Sem5 emphysic 217 electromagnet

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Chapter 2. Electrostatics

2.1. The Electrostatic Field

To calculate the force exerted by some electric charges, q1, q2, q3, ... (the source charges) on

another charge Q (the test charge) we can use the principle of superposition. This principle

states that the interaction between any two charges is completely unaffected by the presence of

other charges. The force exerted on Q by q1, q2, and q3 (see Figure 2.1) is therefore equal to the

vector sum of the force F 1 exerted by q1 on Q, the force F 2 exerted by q2 on Q, and the force F 3exerted by q3 on Q.

q3

q1

q2

Q

F2

F3

F1Ftot

Figure 2.1. Superposition of forces.

The force exerted by a charged particle on another charged particle depends on their

separation distance, on their velocities and on their accelerations. In this Chapter we will

consider the special case in which the source charges are stationary.

The electric field produced by stationary source charges is called and electrostatic field.

The electric field at a particular point is a vector whose magnitude is proportional to the total

force acting on a test charge located at that point, and whose direction is equal to the direction of

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the force acting on a positive test charge. The electric field E , generated by a collection of

source charges, is defined as

E =F

Q

where F is the total electric force exerted by the source charges on the test charge Q. It is

assumed that the test charge Q is small and therefore does not change the distribution of the

source charges. The total force exerted by the source charges on the test charge is equal to

F = F 1 + F 2 + F 3 + ... =1

4pe0

q1Q

r12 ˆ r 1 +

q2Q

r22 ˆ r 2 +

q3Q

r32 ˆ r 3 + ...

Ê Ë Á

ˆ ¯ ˜ =

Q

4pe0

qi

ri2 ˆ r i

i =1

n

Â

The electric field generated by the source charges is thus equal to

E =F

Q=

1

4pe0

qi

ri2 ˆ r i

i =1

n

Â

In most applications the source charges are not discrete, but are distributed continuously over

some region. The following three different distributions will be used in this course:

1. line charge l: the charge per unit length.

2. surface charge s: the charge per unit area.

3. volume charge r: the charge per unit volume.

To calculate the electric field at a point P generated by these charge distributions we have to

replace the summation over the discrete charges with an integration over the continuous charge

distribution:

1. for a line charge: E P ( ) =1

4pe0

ˆ r

r 2 ldlLineÚ

2. for a surface charge: E P ( ) =1

4pe0

ˆ r

r 2 sdaSurface

Ú

3. for a volume charge: E P ( ) =1

4pe0

ˆ r

r 2 rdtVolume

Ú

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Here ˆ r is the unit vector from a segment of the charge distribution to the point P at which we

are evaluating the electric field, and r is the distance between this segment and point P .

Example: Problem 2.2

a) Find the electric field (magnitude and direction) a distance z above the midpoint between two

equal charges q a distance d apart. Check that your result is consistent with what you would

expect when z » d.

b) Repeat part a), only this time make he right-hand charge -q instead of +q.

z

d/2d/2

P

a)

Er El

Etot

z

d/2d/2

P

b)

El

Er

Etot

Figure 2.2. Problem 2.2

a) Figure 2.2a shows that the x components of the electric fields generated by the two point

charges cancel. The total electric field at P is equal to the sum of the z components of the

electric fields generated by the two point charges:

E P ( ) = 21

4pe0

q14

d2 + z 2Ê Ë

ˆ ¯

z

1

4d2 + z 2

ˆ z =1

2pe0

qz

14

d2 + z 2Ê Ë

ˆ ¯

3/ 2 ˆ z

When z » d this equation becomes approximately equal to

E P ( ) @1

2pe0

q

z 2 ˆ z =1

4pe0

2q

z 2 ˆ z

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which is the Coulomb field generated by a point charge with charge 2q.

b) For the electric fields generated by the point charges of the charge distribution shown in

Figure 2.2b the z components cancel. The net electric field is therefore equal to

E P ( ) = 21

4pe0

q14

d2 + z 2Ê Ë

ˆ ¯

d2

1

4d2 + z 2

ˆ x =1

4pe0

qd

14

d2 + z 2Ê Ë

ˆ ¯

3/ 2 ˆ x


Find the electric field a distance z above the center of a circular loop of radius r which carries

a uniform line charge l.

r

z

P

dEldEr

2dEz

Figure 2.3. Problem 2.5.

