 = q2 then null point will be formed at the mid point of the line joining the charges.
b) If two unlike charges q1 and q2 (q1 < q2 in magnitude ) are separated by a distance ‘d’ as shown in the figure then null point N will be formed outside the system of charges and on the line joining them and nearer to charge of smaller magnitude.
\(x=\frac{d}{\sqrt{\frac{{{q}_{2}}}{{{q}_{1}}}}-1}\)
In the above case, if q1 = q2 then there is no possibility for the null point.
c) In the above formulae, the magnitudes of q1 and q2 should be taken without sign.
CONCEPTUAL UNDERSTANDING:
1) When a dielectric material is introduced between the plates of a charged condenser, after disconnecting the battery the electric field between the plates
Ans: decreases
2)The pair of particles which have same acceleration in a uniform electric field is
Ans: Deutron and alpha particle.
3)An electron is projected with certain velocity into an electric field in a direction opposite to the field. Then it is
Ans: accelerated
Application ability:
Ans: \(10\sqrt{2}\)
2. The breakdown electric intensity for air is V/m. The maximum charge that can be held by a sphere of radius 1 mm is
Ans: 0.33 nC
Solved problems:
Problem 1 :
Find the electric field caused by a disk of radius R with a uniform positive surface charge density s at a point along the axis of the disk a distance x from its centre.
Solution : Consider a disc of surface charge density `s'. Let us calculate the electric field due to a ring of charge situated at a distance r, from the centre and having a thickness, dr. Using the result of the previous question, we can say, \(dE=\frac{kx\,dq}{{{\left( {{x}^{2}}+{{r}^{2}} \right)}^{{}^{3}/{}_{2}}}}\) |
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Now, the area of the ring, dS = 2pr. dr, Þ dq = s2prdr,
Thus
\(\begin{align} & {{\left| E \right|}_{^{p}}}=\int{dE=kx}\int\limits_{O}^{R}{\frac{\sigma 2\pi rdr}{{{\left( {{x}^{2}}+{{r}^{2}} \right)}^{3/2}}}=\frac{1}{4\pi {{\in }_{^{o}}}}x.\sigma \pi \int\limits_{O}^{R}{\frac{2r}{{{\left( {{x}^{2}}+{{r}^{2}} \right)}^{3/2}}}dr}} \\ & E=\frac{\sigma }{2{{\in }_{o}}}\left[ 1-\frac{x}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{1/2}}} \right] \\ \end{align}\)
Also, as \(R\to \infty ,E=\frac{\sigma }{2{{\in }_{o}}},\) which is the electric field in front of an infinite plane sheet of charge.
Problem 2: A solid spherical region having a spherical cavity whose diameter ‘R’ is equal to the radius of the spherical region, has a total charge ‘Q’. Find the electric field and potential at a point P as shown. |
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Solution : Charge density r =\(\frac{Q}{\frac{4}{3}\pi \left[ {{R}^{3}}-{{\left( \frac{R}{2} \right)}^{3}} \right]}\) =\(\frac{6Q}{7\pi {{R}^{3}}}\) Using the superposition principle, V = \(\frac{\rho \left( \frac{4}{3}\pi {{R}^{3}} \right)}{4\pi {{\varepsilon }_{0}}x}-\frac{\rho \left[ \frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}} \right]}{4\pi {{\varepsilon }_{0}}\left( x-\frac{R}{2} \right)}\) or V = \(\frac{\rho {{R}^{3}}}{3{{\varepsilon }_{0}}}\left[ \frac{7x-4R}{4x\left( 2x-R \right)} \right]\) |
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Similarly E = \(\frac{\rho \left( \frac{4}{3}\pi {{R}^{3}} \right)}{4\pi {{\varepsilon }_{0}}{{x}^{2}}}-\frac{\rho \left[ \frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}} \right]}{4\pi {{\varepsilon }_{0}}{{\left( x-\frac{R}{2} \right)}^{2}}}\) or E = \(\frac{\rho {{R}^{3}}}{3{{\varepsilon }_{0}}}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{R{{\left( 2x-R \right)}^{2}}} \right]\)
Illustration 1:
What is the electric field at any point on the axis of a charged rod of length `L' and linear charge density`l'? The point is separated from the nearer end by a.
Solution :
Consider an element, dx at a distance, x from the point, P, where we seek to find the electric field. The elemental charge, dq =ldx
\(\text{Then, }dE=k.\frac{\lambda dx}{{{x}^{2}}}\,\,\,\,or\,\,\,\,E=k\lambda \int\limits_{a}^{a+L}{\frac{1}{{{x}^{2}}}dx=k\lambda }\left[ -\frac{1}{x} \right]_{a}^{a+L}=k\lambda \left[ \frac{-1}{a+L}+\frac{1}{a} \right]\)
\(Thus,\,\,E=\frac{\lambda }{4\pi {{\in }_{o}}}\left[ \frac{1}{a}-\frac{1}{L+a} \right]\) [ k = \(\frac{1}{4\pi {{\varepsilon }_{0}}}\) ]
Illustration 2 :
A ring shaped conductor with radius R carries a total charge q uniformly distributed around it as shown in the figure, find the electric field at a point P that lies on the axis of the ring at a distance x from its centre.
Solution:
Consider a differential element of the ring of length ds. Charge on this element is dq = \(\left( \frac{q}{2\pi R} \right)ds\)
This element sets up a differential electric field d \(\overline{E}\)at point P.
The resultant field\(\vec{E}\) at P is found by integrating the effects of all the elements that make up the ring. From symmetry this resultant field must lie along the right axis. Thus, only the component of d \(\vec{E}\) parallel to this axis contributes to the final result.
\(\vec{E}\) = \(\int{d\vec{E}}\)Þ \(\vec{E}\) = \(\int{d\vec{E}}\) cosq
dE = \(\frac{1}{4\pi {{\in }_{0}}}\frac{dq}{{{r}^{2}}}\,=\,\frac{1}{4\pi \varepsilon {{\in }_{0}}}\left( \frac{q\,ds}{2\pi R} \right)\,.\,\frac{1}{\left( {{R}^{2}}+{{x}^{2}} \right)}\)
cosq = \(\frac{x}{{{({{R}^{2}}+{{x}^{2}})}^{1/2}}}\)
To find the total x-component Ex of the field at P, we integrate this expression over all segment of the ring.
. Ex = \(\int{d\vec{E}}\) cosq = \(\frac{1}{4\pi {{\in }_{0}}}\frac{qx}{2\pi R.\,{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}\int{ds}\)
The integral is simply the circumference of the ring = 2pR
E = \(\frac{1}{4\pi {{\in }_{0}}}\frac{qx}{{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}\)
As q is positive charge, field is directed away from the centre of the ring, along its axis.
Electric filed at the axis of a circular uniformly charged ring:
Intensity of electric field at a point P that lies on the axis of the ring at a distance x from its centre is
\(E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qx}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}}\)
where \(\left\{ \cos \theta =\frac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}} \right\}\)
Where R is the radius of the ring. From the above expression E = 0 at the centre of the ring.
E will be maximum when \(\frac{dE}{dx}=0\).
Differentating E w.r.t x and putting it equal to zero we get
\(x=\frac{R}{\sqrt{2}}\) and \({{E}_{\max }}=\frac{2}{3\sqrt{3}}\left( \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{R}^{2}}} \right)\)
Electric field due to a Charged Spherical Conductor(Spherical Shell ):
‘q’ amount of charge be uniformly distributed over a spherical shell of radius ‘r’
\(\sigma =\)Surface charge density,
\(\sigma =\frac{q}{4\pi {{R}^{2}}}\)
Ø When point ‘P’ lies outside the shell :
\(E=\frac{1}{4\pi {{\in }_{0}}}\times \frac{q}{{{r}^{2}}}\)
Ø This is the same expression as obtained for electric field at a point due to a point charge. Hence a charged spherical shell behave as a point charge concentrated at the centre of it.
\(E=\frac{1}{4\pi {{\in }_{0}}}\frac{\sigma .4\pi {{R}^{2}}}{{{r}^{2}}}\,\,\,\,\,\because \sigma =\frac{q}{4\pi {{r}^{2}}}\)
\(E=\frac{\sigma .{{R}^{2}}}{{{\in }_{0}}{{r}^{2}}}\)
When point ‘P’ lies on the shell :
\(E=\frac{\sigma }{{{\in }_{0}}}\)
When Point ‘P’ lies inside the shell:
E = 0
Note : The field inside the cavity is always zero this is known as elctro static shielding
Electric filed due to a Uniformly charged non – conducting sphere:
Electric field intensity due to a uniformly charged non-conducting sphere of charge Q, of radius R at a distance r from the centre of the sphere
q is the amount of charge be uniformly distributed over a solid sphere of radius R.
\(\rho \)= Volume charge density
\(\rho =\frac{q}{\frac{4}{3}\pi {{R}^{3}}}\)
When point ‘P’ lies inside sphere :
\(E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qr}{{{R}^{3}}}\)for \(r<R\)
\(E=\frac{\rho .r}{3{{\in }_{0}}}\)
When point ‘P’ lies on the sphere:
. \(E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{R}^{2}}}\)
\(E=\frac{\rho .\,\,R}{3{{\in }_{0}}}\)
When point ‘P’ lies outside the sphere:
\(E=\frac{1}{4\pi {{\in }_{0}}}\frac{q}{{{r}^{2}}}\)
\(E=\frac{\rho .\,\,{{R}^{3}}}{3{{\in }_{0}}{{r}^{2}}}\)
Electric Field due to a charged Disc:
Electric field due to a uniformly charged disc with surface charge density of radius at a distance x from the centre of the disc is
\(E=\frac{\sigma }{2{{\varepsilon }_{0}}}\left[ 1-\frac{x}{\sqrt{{{x}^{2}}+{{R}^{2}}}} \right]\)
If Q is the total charge on the disc, then
\(E=\frac{2Q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\left[ 1-\frac{x}{\sqrt{{{x}^{2}}+{{R}^{2}}}} \right]\)
CONCEPTUAL UNDERSTANDING:
1) A uniformly charged thin spherical shell of radius R carries uniform surface charge density of per unit area. It is made of two hemispherical shells, held together by pressing them with force F.F is proportional to
Ans: \(\frac{1}{{{\varepsilon }_{o}}}{{\sigma }^{2}}{{R}^{2}}\)
2) An infinite number of charges each of magnitude q are placed on x - axis at distances of 1,2, 4, 8, ... meter from the origin. The intensity of the electric field at origin is
Ans : \(\frac{q}{3\pi {{\varepsilon }_{0}}}\)
Application ability:
1)
A non-conducting ring of radius 0.5 m carries of total charge of 1.11x10-10c distributed non-uniformly on its circumference producing an electric field E everywhere in space. The value of the integral \(\int\limits_{l=\infty }^{l=0}{-\overrightarrow{E}}.d\overrightarrow{l}\) (l=0 being centre of the ring) in volts is
Ans : +2
Solved problems:
Problem 1 :
Find the electric field caused by a disk of radius R with a uniform positive surface charge density s at a point along the axis of the disk a distance x from its centre.
Solution : Consider a disc of surface charge density `s'. Let us calculate the electric field due to a ring of charge situated at a distance r, from the centre and having a thickness, dr. Using the result of the previous question, we can say, \(dE=\frac{kx\,dq}{{{\left( {{x}^{2}}+{{r}^{2}} \right)}^{{}^{3}/{}_{2}}}}\) |
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Now, the area of the ring, dS = 2pr. dr, Þ dq = s2prdr,
Thus
\({{\left| E \right|}_{^{p}}}=\int{dE=kx}\int\limits_{O}^{R}{\frac{\sigma 2\pi rdr}{{{\left( {{x}^{2}}+{{r}^{2}} \right)}^{3/2}}}=\frac{1}{4\pi {{\in }_{^{o}}}}x.\sigma \pi \int\limits_{O}^{R}{\frac{2r}{{{\left( {{x}^{2}}+{{r}^{2}} \right)}^{3/2}}}dr}}\)
\(E=\frac{\sigma }{2{{\in }_{o}}}\left[ 1-\frac{x}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{1/2}}} \right]\)
Also, as \(R\to \infty ,E=\frac{\sigma }{2{{\in }_{o}}},\) which is the electric field in front of an infinite plane sheet of charge.
Problem 2: A solid spherical region having a spherical cavity whose diameter ‘R’ is equal to the radius of the spherical region, has a total charge ‘Q’. Find the electric field and potential at a point P as shown. |
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Solution : Charge density r =\(\frac{Q}{\frac{4}{3}\pi \left[ {{R}^{3}}-{{\left( \frac{R}{2} \right)}^{3}} \right]}\) =\(\frac{6Q}{7\pi {{R}^{3}}}\) Using the superposition principle, V = \(\frac{\rho \left( \frac{4}{3}\pi {{R}^{3}} \right)}{4\pi {{\varepsilon }_{0}}x}-\frac{\rho \left[ \frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}} \right]}{4\pi {{\varepsilon }_{0}}\left( x-\frac{R}{2} \right)}\) or V = \(\frac{\rho {{R}^{3}}}{3{{\varepsilon }_{0}}}\left[ \frac{7x-4R}{4x\left( 2x-R \right)} \right]\) |
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Similarly E = \(\frac{\rho \left( \frac{4}{3}\pi {{R}^{3}} \right)}{4\pi {{\varepsilon }_{0}}{{x}^{2}}}-\frac{\rho \left[ \frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}} \right]}{4\pi {{\varepsilon }_{0}}{{\left( x-\frac{R}{2} \right)}^{2}}}\) or E = \(\frac{\rho {{R}^{3}}}{3{{\varepsilon }_{0}}}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{R{{\left( 2x-R \right)}^{2}}} \right]\)
Higher order thinking :
Solution: The spherically symmetric distribution of charge means that the charge density at any point depends only on the distance of the point from the centre and not on the direction. Secondly, the object can not be a conductor, or else the excess charge will reside on its surface. |
Now, apply Gauss’s law to a spherical Gaussian surface of radius r (r > R for point A)
eo \(\oint{\overrightarrow{\,E\,}}.d\overrightarrow{s}={{q}_{en}}\) Þ \({{\varepsilon }_{o}}E(4\pi {{r}^{2}})\) = q
E = \(\frac{1}{4\pi {{\in }_{0}}}\frac{q}{{{r}^{2}}}\) where q is the total charge
For point B (r<R)
eo\(\oint{\overrightarrow{E\,}}.d\overrightarrow{s}\) = \({{\varepsilon }_{o}}E(4\pi {{r}^{2}})\) = q
q’ = \(\frac{q\frac{4}{3}\pi {{r}^{3}}}{\frac{4}{3}\pi {{R}^{3}}}\,\,=\,\,q{{\left( \frac{r}{R} \right)}^{3}}\)
E = \(\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q\,{{\left( \frac{r}{R} \right)}^{3}}}{{{r}^{2}}}\,=\,\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qr}{{{R}^{3}}}\)
Illustration 2 : Figure shows a section of an infinite rod of charge having linear charge density l which is constant for all points on the line. Find electric field E at a distance r from the line. |
\({{\varepsilon }_{o}}\oint{\vec{E}.\,d\vec{s}}={{q}_{in}}\)
eo \(\left[ \int{\vec{E}.d\vec{s}+\int{\vec{E}.d\vec{s}}} \right]\) = qin
Cylindrical Plane Surface
eo E (2prl) + E.ds. cos 90o = ll
E = \(\frac{\lambda l}{{{\varepsilon }_{o}}2\pi rl}=\frac{\lambda }{2\pi {{\varepsilon }_{0}}r}\)
The direction of \(\overrightarrow{E}\) is radially outward for a line of positive charge.
a) A charged body of mass ‘m’ and charge ‘q’ is initially at rest in a uniform electric field of intensity E.
