Chapter 2

Electrostatics

2.1 Coulomb’s law

Coulomb’s law1 (1785) (Cavendish 1772: unpublished )

The force exerted on q by q′

~F = kqq′

R2R , (2.1)

~R = ~r − ~r′ (2.2)

R =~R

R2, R = |~R| =

√~R · ~R (2.3)

q′ q

O

~r~r′

~R

2 Dimension、system of units、constant of proportionality k :

Dimension: Every physical quantity has a dimension, made up of a com-

bination of of M =mass, L =length and T =time. We denote the dimension

of A as [A].

Examples： velocity [v] = L/T、acceleration [a] = L/T 2、force [F ] = [ma] =

ML/T 2、energy [E] = [mv2] = ML2/T 2, etc.

Unit: Unit is the standard by which one measures a quantity having a

definite dimension.

1Columbus in italian. (The real name is Colombo.)

1

Examples: Unit of length L : m, cm, etc.

MKS system of units: In this system one meaures (L,M, T ) by (m,Kg, s)

Remark: In the case of E & M, there is more freedom in the choice of units

compared to the case of mechanics.

Dimensions of k and q: From the Coulomb’s law, one has

[kqq′] = [kq2] = [FR2] =ML3

T 2. (2.4)

From this relation we can only fix the dimension and the unit of the com-

bination kq2. Those of individual k and q are not uniquely determined. In

the following, we will use the so-called MKSA system .

Introduce independent unit for current

In the MKS system, first we define the unit of electric current as “Ampere”

and then fix the unit of charge called “Coulomb”. Concretely, this is done in

the following way:

• Prepare two parallel currents 1m apart and let the same strength of

currents go through as in the figure. Then if the Lorentz force2 exerted

on the 1m segment is 2× 10−7N , we define the amount of the current

as 1A (Ampere)3

d = 1m

l = 1m

~F1A 1A

• Unit of electric charge C(Coulomb) is defined as

C = A · s (2.5)

2We will explain the Lorentz force in detail later.3The general formula is

F = km2I1I2

dl, km =

k

c2= 10−7

2

• In this case the unit of k becomes [k] = Nm2/C2. As for the actual

value of k, experiments tell us that

k = 10`7c2 = 8.988× 109 (2.6)

c = speed of light = 2.998× 108 m/s (2.7)

The reason why the speed of light appears here will be clarified when

we deal with the electromagnetic wave.

Further we denote k defined this way as

k =1

4πε0

(2.8)

ε0 = dielectric constant of the vacuum (2.9)

To really understand the meaning of this constant, we must learn the

physics of dielectric material. This is an advanced topic, which we will

not be covered in this course.

Exercise 2.1 Actually the“Coulomb” is a tremendously large unit. (The

amount discharged in a lightening is on the order of a few Coulombs )

When two bodies with mass 1 Kg and charge 1C each are placed L meters

apart, the electrostatic force exerted on one of them turned out to be equal

to the gravitational force exereted on the surface of the Earth. What is the

value of L ?

Exercise 2.2 What is the ratio of the gravitational force and the Coulomb

force for a proton ? The mass, the the charge of the proton and the Newton’s

gravitational constant are given by

mp = 1.672× 10−27 kg , ep = 1.602× 10−19 C (2.10)

G = 6.67× 10−11 Nm2/Kg2 (2.11)

3

2.2 Concept of electric field and superposi-

tion principle: Coulomb’s law for contin-

uous distribution of charges

Consider the situation where a charge q is receiving Coulomb forces due to

other charges. Take the point of view that the charges first create an electric

field at the position of q and then it exerts the force on q. This is expressed

as

~F (~r) = q ~E(~r) . (2.12)

Then since we know that the forces can be superposed, we must conclude

that the electric fields can also be superposed. This is called the superpo-

sition principle for electric fieds. If we apply this to the fields created

by each individual charge, we obtain the formula

q

qN

q2

q1

~E

~E(~r) =1

4πε0

N∑i=1

qi

R3i

~Ri (2.13)

~Ri ≡ ~r − ~ri (2.14)

2 Dimension and unit of E :

From F = qE, we find that the unit of E is [E] = N/C

Using E = (1/4πε0)q/R2, the dimension of E must be

4

[E] =1

[ε0]

[Q

L2

](2.15)

This formula is extremely useful in checking the answer for all the problems

for computing the electric field. If E is not of this form, one must have made

a mistake !

2 Electric field due to continuous distribution of charges :

Since the charges are carried by electrons and protons, classically the distri-

bution of charges is strictly speaking discrete. However, for a macroscopic

body with an enormous number (∼ 6 × 1023) of atoms it is possible and

preferable to regard the distribution as continuous.

ρ(~r) = electric charge density (2.16)

ρ(~r)dV = amount of charge in the volume dV

in the vicinity of ~r (2.17)

dV = dxdydz = d3r (2.18)

From this we see that to deal with a continuous distribution, all we have

to do is to make the following replacements in the formulas for the discrete

distribution

qi ⇒ ρ(~r′)d3r′ (2.19)∑

i

⇒∫

(2.20)

In this way we easily obtain the formula

~E(~r) =1

4πε0

∫ρ(~r′)

|~r − ~r′|3 (~r − ~r′)d3r′ (2.21)

5

2.3 Simple applications of Coulomb’s law

Example 1. Electric field due to uniform distribution of charges

on a plane

Consider the situation where charges are uniformly distributed on the y-z

plane with the area density = σ. (Area density = charges per unit area).

From the uniformity, without loss of generality we can choose the position

at which to measure the electric field to be on the x axis.

First we compute the electric field created by the charges in the infinitesimal

area dy′dz′, namely σdy′dz′, located at ~r′. (See the figure in the next page.)

2 Symmetry and dimensional analysis :

It is always recommended to guess the answer.

For this purpose, consideration of the symmetry of the problem and the

dimensional analysis is of utmost importance, as we shall see.

From the symmetry, the only non-vanishing component is Ex perpendicular

to the plane.

The dimension of σ is [σ] = Q/L2. This is precisely the dimension of ε0E

Therefore we should have

Ex = cσ

ε0

(2.22)

where c is dimensionless. Since x is the only quantity having the dimension

of length, c can only be a constant (one cannot form a ratio of x with

another length). Consequently, Ex cannot depend on x! Thus, all we have

to compute is the value of the constant c.