Each segment of the loop is located at the same distance from P (see Figure 2.3). The

magnitude of the electric field at P due to a segment of the ring of length dl is equal to

dE =1

4pe0

ldl

r 2 + z 2

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When we integrate over the whole ring, the horizontal components of the electric field cancel.

We therefore only need to consider the vertical component of the electric field generated by each

segment:

dEz =z

r 2 + z 2dE =

ldl

4pe0

z

r 2 + z 2( )3/ 2

The total electric field generated by the ring can be obtained by integrating dEz over the whole

ring:

E =l

4pe0

z

r 2 + z 2( )3/ 2 dlRingÚ =

1

4pe0

z

r 2 + z 2( )3/ 2 2pr( )l =1

4pe0

z

r 2 + z 2( )3/ 2 q


Find the electric field a distance z from the center of a spherical surface of radius R, which

carries a uniform surface charge density s. Treat the case z < R (inside) as well as z > R

(outside). Express your answer in terms of the total charge q on the surface.

P

z

rcosq

rsinq

z-rcosq

q


Consider a slice of the shell centered on the z axis (see Figure 2.4). The polar angle of this

slice is q and its width is dq. The area dA of this ring is

dA = 2pr sinq( )rdq = 2pr 2 sinqdq

The total charge on this ring is equal to

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dq = sdA =1

2qsinqdq

where q is the total charge on the shell. The electric field produced by this ring at P can be

calculated using the solution of Problem 2.5:

dE =1

8pe0

q

r

z - r cosqr 2 + z 2 - 2zr cosq( )3/ 2 r sinqdq

The total field at P can be found by integrating dE with respect to q:

E =1

8pe0

q

r

z - r cosqr 2 + z 2 - 2zr cosq( )3/ 2 r sinqdq

0

p

Ú =

=1

8pe0

q

r

z - r cosqr 2 + z 2 - 2zr cosq( )3/ 2 d r cosq( )

0

p

Ú =1

8pe0

q

r

z - y

r 2 + z 2 - 2zy( )3/ 2 dy-r

r

Ú

This integral can be solved using the following relation:

z - y

r 2 + z 2 - 2zy( )3/ 2 = -d

dz

1

r 2 + z 2 - 2zy

Substituting this expression into the integral we obtain:

E = -1

8pe0

qr

ddz

1

r 2 + z 2 - 2zydy

-r

r

Ú = -1

8pe0

qr

ddz

r 2 + z 2 - 2zy

-z-r

r

=

= -1

8pe0

q

r

d

dz

r + z( ) - r - z

z

Ï Ì Ó

¸ ˝ ˛

Outside the shell, z > r and consequently the electric field is equal to

E = -1

8pe0

q

r

d

dz

r + z( ) - z - r( )z

= -1

4pe0

qd

dz

1

z=

1

4pe0

q

z 2

Inside the shell, z < r and consequently the electric field is equal to

E = -1

8pe0

q

r

d

dz

r + z( ) - r - z( )z

= -1

4pe0

q

r

d

dz1 = 0

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Thus the electric field of a charged shell is zero inside the shell. The electric field outside the

shell is equal to the electric field of a point charge located at the center of the shell.

2.2. Divergence and Curl of Electrostatic Fields

The electric field can be graphically represented using field lines. The direction of the field

lines indicates the direction in which a positive test charge moves when placed in this field. The

density of field lines per unit area is proportional to the strength of the electric field. Field lines

originate on positive charges and terminate on negative charges. Field lines can never cross

since if this would occur, the direction of the electric field at that particular point would be

undefined. Examples of field lines produced by positive point charges are shown in Figure 2.5.

a) b)

Figure 2.5. a) Electric field lines generated by a positive point charge with charge q. b)Electric field lines generated by a positive point charge with charge 2q.