The force acting on it, F = Eq
Here the direction of F is in the direction of field if ‘q’ is + ve and opposite to the field if ‘q’ is –ve.
b) The body travels in straight line path with uniform acceleration,\(a=\frac{F}{m}=\frac{Eq}{m}\) initial velocity, u= 0.
At an instant of time t ,
c) Its final velocity , \(v=u+at=\left( \frac{Eq}{m} \right)t\)
d) Displacement \(s=ut+\frac{1}{2}a{{t}^{2}}=\frac{1}{2}\left( \frac{Eq}{m} \right){{t}^{2}}\)
e) Momentum, \(P=mv=(Eq)t\)
f) Kinetic energy, K.E =\(\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\left( \frac{{{E}^{2}}{{q}^{2}}}{m} \right){{t}^{2}}\)
g) When a charged particle enters perpendicularly into a uniform electric field of intensity E with a velocity u then it describes parabolic path as shown in figure
Along the horizontal direction, there is no acceleration and hence x = ut.
Along the vertical direction, acceleration, \(a=\frac{F}{m}=\frac{Eq}{m}\) (here gravitational force is not considered)
h) Hence vertical displacement , \(y=\frac{1}{2}\left( \frac{Eq}{m} \right){{t}^{2}}\)
\(y=\frac{1}{2}\left( \frac{qE}{m} \right){{\left( \frac{x}{u} \right)}^{2}}\) \(=\left( \frac{qE}{2m{{u}^{2}}} \right){{x}^{2}}\)
i) At any instant of time t, horizontal component of velocity , vx = u.
j) vertical componet of velocity
\({{v}_{y}}=at=\left( \frac{Eq}{m} \right)t\)
\(v=\left| {\bar{v}} \right|=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{{{u}^{2}}+\frac{{{E}^{2}}{{q}^{2}}{{t}^{2}}}{{{m}^{2}}}}\)
Null point (or) Neutral Point :
The point where the resultant electric field intensity becomes zero is called a null point.
a) If two like charges q1 and q2 are separated by a distance ‘d’ as shown in the figure then null point ‘N’ will be formed in between them and nearer to charge of smaller magnitude.
\(x=\frac{d}{\sqrt{\frac{{{q}_{2}}}{{{q}_{1}}}}+1}\) x is from .
In the above case, if q1 = q2 then null point will be formed at the mid point of the line joining the charges.
b) If two unlike charges q1 and q2 (q1 < q2 in magnitude ) are separated by a distance ‘d’ as shown in the figure then null point N will be formed outside the system of charges and on the line joining them and nearer to charge of smaller magnitude.
\(x=\frac{d}{\sqrt{\frac{{{q}_{2}}}{{{q}_{1}}}}-1}\)
In the above case, if q1 = q2 then there is no possibility for the null point.
c) In the above formulae, the magnitudes of q1 and q2 should be taken without sign.
CONCEPTUAL UNDERSTANDING:
Ans : Vertical velocity changes but horizontal velocity remains constant
Ans : parabola
Ans : proportional to its specific charge
Application ability:
1 )A pendulum bob of mass m carrying a charge q is at rest in a uniform horizontal electric field of intensity E. The tension in the thread is
Ans:
Electric Dipole:
An electric dopole is a pair of equal and opposite point charge and seperated by a distance 2a. The line connecting the two charges defines adirection in space. The direction from to is said to be axis of the dipole.
The total charge of the electric dipole is zero.
(1) Charges (+q) and(-q) are called the poles of the dipole. (2) = the displacement vector from -ve charge to +ve charge. (3) = the dipole moment = \(q\vec{l}\) . |
(4) The straight line joining the two poles is called axial line.
(5) Perpendicular bisector of ‘ is called equatorial line.
Dipole moment\(\left( \overrightarrow{p} \right)\) :
It is defined as the product of magnitude of either charge and the distance of separation between the two charges.
\(\overrightarrow{p}=q\left( 2\overrightarrow{a} \right)\)
Dipole moment \(\overrightarrow{p}\) always points from -q to +q
The electric dipole moment is a measure of the separation of positive and negative electrical charges within a system, that is, a measure of the system's overall polarity.
The SI units for electric dipole moment are coulomb-meter (C.m); however, the most common unit is the debye (D).
Electric field due to a Dipole at a point lying on the Axil Line (End on position):
\({{E}_{axial}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2pr}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}} \) (from negative to positive charge)
In case of a short dipole (r>>a).
\({{E}_{axial}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2p}{{{r}^{3}}}\)
Electric filed due to a dipole at a point lying on the equatorial line (Broad side on position):
\( {{E}_{equatorial}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{p}{{{\left( {{r}^{2}}+{{a}^{2}} \right)}^{{}^{3}/{}_{2}}}}\)
{ from positive to negative charge}
In case a short dipole (r>>a),
\({{E}_{equatorial}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{p}{{{r}^{3}}}\)
Field due to dipole:
Electric filed due to a short dipole at any point p(r, \(\theta \)):
\(E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{p}{{{r}^{3}}}\sqrt{1+3{{\cos }^{2}}\theta }\)
\(\tan \beta =\frac{{{E}_{\theta }}}{{{E}_{r}}}=\frac{1}{2}\tan \theta \) \(\Rightarrow \beta ={{\tan }^{-1}}\left[ \frac{1}{2}\left( \tan \theta \right) \right]\)
Torque and workdone on dipole:
Torque on a dipole placed in a uniform electric filed:
The torque due to the force on the positive charge about a point O is given by Fa sin .The torque on the negative charge about O is also Fa sin
\(\tau =2Fa\sin \theta \)
\(\Rightarrow \tau =2aqE\sin \theta \Rightarrow \tau =pE\sin \theta \)
\(\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E}\)
An object with an electric dipole moment is subject to a torque τ when placed in an external electric field. The torque tends to align the dipole with the field.
Potential energy of a dipole placed in a uniform electric field:
Work done by an external agent to rotate the dipole from an angle \({{\theta }_{0}}\) to \(\theta \)with the field is
\(W=PE\left( \cos {{\theta }_{0}}-\cos \theta \right)\)
The potential energy stored of a dipole in uniform electric field is \(U=-PE\,\cos \theta \)
This is equivalent to the dot product of the vectors \(\overrightarrow{p}\,and\,\overrightarrow{E}.\)
\(U=-\overrightarrow{p}.\overrightarrow{E}=\left( {{p}_{x}}{{E}_{x}}+{{p}_{y}}{{E}_{y}}+{{p}_{z}}{{E}_{z}} \right)\)
Angular SHM of Dipole in uniform electric field:
When a dipole is suspended in a uniform field, it will align itself parallel to the field.
Now if it is given a small angular displacement \(\theta \)about its equilibrium position then restoring couple will be
\(C=-pE\sin \theta =-pE\theta \,\,\left( as\,\,\sin \,\theta =\theta \right)\)
\(\Rightarrow I\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=-pE\theta \,\)
\(\Rightarrow \frac{{{d}^{2}}\theta }{d{{t}^{2}}}=-{{\omega }^{2}}\theta \) with \({{\omega }^{2}}=\frac{pE}{I}\)
Hence motion is simple harmonic.
Time period of oscillation is \(T=2\pi \sqrt{\frac{I}{pE}}\)
CONCEPTUAL UNDERSTANDING:
Ans: 0
Ans : \(\frac{1}{{{r}^{3}}}\)
Ans : neither a force nor a torque
Application ability:
Ans : \(2\times {{10}^{-19}}J\)
Ans : \(\sqrt{3}ql\) along perpendicular bisector of q - q line
Ans : zero.
Higher order thinking :
Illustration 1:
Two tiny spheres, each of of mass M, and charges +q and -q respectively, are connected by a massless rod of length, L. They are placed in a uniform electric field at an angle q with the\(\vec{E}\) (q » 0o). Calculate the minimum time in which the system aligns itself parallel to the \(\vec{E}\).
Solution :
t = pE sin q, (as q ® 0,sin q ® q)
Þ t = -(pE)q (If we assume angular displacement to be anti-clockwise, torque is clockwise)
Þ a = \(-\left( \frac{pE}{I} \right)\theta \) = -w2q
As torque is proportional to 'q' and oppositely directed, there will be an S.H.M.
Here, p = q.L.and moment of inertia, I = M (L/2)2 + M (L/2)2 = ML2/2
\(\therefore \text{ time }\,\text{period}\,\text{ T = }2\pi \sqrt{\frac{I}{pE}}\) The minimum time required to align itself is \(\frac{T}{4}\) |
ELECTRIC POTENTIAL :
i) Electric potential at a point in an electric field is the amount of workdone in bringing a unit positive charge from infinity to that point along any arbitrary path against the intensity of electric field.
If W is the workdone in bringing a test charge q0 from infinity to a point then the potential at that point, V=W/q0
ii) Electric potential at a point is numerically equal to the electric potential energy of a unit positive charge placed at that point.
iii) Electric potential is a scalar quantity. Its S.I. unit is volt.
iv) Electric potential is analogous quantity to temperature in heat and pressure in hydrostatics.
Electric potential due to a point charge :
a) In air or free space, the electric potential at a distance ‘d’ from a point charge q is given by,\(V=\frac{1}{4\pi {{\in }_{o}}}\,\,\frac{q}{d}\) .
b) In a medium ,\({{V}^{1}}=\frac{1}{4\pi {{\in }_{o}}K}\,\,\frac{q}{d}\) .
By the presence of a medium, the potential decreases.
c) Due to an isolated positive charge, the potential is positive.
d) Due to an isolated negative charge, the potential is negative.
vi) Principle of superposition :
Due to several point charges, the net potential at a point is obtained by the principle of superposition.
The electric potential at a point due to a group of point charges q1 , q2 , ....., qn , which are at distances r1 , r2 , r3 , .......rn from the point is given by
V = V1 + V2 + .....+ Vn = \(\frac{1}{4\pi {{\varepsilon }_{0}}}\sum\limits_{i=1}^{n}{\frac{{{q}_{i}}}{{{r}_{i}}}}\)
The above sum is the algebraic sum but not the vector sum.
CONCEPTUAL UNDERSTANDING:
1) Potential at the point of a pointed conductor is
Ans: same as at any other point
2) Electric potential at some point in space is zero. Then at that point
Ans: electric intensity may or may not be zero
Application ability:
1) If \(4\times {{10}^{20}}eV\) is required to move a charge of 0.25 coulomb between two points, the potential difference between these two points is
Ans: \(\frac{1}{256}\) volt
2) An infinite number of charges each equal to 'q' are placed along the X-axis at x = 1, x = 2, x = 4, x = 8 ..... The potential at the point x = 0 due to this set of charges is
Ans: \(\frac{2Q}{4\pi {{\in }_{o}}}\)
Solved problems:
Problem 1 :
Three charges q1 , q2 and q3 are located at the vertices of an equilateral triangle of side a. Find the electric potential energy of the system.
Solution:
Taking datum at infinity. We can assume that the charges are brought one by one.
W = DVq
In absence of any charge
VA = 0,
Þ W1 = (VA - V¥)q1 = 0 (1)
W2 = (VB - V¥)q2
= \(\left( \frac{K{{q}_{1}}}{a}-0 \right){{q}_{2}}=\frac{K{{q}_{1}}{{q}_{2}}}{a}\) (2)
W3 = (Vc - V¥)q3
= \(\left( \frac{K{{q}_{1}}}{a}+\frac{K{{q}_{2}}}{a}-0 \right){{q}_{3}}\) = \(\frac{K{{q}_{1}}{{q}_{3}}}{a}+\frac{K{{q}_{1}}{{q}_{3}}}{a}\) (3) W = W1 + W2 + W3 Þ \(W=\frac{K{{q}_{1}}{{q}_{2}}}{a}+\frac{K{{q}_{2}}{{q}_{3}}}{a}+\frac{K{{q}_{1}}{{q}_{3}}}{a}\) |
|
ØTwo charges and are separated by a distance 'd'. The P.E. of the system of charges is
\(U=\frac{1}{4\pi {{\in }_{0}}}.\frac{{{Q}_{1}}{{Q}_{2}}}{d}\) from \(\text{U=W=Vq}\)
Ø Three charges \({{Q}_{1}},{{Q}_{2}},{{Q}_{3}}\) are placed at the three vertices of an equilateral triangle of side 'a'. The P.E. of the system of charges is
\(U=\frac{1}{4\pi {{\in }_{0}}}\left[ \frac{{{Q}_{1}}{{Q}_{2}}}{a}+\frac{{{Q}_{2}}{{Q}_{3}}}{a}+\frac{{{Q}_{3}}{{Q}_{1}}}{a} \right]\) or
\(U=\frac{1}{4\pi {{\in }_{0}}}\frac{\sum{{{Q}_{1}}{{Q}_{2}}}}{a}\)
Ø A charged particle of charge is held at rest at a distance 'd' from a stationary charge . When the charge is released, the K.E. of the charge at infinity is \(\frac{1}{4\pi {{\in }_{0}}}.\frac{{{Q}_{1}}{{Q}_{2}}}{d}\)
Ø If two like charges are brought closer, P.E of the system increases.
Ø If two unlike chargtes are brought closer, P.E of the system decreses.
For an attractive system U is always NEGATIVE..
For a repulsive system U is always POSITIVE.
For a stable system U is MINIMUM.
i.e. \(F=-\frac{dU}{dx}\) = 0 (for stable system)
Equilibrium of charges:
i) A charge is said to be in equilibrium, if net force acting on it is zero. A system of charges is said to be in equilibrium if each charge is separately in equilibrium.
Equilibrium can be divided into following types.
a) Stable equilibrium : After displacing a charged particle from its equilibrium position, if it returns back then it is said to be in stable equilibrium.
b) Unstable equilibrium : After displacing a charged particle from its equilibrium position, if it never returns back then it is said to be in unstable equilibrium
c) Neutral equilibrium : After displacing a charged particle from its equilibrium position if it neither comes back, nor moves away but remains in the position in which it was kept, is said to be in neutral equilibrium.
ii) Different cases of equilibrium of charges:
a) Freely floating charge in air
If a charged body or oil drop of mass m and charge q is in equilibrium in an electric field of intensity E then its weight is balanced by electrostatic force.
In equilibrium
QE = mg
\(\Rightarrow E=\frac{mg}{Q}\)
b) Suspension of charge from string
If a charged body is suspended from a string in horizontal electric field and is in equilibrium then
\(T\,\cos \,\theta \,\,=\,\,mg\) and \(T\,\sin \,\theta \,\,=\,\,Eq\)
\(\therefore \,Tan\,\theta =\frac{Eq}{mg}\) and \(T=\sqrt{{{(Eq)}^{2}}+{{(mg)}^{2}}}\)
c) System of three collinear charges
In the following figure three charges Q1,Q and Q2 are kept along a straight line.