2 Actual computation :

6

Parametrize the positions as

~r′ = (0, y′, z′) , ~r = (x, 0, 0) ,

q qq ~r − ~r′ = (x,−y′,−z′) , (~r − ~r′)2 = x2 + y′2 + z′2 (2.23)

~r

z

y

σdy′dz′

~r′

x

Thus from the basic formula for the electric field,

Ex

Ey

Ez

=

σ

4πε0

∫dy′dz′

1

(x2 + y′2 + z′2)3/2

x−y′

−z′

The components with y′ and z′ multiplied vanish since they are integrals

over odd functions from −∞ to ∞. (This was obvious from the symmetry

consideration.) Therefore we are left with one integral to perform:

Ey = Ez = 0

Ex =σx

4πε0

∫dy′dz′

1

(x2 + y′2 + z′2)3/2

2 Idea of Scaling :

To compute this type of integral with ease, it is extremely useful to use the

scaling consideration and extract the dependence on x. Since x is the

only quantity having the dimension of length, we can express y′ and z′ in

units of x. Then the x dependence is trivially extracted and the rest is just

a dimensionless integral giving a number.

y′ = xu , z′ = xv∫dy′dz′

1

(x2 + y′2 + z′2)3/2=

C

x

where C =

∫dudv

(1 + u2 + v2)3/2= number

7

C can be computed easily with the use of the polar coordinate:

u

v

a

da

(u, v)

φ

u = a cos φ , v = a sin φ

C =

∫ ∞

0

2πada

(1 + a2)3/2

= π

∫ ∞

0

db

(1 + b)3/2(b ≡ a2)

= 2π (2.24)

In this way, we obtain the simple result

Ex(~r) =σx

4πε0

· 2π

x=

σ

2ε0

= constant (2.25)

Ey = Ez = 0 (2.26)

We will later rederive this result using a beautiful and powerful formula

called the Gauss’ law.

Exercise 2.3 We wish to find the electric field produced by the charge

uniformly distributed along the z-axis with the line density (charge per unit

length) τ .

(1) First guess the form of the answer from the symmetry consideration and

the dimensional analysis.

(2) Find the electric field by actual computation

8

Er(r)r

Example 2. Electric field produced by the uniform charge distribu-

tion on the surface of a sphere

Radius of the sphere = a. Total amout of charge = Q.

Then the area density is σ = Q/(4πa2).

From the symmetry we only have the normal component En only 。From the previous consideration, the form of En must be

En =1

ε0

Q

L2

where L2 = r2, a2, ar,etc. Noticing that the dimensionless ratio a/r is ava-

ialbe, in general it can be of the form

En =1

4πε0

Q

r2f(a/r)

where f(x) is an arbitrary function. However, since the sphere appears point-

like when looked at from infinity ( r → ∞), the function must have the

property f(0) = 1.

The direct calculation using the Coulomb’s law is rather complicated and we

will not discuss it in this course. However, the result is astonishingly simple:

En(~r) =

Q

4πε0r2 r > a

0 r < a(2.27)

9

(discontinuous at r = a)

For r > a, the answer is the same as for the case where a point

charge Q is located at the origin. On the other hand, inside the

sphere, the field cancels out completely.

This feature will be very easily deduced from the powerful Gauss’ law, to be

explained in the next section.

2.4 Gauss’ law and Maxwell’s first equation

As we have seen, the Coulomb’s law is applicable to general distribution of

charges but its direct application often requires rather involved calculations.

Also it is expressed as a law at a distance. In order to capture the essence of

the Coulomb’s law in the form of the Maxwell’s first equation for the electric

field, we shall now develop a new perspective.

2 Basic picture :

When constructing his theory of E & M, Maxwell made essential use of the

analogy with the fluid dynamics. One such viewpoint is to regard the

charge as the source of the fluid and focus on the conservation of the

fluid as it flows across a closed surface. More specifically, we can derive

the extremely powerful Gauss’ law by considering the following situation:

q

q

2.4.1 Gauss’ law

2 Case where the charge is inside a closed surface:

10

We define an infinitesimal electric flux as the product of the infinitesimal

area dS and the normal component of the electric field En = ~E · n at that

point. The total flux is then given by

Φe =

∫EndS =

∫~E · ndS . (2.28)

q

dΩ

~E

En

n

r

r2dΩ

Now consider a unit sphere around the charge q, and

define the solid angle dΩ as the infinitesimal area cut

out from this sphere by the cone as in the figure. (Note

that the ordinary angle is measured by the length of the

arc.)

θ

θ radian1

From this definition we get

∫dΩ = 4π

= surface area of a unit sphere

dS

r2dΩ

θ

Also, from the figure it is easy to see that

r2dΩ = dS cos θ → dS =r2dΩ

cos θ(2.29)

Since En = E cos θ ( E ≡ | ~E|), we can write

Φe =

∫E(~r)r2dΩ (2.30)

Now apply the Coulomb’s law. We get

Φe =

∫q

4πε0r2r2dΩ =

q

4πε0

∫dΩ

=q

4πε0

4π =q

ε0

q qq∫

~E · ndS =q

ε0

(2.31)

11

This says that the total flux is proportional to the charge inside. Note that

for this conclusion the inverse square law E ∼ 1/r2 is of crucial im-

portance.

2 Case where the charge is outside the closed surface :

dS1

~E(1)

~E(2)

dS2

q

Remembering that n is defined to be the outward normal, the flux is com-

puted as

∫EndS =

∫E(1)

n dS1 −∫

E(2)n dS2

= 0 ← integrals over the same solid angle

For the general distribution of charges, we can apply the superposition principle

and obtain the following result：

Gauss’ law= Integral form of the Maxwell’s first equation

∫

∂V

~E · ndS =1

ε0

∑i

qi =1

ε0

(total charge inside V )

=1

ε0

∫

V

ρ(~r)d3r

∂V = the boundary of V

12

2.4.2 Calculation of the electric field using Gauss’ law

Not only is the Gauss’ law beautiful in form and easy to understand phys-

ically, it can be an extremely powerful tool for computing the electric field

itself when the system has an appropriate symmetry.