The flux of electric field lines through any surface is proportional to the number of field lines

passing through that surface. Consider for example a point charge q located at the origin. The

electric flux F E through a sphere of radius r, centered on the origin, is equal to

F E = E ∑ da Surface

Ú =1

4pe0

q

r 2 ˆ r Ê Ë

ˆ ¯ ∑ r 2 sinqdqdfˆ r ( )

SurfaceÚ =

q

e0

Since the number of field lines generated by the charge q depends only on the magnitude of the

charge, any arbitrarily shaped surface that encloses q will intercept the same number of field

lines. Therefore the electric flux through any surface that encloses the charge q is equal to q / e0 .

Using the principle of superposition we can extend our conclusion easily to systems containing

more than one point charge:

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Ú = E i ∑ da Surface

Úi

Â =1

e0

qii

Â

We thus conclude that for an arbitrary surface and arbitrary charge distribution

E ∑ da Surface

Ú =Qenclosed

e0

where Qenclosed is the total charge enclosed by the surface. This is called Gauss's law. Since

this equation involves an integral it is also called Gauss's law in integral form.

Using the divergence theorem the electric flux F E can be rewritten as


Ú = — ∑ E ( )dtVolume

Ú

We can also rewrite the enclosed charge Qencl in terms of the charge density r:

Qenclosed = rdtVolume

Ú

Gauss's law can thus be rewritten as

— ∑ E ( )dtVolume

Ú =1

e0

rdtVolume

Ú

Since we have not made any assumptions about the integration volume this equation must hold

for any volume. This requires that the integrands are equal:

— ∑ E =re0

This equation is called Gauss's law in differential form.

Gauss's law in differential form can also be obtained directly from Coulomb's law for a

charge distribution r r '( ) :

E r '( ) =1

4pe0

Dˆ r

Dr( )2 r r '( )dt 'Volume

Ú

where Dr = r - r '. The divergence of E r is equal to

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— ∑ E r ( ) =1

4pe0

— ∑Dˆ r

Dr( )2

Ê Ë Á

ˆ ¯ ˜ r r '( )dt '

VolumeÚ =

1

4pe0

4pd 3 r - r '( )r r '( )dt 'Volume

Ú =r r ( )

e0

which is Gauss's law in differential form. Gauss's law in integral form can be obtained by

integrating — ∑ E r ( ) over the volume V:

— ∑ E r ( )( )dtVolume

Ú = E ∑ da Surface

Ú = F E =r r ( )

e0

dtVolume

Ú =QEnclosed

e0


If the electric field in some region is given (in spherical coordinates) by the expression

E r ( ) =Aˆ r + B sinq cosf ˆ f

r

where A and B are constants, what is the charge density r?

The charge density r can be obtained from the given electric field, using Gauss's law in

differential form:

r = e0 — ∑ E ( ) = e0

1

r 2

∂∂r

r 2Er( ) +1

r sinq∂

∂qsinqEq( ) +

1

r sinq∂

∂fEf( )Ê

Ë Á ˆ ¯ ˜ =

= e0

1

r 2

∂∂r

Ar( ) +1

r sinq∂

∂fB sinq cosf

rÊ Ë

ˆ ¯

Ê Ë Á

ˆ ¯ ˜ = e0

A

r 2 - e0

B

r 2 sinf

2.2.1. The curl of E

Consider a charge distribution r(r). The electric field at a point P generated by this charge

distribution is equal to

E r ( ) =1

4pe0

Dˆ r

Dr( )2 r r '( )Ú dt '

where Dr = r - r '. The curl of E is equal to

— ¥ E r ( ) =1

4pe0

— ¥Dˆ r

Dr( )2

Ê Ë Á

ˆ ¯ ˜ r r '( )Ú dt '

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However, — ¥ ˆ r / r 2 = 0 for every vector r and we thus conclude that

— ¥ E r ( ) = 0

2.2.2. Applications of Gauss's law

Although Gauss's law is always true it is only a useful tool to calculate the electric field if the

charge distribution is symmetric:

1. If the charge distribution has spherical symmetry, then Gauss's law can be used with

concentric spheres as Gaussian surfaces.

2. If the charge distribution has cylindrical symmetry, then Gauss's law can be used with

coaxial cylinders as Gaussian surfaces.

3. If the charge distribution has plane symmetry, then Gauss's law can be used with pill boxes

as Gaussian surfaces.