Charge Q will be in equilibrium if and only if (Force applied by charge Q1) = (Force applied by charge Q2)
i.e\(\frac{1}{4\pi {{\in }_{0}}}\frac{{{Q}_{1}}Q}{x_{1}^{2}}=\frac{1}{4\pi {{\in }_{0}}}\frac{{{Q}_{2}}Q}{x_{2}^{2}}dx\) \(\Rightarrow \frac{{{Q}_{1}}}{{{Q}_{2}}}={{\left( \frac{{{x}_{1}}}{x_{2}^{{}}} \right)}^{2}}\)
This is necessary condition for Q to be in equilibrium. If all the three charges (Q1, Q and Q2) are similar, Q will be in stable equilibrium.
d) Two identical charged spheres of same mass are suspended by strings of same length from same point .
a) They arrange themselves and come to equilibrium as shown in figure (stable equilibrium).
\(F=\frac{1}{4\pi {{\in }_{o}}}\,\,\frac{{{q}^{2}}}{{{x}^{2}}}\) , \(\frac{F}{\ell \,\sin \theta }\,\,=\,\,\frac{W}{\ell \cos \theta }\,\,=\,\,\frac{T}{\ell }\)
b) If the arrangement is immersed in a non conducting liquid of dielectric constant K if the distance between the bobs remains same then is constant.
\(\frac{F}{{{F}^{1}}}=\frac{W}{{{W}^{1}}}\)
\(\therefore \,\,K=\frac{mg}{mg\left( 1-\frac{{{d}_{\ell }}}{{{d}_{B}}} \right)}=\frac{{{d}_{B}}}{{{d}_{B}}-{{d}_{\ell }}}\)
c) If the above arrangement is kept in an artificial satellite \(({{W}^{1}}=0)\) then \(\theta =180\) º and Tension in each string,
\(T=\frac{1}{4\pi {{\in }_{o}}}\,\,\frac{{{q}^{2}}}{{{(2\ell )}^{2}}}\)
CONCEPTUAL UNDERSTANDING:
Ans: Electric field is normal to the surface of the cube
Application ability:
1)Three charges each \(20\mu C\) are placed at the corners of an equilateral triangle of side \(0.4m\) . The potential energy of the system is
Ans: \(27J\)
2)A charge \(-2\mu C\) at the origin, \(-1\mu C\) at\(-1\mu C\) and \(1\mu C\) at \(-7cm\) are placed on \(X-\) axis. The mutual potential energy of the system is
Ans: \(-0.064J\)
Solved problems:
Problem 1 :
Two point charges each of charge +q are fixed at (+a, 0) and (-a, 0). Another positive point charge q placed at the origin is free to move along x- axis.
The charge q at origin in equilibrium will have |
Solution :
The net force on q at origin is
\(\vec{F}={{\vec{F}}_{1}}+{{\vec{F}}_{2}}\)
= \(\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{{{r}^{2}}}\hat{i}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{{{r}^{2}}}(-\hat{i})=\hat{0}\)
The P.E. of the charge q in between the extreme charges at a distance x from the origin along +ve x axis is
U = \(\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{\left( a-x \right)}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{(a+x)}\) = \(\frac{1}{4\pi {{\varepsilon }_{0}}}.{{q}^{2}}\left[ \frac{1}{a-x}+\frac{1}{a+x} \right]\) .
\(\frac{dU}{dx}=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}}\left[ -\frac{1}{{{(a-x)}^{2}}}+\frac{1}{{{(a+x)}^{2}}} \right]\)
For U to be minimum
\(\frac{dU}{dx}=0,\) \(\frac{{{d}^{2}}U}{d{{x}^{2}}}>0,\)
Þ (a-x)2 = (a + x)2 Þ a + x = ± (a – x)
Þ x = 0, because other solution is relevant.
Thus the charged particle at the origin will have minimum force and minimum P.E.
\ (D).
Problem 2 :
Three particles, each of mass 1 gm and carrying a charge, q, are suspended from a common point by insulated massless strings, each 100 cm long. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side 3 cm, calculate the charge, q on each particle (take g= 10ms-2)
\({{\vec{F}}_{A}}={{\vec{F}}_{AB}}+{{\vec{F}}_{AC}}\)
Solution:
\(=\left[ \frac{1}{4\pi {{\in }_{o}}}\frac{{{q}^{_{2}}}}{{{a}^{2}}} \right]2\cos \left[ \frac{{{60}^{o}}}{2} \right]\text{ in the direction D to A}\)
\(=2\left[ \frac{1}{4\pi {{\in }_{^{o}}}}\frac{{{q}^{2}}}{{{a}^{_{2}}}} \right]\frac{\sqrt{3}}{2}\) in the direction D to A
For equilibrium
Tcosq = mg.....(1) & Tsinq=FA..... (2)
Divide (2) by (1), we get; tanq = \(\frac{{{F}_{A}}}{mg}\) Where , tanq » sinq \(\frac{AD}{AO}=\frac{\sqrt{3}\times {{10}^{-2}}}{1}\) (AD = 2/3 of the median length, as D is the centroid) \(\begin{align} & or2\left[ \frac{1}{4\pi {{\in }_{o}}} \right]\frac{{{q}^{_{2}}}}{{{a}^{2}}}\frac{\sqrt{3}}{2}={{10}^{_{-3}}}\times (\sqrt{3}\times {{10}^{-2}})\, \\ & or\,q=\sqrt{10}\times {{10}^{-9}}C \\ & \therefore q=3.16\times {{10}^{-9}}C \\ \end{align} \) |
In Chapters 6 and 8 (Class XI), the notion of potential energy was introduced. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, that work gets stored as potential energy of the body. When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces.
Coulomb force between two (stationary) charges is also a conservative force. This is not surprising, since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law. Thus, like the potential energy of a mass in a gravitational field, we can define electrostatic potential energy of a charge in an electrostatic field.
Consider an electrostatic field E due to some charge configuration. First, for simplicity, consider the field E due to a charge Q placed at the origin. Now, imagine that we bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. With reference to Fig. 2.1, this will happen if Q and q are both positive or both negative. For definiteness, let us take Q, q > 0.
FIGURE 2.1 A test charge q (> 0) is moved from the point R to the point P against the repulsive force on it by the charge Q (> 0) placed at the origin.
Two remarks may be made here. First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the charge q from R to P, we apply an external force Fext just enough to counter the repulsive electric force FE (i.e, Fext= –FE).
This means there is no net force on or acceleration of the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed. In this situation, work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. If the external force is removed on reaching P, the electric force will take the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is conserved.
Thus, work done by external forces in moving a charge q from R to P is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGxb % WaaSbaaSqaaiaadkfacaWGqbaabeaakiabg2da9maapehabaGaamOr % amaaBaaaleaacaWGLbGaamiEaiaadshaaeqaaOGaaiOlaiaadsgaca % WGYbaaleaacaWGsbaabaGaamiuaaqdcqGHRiI8aaGcbaGaeyypa0Ja % eyOeI0Yaa8qCaeaacaWGgbWaaSbaaSqaaiaadweaaeqaaOGaaiOlai % aadsgacaWGYbaaleaacaWGsbaabaGaamiuaaqdcqGHRiI8aOGaaGPa % VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 % caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua % aGPaVlaaykW7caaMc8UaaGPaVlaaykW7daqadaqaaiaaikdacaGGUa % GaaGymaaGaayjkaiaawMcaaaaaaa!8D16! \begin{array}{l} {W_{RP}} = \int\limits_R^P {{F_{ext}}.dr} \\ = - \int\limits_R^P {{F_E}.dr} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {2.1} \right) \end{array}\)
This work done is against electrostatic repulsive force and gets stored as potential energy.
At every point in electric field, a particle with charge q possesses a certain electrostatic potential energy, this work done increases its potential energy by an amount equal to potential energy difference between points R and P.
Thus, potential energy difference
\(\Delta\)U = UP + UR = WRP (2.2)
( Note here that this displacement is in an opposite sense to the electric force and hence work done by electric field is negative, i.e., –WRP .)
Therefore, we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating ) charge q from one point to another for electric field of any arbitrary charge configuration.
Two important comments may be made at this stage:
(1) The right side of Eq. (2.2) depends only on the initial and final positions of the charge. It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force. The concept of the potential energy would not be meaningful if the work depended on the path. The path-independence of work done by an electrostatic field can be proved using the Coulomb’s law. We omit this proof here.
(2) Equation (2.2) defines potential energy difference in terms of the physically meaningful quantity work. Clearly, potential energy so defined is undetermined to within an additive constant.What this means is that the actual value of potential energy is not physically significant; it is only the difference of potential energy that is significant. We can always add an arbitrary constant \(\alpha\) to potential energy at every point, since this will not change the potential energy difference:
(UP + \(\alpha\) ) - (UR + \(\alpha\) ) = UP + UR
Put it differently, there is a freedom in choosing the point where potential energy is zero. A convenient choice is to have electrostatic potential energy zero at infinity. With this choice, if we take the point R at infinity, we get from Eq. (2.2)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vamaaBa % aaleaacqGHEisPcaWGqbaabeaakiabg2da9iaadwfadaWgaaWcbaGa % amiuaaqabaGccqGHsislcaWGvbWaaSbaaSqaaiabg6HiLcqabaGccq % GH9aqpcaWGvbWaaSbaaSqaaiaadcfaaeqaaaaa!4288! {W_{\infty P}} = {U_P} - {U_\infty } = {U_P}\) (2.3)
Since the point P is arbitrary, Eq. (2.3) provides us with a definition of potential energy of a charge q at any point. Potential energy of charge q at a point (in the presence of field due to any charge configuration) is the work done by the external force (equal and opposite to the electric force) in bringing the charge q from infinity to that point.
ELECTROSTATIC POTENTIAL
Consider any general static charge configuration. We define potential energy of a test charge q in terms of the work done on the charge q. This work is obviously proportional to q, since the force at any point is qE, where E is the electric field at that point due to the given charge configuration. It is, therefore, convenient to divide the work by the amount of charge q, so that the resulting quantity is independent of q. In other words, work done per unit test charge is characteristic of the electric field associated with the charge configuration. This leads to the idea of electrostatic potential V due to a given charge configuration. From Eq. (2.1), we get:
Work done by external force in bringing a unit positive charge from point R to P
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaam % OvamaaBaaaleaacaWGsbaabeaakiabgkHiTiaadAfadaWgaaWcbaGa % amOuaaqabaGcdaqadaqaaiabg2da9maalaaabaGaamyvamaaBaaale % aacaWGqbaabeaakiabgkHiTiaadwfadaWgaaWcbaGaamOuaaqabaaa % keaacaWGXbaaaaGaayjkaiaawMcaaaaa!4407! = {V_R} - {V_R}\left( { = \frac{{{U_P} - {U_R}}}{q}} \right)\) (2.4)
where VP and VR are the electrostatic potentials at P and R, respectively. Note, as before, that it is not the actual value of potential but the potential difference that is physically significant. If, as before, we choose the potential to be zero at infinity, Eq. (2.4) implies:
Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point.
In other words, the electrostatic potential (V ) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point.
The qualifying remarks made earlier regarding potential energy also apply to the definition of potential. To obtain the work done per unit test charge, we should take an infinitesimal test charge \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqgaaa!379B! \delta \)q, obtain the work done \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqgaaa!379B! \delta \)W in bringing it from infinity to the point and determine the ratio \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqgaaa!379B! \delta \)W/\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqgaaa!379B! \delta \)q. Also, the external force at every point of the path is to be equal and opposite to the electrostatic force on the test charge at that point.
FIGURE 2.2 Work done on a test charge q by the electrostatic field due to any given charge configuration is independent of the path, and depends only on its initial and final positions.
POTENTIAL DUE TO A POINT CHARGE
Consider a point charge Q at the origin (Fig. 2.3).
FIGURE 2.3 Work done in bringing a unit positive test charge from infinity to the point P, against the repulsive force of charge Q (Q > 0), is the potential at P due to the charge Q.
For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P. For Q > 0, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to the point P.
At some intermediate point \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmiuayaafa % aaaa!36D7! P'\) on the path, the electrostatic force on a unit positive charge is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGrbGaey41aqRaaGymaaqaaiaaisdacqaHapaCcqaH1oqzdaWgaaWc % baGaaGimaaqabaGcceWGYbGbauaadaahaaWcbeqaaiaaikdaaaaaaO % GabmOCayaajyaafaaaaa!41C8! \frac{{Q \times 1}}{{4\pi {\varepsilon _0}{{r'}^2}}}\hat r'\) (2.5)
where \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOCayaajy % aafaaaaa!3708! \hat r'\) is the unit vector along \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiqadc % fagaqbaaaa!37AB! OP'\). Work done against this force from \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOCayaafa % aaaa!36F9! r'\) to \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOCayaafa % Gaey4kaSIaeyiLdqKabmOCayaafaaaaa!3A45! r' + \Delta r'\) is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % 4vaiabg2da9iabgkHiTmaalaaabaGaamyuaaqaaiaaisdacqaHapaC % cqaH1oqzdaWgaaWcbaGaaGimaaqabaGcceWGYbGbauaadaahaaWcbe % qaaiaaikdaaaaaaOGaeyiLdqKabmOCayaafaaaaa!4484! \Delta W = - \frac{Q}{{4\pi {\varepsilon _0}{{r'}^2}}}\Delta r'\) (2.6)
The negative sign appears because for \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKabm % OCayaafaaaaa!3860! \Delta r'\) < 0, \(\Delta\)W is positive. Total work done (W) by the external force is obtained by integrating Eq. (2.6) from \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOCayaafa % Gaeyypa0JaeyOhIuQaaGPaVlaaykW7caWG0bGaam4BaiaaykW7caaM % c8UabmOCayaafaGaeyypa0JaamOCaaaa!4489! r' = \infty \,\,to\,\,r' = r\),
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vaiabg2 % da9iabgkHiTmaapehabaWaaSaaaeaacaWGrbaabaGaaGinaiabec8a % Wjabew7aLnaaBaaaleaacaaIWaaabeaakiqadkhagaqbamaaCaaale % qabaGaaGOmaaaaaaGccaWGKbGabmOCayaafaaaleaacqGHEisPaeaa % caWGYbaaniabgUIiYdaaaa!4770! W = - \int\limits_\infty ^r {\frac{Q}{{4\pi {\varepsilon _0}{{r'}^2}}}dr'} \) (2.7)
This, by definition is the potential at P due to the charge Q
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaabm % aabaGaamOCaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamyuaaqa % aiaaisdacqaHapaCcqaH1oqzdaWgaaWcbaGaaGimaaqabaGccaWGYb % aaaaaa!4146! V\left( r \right) = \frac{Q}{{4\pi {\varepsilon _0}r}}\) (2.8)
Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are in the same direction.] Finally, we note that Eq. (2.8) is consistent with the choice that potential at infinity be zero.
Figure (2.4) shows how the electrostatic potential ( \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOhIukaaa!3767! \infty \) 1/r ) and the electrostatic field (\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOhIukaaa!3767! \infty \) 1/r 2 ) varies with r.