Condition:

Symmetry of the system allows one to take the closed surface ∂V

in such a way that En on it is constant. In such a case, En can be

taken outside the integral and can be computed.

Specifically,

∫

∂V

~E · ndS = En

∫

∂V

dS = EnS =Q

ε0

q qq En =Q

ε0S(2.32)

Example 1. Electric field due to the uniform charge on the surface

of a sphere

We already said that we can compute the electric field for this case using the

Coulomb’s law. However, the computation can be enourmously simplified if

we use the Gauss’ law. From symmetry, only the radial component can be

non-vanishing and moreover it depends only on r not on the angle. Denoting

this as E(r), Gauss’ law tells us

∫

S

EndS = E(r)4πr2

=

Q/ε0 if r > a0 if r < a

q qq E(r) =

Q

4πε0r2 if r > a

0 if r < a

Qa

r

~E(~r)

Exercise 2.4 Find the electric field produced by the uniformly charged

sphere ( including the interior). In particular what is the field inside ?

13

Example 2. Electric field produced by the uniformly distributed

charge along an infinitely long straight line

Let the line-density be τ . In the right figure, E points

along the radial direction by symmetry and therefore

there is no contributions from the upper and the lower

surfaces. Therefore we only need to consider the integral

over the side of the cylinder. Gauss’ law then dictates

ε0

∫EndS = ε0E(r)2πrl = lτ

q qq E(r) =1

2πε0

τ

r(2.33)

電荷 τ l

rEr

l

So the electric field decreases only as 1/r. The intuitive reason for

this behavior is as follows. Since the system is uniform in the z direction,

this direction can be ignored and effectively the problem becomes that of

computing the field due to a point charge at the origin on a plane. ( Indeed

the form of the answer does not contain the length l along z direction.) Thus

the integral for the Gauss’ law gives us the circumference 2πr instead of the

area. This shows that due to the inherently two-dimensional nature, the

electric flux can diverge only on a plane and hence the field does not die

away as quickly as in 3 dimensions.

Exercise 2.5 Using the Gauss’ law, find the electric field produced by

the charge distributed uniformly on the y-z plane.

Exercise 2.6 Consider the situation where the electric charge is uni-

formly distributed, with area density σ, over the side of an infinitely long

cyliner of radius R. Compute the electric field produced (inside and outside

the cylinder) using the Gauss’ law.

14

2.4.3 Gauss’ theorem

Gauss’ law gives a relation between an area integral and a volume integra.

We will now derive the differential form of the Maxwell’s first equation

by extracting a law which holds locally at every point. What allows us to

do this is the Gauss’ theorem, a mathematical identity. (Later, we will

encouter a similar theorem called “Stokes’ theorem”. ) The essence of the

theorem is the conversion of the area integral into a volume integral.

2 Gauss’ theorem :

Let ~E(~x) be a vector field, V be a finite 3-dimensional region and ∂V be its

boundary. Then, the following identity holds:

∫

∂V

~E · ndS =

∫

V

~∇ · ~~E d3x (2.34)

~∇ · ~~E =∂Ex

∂x+

∂Ey

∂y+

∂Ez

∂z

=∑

i

∂iEi

Proof：

(x, y, z)

Ex(x + ∆x, y′, z′)

Ex(x, y′, z′)

∆z

∆y

∆x

S1

S2

Decompose the region into small rectangular parallelepiped’s

Focus on the component in the x direction. The flux going out in the x

15

direction from this parallelopiped is

Φx =

∫

S′1

Ex(x + ∆x, y′, z′)dy′dz′

−∫

S1

Ex(x, y′, z′)dy′dz′

Expanding y′, z′ around y, z and keeping only the largest contribution

Φx ' [Ex(x + ∆x, y, z)− Ex(x, y, z)] ∆y∆z

=∂Ex

∂x(x, y, z)∆x∆y∆z (2.35)

Similarly, we add the flux going out in the y, z directions. Then we get, for

this parallelepipied,∫

~E · ndS =

(∂Ex

∂x+

∂Ey

∂y+

∂Ez

∂z

)∆V (2.36)

Thus the theorem is proved for this small parallelepiped.

Now we put together the contributions from these infinitesimal paral-

lelpiped to recover the entire region. Then, except for the contrigution

of the surface of the entire region, all the contributions on the left hand

side cancel with each other. Thus the Gauss’ theorem holds for arbitrary V .

//

2.4.4 Maxwell’s first equation: differential form

We now apply the Gauss’ theorem to the LHS of the Gauss’ law. This gives∫

∂V

~E · ndS =

∫

V

~∇ · ~~Ed3x =1

ε0

∫

V

ρd3x (2.37)

Since this equality is valid for arbitrary V , the integrand must be equal.

In this way, we obtain the Maxwell’s first equation:

~∇ · ~E =ρ

ε0

(2.38)

16

Remember that this law came from the Coulomb’s law. Conversely, we

will show, after introducing the concept of electrostatic potential, that the

Coulomb’s law can be obtained as the solution of this equation regarded as

the differential equation for ~E.

Coulomb

Gauss Maxwell 1st eq

2.5 Potential energy and the electric poten-

tial

2.5.1 Reviews of the “work” and the potential energy

2 work and conservative force :

Consider the electric Lorentz force ~F = q ~E acting on a charge q. Then the

work done by this force in bringing the charge from P0 to P along a path is

given by

W =

∫ P

P0

~F · d~x =

∫ t

t0

~F (~x(t)) · d~x

dtdt

(2.39)

If this work does not depend on the path, ~F is said to

be a conservative force. In such a case, by fixing the

starting point P0 once and for all W becomes a single-

valued function of the other end point P .

P0

P~F

d~x

This quantity is written as −U(P ), where U(P ) is called the potential en-

17

ergy . If we denote the position of P as ~x, we have thus

U(~x) =

∫ ~x

(−~F ) · d~x (2.40)

Now we want to solve ~F in terms of U . To do this, take the time derivative

of both sides. The one for the LHS is

dU

dt=

∂U

∂x

dx

dt+

∂U

∂y

dy

dt+

∂U

∂z

dz

dt

= ~∇U · d~x

dt(2.41)

On the other hand the one for the RHS is

d

dt

∫ t

(−~F (~x(t))) · d~x

dtdt = −~F · d~x

dt(2.42)

Since the path of the integration is arbitrary, d~x/dt is an arbitrary function.