Use Gauss's law to find the electric field inside a uniformly charged sphere (charge density r)

of radius R.

The charge distribution has spherical symmetry and consequently the Gaussian surface used

to obtain the electric field will be a concentric sphere of radius r. The electric flux through this

surface is equal to


Ú = 4pr 2E r( )

The charge enclosed by this Gaussian surface is equal to

QEnclosed =4

3pr 3r

Applying Gauss's law we obtain for the electric field:

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E r( ) =1

4pr 2

QEnclosed

e0

=1

4pr 2

43

pr 3r

e0

=r

3e0

r


Find the electric field inside a sphere which carries a charge density proportional to the

distance from the origin: r = k r, for some constant k.

The charge distribution has spherical symmetry and we will therefore use a concentric sphere

of radius r as a Gaussian surface. Since the electric field depends only on the distance r, it is

constant on the Gaussian surface. The electric flux through this surface is therefore equal to


Ú = 4pr 2E r( )

The charge enclosed by the Gaussian surface can be obtained by integrating the charge

distribution between r' = 0 and r' = r:

QEnclosed = r r '( )dtVolume

Ú = kr ' 4pr '2( )dr '0

r

Ú = pkr4

Applying Gauss's law we obtain:

F E = 4pr 2E r( ) =QEnclosed

e0

=pkr4

e0

or

E r( ) =

pkr4

e0

Ê Ë Á

ˆ ¯ ˜

4pr 2 =1

4e0

kr2


A long coaxial cable carries a uniform (positive) volume charge density r on the inner

cylinder (radius a), and uniform surface charge density on the outer cylindrical shell (radius b).

The surface charge is negative and of just the right magnitude so that the cable as a whole is

neutral. Find the electric field in each of the three regions: (1) inside the inner cylinder (r < a),

(2) between the cylinders (a < r < b), (3) outside the cable (b < r).

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The charge distribution has cylindrical symmetry and to apply Gauss's law we will use a

cylindrical Gaussian surface. Consider a cylinder of radius r and length L. The electric field

generated by the cylindrical charge distribution will be radially directed. As a consequence,

there will be no electric flux going through the end caps of the cylinder (since here E da ). The

total electric flux through the cylinder is equal to


Ú = 2prLE r( )

The enclosed charge must be calculated separately for each of the three regions:

1. r < a: QEnclosed = pr 2Lr

2. a < r < b: QEnclosed = pa2Lr

3. b < r: QEnclosed = 0

Applying Gauss's law we find

E r( ) =1

2prL

QEnclosed

e0

Substituting the calculated Qencl for the three regions we obtain

1. r < a: E r( ) =1

2prL

QEnclosed

e0

=1

2prL

pr 2Lre0

=1

2e0

rr .

2. a < r < b: E r( ) =1

2prL

QEnclosed

e0

=1

2prL

pa2Lre0

=1

2e0

a2

rr

3. b < r E r( ) =1

2prL

QEnclosed

e0

= 0


Two spheres, each of radius R and carrying uniform charge densities of +r and -r ,

respectively, are placed so that they partially overlap (see Figure 2.6). Call the vector from the

negative center to the positive center s . Show that the field in the region of overlap is constant

and find its value.

To calculate the total field generated by this charge distribution we use the principle of

superposition. The electric field generated by each sphere can be obtained using Gauss' law (see

- 13 -

Problem 2.12). Consider an arbitrary point in the overlap region of the two spheres (see Figure

2.7). The distance between this point and the center of the negatively charged sphere is r-. The

distance between this point and the center of the positively charged sphere is r+. Figure 2.7

shows that the vector sum of s and r + is equal to r - . Therefore,

r + - r - = -s

The total electric field at this point in the overlap region is the vector sum of the field due to the

positively charged sphere and the field due to the negatively charged sphere:

E tot =r3e0

r + - r -( )

-

+s


-

+s

E+

E-Etot

q

Figure 2.7. Calculation of Etot.

- 14 -

The minus sign in front of r - shows that the electric field generated by the negatively charged

sphere is directed opposite to r - . Using the relation between r + and r - obtained from Figure 2.7

we can rewrite E tot as

E tot = -r3e0

s

which shows that the field in the overlap region is homogeneous and pointing in a direction

opposite to s .