FIGURE 2.4 Variation of potential V with r [in units of (Q/\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGinaiabec % 8aWjabew7aLnaaBaaaleaacaaIWaaabeaaaaa!3AFE! 4\pi {\varepsilon _0}\) ) m-1] (blue curve) and field with r [in units of (Q/\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGinaiabec % 8aWjabew7aLnaaBaaaleaacaaIWaaabeaaaaa!3AFE! 4\pi {\varepsilon _0}\) ) m-2] (black curve) for a point charge Q.
EXAMPLE 1
a. Calculate the potential at a point P due to a charge of 4 × 10–7C located 9 cm away.
b. Hence obtain the work done in bringing a charge of 2 × 10–9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought?
SOLUTION
a. \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaisdacqaHapaCcqaH1oqzdaWgaaWc % baGaaGimaaqabaaaaOWaaSaaaeaacaWGrbaabaGaamOCaaaacqGH9a % qpcaaI5aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiaaiMdaaaGc % caWGobGaamyBamaaCaaaleqabaGaaGOmaaaakiaadoeadaahaaWcbe % qaaiabgkHiTiaaikdaaaGccqGHxdaTdaWcaaqaaiaaisdacqGHxdaT % caaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaG4naaaakiaadoeaae % aacaaIWaGaaiOlaiaaicdacaaI5aGaamyBaaaacqGH9aqpcaaI0aGa % ey41aqRaaGymaiaaicdadaahaaWcbeqaaiaaisdaaaGccaWGwbaaaa!5F59! V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{r} = 9 \times {10^9}N{m^2}{C^{ - 2}} \times \frac{{4 \times {{10}^{ - 7}}C}}{{0.09m}} = 4 \times {10^4}V\)
b. \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vaiabg2 % da9iaadghacaWGwbGaeyypa0JaaGOmaiabgEna0kaaigdacaaIWaWa % aWbaaSqabeaacqGHsislcaaI5aaaaOGaam4qaiabgEna0kaaisdacq % GHxdaTcaaIXaGaaGimamaaCaaaleqabaGaaGinaaaakiaadAfacqGH % 9aqpcaaI4aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTi % aaiwdaaaGccaWGkbaaaa!51DD! W = qV = 2 \times {10^{ - 9}}C \times 4 \times {10^4}V = 8 \times {10^{ - 5}}J\)
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.
POTENTIAL DUE TO AN ELECTRIC DIPOLE
As we learnt in the last chapter, an electric dipole consists of two charges q and –q separated by a (small) distance 2a. Its total charge is zero. It is characterised by a dipole moment vector p whose magnitude is q × 2a and which points in the direction from –q to q (Fig. 2.5).
FIGURE 2.5 Quantities involved in the calculation of potential due to a dipole.
We also saw that the electric field of a dipole at a point with position vector r depends not just on the magnitude r, but also on the angle between r and p. Further, the field falls off, at large distance, not as 1/r 2 (typical of field due to a single charge) but as 1/r3. We, now, determine the electric potential due to a dipole and contrast it with the potential due to a single charge. As before, we take the origin at the centre of the dipole. Now we know that the electric field obeys the superposition principle. Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. Thus, the potential due to the dipole is the sum of potentials due to the
charges q and –q
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaisdacqaHapaCcqaH1oqzdaWgaaWc % baGaaGimaaqabaaaaOWaaeWaaeaadaWcaaqaaiaadghaaeaacaWGYb % WaaSbaaSqaaiaaigdaaeqaaaaakiabgkHiTmaalaaabaGaamyCaaqa % aiaadkhadaWgaaWcbaGaaGOmaaqabaaaaaGccaGLOaGaayzkaaaaaa!4607! V = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{q}{{{r_1}}} - \frac{q}{{{r_2}}}} \right)\) (2.9)
where r1 and r2 are the distances of the point P from q and –q, respectively.
Now, by geometry,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGYb % Waa0baaSqaaiaaigdaaeaacaaIYaaaaOGaeyypa0JaamOCamaaCaaa % leqabaGaaGOmaaaakiabgUcaRiaadggadaahaaWcbeqaaiaaikdaaa % GccqGHsislcaaIYaGaamyyaiaadkhaciGGJbGaai4BaiaacohacqaH % 4oqCaeaacaWGYbWaa0baaSqaaiaaikdaaeaacaaIYaaaaOGaeyypa0 % JaamOCamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadggadaahaaWc % beqaaiaaikdaaaGccqGHRaWkcaaIYaGaamyyaiaadkhaciGGJbGaai % 4BaiaacohacqaH4oqCcaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa % ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG % PaVlaaykW7caaMc8UaaGPaVlaaykW7daqadaqaaiaaikdacaGGUaGa % aGymaiaaicdaaiaawIcacaGLPaaaaaaa!76E3! \begin{array}{l} r_1^2 = {r^2} + {a^2} - 2ar\cos \theta \\ r_2^2 = {r^2} + {a^2} + 2ar\cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {2.10} \right) \end{array}\)
We take r much greater than a ( r >> a ) and retain terms only upto the first order in a/r
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGYb % Waa0baaSqaaiaaigdaaeaacaaIYaaaaOGaeyypa0JaamOCamaaCaaa % leqabaGaaGOmaaaakmaabmaabaGaaGymaiabgkHiTmaalaaabaGaaG % OmaiaadggaciGGJbGaai4BaiaacohacqaH4oqCaeaacaWGYbaaaiab % gUcaRmaalaaabaGaamyyamaaCaaaleqabaGaaGOmaaaaaOqaaiaadk % hadaahaaWcbeqaaiaaikdaaaaaaaGccaGLOaGaayzkaaaabaGaeyyr % IaKaamOCamaaCaaaleqabaGaaGOmaaaakmaabmaabaGaaGymaiabgk % HiTmaalaaabaGaaGOmaiaadggaciGGJbGaai4BaiaacohacqaH4oqC % aeaacaWGYbaaaaGaayjkaiaawMcaaiaaykW7caaMc8UaaGPaVlaayk % W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa % VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 % caaMc8UaaGPaVlaaykW7caaMc8+aaeWaaeaacaaIYaGaaiOlaiaaig % dacaaIXaaacaGLOaGaayzkaaaaaaa!8DF7! \begin{array}{l} r_1^2 = {r^2}\left( {1 - \frac{{2a\cos \theta }}{r} + \frac{{{a^2}}}{{{r^2}}}} \right)\\ \cong {r^2}\left( {1 - \frac{{2a\cos \theta }}{r}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {2.11} \right) \end{array}\)
Similarly,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCamaaDa % aaleaacaaIYaaabaGaaGOmaaaakiabgwKiajaadkhadaahaaWcbeqa % aiaaikdaaaGcdaqadaqaaiaaigdacqGHRaWkdaWcaaqaaiaaikdaca % WGHbGaci4yaiaac+gacaGGZbGaeqiUdehabaGaamOCaaaaaiaawIca % caGLPaaaaaa!4611! r_2^2 \cong {r^2}\left( {1 + \frac{{2a\cos \theta }}{r}} \right)\) (2.12)
Using the Binomial theorem and retaining terms upto the first order in a/r ; we obtain
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaamOCamaaBaaaleaacaaIXaaabeaaaaGccqGHfjcqdaWc % aaqaaiaaigdaaeaacaWGYbaaamaabmaabaGaaGymaiabgkHiTmaala % aabaGaaGOmaiaadggaciGGJbGaai4BaiaacohacqaH4oqCaeaacaWG % YbaaaaGaayjkaiaawMcaamaaCaaaleqabaGaeyOeI0IaaGymaiaac+ % cacaaIYaaaaOGaeyyrIa0aaSaaaeaacaaIXaaabaGaamOCaaaadaqa % daqaaiaaigdacqGHRaWkdaWcaaqaaiaadggaaeaacaWGYbaaaiGaco % gacaGGVbGaai4CaiabeI7aXbGaayjkaiaawMcaaaaa!55E0! \frac{1}{{{r_1}}} \cong \frac{1}{r}{\left( {1 - \frac{{2a\cos \theta }}{r}} \right)^{ - 1/2}} \cong \frac{1}{r}\left( {1 + \frac{a}{r}\cos \theta } \right)\) (2.13(a))
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaamOCamaaBaaaleaacaaIYaaabeaaaaGccqGHfjcqdaWc % aaqaaiaaigdaaeaacaWGYbaaamaabmaabaGaaGymaiabgkHiTmaala % aabaGaaGOmaiaadggaciGGJbGaai4BaiaacohacqaH4oqCaeaacaWG % YbaaaaGaayjkaiaawMcaamaaCaaaleqabaGaeyOeI0IaaGymaiaac+ % cacaaIYaaaaOGaeyyrIa0aaSaaaeaacaaIXaaabaGaamOCaaaadaqa % daqaaiaaigdacqGHsisldaWcaaqaaiaadggaaeaacaWGYbaaaiGaco % gacaGGVbGaai4CaiabeI7aXbGaayjkaiaawMcaaaaa!55EC! \frac{1}{{{r_2}}} \cong \frac{1}{r}{\left( {1 - \frac{{2a\cos \theta }}{r}} \right)^{ - 1/2}} \cong \frac{1}{r}\left( {1 - \frac{a}{r}\cos \theta } \right)\) (2.13(b))
Using Eqs. (2.9) and (2.13) and p = 2qa, we get
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaamyCaaqaaiaaisdacqaHapaCcqaH1oqzdaWgaaWc % baGaaGimaaqabaaaaOWaaSaaaeaacaaIYaGaamyyaiGacogacaGGVb % Gaai4CaiabeI7aXbqaaiaadkhadaahaaWcbeqaaiaaikdaaaaaaOGa % eyypa0ZaaSaaaeaacaWGWbGaci4yaiaac+gacaGGZbGaeqiUdehaba % GaaGinaiabec8aWjabew7aLnaaBaaaleaacaaIWaaabeaakiaadkha % daahaaWcbeqaaiaaikdaaaaaaaaa!539A! V = \frac{q}{{4\pi {\varepsilon _0}}}\frac{{2a\cos \theta }}{{{r^2}}} = \frac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\) (2.14)
Now,p cos\(\theta\) = p\(\hat r\)
where \(\hat r\) is the unit vector along the position vector OP.
The electric potential of a dipole is then given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaisdacqaHapaCcqaH1oqzdaWgaaWc % baGaaGimaaqabaaaaOWaaSaaaeaacaWGWbGaaiOlaiqadkhagaqcaa % qaaiaadkhadaahaaWcbeqaaiaaikdaaaaaaOGaaGPaVlaacUdacaaM % c8UaaGPaVpaabmaabaGaamOCaiabg6da+iabg6da+iaadggaaiaawI % cacaGLPaaaaaa!4D32! V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{p.\hat r}}{{{r^2}}}\,;\,\,\left( {r > > a} \right)\) (2.15)
Equation (2.15) is, as indicated, approximately true only for distances large compared to the size of the dipole, so that higher order terms in a/r are negligible. For a point dipole p at the origin, Eq. (2.15) is, however, exact.
From Eq. (2.15), potential on the dipole axis ( \(\theta\) = 0,\(\pi\) ) is given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9iabgglaXoaalaaabaGaaGymaaqaaiaaisdacqaHapaCcqaH1oqz % daWgaaWcbaGaaGimaaqabaaaaOWaaSaaaeaacaWGWbaabaGaamOCam % aaCaaaleqabaGaaGOmaaaaaaGccaaMc8oaaa!441C! V = \pm \frac{1}{{4\pi {\varepsilon _0}}}\frac{p}{{{r^2}}}\,\) (2.16)
(Positive sign for \(\theta\) = 0, negative sign for \(\theta\) = \(\pi\).) The potential in the equatorial plane (\(\theta\) = \(\pi\)/2) is zero.
The important contrasting features of electric potential of a dipole from that due to a single charge are clear from Eqs. (2.8) and (2.15):
(1) The potential due to a dipole depends not just on r but also on the angle between the position vector r and the dipole moment vector p. (It is, however, axially symmetric about p. That is, if you rotate the position vector r about p, keeping \(\theta\) fixed, the points corresponding to P on the cone so generated will have the same potential as at P.)
(2) The electric dipole potential falls off, at large distance, as 1/r 2, not as 1/r, characteristic of the potential due to a single charge. (You can refer to the Fig. 2.5 for graphs of 1/r 2 versus r and 1/r versus r, drawn there in another context.)
POTENTIAL DUE TO A SYSTEM OF CHARGES
Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…, rn relative to some origin (Fig. 2.6). The potential V1 at P due to the charge q1 is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaaIXaaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaisda % cqaHapaCcqaH1oqzdaWgaaWcbaGaaGimaaqabaaaaOWaaSaaaeaaca % WGXbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamOCamaaBaaaleaacaaI % XaGaamiuaaqabaaaaaaa!434F! {V_1} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}}}{{{r_{1P}}}}\)
FIGURE 2.6 Potential at a point due to a system of charges is the sum of potentials due to individual charges.
where r1P is the distance between q1 and P.
Similarly, the potential V2 at P due to q2 and V3 due to q3 are given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaaIYaaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaisda % cqaHapaCcqaH1oqzdaWgaaWcbaGaaGimaaqabaaaaOWaaSaaaeaaca % WGXbWaaSbaaSqaaiaaikdaaeqaaaGcbaGaamOCamaaBaaaleaacaaI % YaGaamiuaaqabaaaaOGaaiilaiaadAfadaWgaaWcbaGaaG4maaqaba % GccqGH9aqpdaWcaaqaaiaaigdaaeaacaaI0aGaeqiWdaNaeqyTdu2a % aSbaaSqaaiaaicdaaeqaaaaakmaalaaabaGaamyCamaaBaaaleaaca % aIZaaabeaaaOqaaiaadkhadaWgaaWcbaGaaG4maiaadcfaaeqaaaaa % aaa!516B! {V_2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_2}}}{{{r_{2P}}}},{V_3} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_3}}}{{{r_{3P}}}}\)
where r2P and r3P are the distances of P from charges q2 and q3, respectively; and so on for the potential due to other charges. By the
superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9iaadAfadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGwbWaaSba % aSqaaiaaikdaaeqaaOGaey4kaSIaaiOlaiaac6cacaGGUaGaaiOlai % aac6cacaGGUaGaaiOlaiabgUcaRiaadAfadaWgaaWcbaGaamOBaaqa % baaaaa!44EE! V = {V_1} + {V_2} + ....... + {V_n}\) (2.17)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIXaaabaGaaGinaiabec8aWjabew7aLnaaBaaaleaacaaI % WaaabeaaaaGcdaqadaqaamaalaaabaGaamyCamaaBaaaleaacaaIXa % aabeaaaOqaaiaadkhadaWgaaWcbaGaaGymaiaadcfaaeqaaaaakiab % gUcaRmaalaaabaGaamyCamaaBaaaleaacaaIYaaabeaaaOqaaiaadk % hadaWgaaWcbaGaaGOmaiaadcfaaeqaaaaakiabgUcaRiaac6cacaGG % UaGaaiOlaiaac6cacaGGUaGaey4kaSYaaSaaaeaacaWGXbWaaSbaaS % qaaiaad6gaaeqaaaGcbaGaamOCamaaBaaaleaacaWGUbGaamiuaaqa % baaaaaGccaGLOaGaayzkaaaaaa!5310! = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_1}}}{{{r_{1P}}}} + \frac{{{q_2}}}{{{r_{2P}}}} + ..... + \frac{{{q_n}}}{{{r_{nP}}}}} \right)\) (2.18)
If we have a continuous charge distribution characterised by a charge density \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdihaaa!37B6! \rho \) (r), we divide it, as before, into small volume elements each of size \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % ODaaaa!3858! \Delta v\) and carrying a charge \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdiNaey % iLdqKaamODaaaa!3A18! \rho \Delta v\). We then determine the potential due to each volume element and sum (strictly speaking , integrate) over all such contributions, and thus determine the potential due to the entire distribution.