With this in minde, compare (2.41) with (2.42). We must then have

~F (~x) = −~∇U(~x) (2.43)

Conversely, when ~F (~x) can be expresed in this form, W does not indeed

depend on the path. The work needed to move a body from ~x0 to ~x against

the conservative force ~F is given by

∫ ~x

~x0

(−~F ) · d~x = U(~x)− U(~x0) (2.44)

This amount of energy is stored as the potential energy.

A simple example: Gravitational potential energy of a body at height h as

measured from the surface of the Earth is mg × h = mgh.

2.5.2 Potential energy and electrostatic potential dueto Coulomb force

The electric field produced at ~x by a charge q located at ~x′ is

~E(~x) =1

4πε0

q

|~x− ~x′|3 (~x− ~x′) (2.45)

18

This can be interpreted as the force exerted on a unit charge at ~x. The

fact that this force is conservative can be easily checked by the calculation

∂

∂xi

1

|~x− ~x′| =∂

∂xi

[(~x− ~x′)2]−1/2 = −1

2[(~x− ~x′)2]−3/2 ∂

∂xi

∑j

(xj − x′j)2

= −1

2[(~x− ~x′)2]−3/2 × 2(xi − x′i) = − xi − x′i

|~x− ~x′|3 (2.46)

So we have the formula

~∇ 1

|~x− ~x′| = − ~x− ~x′

|~x− ~x′|3 (2.47)

This formula will be important later.

Comparing this with the expression for ~E, we immediately see that

~E(~x) = −~∇φ(~x) (2.48)

φ(~x) =1

4πε0

q

|~x− ~x′| + const (2.49)

φ(~x) is called the electrostatic potential. When it is multiplied by a

charge, it acquires the dimension of energy. The constant in the above ex-

pression can be set to zero by defining φ(~x = ∞) = 0. Then, we can write

φ(~x) =1

4πε0

q

|~x− ~x′|

= −∫ ~x

∞~E(~y) · d~y (2.50)

19

This expresses the work done on a unit charge by moving it from ∞ to ~x

against the electric field ~E. Electrostatic potential has the following impor-

tant properties:

1. Superposition of fields −→ superposition (addition) of the potential

2. The electric field is a vector field with 3 components. In contrast, the

electrostatic potential is a “scalar” having only one component and

hence is easier to deal with. Thus it is advantageous to compute the

electrostatic potential first and then compute ~E from ~E = −~∇φ. This

is what we will do from now on.

2 electrostatic potential due to general distribution of charges :

From the superposition principle we have

φ(~x) =1

4πε0

∑i

qi

|~x− ~xi| (2.51)

φ(~x) =1

4πε0

∫ρ(~x′)d3x′

|~x− ~x′| (2.52)

Exercise 2.7 (1) Compute the electrostatic potential due to the charges

uniformly distributed, with line-density τ , along an infinitely long straight

20

line (say z-axis). Take the zero of the potential to be at the point r = a,

where r is the perpendicular distance from the line.

Remark：To avoid a divergence (infinity) in the intermediate steps, take the

length of the straight line to be finite (for example 2l) and then take the limit

l →∞ at the end of the calculation.

(2) Compute the electric field from the electrostatic potential and confirm

that the answer is the same as we got before.

2 Equipotential surface and the direction of the electric field :

If we know the equipotential surface, we can immedi-

ately learn the directions of the electric field by a simple

geometrical consideration.

equipotential surface ⇐⇒ φ(~x) = C = const

q qq 0 = φ(~x + ∆~x)− φ(~x)

= ∆~x · ~∇φ(~x)

⇒ ∆~x · ~E(~x) = 0

(2.53)

∆~x

~x ~x + ∆~x

φ = const

Thus, the electric field is perpendicular to the equipotential surface.

The equipotential surface (or line) corresponds to the isobaric surface (or

line) in the weather chart, and ~E corresponds to the wind velocity vector.

~Eφ = const

21

2.5.3 The equation satisfied by the electrostatic poten-tial: The Poisson equation

As we have seen, one can compute the electric field by first obtaining the

electrostatic potential and then differentiating it. Now if we combine the two

equations we already know, i.e.

~∇ · ~~E =ρ

ε0

(2.54)

~E = −~∇φ (2.55)

we can eliminate the electric field and obtain an equation for the potential φ

~∇ · ~~E = −~∇ · ~∇φ

= −∇2φ =ρ

ε0

(2.56)

This is the Poisson equation

∇2φ = − ρ

ε0

(2.57)

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2= Laplacian (2.58)

In particular, when ρ = 0, it is called the Laplace equation.

2.5.4 General solution of the Poisson equation

Since we can make use of our knowledge of the electrostatic potential due

to the Coulomb force, it is not so difficult to obtain the general solution of

the Poisson equation. Suppose we find, by some means, one solution of the

Poisson equation (usually called a special solution) φ0(x). If we denote the

general solution by φ(x) we have

∇2φ = − ρ

ε0

∇2φ0 = − ρ

ε0

22

We then find that the difference satisfies the Laplace equation and hence

∇2(φ− φ0) = 0

q qq φ(~x) = φ0(~x) + f(~x) (2.59)

f(x) = general solution of the Laplace equation

(2.60)

When an appropriate boundary condition is specified, the solution of the

Laplace equation can be obtained, for simple cases, by methods to be de-

scribed later.

Since the electrostatic potential due to the Coulomb’s force

must be a solution of the Poisson equation the following theorem holds:

Theorem

General solution of the Poisson equation φ(x)

φ(~x) =1

4πε0

∫d3x′

ρ(~x′)

|~x− ~x′| + f(~x) (2.61)

∇2f(~x) = 0 (2.62)

The function f(x) depends on the specific situation of the problem (to be

elaborated later).