2.3. The Electric Potential

The requirement that the curl of the electric field is equal to zero limits the number of vector

functions that can describe the electric field. In addition, a theorem discussed in Chapter 1 states

that any vector function whose curl is equal to zero is the gradient of a scalar function. The

scalar function whose gradient is the electric field is called the electric potential V and it is

defined as

E = -— V

Taking the line integral of — V between point a and point b we obtain

— V ∑ dl a

b

Ú = V b( ) - V a( ) = - E ∑ dl a

b

Ú

Taking a to be the reference point and defining the potential to be zero there, we obtain for V(b)

V b( ) = - E ∑ dl a

b

Ú

The choice of the reference point a of the potential is arbitrary. Changing the reference point of

the potential amounts to adding a constant to the potential:

V ' b( ) = - E ∑ dl a '

b

Ú = - E ∑ dl a '

a

Ú - E ∑ dl a

b

Ú = K + V b( )

where K is a constant, independent of b, and equal to

K = - E ∑ dl a '

a

Ú

However, since the gradient of a constant is equal to zero

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E'= -— V '= - — V = E

Thus, the electric field generated by V' is equal to the electric field generated by V. The physical

behavior of a system will depend only on the difference in electric potential and is therefore

independent of the choice of the reference point. The most common choice of the reference

point in electrostatic problems is infinity and the corresponding value of the potential is usually

taken to be equal to zero:

V b( ) = - E ∑ dl •

b

Ú

The unit of the electrical potential is the Volt (V, 1V = 1 Nm/C).


One of these is an impossible electrostatic field. Which one?

a) E = k xy( )ˆ i + 2yz( )ˆ j + 3xz( )ˆ k [ ]b) E = k y2( )ˆ i + 2xy + z 2( )ˆ j + 2yz( )ˆ k [ ]Here, k is a constant with the appropriate units. For the possible one, find the potential, using the

origin as your reference point. Check your answer by computing — V .

a) The curl of this vector function is equal to

— ¥ E = k∂

∂y3xz( ) -

∂∂z

2yz( )Ê Ë Á

ˆ ¯ ˜ ˆ i + k

∂∂z

xy( ) -∂

∂x3xz( )Ê

Ë ˆ ¯

ˆ j +

k∂

∂x2yz( ) -

∂∂y

xy( )Ê Ë Á

ˆ ¯ ˜ = k -2yˆ i - 3zˆ j - xˆ k ( )

Since the curl of this vector function is not equal to zero, this vector function can not describe an

electric field.

b) The curl of this vector function is equal to

— ¥ E = k∂

∂y2yz( ) -

∂∂z

2xy + z 2( )Ê Ë Á

ˆ ¯ ˜ ˆ i + k

∂∂z

y2( ) -∂

∂x2yz( )Ê

Ë ˆ ¯

ˆ j +

k∂

∂x2xy + z 2( ) -

∂∂y

y2( )Ê Ë Á

ˆ ¯ ˜ = 0

- 16 -

Since the curl of this vector function is equal to zero it can describe an electric field. To

calculate the electric potential V at an arbitrary point (x, y, z), using (0, 0, 0) as a reference point,

we have to evaluate the line integral of E between (0, 0, 0) and (x, y, z). Since the line integral

of E is path independent we are free to choose the most convenient integration path. I will use

the following integration path:

0, 0, 0 Æ x , 0, 0 Æ x , y , 0 Æ x , y , z

The first segment of the integration path is along the x axis:

dl = dxˆ i

and

E ∑ dl = ky2dx = 0

since y = 0 along this path. Consequently, the line integral of E along this segment of the

integration path is equal to zero. The second segment of the path is parallel to the y axis:

dl = dy j

and

E ∑ dl = k 2xy + z 2( )dy = 2kxydy

since z = 0 along this path. The line integral of E along this segment of the integration path is

equal to

E ∑ dl (x, 0,0)

(x,y, 0)

Ú = 2kxydy0

y

Ú = kxy2

The third segment of the integration path is parallel to the z axis:

dl = dz ˆ k

and

E ∑ dl = 2k yz( )dz

The line integral of E along this segment of the integration path is equal to

- 17 -

E ∑ dl (x, y, 0)