We have seen in Chapter 1 that for a uniformly charged spherical shell, the electric field outside the shell is as if the entire charge is concentrated at the centre. Thus, the potential outside the shell is given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaisdacqaHapaCcqaH1oqzdaWgaaWc % baGaaGimaaqabaaaaOWaaSaaaeaacaWGXbaabaGaamOCaaaadaqada % qaaiaadkhacqGHLjYScaWGsbaacaGLOaGaayzkaaaaaa!44CE! V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{r}\left( {r \ge R} \right)\) (2.19(a))
where q is the total charge on the shell and R its radius. The electric field inside the shell is zero. This implies (Section 2.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaisdacqaHapaCcqaH1oqzdaWgaaWc % baGaaGimaaqabaaaaOWaaSaaaeaacaWGXbaabaGaamOuaaaaaaa!3F91! V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{R}\) (2.19(b))
EXAMPLE 2
Two charges 3 × 10–8 C and –2 × 10–8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
SOLUTION
Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin (Fig. 2.7).
Fig. 2.7
Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGinaiabec8aWjabew7aLnaaBaaaleaacaaIWaaabeaa % aaGcdaWadaqaamaalaaabaGaaG4maiabgEna0kaaigdacaaIWaWaaW % baaSqabeaacqGHsislcaaI4aaaaaGcbaGaamiEaiabgEna0kaaigda % caaIWaWaaWbaaSqabeaacqGHsislcaaIYaaaaaaakiabgkHiTmaala % aabaGaaGOmaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsisl % caaI4aaaaaGcbaWaaeWaaeaacaaIXaGaaGynaiabgkHiTiaadIhaai % aawIcacaGLPaaacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOe % I0IaaGOmaaaaaaaakiaawUfacaGLDbaacqGH9aqpcaaIWaaaaa!5DB1! \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{3 \times {{10}^{ - 8}}}}{{x \times {{10}^{ - 2}}}} - \frac{{2 \times {{10}^{ - 8}}}}{{\left( {15 - x} \right) \times {{10}^{ - 2}}}}} \right] = 0\)
where x is in cm. That is,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaaabaGaamiEaaaacqGHsisldaWcaaqaaiaaikdaaeaacaaIXaGa % aGynaiabgkHiTiaadIhaaaGaeyypa0JaaGimaaaa!3E9D! \frac{3}{x} - \frac{2}{{15 - x}} = 0\)
which gives x = 9 cm.
If x lies on the extended line OA, the required condition is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaaabaGaamiEaaaacqGHsisldaWcaaqaaiaaikdaaeaacaWG4bGa % eyOeI0IaaGymaiaaiwdaaaGaeyypa0JaaGimaaaa!3E9D! \frac{3}{x} - \frac{2}{{x - 15}} = 0\)
which gives x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.
EXAMPLE 3
Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively.
Figures 2.8
a. Give the signs of the potential difference VP – VQ; VB – VA.
b. Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B.
c. Give the sign of the work done by the field in moving a small positive charge from Q to P.
d. Give the sign of the work done by the external agency in moving a small negative charge from B to A.
e. Does the kinetic energy of a small negative charge increase or decrease in going from B to A?
SOLUTION
a. As \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaabaaa % aaaaaapeGaeyyhIu7aaSaaaeaacaaIXaaabaGaamOCaaaacaGGSaGa % amOvamaaBaaaleaacaWGqbaabeaakiabg6da+iaadAfadaWgaaWcba % Gaamyuaaqabaaaaa!3FAE! V \propto \frac{1}{r},{V_P} > {V_Q}\) Thus, \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaqadaqaaiaadAfadaWgaaWcbaGaamiuaaqabaGccqGHsislcaWG % wbWaaSbaaSqaaiaadgfaaeqaaaGccaGLOaGaayzkaaaaaa!3C59! \left( {{V_P} - {V_Q}} \right)\) is positive. Also VB is less negative that VA Thus VB > VA or (VB-VA)is positive.
b. A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)A > (P.E.)B and hence sign of potential energy differences is positive.
c. In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.
d. In moving a small negative charge from B to A work has to be done by the external agency. It is positive.
e. Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.
EQUIPOTENTIAL SURFACES
An equipotential surface is a surface with a constant value of potential at all points on the surface. For a single charge q, the potential is given by Eq. (2.8):
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGwbGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGinaiabec8aWjab % ew7aLnaaBaaaleaacaaIWaaabeaaaaGcdaWcaaqaaiaadghaaeaaca % WGYbaaaaaa!3FD1! V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{r}\)
This shows that V is a constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge.
Now the electric field lines for a single charge q are radial lines starting from or ending at the charge, depending on whether q is positive or negative. Clearly, the electric field at every point is normal to the equipotential surface passing through that point. This is true in general: for any charge configuration, equipotential surface through a point is normal to the electric field at that point. The proof of this statement is simple.
If the field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done. But this is in contradiction to the definition of an equipotential surface: there is no potential difference between any two points on the surface and no work is required to move a test charge on the surface. The electric field must, therefore, be normal to the equipotential surface at every point. Equipotential surfaces offer an alternative visual picture in addition to the picture of electric field lines around a charge configuration.
FIGURE 2.9 For a single charge q a. equipotential surfaces are spherical surfaces centred at the charge, and b. electric field lines are radial, starting from the charge if q > 0.
For a uniform electric field E, say, along the x -axis, the equipotential surfaces are planes normal to the x -axis, i.e., planes parallel to the y-z plane (Fig. 2.10). Equipotential surfaces for (a) a dipole and (b) two identical positive charges are shown in Fig. 2.11.
FIGURE 2.10 Equipotential surfaces for a uniform electric field
FIGURE 2.11 Some equipotential surfaces for (a) a dipole, (b) two identical positve charges.
Relation between field and potential
Consider two closely spaced equipotential surfaces A and B (Fig. 2.12) with potential values V and V + \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqaH0oazaaa!37BB! \delta \)V, where \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqaH0oazaaa!37BB! \delta \) V is the change in V in the direction of the electric field E. Let P be a point on the surface B. \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqaH0oazaaa!37BB! \delta \) l is the perpendicular distance of the surface A from P. Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is |E|\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqaH0oazaaa!37BB! \delta \) l.
FIGURE 2.12 From the potential to the field.
This work equals the potential difference VA - VB
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaqaaaaa % aaaaWdbmaaemaabaGaamyraaGaay5bSlaawIa7aiabes7aKjaadYga % cqGH9aqpcaWGwbGaeyOeI0YaaeWaaeaacaWGwbGaey4kaSIaeqiTdq % MaamOvaaGaayjkaiaawMcaaiabg2da9iabgkHiTiabes7aKjaadAfa % aeaacaWGPbGaaiOlaiaadwgacaGGUaGaaiilaiaaykW7caaMc8UaaG % PaVlaaykW7daabdaqaaiaadweaaiaawEa7caGLiWoacqGH9aqpcqGH % sisldaWcaaqaaiabes7aKjaadAfaaeaacqaH0oazcaWGSbaaaiaayk % W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa % VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 % caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl % aaykW7caaMc8+aaeWaaeaacaaIYaGaaiOlaiaaikdacaaIWaaacaGL % OaGaayzkaaaaaaa!9931! \begin{array}{l} \left| E \right|\delta l = V - \left( {V + \delta V} \right) = - \delta V\\ i.e.,\,\,\,\,\left| E \right| = - \frac{{\delta V}}{{\delta l}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {2.20} \right) \end{array}\)
Since \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqaH0oazaaa!37BB! \delta \)V is negative, \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqaH0oazaaa!37BB! \delta \)V = –\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHsisldaabdaqaaiabes7aKjaadAfaaiaawEa7caGLiWoaaaa!3CA5! \left| {\delta V} \right|\). we can rewrite Eq (2.20) as
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaabdaqaaiaadweaaiaawEa7caGLiWoacqGH9aqpcqGHsisldaWc % aaqaaiabes7aKjaadAfaaeaacqaH0oazcaWGSbaaaiabg2da9iabgU % caRmaalaaabaWaaqWaaeaacqaH0oazcaWGwbaacaGLhWUaayjcSdaa % baGaeqiTdqMaamiBaaaaaaa!4B4B! \left| E \right| = - \frac{{\delta V}}{{\delta l}} = + \frac{{\left| {\delta V} \right|}}{{\delta l}}\) (2.21)
We thus arrive at two important conclusions concerning the relation between electric field and potential:
1. Electric field is in the direction in which the potential decreases steepest.
2. Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.
POTENTIAL ENERGY OF A SYSTEM OF CHARGES
Consider first the simple case of two charges q1 and q2 with position vector r1 and r2 relative to some origin. Let us calculate the work done (externally) in building up this configuration. This means that we consider the charges q1 and q2 initially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge q1 is brought from infinity to the point r1. There is no external field against which work needs to be done, so work done in bringing q1 from infinity to r1 is zero. This charge produces a potential in space given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGwbWaaSbaaSqaaiaaigdaaeqaaOGaeyypa0ZaaSaaaeaacaaI % XaaabaGaaGinaiabec8aWjabew7aLnaaBaaaleaacaaIWaaabeaaaa % GcdaWcaaqaaiaadghadaWgaaWcbaGaaGymaaqabaaakeaacaWGYbWa % aSbaaSqaaiaaigdacaWGqbaabeaaaaaaaa!436F! {V_1} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}}}{{{r_{1P}}}}\)
where r1P is the distance of a point P in space from the location of q1. From the definition of potential, work done in bringing charge q2 from infinity to the point r2 is q2 times the potential at r2 due to q1:
work done on q2 = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaqaaiaaigdaaeaacaaI0aGaeqiWdaNaeqyTdu2aaSbaaSqa % aiaaicdaaeqaaaaakmaalaaabaGaamyCamaaBaaaleaacaaIXaaabe % aakiaadghadaWgaaWcbaGaaGOmaaqabaaakeaacaWGYbWaaSbaaSqa % aiaaigdacaaIYaaabeaaaaaaaa!426C! \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r_{12}}}}\)
Since electrostatic force is conservative, this work gets stored in the form of potential energy of the system. Thus, the potential energy of a system of two charges q1 and q2 is
U = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaqaaiaaigdaaeaacaaI0aGaeqiWdaNaeqyTdu2aaSbaaSqa % aiaaicdaaeqaaaaakmaalaaabaGaamyCamaaBaaaleaacaaIXaaabe % aakiaadghadaWgaaWcbaGaaGOmaaqabaaakeaacaWGYbWaaSbaaSqa % aiaaigdacaaIYaaabeaaaaaaaa!426C! \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r_{12}}}}\) (2.22)
Obviously, if q2 was brought first to its present location and q1 brought later, the potential energy U would be the same. More generally, the potential energy expression, Eq. (2.22), is unaltered whatever way the charges are brought to the specified locations, because of path-independence of work for electrostatic force.
Equation (2.22) is true for any sign of q1 and q2. If q1q2 > 0, potential energy is positive. This is as expected, since for like charges (q1q2 > 0), electrostatic force is repulsive and a positive amount of work is needed to be done against this force to bring the charges from infinity to a finite distance apart. For unlike charges (q1 q2 < 0), the electrostatic force is attractive. In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity. In other words, a negative amount of work is needed for the reverse path (from infinity to the present locations), so the potential energy is negative.
FIGURE 2.13 Potential energy of a system of charges q1 and q2 is directly proportional to the product of charges and inversely to the distance between them.
Equation (2.22) is easily generalised for a system of any number of point charges. Let us calculate the potential energy of a system of three charges q1, q2 and q3 located at r1, r2, r3, respectively. To bring q1 first from infinity to r1, no work is required. Next we bring q2 from infinity to r2. As before, work done in this step is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGXbWaaSbaaSqaaiaaikdaaeqaaOGaamOvamaaBaaaleaacaaI % XaaabeaakmaabmaabaGaamOCamaaBaaaleaacaaIYaaabeaaaOGaay % jkaiaawMcaaiabg2da9maalaaabaGaaGymaaqaaiaaisdacqaHapaC % cqaH1oqzdaWgaaWcbaGaaGimaaqabaaaaOWaaSaaaeaacaWGXbWaaS % baaSqaaiaaigdaaeqaaOGaamyCamaaBaaaleaacaaIYaaabeaaaOqa % aiaadkhadaWgaaWcbaGaaGymaiaaikdaaeqaaaaaaaa!4A98! {q_2}{V_1}\left( {{r_2}} \right) = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r_{12}}}}\) (2.23)
The charges q1 and q2 produce a potential, which at any point P is given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGwbWaaSbaaSqaaiaaigdacaGGSaGaaGOmaaqabaGccqGH9aqp % daWcaaqaaiaaigdaaeaacaaI0aGaeqiWdaNaeqyTdu2aaSbaaSqaai % aaicdaaeqaaaaakmaabmaabaWaaSaaaeaacaWGXbWaaSbaaSqaaiaa % igdaaeqaaaGcbaGaamOCamaaBaaaleaacaaIXaGaamiuaaqabaaaaO % Gaey4kaSYaaSaaaeaacaWGXbWaaSbaaSqaaiaaikdaaeqaaaGcbaGa % amOCamaaBaaaleaacaaIYaGaamiuaaqabaaaaaGccaGLOaGaayzkaa % aaaa!4C06! {V_{1,2}} = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_1}}}{{{r_{1P}}}} + \frac{{{q_2}}}{{{r_{2P}}}}} \right)\) (2.24)
Work done next in bringing q3 from infinity to the point r3 is q3 times V1, 2 at r3
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGXbWaaSbaaSqaaiaaiodaaeqaaOGaamOvamaaBaaaleaacaaI % XaGaaiilaiaaikdaaeqaaOWaaeWaaeaacaWGYbWaaSbaaSqaaiaaio % daaeqaaaGccaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaIXaaabaGa % aGinaiabec8aWjabew7aLnaaBaaaleaacaaIWaaabeaaaaGcdaqada % qaamaalaaabaGaamyCamaaBaaaleaacaaIXaaabeaakiaadghadaWg % aaWcbaGaaG4maaqabaaakeaacaWGYbWaaSbaaSqaaiaaigdacaaIZa % aabeaaaaGccqGHRaWkdaWcaaqaaiaadghadaWgaaWcbaGaaGOmaaqa % baGccaWGXbWaaSbaaSqaaiaaiodaaeqaaaGcbaGaamOCamaaBaaale % aacaaIYaGaaG4maaqabaaaaaGccaGLOaGaayzkaaaaaa!5504! {q_3}{V_{1,2}}\left( {{r_3}} \right) = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_1}{q_3}}}{{{r_{13}}}} + \frac{{{q_2}{q_3}}}{{{r_{23}}}}} \right)\) (2.25)
The total work done in assembling the charges at the given locations is obtained by adding the work done in different steps [Eq. (2.23) and Eq. (2.25)],
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGvbGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGinaiabec8aWjab % ew7aLnaaBaaaleaacaaIWaaabeaaaaGcdaqadaqaamaalaaabaGaam % yCamaaBaaaleaacaaIXaaabeaakiaadghadaWgaaWcbaGaaGOmaaqa % baaakeaacaWGYbWaaSbaaSqaaiaaigdacaaIYaaabeaaaaGccqGHRa % WkdaWcaaqaaiaadghadaWgaaWcbaGaaGymaaqabaGccaWGXbWaaSba % aSqaaiaaiodaaeqaaaGcbaGaamOCamaaBaaaleaacaaIXaGaaG4maa % qabaaaaOGaey4kaSYaaSaaaeaacaWGXbWaaSbaaSqaaiaaikdaaeqa % aOGaamyCamaaBaaaleaacaaIZaaabeaaaOqaaiaadkhadaWgaaWcba % GaaGOmaiaaiodaaeqaaaaaaOGaayjkaiaawMcaaaaa!54AF! U = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_1}{q_2}}}{{{r_{12}}}} + \frac{{{q_1}{q_3}}}{{{r_{13}}}} + \frac{{{q_2}{q_3}}}{{{r_{23}}}}} \right)\) (2.26)
Again, because of the conservative nature of the electrostatic force (or equivalently, the path independence of work done), the final expression for U, Eq. (2.26), is independent of the manner in which the configuration is assembled. The potential energy is characteristic of the present state of configuration, and not the way the state is achieved.