2 Check:

Let us check that the Coulomb potential actually satisfies the Poisson equa-

23

tion. If we apply ∇2 and compute naively, we find

~∇ 1

|~x− ~x′| = − ~x− ~x′

|~x− ~x′|3 (2.63)

q qq ~∇ · ~∇ 1

|~x− ~x′| =−~∇ · (~x− ~x′)

|~x− ~x′|3 − (~x− ~x′) · ~∇ 1

|~x− ~x′|3

= − 3

|~x− ~x′|3 − (~x− ~x′) · (−3)(~x− ~x′)

|~x− ~x′|5= 0! (2.64)

So it appears that one cannot account for the source term of the Poisson

equaiton. Although very subtle, there is actually an implicit assumption

made in the calculation above. Since the denominator involves the factor

|~x − ~x′| which vanishes for ~x′ = ~x, we have assumed ~x 6= ~x′. However, since

we have to integrate over ~x′ over the entire space to get φ(x), there must be

an instance where ~x′ coincides with ~x. At that point the calculation above

does not apply.

Therefore we have to consider carefully how we should perform the inte-

gral over ~x′ in the small neighborhood of ~x.

So consider a small sphere around ~x of radius ε and try to perform an

integral inside this region. But we wish to avoid integrating over the point

~x = ~x′. How can we achieve such a thing ? There is a clever trick. Recall the

Gauss’ theorem which converts between the volume integral and

the surface integral. Since the surface of the sphere never touches the

center, we can avoid the problem by replacing the integral inside the sphere

by an integral over its surface.

What we want to compute is

∇2φ(~x) =1

4πε0

∫

|~x′−~x|≤ε

d3x′ρ(~x′)∇2 1

|~x′ − ~x|(2.65)

=1

4πε0

∫

|~x′−~x|≤ε

d3x′ρ(~x′)~∇ · ~∇ 1

|~x− ~x′|(2.66)

ε

~x

~x′n

If we take ε to be sufficiently small, we can approximate ρ(~x′) ' ρ(~x). Also,

24

we can write ~∇· ~∇(1/|~x− ~x′|) = ~∇′ · ~∇′(1/|~x− ~x′|). By using the formula we

derived before (which is completely valid since ~x′ 6= ~x here) i.e. ~∇′(1/|~x −~x′|) = (~x − ~x′)/|~x − ~x′|3, we can compute the integral in the following way

using the Gauss’ theorem:

∇2φ(~x) =ρ(~x)

4πε0

∫

|~x′−~x|≤ε

d3x′~∇′ ·(

~x− ~x′

|~x− ~x′|3

)

=ρ(~x)

4πε0

∫

|~x′−~x|=ε

ε2dΩn · (~x− ~x′)

ε3← Gauss’ theorem

= −ρ(~x)

4πε0

∫dΩ = −ρ(~x)

ε0

(2.67)

Thus, we do recover the source term and check that the Coulomb potential

is indeed a solution of the Poisson bracket.

Exercise 2.8 Suppose the electrostatic potential is of the so-called Yukawa

form

φ(r) =e−kr

r,

(k > 0 , r =

√x2 + y2 + z2

)

(i) Find the electric field ~E(~x).

(ii) Using the Gauss’ law, find the total amount of charge inside the sphere

of radius R, including the origin.

(iii) Find the charge inside an infinitesimal neighborhood of the origin. Also

find the total charge present in the entire space.

(iv) Find the charge density outside the origin .

(v) Find the total amount of charge in the entire space excluding the origin.

（Use the formula∫∞

0drre−kr = 1/k2 if needed）Compare the result with that

of (iii) and discuss the consistency.

25

2.6 The electrostatic potential and the elec-

tric field produced an electric dipole

The superposition of the electric field and the potential is one of the cru-

cial principles of E & M. It in turn means that one can decompose the

general electric field and the potential into a sum of some “basic

configurations of charges”. What then should be considered as such ba-

sic configurations ? Let us begin with some intuitively plausible answer and

then justify it more logically.

2 1. A point charge at the origin:

This should obviously be fundamental. The electrostatic potential it pro-

duces is

φ0(~r) =1

4πε0

q

r(2.68)

2 2. Electric dipole :

Consider a pair of charges of equal magnitude with opposite sign closely

located to each other near the origin.

∆r = d cos θ

r

r + ∆rθ

q

−q

d

If we assume d ¿ r、hence ∆r ¿ r, the electrostatic potential is given by

φ1(r) =1

4πε0

(q

r− q

r + ∆r

)

=1

4πε0

q∆r

r(r + ∆r)

' 1

4πε0

qd cos θ

r2(2.69)

Now if we take the limit d → 0 with p ≡ qd held fixed the system becomes

the so-called an electric dipole. Its potential is given by

26

φ1(~r) =1

4πε0

p cos θ

r2

=1

4πε0

~p · rr2

∼ 1

r2

Remarks： Here θ is the angle betwee ~p and ~r. ( Not the angle from the

y-axis.)

In the second expression, ~p is a vector of magnitude p in the direction from

−q to q and is called the electric dipole moment (vector).

2 Electric field produced by an electric dipole:

To get the profile of the electric field, we should first draw the equipotential

surface (↔ r2 ∝ cos θ) and then draw the electric field perpendicular to this

surface. Since there is an obvious axial symmetry, it suffices to consider the

projection to the z-y plane.

z

y

θ = 0 line

θ = π line

~p

cos θ > 0

cos θ < 0

Precise expression of the electric field becomes

~E1(~r) =1

4πε0

1

r3(3(~p · r)r − ~p) (2.70)

Exercise 2.9 Compute the field ~E1 from the potential φ1.

27

2.7 System of conductors and the charge dis-

tribution, the electrostatic potential and

the electric field produced by them

Matter can be classified according to its electrical conductivity roughly into

two categories

• Electric conductor: Example metal（conduction electrons）、Elec-

trolytic solution（positive and negative ions)

• Insulator, or dielectric media (no freely moving charge carriers

inside)

Electrostatiscs of conductors is quite important practically as well as theo-

retically.

2.7.1 Static properties of a conductor

Static = no flow of electric current

Characteristic properties

1. No electric field in a conductor：← If there is, then a current must

flow and it is not static.

2. No charge density inside a conductor：← If there is a charge

distribution, then from the Gauss’ law it produces an electric field.

3. Therefore charge and electric field can exist only onthe surface

of a conductor .