(x,y, z)

Ú = 2k yz( )dz0

z

Ú = kyz2

The electric potential at (x, y, z) is thus equal to

V x,y,z( ) = - E ∑ dl (0,0,0)

(x, 0,0)

Ú - E ∑ dl (x, 0,0)

(x, y, 0)

Ú - E ∑ dl (x, y, 0)

(x, y, z)

Ú =

= 0 - kxy2 - kyz2 = -k xy2 + yz2( )

The answer can be verified by calculating the gradient of V:

— V =∂V

∂xˆ i +

∂V

∂yˆ j +

∂V

∂zˆ k = -k y2ˆ i + 2xy + z 2( ) ˆ j + 2yz( ) ˆ k ( ) = -E

which is the opposite of the original electric field E .

The advantage of using the electric potential V instead of the electric field is that V is a scalar

function. The total electric potential generated by a charge distribution can be found using the

superposition principle. This property follows immediately from the definition of V and the fact

that the electric field satisfies the principle of superposition. Since

E = E 1 + E 2 + E 3 + .....

it follows that

V = - E ∑ dl •

b

Ú = - E 1 ∑ dl •

b

Ú - E 2 ∑ dl •

b

Ú - E 3 ∑ dl •

b

Ú - ... = V1 + V2 + V3 + ....

This equation shows that the total potential at any point is the algebraic sum of the potentials at

that point due to all the source charges separately. This ordinary sum of scalars is in general

easier to evaluate then a vector sum.


Suppose the electric potential is given by the expression

V r ( ) = Ae-lr

r

for all r (A and l are constants). Find the electric field E r ( ) , the charge density r r ( ) , and the

total charge Q.

- 18 -

The electric field E r ( ) can be immediately obtained from the electric potential:

E r ( ) = - — V r ( ) = -∂

∂rA

e-lr

r

Ê Ë Á

ˆ ¯ ˜ ˆ r = lA

e-lr

r+ A

e-lr

r 2

Ê Ë Á

ˆ ¯ ˜ ˆ r

The charge density r r ( ) can be found using the electric field E r ( ) and the following relation:

— ∑ E r ( ) =r r ( )

e0

This expression shows that

r r ( ) = e0 — ∑ E r ( )[ ]

Substituting the expression for the electric field E r ( ) we obtain for the charge density r r ( ) :

r r ( ) = e0A — ∑ 1 + lr( )e-lr ˆ r

r 2

Ê Ë Á

ˆ ¯ ˜

È

Î Í

˘

˚ ˙ =

= e0A 1 + lr( )e-lr — ∑ˆ r

r 2

Ê Ë Á

ˆ ¯ ˜ +

ˆ r

r 2 ∑ — 1 + lr( )e-lr( )È

Î Í

˘

˚ ˙ =

= e0A 4p 1 + lr( )e-lrd 3 r ( ) -l2e-lr

r

È

Î Í

˘

˚ ˙ = e0A 4pd 3 r ( ) -

l2e-lr

r

È

Î Í

˘

˚ ˙

The total charge Q can be found by volume integration of r r ( ) :

Qtot = r r ( )dtVolume

Ú = e0A 4pd 3 r ( ) -l2e-lr

r

È

Î Í

˘

˚ ˙ 4pr 2dr

0

•

Ú =

= 4pe0A 4pd 3 r ( )r 2dr0

•

Ú - rl2e-lr dr0

•

ÚÈ Î Í

˘ ˚ ˙ =

= -4pe0A rl2e-lr dr0

•

Ú

The integral can be solved easily:

re -lr dr0

•

Ú = -d

dle-lr dr

0

•

Ú = -d

dl1

lÊ Ë

ˆ ¯ =

1

l2

The total charge is thus equal to

- 19 -

Qtot = -4pe0A

The charge distribution r r ( ) can be directly used to obtained from the electric potential

V r ( )

r r ( ) = e0 — ∑ E r ( )[ ] = -e0 — ∑ — V r ( )[ ] = -e0 — 2V r ( )