FIGURE 2.14 Potential energy of a system of three charges is given by Eq. (2.26), with the notation given in the figure.
EXAMPLE 4
Four charges are arranged at the corners of a square ABCD of side d, as shown in Fig. 2.15.(a) Find the work required to put together this arrangement. (b) A charge q0 is brought to the centre E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this?
Fig. 2.15.
SOLUTION
(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps:
(1) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.
(2) Work needed to bring –q to B when +q is at A. This is given by (charge at B) × (electrostatic potential at B due to charge +q at A)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaey % OeI0IaamyCaiabgEna0oaabmaabaWaaSaaaeaacaWGXbaabaGaaGin % aiabec8aWjabew7aLnaaBaaaleaacaaIWaaabeaakiaadsgaaaaaca % GLOaGaayzkaaGaeyypa0JaeyOeI0YaaSaaaeaacaWGXbWaaSbaaSqa % aiaaikdaaeqaaaGcbaGaaGinaiabec8aWjabew7aLnaaBaaaleaaca % aIWaaabeaakiaadsgaaaaaaa!4D66! = - q \times \left( {\frac{q}{{4\pi {\varepsilon _0}d}}} \right) = - \frac{{{q_2}}}{{4\pi {\varepsilon _0}d}}\)
(3) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at C) × (potential at C due to charges at A and B)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpcqGHRaWkcaWGXbGaey41aq7aaeWaaeaadaWcaaqaaiaadghaaeaa % caaI0aGaeqiWdaNaeqyTdu2aaSbaaSqaaiaaicdaaeqaaOGaamizaa % aacqGHRaWkdaWcaaqaaiabgkHiTiaadghadaWgaaWcbaGaaGOmaaqa % baaakeaacaaI0aGaeqiWdaNaeqyTdu2aaSbaaSqaaiaaicdaaeqaaO % GaamizaaaaaiaawIcacaGLPaaaaeaacqGH9aqpdaWcaaqaaiabgkHi % TiaadghadaWgaaWcbaGaaGOmaaqabaaakeaacaaI0aGaeqiWdaNaeq % yTdu2aaSbaaSqaaiaaicdaaeqaaOGaamizaaaadaqadaqaaiaaigda % cqGHsisldaWcaaqaaiaaigdaaeaadaGcaaqaaiaaikdaaSqabaaaaa % GccaGLOaGaayzkaaaaaaa!5C01! \begin{array}{l} = + q \times \left( {\frac{q}{{4\pi {\varepsilon _0}d}} + \frac{{ - {q_2}}}{{4\pi {\varepsilon _0}d}}} \right)\\ = \frac{{ - {q_2}}}{{4\pi {\varepsilon _0}d}}\left( {1 - \frac{1}{{\sqrt 2 }}} \right) \end{array}\)
(4) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) × (potential at D due to charges at A, B and C)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpcqGHsislcaWGXbGaey41aq7aaeWaaeaadaWcaaqaaiabgUcaRiaa % dghaaeaacaaI0aGaeqiWdaNaeqyTdu2aaSbaaSqaaiaaicdaaeqaaO % GaamizaaaacqGHRaWkdaWcaaqaaiabgkHiTiaadghaaeaacaaI0aGa % eqiWdaNaeqyTdu2aaSbaaSqaaiaaicdaaeqaaOGaamizamaakaaaba % GaaGOmaaWcbeaaaaGccqGHRaWkdaWcaaqaaiaadghaaeaacaaI0aGa % eqiWdaNaeqyTdu2aaSbaaSqaaiaaicdaaeqaaOGaamizaaaaaiaawI % cacaGLPaaaaeaacqGH9aqpdaWcaaqaaiabgkHiTiaadghadaahaaWc % beqaaiaaikdaaaaakeaacaaI0aGaeqiWdaNaeqyTdu2aaSbaaSqaai % aaicdaaeqaaOGaamizaaaadaqadaqaaiaaikdacqGHsisldaWcaaqa % aiaaigdaaeaadaGcaaqaaiaaikdaaSqabaaaaaGccaGLOaGaayzkaa % aaaaa!64C2! \begin{array}{l} = - q \times \left( {\frac{{ + q}}{{4\pi {\varepsilon _0}d}} + \frac{{ - q}}{{4\pi {\varepsilon _0}d\sqrt 2 }} + \frac{q}{{4\pi {\varepsilon _0}d}}} \right)\\ = \frac{{ - {q^2}}}{{4\pi {\varepsilon _0}d}}\left( {2 - \frac{1}{{\sqrt 2 }}} \right) \end{array}\)
Add the work done in steps (i), (ii), (iii) and (iv). The total work required is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpdaWcaaqaaiabgkHiTiaadghadaahaaWcbeqaaiaaikdaaaaakeaa % caaI0aGaeqiWdaNaeqyTdu2aaSbaaSqaaiaaicdaaeqaaOGaamizaa % aadaGadaqaamaabmaabaGaaGimaaGaayjkaiaawMcaaiabgUcaRmaa % bmaabaGaaGymaaGaayjkaiaawMcaaiabgUcaRmaabmaabaGaaGymai % abgkHiTmaalaaabaGaaGymaaqaamaakaaabaGaaGOmaaWcbeaaaaaa % kiaawIcacaGLPaaacqGHRaWkdaqadaqaaiaaikdacqGHsisldaWcaa % qaaiaaigdaaeaadaGcaaqaaiaaikdaaSqabaaaaaGccaGLOaGaayzk % aaaacaGL7bGaayzFaaaabaGaeyypa0ZaaSaaaeaacqGHsislcaWGXb % aabaGaaGinaiabec8aWjabew7aLnaaBaaaleaacaaIWaaabeaakiaa % dsgaaaWaaeWaaeaacaaI0aGaeyOeI0YaaOaaaeaacaaIYaaaleqaaa % GccaGLOaGaayzkaaaaaaa!6006! \begin{array}{l} = \frac{{ - {q^2}}}{{4\pi {\varepsilon _0}d}}\left\{ {\left( 0 \right) + \left( 1 \right) + \left( {1 - \frac{1}{{\sqrt 2 }}} \right) + \left( {2 - \frac{1}{{\sqrt 2 }}} \right)} \right\}\\ = \frac{{ - q}}{{4\pi {\varepsilon _0}d}}\left( {4 - \sqrt 2 } \right) \end{array}\)
The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
(Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)
(b) The extra work necessary to bring a charge q0 to the point E when the four charges are at A, B, C and D is q0 × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence, no work is required to bring any charge to point E.
POTENTIAL ENERGY IN AN EXTERNAL FIELD
Potential energy of a single charge
In Section 2.7, the source of the electric field was specified – the charges and their locations - and the potential energy of the system of those charges was determined. In this section, we ask a related but a distinct question. What is the potential energy of a charge q in a given field? This question was, in fact, the starting point that led us to the notion of the electrostatic potential (Sections 2.1 and 2.2). But here we address this question again to clarify in what way it is different from the discussion in Section 2.7.
The main difference is that we are now concerned with the potential energy of a charge (or charges) in an external field. The external field E is not produced by the given charge(s) whose potential energy we wish to calculate. E is produced by sources external to the given charge(s).The external sources may be known, but often they are unknown or unspecified; what is specified is the electric field E or the electrostatic potential V due to the external sources. We assume that the charge q does not significantly affect the sources producing the external field. This is true if q is very small, or the external sources are held fixed by other unspecified forces. Even if q is finite, its influence on the external sources may still be ignored in the situation when very strong sources far away at infinity produce a finite field E in the region of interest. Note again that we are interested in determining the potential energy of a given charge q (and later, a system of charges) in the external field; we are not interested in the potential energy of the sources producing the external electric field. The external electric field E and the corresponding external potential V may vary from point to point. By definition, V at a point P is the work done in bringing a unit positive charge from infinity to the point P.
(We continue to take potential at infinity to be zero.) Thus, work done in bringing a charge q from infinity to the point P in the external field is qV. This work is stored in the form of potential energy of q. If the point P has position vector r relative to some origin, we can write:
Potential energy of q at r in an external field
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaam % yCaiaadAfadaqadaqaaiaadkhaaiaawIcacaGLPaaaaaa!3B4D! = qV\left( r \right)\) (2.27)
where V(r) is the external potential at the point r.
Thus, if an electron with charge q = e = 1.6×10–19 C is accelerated by a potential difference of \(\Delta\)V = 1 volt, it would gain energy of q\(\Delta\)V = 1.6 × 10–19J. This unit of energy is defined as 1 electron volt or 1eV, i.e., 1 eV=1.6 × 10–19J. The units based on eV are most commonly used in atomic, nuclear and particle physics, (1 keV = 103eV = 1.6 × 10–16J, 1 MeV
= 106eV = 1.6 × 10–13J, 1 GeV = 109eV = 1.6 × 10–10J and 1 TeV = 1012eV
= 1.6 × 10–7J). [This has already been defined on Page 117, XI Physics Part I, Table 6.1.]
Potential energy of a system of two charges in an external field
Next, we ask: what is the potential energy of a system of two charges q1 and q2 located at r1 and r2, respectively, in an external field? First, we calculate the work done in bringing the charge q1 from infinity to r1. Work done in this step is q1 V(r1), using Eq. (2.27). Next, we consider the work done in bringing q2 to r2. In this step, work is done not only against the external field E but also against the field due to q1.
Work done on q2 against the external field
= q2 V (r2)
Work done on q2 against the field due to q1
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaWGXbWaaSbaaSqaaiaaigdaaeqaaOGaamyCamaaBaaaleaa % caaIYaaabeaaaOqaaiaaisdacqaHapaCcqaH1oqzdaWgaaWcbaGaaG % imaaqabaGccaWGYbWaaSbaaSqaaiaaigdacaaIYaaabeaaaaaaaa!4287! = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r_{12}}}}\)
where r12 is the distance between q1 and q2. We have made use of Eqs. (2.27) and (2.22). By the superposition principle for fields, we add up the work done on q2 against the two fields (E and that due to q1):
Work done in bringing q2 to r2
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaam % yCamaaBaaaleaacaaIYaaabeaakiaadAfadaqadaqaaiaadkhadaWg % aaWcbaGaaGOmaaqabaaakiaawIcacaGLPaaacqGHRaWkdaWcaaqaai % aadghadaWgaaWcbaGaaGymaaqabaGccaWGXbWaaSbaaSqaaiaaikda % aeqaaaGcbaGaaGinaiabec8aWjabew7aLnaaBaaaleaacaaIWaaabe % aakiaadkhadaWgaaWcbaGaaGymaiaaikdaaeqaaaaaaaa!499E! = {q_2}V\left( {{r_2}} \right) + \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r_{12}}}}\) (2.28)
Thus,
Potential energy of the system = the total work done in assembling the configuration
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaam % yCamaaBaaaleaacaaIXaaabeaakiaadAfadaqadaqaaiaadkhadaWg % aaWcbaGaaGymaaqabaaakiaawIcacaGLPaaacqGHRaWkcaWGXbWaaS % baaSqaaiaaikdaaeqaaOGaamOvamaabmaabaGaamOCamaaBaaaleaa % caaIYaaabeaaaOGaayjkaiaawMcaaiabgUcaRmaalaaabaGaamyCam % aaBaaaleaacaaIXaaabeaakiaadghadaWgaaWcbaGaaGOmaaqabaaa % keaacaaI0aGaeqiWdaNaeqyTdu2aaSbaaSqaaiaaicdaaeqaaOGaam % OCamaaBaaaleaacaaIXaGaaGOmaaqabaaaaaaa!50B3! = {q_1}V\left( {{r_1}} \right) + {q_2}V\left( {{r_2}} \right) + \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r_{12}}}}\) (2.29)
EXAMPLE 2.5
a. Determine the electrostatic potential energy of a system consisting of two charges 7 \(\mu\)C and –2 \(\mu\)C (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively.
b. How much work is required to separate the two charges infinitely away from each other?
c. Suppose that the same system of charges is now placed in an external electric field E = A (1/r 2); A = 9 × 105 NC–1 m2. What would the electrostatic energy of the configuration be?
SOLUTION
a. \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyvaiabg2 % da9maalaaabaGaaGymaaqaaiaaisdacqaHapaCcqaH1oqzdaWgaaWc % baGaaGimaaqabaaaaOWaaSaaaeaacaWGXbWaaSbaaSqaaiaaigdaae % qaaOGaamyCamaaBaaaleaacaaIYaaabeaaaOqaaiaadkhaaaGaeyyp % a0JaaGyoaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacaaI5aaaaO % Gaey41aq7aaSaaaeaacaaI3aGaey41aq7aaeWaaeaacqGHsislcaaI % YaaacaGLOaGaayzkaaGaey41aqRaaGymaiaaicdadaahaaWcbeqaai % abgkHiTiaaigdacaaIYaaaaaGcbaGaaGimaiaac6cacaaIXaGaaGio % aaaacqGH9aqpcqGHsislcaaIWaGaaiOlaiaaiEdacaWGkbaaaa!5F08! U = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{r} = 9 \times {10^9} \times \frac{{7 \times \left( { - 2} \right) \times {{10}^{ - 12}}}}{{0.18}} = - 0.7J\)
b. W = U2 – U1 = 0 – U = 0 – (–0.7) = 0.7 J.
c. The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGXb % WaaSbaaSqaaiaaigdaaeqaaOGaamOvamaabmaabaGaamOCamaaBaaa % leaacaaIXaaabeaaaOGaayjkaiaawMcaaiabgUcaRiaadghadaWgaa % WcbaGaaGOmaaqabaGccaWGwbWaaSbaaSqaaiaaigdaaeqaaOWaaeWa % aeaacaWGYbWaaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaGaey % ypa0JaamyqamaalaaabaGaaG4naiabeY7aTjaadoeaaeaacaaIWaGa % aiOlaiaaicdacaaI5aGaamyBaaaacqGHRaWkcaWGbbWaaSaaaeaacq % GHsislcaaIYaGaeqiVd0Maam4qaaqaaiaaicdacaGGUaGaaGimaiaa % iMdacaWGTbaaaiabgkHiTiaaicdacaGGUaGaaG4naiaadQeaaeaacq % GH9aqpcaaI3aGaaGimaiabgkHiTiaaikdacaaIWaGaeyOeI0IaaGim % aiaac6cacaaI3aGaeyypa0JaaGinaiaaiMdacaGGUaGaaG4maiaadQ % eaaaaa!6794! \begin{array}{l} {q_1}V\left( {{r_1}} \right) + {q_2}{V_1}\left( {{r_2}} \right) = A\frac{{7\mu C}}{{0.09m}} + A\frac{{ - 2\mu C}}{{0.09m}} - 0.7J\\ = 70 - 20 - 0.7 = 49.3J \end{array}\)
Potential energy of a dipole in an external field
Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform electric field E, as shown in Fig. 2.16.