Moreover, in order to avoid a current flow, the electric field must

be perpendicular to the surface. According to Gauss’ law, its

magnitude is

En =σ

ε0

(2.71)

28

En

dS

EndS = σε0dS

Since the electric field is produce only on oneside (i.e. outside of the

conductor), its magnitude is twice as large compared to the case where

it exists on both sides.

The charges distribute themselves so that ~E = 0 inside the conductor.

Put it another way, the distribution is such that the electrostatic

potential is constant inside and on the surface.

4. When on places a conductor in an external electric field, even if the

conductor does not have an overall charge, some charge distribution get

induced on the surface. This phenomenon is called the electrostatic

induction.

⊕+ + +

− − −=

+ + +

− − −~E = 0

We can summarize the above three properties in one figure

En = σε0

~E = 0, ρ = 0φ = const

φ = const

The essential characteristic is “conductor=equipotential body”

29

2.7.2 How to compute the electrostatic field around aconductor

Special feature of a conductor:

So far, we have been computing the electrostatic potential φ(~r) and the

electric field from a given charge density ρ(~r). However, in the case involving

a conductor, we have to devise a new method since the charge distribution

on the surface of the conductor is not apriori known.

2 How to set up the problem :

The electrostatic field around a conductor is determined from the following

principle.

1. We know the answer inside the conductor, namely

ρ = ~E = −~∇φ = 0

→ φ(~r) = c = const.

2. On the boundary of the conductor the potential is constant, called the

Dirichlet boundary condition

φ = c = const. (2.72)

3. We then must solve the Poisson equation for the potential in the outside

region with this boundary condition

−∇2φ =ρ

ε0

(2.73)

(ρ denotes the charge density outside)

In summary, what we have to do is to solve the so-called boundary value

problem restricted to the outer region. In this regard, we can forget

about the fact that we have a conductor in the inner region.

30

Now recall that we already have the general solution of the Poisson equation

(2.74):

φ(~x) =1

4πε0

∫d3x′

ρ(~x′)

|~x− ~x′| + f(~x) (2.74)

∇2f(~x) = 0 (2.75)

The problem here is how to determine the fucntion f(~x). As we empha-

sized already, the crucial requirement is that φ(~x) = c on the boundary. So

we must choose f(~x) in such a way that this is realized. Moreover we must

make sure that f(~x) satisfies the Laplace equation. Apparently this requires

the knowledge of the general solution of the Laplace equation, which is in

general a difficult problem.

Fortunately, however, when the shape of the conductor is simple enough,

we can easily find the solution by an ingeneous method called Kelvin’s

method of images without invoking the detailed knowledge of the general

solution of the Laplace equation.

2.7.3 Kelvin’s method of images

Let us explain this method by way of examples.

Example 1.

　Consider the situation where a charge q is placed

at a certain distance from the infinite plane, say

y-z plane, which forms the surface of a conduc-

tor and is grounded, as in the figure. We wish

to find the electrostatic potential at an arbitrary

point P (x, y, z). Here grounding means that the

potential is set equal to the one for the earth, which

can be regarded as an “infinitely large conductor”

with φ = 0.

y-z plane P

q−q

rr′

φ = 0

image charge

grounded

Clearly, one obvious solution of the Poisson equation in the outside region is

the Coulomb potential given by

φ0(P ) =1

4πε0

q

r

31

However, clearly it does not satisfy the boundary condition that φ = 0 on

the surface of the conductor, i.e. the y-z plane. To remedy this situation,

we place a virtual so-called image charge −q as in the figure. Then the

potential created by this charge is

φ1 =1

4πε0

−q

r′(2.76)

Adding this to the Coulomb potential, we obtain

φ =1

4πε0

q

(1

r− 1

r′

)

Evidently, this satisfies the condition that φ = 0 on the boundary.

Why are we allowed to put such an additional contribution ?

Note that the additional potential (2.76) is of Coulombic in form and,

since the virtual charge is placed inside the conductor, r0 never vanishes

outside. In such a situation、as we checked before, it satisfies the Laplace

equation. So we have obtained the appropriate solution f(~x) = φ1(~x) of the

Laplace equation by the trick of representing it as the potential produced by

a virtual image charge.

2 Charge distribution on the surface of the conductor :

Having obtained the electrostatic potential in the outside region, we can

easily compute the charge distribution on the surface of the conductor σ(y, z).

We already derived the relevant formula, namely Ex = σ/ε0. So all we have

to do is to compute Ex = −∂φext/∂x. Let us coordinatize the problem as

follows

R2 ≡ y2 + z2

r =√

(x− d)2 + R2

r′ =√

(x + d)2 + R2

φ =q

4πε0

(1√

(x− d)2 + R2− 1√

(x + d)2 + R2

)(for x ≥ 0)

φ = 0 (for x < 0)

32

Performing the differetiation,

Ex = − ∂

∂xφext

= − q

4πε0

( −x + d

((x− d)2 + R2)3/2+

x + d

((x + d)2 + R2)3/2

)

Setting x = 0, we get the electric field on the surface as

Ex(0, y, z) = − q

4πε0

2d

(d2 + R2)3/2

=σ(y, z)

ε0

Thus the surface charge density takes the form

σ(y, z) =−q

2π

d

(d2 + R2)3/2

Further, let us find the total charge induced on the surface:

Qind =

∫dy dz σ(y, z)

= 2π

(−qd

2π

) ∫ ∞

0

RdR

(d2 + R2)3/2

= −qd1

d

∫ ∞

0

udu

(u2 + 1)3/2

︸ ︷︷ ︸1

← R = ud

= −q (2.77)

Thus we find that the total induced charge is precisely equal to that of the

image charge.

Exercise 2.10 Derive this simple fact by using the Gauss’ law.

Exercise 2.11 Consider the configuration where a charge is placed at

(b, 0, a) in the empty region surrounded by the surfaces of a conductor as in

the figure, namely the planes given by z = 0, x ≥ 0 and x = 0, z ≥ 0, as in

the figure.

33

b

ax

zq

1. Applying the method of images, find the electrostatic potential at an

arbitrary point (x, y, z) inside the empty region x ≥ 0, z ≥ 0.

2. When a = b, find the charge density induced on the surface of the con-

ductor. Also find the total induced charge. Use the following formulas

as needed.∫

dx

(x2 + c2)3/2=

1

c2

x√x2 + c2∫

dx

x2 + c2=

1

ctan−1 x

c

2.8 Concept of the energy of the electric field

In this subsection we compute the electrostatic potential energy of the basic

charge distributions and learn that they can be interpreted as the energy

stored in the electric field itself.

2.8.1 Electrostatic energy

As we already mentioned, the product of the charge and the electrostatic

potential has the meaning of the potential energy. Below we compute the

concrete form for simple charge distributions.

34

1. Energy of the system of two point charges:

U =1

4πε0

q1q2

r12

(2.78)

2. The case of three charges

U =1

4πε0

(q1q2

r12

+q2q3

r23

+q3q1

r31

)

=1

4πε0

1

2

∑

i6=j

qiqj

rij

(2.79)

Here the factor of 1/2 corrects the double-counting

of qiqj/rij and qjqi/rji.

r12

r12

r23

r31

q1

q1

q2

q2

q3

3. The case of n charges： Generalizing the case of 3 charges, we imme-

diately get

U =1

4πε0

1

2

∑

i 6=j

qiqj

rij

(2.80)

4. Expression in terms of the electrostatic potential produced by qj (j 6=i) at the position of qi ： Since the electrostatic potential is

φi =1

4πε0

∑

j 6=i

qj

rij

U above can be reexpressed as

U =1

2

n∑i=1

qiφi (2.81)

5. The case of continuous distribution of charges：

qi −→ ρ(~x)d3x

φi −→ φ(~x)∑

i

−→∫

d3x

35

This leads to

U =1

2

∫d3xρ(~x)φ(~x) (2.82)

6. When the charges are on the surface of conductors：Since φ =const on each conductor, the expression

for the electrostatic energy simplifies to

U =1

2

∑i

∫

Vi

d3x ρ(~x)φi

=1

2

∑φi

∫

Vi

d3x ρ(~x)

=1

2

∑Qiφi (2.83)

Here Vi = volume of the i-th conductor

Qi = total charge on the i-th conductor

This result shows that we can effectively think of

each conductor as a charged particle。

Q1

Q2

Q3

φ2

φ3

φ1

V1

V2

V3

36

2.8.2 Electrostatic energy of simple systems

1. Conducting sphere： From Gauss’ law, the electric

field outside is exactly the same as for a point particle

at the center with total charge Q. Therefore

electrostatic potential φ =1

4πε0

Q

a

Therefore U =1

2Qφ =

1

8πε0

Q2

a

2. Uniformly charged sphere： Denote the constant

charge density as ρ:

ρ =Q

43πa3

= const.

First, from the Gauss’ law, the electric field can be com-

puted as

Q

a

Q

a

Outside (r ≥ a) En(r) =1

4πε0

Q

r2=

ρa3

3ε0

1

r2

Inside (r ≤ a) En(r) =ρ

4πε0

(4/3)πr3

r2=

ρr

3ε0

(2.84)

From these result, the electrostatic potential is obtained as

φ(r) = −∫ r

∞En(r′)dr′ = −

∫ a

∞

ρa3

3ε0

dr

r2−

∫ r

a

ρr

3ε0

dr

=ρa3

3ε0a− ρ

6ε0

(r2 − a2) =ρ

6ε0

(3a2 − r2) (2.85)

a

φ

r0

ρa2

3ε0

ρa2

2ε0

37

The total energy is

U =1

2

∫ρφd3x =

ρ2

12ε0

∫ a

0

(3a2 − r2)4πr2dr

=4πρ2a5

15ε0

(2.86)

Dimensional analysis:

φ ∼ EL ∼ Q

ε0L(2.87)

U ∼ Qφ ∼ Q2

ε0L(2.88)

On the other hand ρ2 ∼(

Q

L3

)2

=Q2

L6

q qq U ∼ 1

ε0

ρ2L5 (2.89)

Comment：Very roughly one can regard a nuleus as a uniformly charged

sphere where ρ is constant and independent of the radius. Then

Mass M ∝ a3

U ∝ a5

q qq U ∝ M5/3 (2.90)

This shows that the electrostatic energy increases as M5/3. This ex-

plains the fact that the nucleus of heavy element is easy to undergo

fission.

38

2.8.3 Interpretation of the electrostatic energy as theenergy of the electric field

Let us try to express the electrostatic energy U directly in terms of the

electric field ~E:

U =1

2

∫d3xρφ

Substitute ρ = ε0~∇ · ~E

Then U =ε0

2

∫d3xφ~∇ · ~E

=ε0

2

∫d3x

(~∇ · ( ~Eφ)− ~E · ~∇φ

)

=ε0

2

∫

∞dSEnφ +

ε0

2

∫d3x~E · ~E

In the first integral, En and φ behave at infinity at most like

En ∼ 1

r2, φ ∼ 1

r

Therefore this interal vanishes. Therefore

U =ε0

2

∫d3x| ~E|2

u(~x) = Energy density of the electric field =ε0

2| ~E|2

2.9 electrostatic potential of a system of con-

ductors and the concept of electric capac-

ity

As we stressed, a conductor can be characterized as an equipotential body.

We now point out that, due to the superposition principle, there is a simple

39

relation between the electrostatic potential of a conductor and the charge

on the conductor.

2 Case of a single conductor :

Let us derive a general relation between the electrostatic potential φ and the

charge Q of a conductor.

For this purpose we recall how φ is determined. When

there is no true charge outside, the properties of φ for

the conductor are

(i) In the space outside ∇2φ = 0

(ii) Boundary condition at ∞ (convention) φ = 0

(iii) Boundary on the surface φ = φc = const

φC

These conditions determine φ uniquely. From such φ, we can compute the

electric field En on the surface and this in turn will dictate the total charge

on the surface in the following way:

En = n · (−~∇φ) (2.91)

Q = ε0

∫dSEn (2.92)

The crucial point is that all these relations are linear. This is due to

the superposition principle of electric field and the electrostatic potential.

Specifically this means that if φ is a solution of the Laplace equation, then

λφ, where λ is an arbitrary constant, is also a solution. In this case, the

boundary condition on the surface is scaled by λ. Then clearly the

charge on the conductor is scaled by the same factor λ. That is

φ −→ λφ

Q −→ λQ

Evidently this means that φ and Q must be proportional to each other. This

is exprssed as

40

Q = Cφc (2.93)

The constant of proportionality C is called the electric capacity of the

conductor.