FIGURE 2.16 Potential energy of a dipole in a uniform external field.
As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiXdqhaaa!37BB! \tau \) given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiXdqhaaa!37BB! \tau \) = p \(\times\) E
which will tend to rotate it (unless p is parallel or antiparallel to E). Suppose an external torque \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiXdq3aaS % baaSqaaiaadwgacaWG4bGaamiDaaqabaaaaa!3AC7! {\tau _{ext}}\) is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle \(\theta\)0 to angle \(\theta\)1 at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque will be given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGxb % Gaeyypa0Zaa8qCaeaacaWG0bWaaSbaaSqaaiaadwgacaWG4bGaamiD % aaqabaGcdaqadaqaaiabeI7aXbGaayjkaiaawMcaaiaadsgacqaH4o % qCcqGH9aqpdaWdXbqaaiaadchacaWGfbGaci4CaiaacMgacaGGUbGa % eqiUdeNaamizaiabeI7aXbWcbaGaeqiUde3aaSbaaWqaaiaaicdaae % qaaaWcbaGaeqiUde3aaSbaaWqaaiaaigdaaeqaaaqdcqGHRiI8aaWc % baGaeqiUde3aaSbaaWqaaiaaicdaaeqaaaWcbaGaeqiUde3aaSbaaW % qaaiaaigdaaeqaaaqdcqGHRiI8aaGcbaGaeyypa0JaamiCaiaadwea % daqadaqaaiGacogacaGGVbGaai4CaiabeI7aXnaaBaaaleaacaaIWa % aabeaakiabgkHiTiGacogacaGGVbGaai4CaiabeI7aXnaaBaaaleaa % caaIXaaabeaaaOGaayjkaiaawMcaaaaaaa!6B54! \begin{array}{l} W = \int\limits_{{\theta _0}}^{{\theta _1}} {{t_{ext}}\left( \theta \right)d\theta = \int\limits_{{\theta _0}}^{{\theta _1}} {pE\sin \theta d\theta } } \\ = pE\left( {\cos {\theta _0} - \cos {\theta _1}} \right) \end{array}\)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyvamaabm % aabaGaeqiUdehacaGLOaGaayzkaaGaeyypa0JaamiCaiaadweadaqa % daqaaiGacogacaGGVbGaai4CamaalaaabaGaeqiWdahabaGaaGOmaa % aacqGHsislciGGJbGaai4BaiaacohacqaH4oqCaiaawIcacaGLPaaa % cqGH9aqpcaWGWbGaamyraiGacogacaGGVbGaai4CaiabeI7aXjabg2 % da9iabgkHiTiaadchacaGGUaGaamyraaaa!54E1! U\left( \theta \right) = pE\left( {\cos \frac{\pi }{2} - \cos \theta } \right) = pE\cos \theta = - p.E\)
This expression can alternately be understood also from Eq. (2.29). We apply Eq. (2.29) to the present system of two charges +q and –q. The potential energy expression then reads
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyvayaafa % WaaeWaaeaacqaH4oqCaiaawIcacaGLPaaacqGH9aqpcaWGXbWaamWa % aeaacaWGwbWaaeWaaeaacaWGYbWaaSbaaSqaaiaaigdaaeqaaaGcca % GLOaGaayzkaaGaeyOeI0IaamOvamaabmaabaGaamOCamaaBaaaleaa % caaIYaaabeaaaOGaayjkaiaawMcaaaGaay5waiaaw2faaiabgkHiTm % aalaaabaGaamyCamaaBaaaleaacaaIYaaabeaaaOqaaiaaisdacqaH % apaCcqaH1oqzdaWgaaWcbaGaaGimaaqabaGccqGHxdaTcaaIYaGaam % yyaaaaaaa!533F! U'\left( \theta \right) = q\left[ {V\left( {{r_1}} \right) - V\left( {{r_2}} \right)} \right] - \frac{{{q_2}}}{{4\pi {\varepsilon _0} \times 2a}}\)
Here, r1 and r2 denote the position vectors of +q and –q. Now, the potential difference between positions r1 and r2 equals the work done in bringing a unit positive charge against field from r2 to r1. The displacement parallel to the force is 2a cos\(\theta\). Thus, [V(r1)–V (r2)] = –E × 2a cos\(\theta\) , We thus obtain
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyvayaafa % WaaeWaaeaacqaH4oqCaiaawIcacaGLPaaacqGH9aqpcqGHsislcaWG % WbGaamyraiGacogacaGGVbGaai4CaiabeI7aXjabgkHiTmaalaaaba % GaamyCamaaBaaaleaacaaIYaaabeaaaOqaaiaaisdacqaHapaCcqaH % 1oqzdaWgaaWcbaGaaGimaaqabaGccqGHxdaTcaaIYaGaamyyaaaacq % GH9aqpcqGHsislcaWGWbGaaiOlaiaadweacqGHsisldaWcaaqaaiaa % dghadaWgaaWcbaGaaGOmaaqabaaakeaacaaI0aGaeqiWdaNaeqyTdu % 2aaSbaaSqaaiaaicdaaeqaaOGaey41aqRaaGOmaiaadggaaaaaaa!5E1A! U'\left( \theta \right) = - pE\cos \theta - \frac{{{q_2}}}{{4\pi {\varepsilon _0} \times 2a}} = - p.E - \frac{{{q_2}}}{{4\pi {\varepsilon _0} \times 2a}}\) (2.34)
We note that \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyvayaafa % WaaeWaaeaacqaH4oqCaiaawIcacaGLPaaaaaa!3A1B! U'\left( \theta \right)\) differs from U(\(\theta\)) by a quantity which is just a constant for a given dipole. Since a constant is insignificant for potential energy, we can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32).
We can now understand why we took \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUde3aaS % baaSqaaiaaicdaaeqaaOGaeyypa0JaeqiWdaNaai4laiaaikdaaaa!3CCE! {\theta _0} = \pi /2\) In this case, the work done against the external field E in bringing +q and – q are equal and opposite and cancel out, i.e., q [V (r1) – V (r2)]=0.
ELECTROSTATICS OF CONDUCTORS
Conductors and insulators were described briefly in Chapter 1. Conductors contain mobile charge carriers. In metallic conductors, these charge carriers are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal. The free electrons form a kind of ‘gas’; they collide with each other and with the ions, and move randomly in different directions. In an external electric field, they drift against the direction of the field. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. In electrolytic conductors, the charge carriers are both positive and negative ions; but the situation in this case is more involved – the movement of the charge carriers is affected both by the external electric field as also by the so-called chemical forces (see Chapter 3). We shall restrict our discussion to metallic solid conductors. Let us note important results regarding electrostatics of conductors.
1. Inside a conductor, electrostatic field is zero
Consider a conductor, neutral or charged. There may also be an external electrostatic field. In the static situation, when there is no current inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor. This fact can be taken as the defining property of a conductor. A conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift. In the static situation, the free charges have so distributed themselves that the electric field is zero everywhere inside. Electrostatic field is zero inside a conductor.
2. At the surface of a charged conductor, electrostatic field must be normal to the surface at every point
If E were not normal to the surface, it would have some non-zero component along the surface. Free charges on the surface of the conductor would then experience force and move. In the static situation, therefore, E should have no tangential component. Thus electrostatic field at the surface of a charged conductor must be normal to the surface at every point. (For a conductor without any surface charge density, field is zero even at the surface.) See result 5.
3. The interior of a conductor can have no excess charge in the static situation
A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. When the conductor is charged, the excess charge can reside only on the surface in the static situation. This follows from the Gauss’s law. Consider any arbitrary volume element v inside a conductor. On the closed surface S bounding the volume element v, electrostatic field is zero. Thus the total electric flux through S is zero. Hence, by Gauss’s law, there is no net charge enclosed by S. But the surface S can be made as small as you like, i.e., the volume v can be made vanishingly small. This means there is no net charge at any point inside the conductor, and any excess charge must reside at the surface.
4. Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface
This follows from results 1 and 2 above. Since E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface. That is, there is no potential difference between any two points inside or on the surface of the conductor. Hence, the result. If the conductor is charged, electric field normal to the surface exists; this means potential will be different for the surface and a point just outside the surface.
In a system of conductors of arbitrary size, shape and charge configuration, each conductor is characterised by a constant value of potential, but this constant may differ from one conductor to the other.
5. Electric field at the surface of a charged conductor
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyraiabg2 % da9maalaaabaGaeq4WdmhabaGaeqyTdu2aaSbaaSqaaiaaicdaaeqa % aaaakiqad6gagaqcaaaa!3D33! E = \frac{\sigma }{{{\varepsilon _0}}}\hat n\)
COMBINATION OF CAPACITORS
We can combine several capacitors of capacitance C1, C2,…, Cn to obtain a system with some effective capacitance C. The effective capacitance depends on the way the individual capacitors are combined. Two simple possibilities are discussed below.
Capacitors in series
Figure 2.26 shows capacitors C1 and C2 combined in series.
FIGURE 2.26 Combination of two capacitors in series.
The proof clearly goes through for any number of capacitors arranged in a similar way. Equation (2.55), for n capacitors arranged in series, generalises to
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9iaadAfadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGwbWaaSba % aSqaaiaaikdaaeqaaOGaey4kaSIaaiOlaiaac6cacaGGUaGaaiOlai % aac6cacqGHRaWkcaWGwbWaaSbaaSqaaiaad6gaaeqaaOGaeyypa0Za % aSaaaeaacaWGrbaabaGaam4qamaaBaaaleaacaaIXaaabeaaaaGccq % GHRaWkdaWcaaqaaiaadgfaaeaacaWGdbWaaSbaaSqaaiaaikdaaeqa % aaaakiabgUcaRiaac6cacaGGUaGaaiOlaiabgUcaRmaalaaabaGaam % yuaaqaaiaadoeadaWgaaWcbaGaamOBaaqabaaaaaaa!5162! V = {V_1} + {V_2} + ..... + {V_n} = \frac{Q}{{{C_1}}} + \frac{Q}{{{C_2}}} + ... + \frac{Q}{{{C_n}}}\) (2.59)
Following the same steps as for the case of two capacitors, we get the general formula for effective capacitance of a series combination of n capacitors:
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaam4qaaaacqGH9aqpdaWcaaqaaiaaigdaaeaacaWGdbWa % aSbaaSqaaiaaigdaaeqaaaaakiabgUcaRmaalaaabaGaaGymaaqaai % aadoeadaWgaaWcbaGaaGOmaaqabaaaaOGaey4kaSYaaSaaaeaacaaI % XaaabaGaam4qamaaBaaaleaacaaIZaaabeaaaaGccqGHRaWkcaGGUa % GaaiOlaiaac6cacqGHRaWkdaWcaaqaaiaaigdaaeaacaWGdbWaaSba % aSqaaiaad6gaaeqaaaaaaaa!486E! \frac{1}{C} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}} + ... + \frac{1}{{{C_n}}}\)(2.60)
Capacitors in parallel
Figure 2.28 (a) shows two capacitors arranged in parallel. In this case, the same potential difference is applied across both the capacitors. But the plate charges (±Q1) on capacitor 1 and the plate charges (±Q2) on the
FIGURE 2.28 Parallel combination of (a) two capacitors, (b) n capacitors
Q1 = C1V, Q2 = C2V (2.61)
The equivalent capacitor is one with charge
Q = Q1 + Q2 (2.62)
and potential difference V.
Q = CV = C1V + C2V (2.63)
The effective capacitance C is, from Eq. (2.63),
C = C1 + C2 (2.64)
The general formula for effective capacitance C for parallel combination of n capacitors [Fig. 2.28 (b)] follows similarly
Q = Q1 + Q2 + ... + Qn (2.65)
i.e., CV = C1V + C2V + ... CnV (2.66)
which gives
C = C1 + C2 + ... Cn (2.67)
ENERGY STORED IN A CAPACITOR
A capacitor, as we have seen above, is a system of two conductors with charge Q and –Q. To determine the energy stored in this configuration, consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by bit, so that at the end, conductor 1 gets charge Q. By charge conservation, conductor 2 has charge –Q at the end (Fig 2.30 ).
FIGURE 2.30 (a) Work done in a small step of building charge on conductor 1 from \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) to \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) + \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\). (b) Total work done in charging the capacitor may be viewed as stored in the energy of electric field between the plates.
In transferring positive charge from conductor 2 to conductor 1, work will be done externally, since at any stage conductor 1 is at a higher potential than conductor 2. To calculate the total work done, we first calculate the work done in a small step involving transfer of an infinitesimal (i.e., vanishingly small) amount of charge. Consider the intermediate situation when the conductors 1 and 2 have charges \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) and
-\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) respectively. At this stage, the potential difference \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! V'\) between conductors 1 to 2 is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\)/C, where C is the capacitance of the system. Next imagine that a small charge \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\) is transferred from conductor 2 to 1. Work done in this step (\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqgaaa!379B! \delta \) W), resulting in charge \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) on conductor 1 increasing to \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\)+ \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\), is given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMaam % 4vaiabg2da9iqadAfagaqbaiabes7aKjqadgfagaqbaiabg2da9maa % laaabaGabmyuayaafaaabaGaam4qaaaacqaH0oazceWGrbGbauaaaa % a!4232! \delta W = V'\delta Q' = \frac{{Q'}}{C}\delta Q'\) (2.68)
Since \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\) can be made as small as we like, Eq. (2.68) can be written as
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMaam % 4vaiabg2da9maalaaabaGaaGymaaqaaiaaikdacaWGdbaaamaadmaa % baWaaeWaaeaaceWGrbGbauaacqGHRaWkcqaH0oazceWGrbGbauaaai % aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHsislceWGrbGb % auaadaahaaWcbeqaaiaaikdaaaaakiaawUfacaGLDbaaaaa!4747! \delta W = \frac{1}{{2C}}\left[ {{{\left( {Q' + \delta Q'} \right)}^2} - {{Q'}^2}} \right]\) (2.69)
EXAMPLE 9
A network of four 10\(\mu\)F capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.)