The unit of C： [C] = Coulomb/Volt = F (Farad). Here, Volt = [E/Q] =

Joule/Coulomb. Just like the unit of charge Coulomb, Farad is actually too

large for practical use. What are commonly used are

µF = 10−6F micro Farad

pF = 10−12F pico Farad

2 How C is determined :

The electric capacity C is determined by the size and the shape of the

conductor. To actually compute C, we should either

(i) compute φc for a given value of Q,

(ii) or instead, solve the Laplace equation with a prescribed φc as the bound-

ary condition and then compute Q.

In this way on can find the relation between Q and φc and hence C.

2 Example: :

Let us compute the electric capacity of a spherical conductor with radius

a = 1m. If the total charge on the conductor is Q, the electrostatic potential

and the capacity are given by

φ =1

4πε0

Q

a

q qq C = 4πε0a =107

c2= 1.1× 10−10F = 110 pF

(2.94)

2 System of two conductors =Condenser :

41

We can generalize the concept of the electric capacity to a system of more

than one conductors.

For simplicity, consider a system made of two conductors. Endow charges Q1

and Q2 to conductors 1 and 2 respectively. Then the electrostatic potential

produced outside is, according to the superposition priniciple, the sum

of the potential due to Q1 and that due to Q2.

Q1 Q2

φ2

φ1

For the case Q2 = 0, φ1, φ2 must be proportional to Q1, which is the only

charge. Therefore

φ1 = D11Q1 , φ2 = D21Q1

Similarly, when Q1 is zero, we have

φ1 = D12Q2 , φ2 = D22Q2

Then the potential for the general case with both charges non-vanishing is

obtained by the superposition of these two results, i.e.

φ1 = D11Q1 + D12Q2

φ2 = D21Q1 + D22Q2

(2.95)

We can summarize the result as

φi =∑

j

DijQj (2.96)

42

The coefficients Dij are sometimes called the coefficients of electrostatic

potential . Solving these equations for Qi, we get

Qi =∑

j

Cijφj (2.97)

Cij are called the coefficients of capacity4. Once Dij or Cij are obtained

we can compute φi from Qi or vice versa.

2 Condenser and its capacity :

In particular, when the two conductors carry

charges related as Q1 = Q = −Q2, the system is

called a condenser. In this case, φ1 and φ2 are

both proportional to Q and hence the potential

difference V = φ1 − φ2 is also proportional

to Q. Therefore one has the well-known relation

Q −Q

φ2

φ1

Q = CV (2.98)

C is called the capacity of the condenser. Note that the potential dif-

ference and hence the capacity of a condenser do not depend on what

conventinal value we assign to the electrostatic potential at ∞.

Exercise 2.12 Express the capacity of a condenser C in terms of the

coefficients of electrostatic potential Dij.4In some cases, the diagonal elements Cii are called the capacities while the off-diagonal

elements Cij , i 6= j are called the coefficients of induction.

43

2 Examples of condensers :

Example １: Parallel plate condenser

Q

−Q

~E d

Area = S

From the figure

σ = Q/S = Surface charge density = finite

E = σ/ε0

V = Ed = σd/ε0 =d

ε0SQ

Therefore we obtain

C = ε0S

d∼ ε0

area

length(2.99)

Dimensional analysis：

Q = Cφc ∼ CQ

ε0L

q qq C ∼ ε0L ∼ ε0L2

L(2.100)

So the smaller the distance between the plates the smaller the potential

difference and the larger the capacity. (But if we take the distance too small,

there occurs an electric discharge ! )

Example ２: Concentric spherical condenser

44

−Q

Qa

b

From Gauss’ law, there is no electric field outside since the total charge

enclosed is Q + (−Q) = 0. This clearly shows that the electric field is non-

zero only in the intermediate region. The field and the potential difference

there is

E(r) =1

4πε0

Q

r2

V =

∫ a

b

(−E(r))dr =Q

4πε0

(1

a− 1

b

)=

Q

4πε0

b− a

ab

Therefore the capacity is

C =4πε0ab

b− a(2.101)

Again, apart from 1/ε0, the dimension is area/distance.

Exercise 2.13 Compute the coefficients of capacity Cij for the concen-

tric spherical condenser as above and show that they have the following

properties:

Cij = Cji i 6= j

Cii ≥ 0

Cij ≤ 0

2.10 Force acting on conductors

2 Example: The force acting on the parallel plate condenser :

45

Consider a parallel plate condenser as in the fig-

ure. Since the two plates are charged oppositely,

Coulomb force acts to attract them to each other.

What is the magnitude of the force ?

One way is to honestly compute the Coulomb force

exerted on each small part of the plates and add

them up. This computation is rather cumbersome.

There is , however, a clever and simpler method of

wide applicability. It is the method of virtual

work invented by D’Alembert in 1743.

Q

−Q

x~Fext

~F~E

If we let the Coulomb force act freely, of course the two plates get pulled

toward each other. To keep the plates still as in the figure, we must exert

an external force ~Fext which is opposite in direction and exactly the same in

magnitude as the Coulomb force, i.e. ~Fext = −~F , so that they balance.

Now imagine that we increase the external force infinitesimally by δFext

and move the plates a small distance δx away from each other. The work5

done by the external force in this process is (Fext + δFext)δx ' Fextδx and

the potential energy of the condenser must increase by this amount. This is

expressed by

Fextδx = −Fδx = δU (2.102)

Dividing both sides by δx we get

F = −δU

δx(2.103)

Note that this relation is exactly the same as the one between the conservative

force and the potential energy.

Let us apply this formula to the system above and compute the force.

5Since we are “imagining” this ideal process to take place, it is called the “virtualwork”.

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We get

U =1

2

∑i

Qiφi =1

2Q(φ1 − φ2) =

1

2QV

V = Ex =σ

ε0

x , σ =Q

S

q qq U =1

2QEx

q qq δU =1

2QEδx

q qq F = −1

2QE = − Q2

2ε0S(2.104)

The negative sign means that the direction of the force is such that the

distance between the plates decreases, i.e. attractive.

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