Fig. 2.29
SOLUTION
(a) In the given network, C1, C2 and C3 are connected in series. The effective capacitance \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4qayaafa % aaaa!36CA! C'\) of these three capacitors is given by \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGabm4qayaafaaaaiabg2da9maalaaabaGaaGymaaqaaiaa % doeadaWgaaWcbaGaaGymaaqabaaaaOGaey4kaSYaaSaaaeaacaaIXa % aabaGaam4qamaaBaaaleaacaaIYaaabeaaaaGccqGHRaWkdaWcaaqa % aiaaigdaaeaacaWGdbWaaSbaaSqaaiaaiodaaeqaaaaaaaa!41E4! \frac{1}{{C'}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)
For C1 = C2 = C3 = 10\(\mu\)F, \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4qayaafa % aaaa!36CA! C'\) = (10/3)\(\mu\)F. The network has \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4qayaafa % aaaa!36CA! C'\) and C4 connected in parallel. Thus, the equivalent capacitance C of the network is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qaiabg2 % da9iqadoeagaqbaiabgUcaRiaadoeadaWgaaWcbaGaaGinaaqabaGc % cqGH9aqpdaqadaqaamaalaaabaGaaGymaiaaicdaaeaacaaIZaaaai % abgUcaRiaaigdacaaIWaaacaGLOaGaayzkaaGaeqiVd0MaamOraiab % g2da9iaaigdacaaIZaGaaiOlaiaaiodacqaH8oqBcaWGgbaaaa!4B4D! C = C' + {C_4} = \left( {\frac{{10}}{3} + 10} \right)\mu F = 13.3\mu F\)
(b) Clearly, from the figure, the charge on each of the capacitors, C1, C2 and C3 is the same, say Q. Let the charge on C4 be \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4qayaafa % aaaa!36CA! C'\). Now, since the potential difference across AB is Q/C1, across BC is Q/C2, across CD is Q/C3 , we have
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGrbaabaGaam4qamaaBaaaleaacaaIXaaabeaaaaGccqGHRaWkdaWc % aaqaaiaadgfaaeaacaWGdbWaaSbaaSqaaiaaikdaaeqaaaaakiabgU % caRmaalaaabaGaamyuaaqaaiaadoeadaWgaaWcbaGaaG4maaqabaaa % aOGaeyypa0JaaGynaiaaicdacaaIWaGaamOvaaaa!43AE! \frac{Q}{{{C_1}}} + \frac{Q}{{{C_2}}} + \frac{Q}{{{C_3}}} = 500V\)
Also, \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % Gaai4laiaadoeadaWgaaWcbaGaaGinaaqabaaaaa!393D! Q'/{C_4}\) = 500 V.
This gives for the given value of the capacitances,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyuaiabg2 % da9iaaiwdacaaIWaGaaGimaiaadAfacqGHxdaTdaWcaaqaaiaaigda % caaIWaaabaGaaG4maaaacqaH8oqBcaWGgbGaeyypa0JaaGymaiaac6 % cacaaI3aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgkHiTiaa % iodaaaGccaWGdbGaaGPaVlaaykW7caaMc8UaaGPaVlaadggacaWGUb % GaamizaiaaykW7caaMc8UabmyuayaafaGaeyypa0JaaGynaiaaicda % caaIWaGaaGPaVlaadAfacqGHxdaTcaaIXaGaaGimaiabeY7aTjaadA % eacqGH9aqpcaaI1aGaaiOlaiaaicdacqGHxdaTcaaIXaGaaGimamaa % CaaaleqabaGaeyOeI0IaaG4maaaakiaaykW7caaMc8Uaam4qaaaa!7031! Q = 500V \times \frac{{10}}{3}\mu F = 1.7 \times {10^{ - 3}}C\,\,\,\,and\,\,Q' = 500\,V \times 10\mu F = 5.0 \times {10^{ - 3}}\,\,C\)
A capacitor, as we have seen above, is a system of two conductors with charge Q and –Q. To determine the energy stored in this configuration, consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by bit, so that at the end, conductor 1 gets charge Q. By charge conservation, conductor 2 has charge –Q the end (Fig 2.30 ).
FIGURE 2.30 (a) Work done in a small step of building charge on conductor 1 from \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) to \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) + \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\). (b) Total work done in charging the capacitor may be viewed as stored in the energy of electric field between the plates.
In transferring positive charge from conductor 2 to conductor 1, work will be done externally, since at any stage conductor 1 is at a higher potential than conductor 2. To calculate the total work done, we first calculate the work done in a small step involving transfer of an infinitesimal (i.e., vanishingly small) amount of charge. Consider the intermediate situation when the conductors 1 and 2 have charges Q and –\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) respectively. At this stage, the potential difference \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! V'\) between conductors 1 to 2 is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\)/C, where C is the capacitance of the system. Next imagine that a small charge \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\) is transferred from conductor 2 to 1. Work done in this step (\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqgaaa!379B! \delta \) W), resulting in charge \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) on conductor 1 increasing to \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\)+ \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\), is given by
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMaam % 4vaiabg2da9iqadAfagaqbaiabes7aKjqadgfagaqbaiabg2da9maa % laaabaGabmyuayaafaaabaGaam4qaaaacqaH0oazceWGrbGbauaaaa % a!4232! \delta W = V'\delta Q' = \frac{{Q'}}{C}\delta Q'\) (2.68)
Since \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\) can be made as small as we like, Eq. (2.68) can be written as
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMaam % 4vaiabg2da9maalaaabaGaaGymaaqaaiaaikdacaWGdbaaamaadmaa % baWaaeWaaeaaceWGrbGbauaacqGHRaWkcqaH0oazceWGrbGbauaaai % aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHsislceWGrbGb % auaadaahaaWcbeqaaiaaikdaaaaakiaawUfacaGLDbaaaaa!4747! \delta W = \frac{1}{{2C}}\left[ {{{\left( {Q' + \delta Q'} \right)}^2} - {{Q'}^2}} \right]\) (2.59)
Equations (2.68) and (2.69) are identical because the term of second order in \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\), i.e., \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaWaaWbaaSqabeaacaaIYaaaaaaa!3966! \delta {Q'^2}\)/2C, is negligible, since \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\) is arbitrarily small. The total work done (W ) is the sum of the small work (\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqgaaa!379B! \delta \)W ) over the very large number of steps involved in building the charge \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMabm % yuayaafaaaaa!387D! \delta Q'\) from zero to Q.
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGxb % Gaeyypa0ZaaabuaeaacqaH0oazcaWGxbaaleaacaWGZbGaamyDaiaa % d2gacaaMc8UaaGPaVlaad+gacaWG2bGaamyzaiaadkhacaaMc8UaaG % PaVlaadggacaWGSbGaamiBaiaaykW7caaMc8Uaam4CaiaadshacaWG % LbGaamiCaiaadohaaeqaniabggHiLdaakeaacqGH9aqpdaaeqbqaam % aalaaabaGaaGymaaqaaiaaikdacaWGdbaaaaWcbaGaam4Caiaadwha % caWGTbGaaGPaVlaaykW7caWGVbGaamODaiaadwgacaWGYbGaaGPaVl % aadggacaWGSbGaamiBaiaaykW7caaMc8Uaam4CaiaadshacaWGLbGa % amiCaiaadohaaeqaniabggHiLdGcdaWadaqaamaabmaabaGabmyuay % aafaGaey4kaSIaeqiTdqMabmyuayaafaaacaGLOaGaayzkaaWaaWba % aSqabeaacaaIYaaaaOGaeyOeI0IabmyuayaafaWaaWbaaSqabeaaca % aIYaaaaaGccaGLBbGaayzxaaGaaGPaVlaaykW7caaMc8UaaGPaVlaa % cIcacaaIYaGaaiOlaiaaiEdacaaIWaGaaiykaaqaaiabg2da9maala % aabaGaaGymaaqaaiaaikdacaWGdbaaamaadmaabaWaaiWaaeaacqaH % 0oazceWGrbGbauaadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIWa % aacaGL7bGaayzFaaGaey4kaSYaaiWaaeaadaqadaqaaiaaikdacqaH % 0oazceWGrbGbauaaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaa % GccqGHsislcqaH0oazceWGrbGbauaacaaIYaaacaGL7bGaayzFaaGa % ey4kaSYaaiWaaeaadaqadaqaaiaaiodacqaH0oazceWGrbGbauaaai % aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHsisldaqadaqa % aiaaikdacqaH0oazceWGrbGbauaaaiaawIcacaGLPaaadaahaaWcbe % qaaiaaikdaaaaakiaawUhacaGL9baacqGHRaWkcaGGUaGaaiOlaiaa % c6cacaGGUaGaey4kaSYaaiWaaeaacaWGrbWaaWbaaSqabeaacaaIYa % aaaOGaeyOeI0YaaeWaaeaacaWGrbGaeyOeI0IaeqiTdqMabmyuayaa % faWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaaacaGL7bGaay % zFaaaacaGLBbGaayzxaaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 % caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caGGOaGaaGOmai % aac6cacaaI3aGaaGymaiaacMcaaeaacqGH9aqpdaWcaaqaaiaaigda % aeaacaaIYaGaam4qaaaadaWadaqaaiaadgfadaahaaWcbeqaaiaaik % daaaGccqGHsislcaaIWaaacaGLBbGaayzxaaGaeyypa0ZaaSaaaeaa % caWGrbGaaGOmaaqaaiaaikdacaWGdbaaaiaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua % aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca % aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa % ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG % PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM % c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk % W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa % VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 % caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua % aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca % aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa % ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG % PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM % c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk % W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa % VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % UaaiikaiaaikdacaGGUaGaaG4naiaaikdacaGGPaaaaaa!CB18! \begin{array}{l} W = \sum\limits_{sum\,\,over\,\,all\,\,steps} {\delta W} \\ = \sum\limits_{sum\,\,over\,all\,\,steps} {\frac{1}{{2C}}} \left[ {{{\left( {Q' + \delta Q'} \right)}^2} - {{Q'}^2}} \right]\,\,\,\,(2.70)\\ = \frac{1}{{2C}}\left[ {\left\{ {\delta {{Q'}^2} - 0} \right\} + \left\{ {{{\left( {2\delta Q'} \right)}^2} - \delta Q'2} \right\} + \left\{ {{{\left( {3\delta Q'} \right)}^2} - {{\left( {2\delta Q'} \right)}^2}} \right\} + .... + \left\{ {{Q^2} - \left( {Q - \delta {{Q'}^2}} \right)} \right\}} \right]\,\,\,\,\,\,\,\,\,\,\,(2.71)\\ = \frac{1}{{2C}}\left[ {{Q^2} - 0} \right] = \frac{{Q2}}{{2C}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2.72) \end{array}\)
The same result can be obtained directly from Eq. (2.68) by integration
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vaiabg2 % da9maalaaabaGaamyuamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikda % caWGdbaaaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaGaam4qai % aadAfadaahaaWcbeqaaiaaikdaaaGccqGH9aqpdaWcaaqaaiaaigda % aeaacaaIYaaaaiaadgfacaWGwbaaaa!4496! W = \frac{{{Q^2}}}{{2C}} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\) (2.73)
This is not surprising since integration is nothing but summation of a large number of small terms.
We can write the final result, Eq. (2.72) in different ways
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIXaaabaGaaGOmaaaadaWcaaqaaiaadgfadaahaaWcbeqa % aiaaikdaaaaakeaacaWGdbaaaiabg2da9maalaaabaWaaiWaaeaaca % WGbbGaeq4WdmhacaGL7bGaayzFaaWaaWbaaSqabeaacaaIYaaaaaGc % baGaaGOmaaaacqGHxdaTdaWcaaqaaiaadsgaaeaacqaH1oqzdaWgaa % WcbaGaaGimaaqabaGccaWGbbaaaaaa!4910! = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{{{{\left\{ {A\sigma } \right\}}^2}}}{2} \times \frac{d}{{{\varepsilon _0}A}}\) (2.74)
The surface charge density \(\sigma\) is related to the electric field E between the plates,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyraiabg2 % da9maalaaabaGaeq4WdmhabaGaeqyTdu2aaSbaaSqaaiaaicdaaeqa % aaaaaaa!3C26! E = \frac{\sigma }{{{\varepsilon _0}}}\)
From Eqs. (2.74) and (2.75) , we get Energy stored in the capacitor
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyvaiabg2 % da9maabmaabaGaaGymaiaac+cacaaIYaaacaGLOaGaayzkaaGaeqyT % du2aaSbaaSqaaiaaicdaaeqaaOGaamyramaaCaaaleqabaGaaGOmaa % aakiabgEna0kaadgeacaWGKbaaaa!43A3! U = \left( {1/2} \right){\varepsilon _0}{E^2} \times Ad\) (2.76)
Note that Ad is the volume of the region between the plates (where electric field alone exists). If we define energy density as energy stored per unit volume of space, Eq (2.76) shows that
Energy density of electric field,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDaiabg2 % da9maabmaabaGaaGymaiaac+cacaaIYaaacaGLOaGaayzkaaGaeqyT % du2aaSbaaSqaaiaaicdaaeqaaOGaamyramaaBaaaleaacaaIYaaabe % aaaaa!3FF2! u = \left( {1/2} \right){\varepsilon _0}{E_2}\) (2.77)
Though we derived Eq. (2.77) for the case of a parallel plate capacitor, the result on energy density of an electric field is, in fact very general and holds true for electric field due to any configuration of charges.
EXAMPLE 10
(a) A 900 pF capacitor is charged by 100 V battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor?
(b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system?
Figure 2.31
SOLUTION
(a)The charge on the capacitor is
Q = CV = 900 × \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaic % dadaahaaWcbeqaaiabgkHiTiaaigdacaaIYaaaaaaa!39FC! {10^{ - 12}}\) F × 100 V = 9 × \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaic % dadaahaaWcbeqaaiabgkHiTiaaigdacaaIYaaaaaaa!39FC! {10^{ - 8}}\) C
The energy stored by the capacitor is
= (1/2) CV2= (1/2) QV
= (1/2) × 9 × \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaic % dadaahaaWcbeqaaiabgkHiTiaaigdacaaIYaaaaaaa!39FC! {10^{ - 8}}\)C × 100 V = 4.5 × \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiaaic % dadaahaaWcbeqaaiabgkHiTiaaigdacaaIYaaaaaaa!39FC! {10^{ - 16}}\) J
b. In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! V'\). The charge on each capacitor is then \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) = C\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! V'\). By charge conservation, \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! Q'\) = Q/2. This implies \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyuayaafa % aaaa!36D8! V'\) = V/2. The total energy of the system is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiabgE % na0oaalaaabaGaaGymaaqaaiaaikdaaaGabmyuayaafaGabmOvayaa % faGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGinaaaacaWGrbGaamOvai % abg2da9iaaikdacaGGUaGaaGOmaiaaiwdacqGHxdaTcaaIXaGaaGim % amaaCaaaleqabaGaeyOeI0IaaGOnaaaakiaadQeaaaa!4A87! 2 \times \frac{1}{2}Q'V' = \frac{1}{4}QV = 2.25 \times {10^{ - 6}}J\)
Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone?
There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.