This equation can be rewritten as

— 2V r ( ) = -r r ( )

e0

and is known as Poisson's equation. In the regions where r r ( ) = 0 this equation reduces to

Laplace's equation:

— 2V r ( ) = 0

The electric potential generated by a discrete charge distribution can be obtained using the

principle of superposition:

Vtot r ( ) = Vi r ( )i =1

n

Â

where Vi r ( ) is the electric potential generated by the point charge qi . A point charge qi located

at the origin will generate an electric potential Vi r ( ) equal to

Vi r ( ) = -1

4pe0

qi

r '2dr'

•

r

Ú =1

4pe0

qi

r

In general, point charge qi will be located at position r i and the electric potential generated by

this point charge at position r is equal to

Vi r ( ) =1

4pe0

qi

r - r i

The total electric potential generated by the whole set of point charges is equal to

Vtot r ( ) =1

4pe0

qi

r - r ii =1

n

Â

- 20 -

To calculate the electric potential generated by a continuous charge distribution we have to

replace the summation over point charges with an integration over the continuous charge

distribution. For the three charge distributions we will be using in this course we obtain:

1. line charge l : Vtot r ( ) =1

4pe0

lr - r '

dl 'LineÚ

2. surface charge s : Vtot r ( ) =1

4pe0

sr - r '

da 'Surface

Ú

3. volume charge r : Vtot r ( ) =1

4pe0

rr - r '

dt 'Volume

Ú


Using the general expression for V in terms of r find the potential at a distance z above the

center of the charge distributions of Figure 2.8. In each case, compute E = -— V . Suppose that

we changed the right-hand charge in Figure 2.8a to -q. What is then the potential at P? What

field does this suggest? Compare your answer to Problem 2.2b, and explain carefully any

discrepancy.

P

z

c)

Rs

P

z

a)

q q

d

P

z

b)

l2L


a) The electric potential at P generated by the two point charges is equal to

- 21 -

V =1

4pe0

q

1

4d2 + z 2

+1

4pe0

q

1

4d2 + z 2

=1

2pe0

q

1

4d2 + z 2

The electric field generated by the two point charges can be obtained by taking the gradient of

the electric potential:

E = -— V = -∂

∂z

1

2pe0

q

1

4d2 + z 2

Ê

Ë

Á Á Á Á

ˆ

¯

˜ ˜ ˜ ˜

ˆ k =1

2pe0

qz

14

d2 + z 2Ê Ë

ˆ ¯

3/ 2ˆ k

If we change the right-hand charge to -q then the total potential at P is equal to zero. However,

this does not imply that the electric field at P is equal to zero. In our calculation we have

assumed right from the start that x = 0 and y = 0. Obviously, the potential at P will therefore not

show an x and y dependence. This however not necessarily indicates that the components of the

electric field along the x and y direction are zero. This can be demonstrated by calculating the

general expression for the electric potential of this charge distribution at an arbitrary point (x,y,z):

V x,y,z( ) =1

4pe0

q

x +12

dÊ Ë

ˆ ¯

2

+ y2 + z 2

+1

4pe0

-q

x -12

dÊ Ë

ˆ ¯

2

+ y2 + z 2

=

=q

4pe0

1

x +1

2d

Ê Ë

ˆ ¯

2

+ y2 + z 2

-q

4pe0

1

x -1

2d

Ê Ë

ˆ ¯

2

+ y2 + z 2

The various components of the electric field can be obtained by taking the gradient of this

expression:

E x x,y,z( ) = -∂V

∂x=

q

4pe0

x +1

2d

Ê Ë

ˆ ¯

x +1

2d

Ê Ë

ˆ ¯

2

+ y2 + z 2Ê Ë Á

ˆ ¯ ˜

3/ 2 -q

4pe0

x -1

2d

Ê Ë

ˆ ¯

x -1

2d

Ê Ë

ˆ ¯

2

+ y2 + z 2Ê Ë Á

ˆ ¯ ˜

3/ 2

E y x,y,z( ) = -∂V

∂y=

q

4pe0

y

x +1

2d

Ê Ë

ˆ ¯

2

+ y2 + z 2Ê Ë Á

ˆ ¯ ˜

3/ 2 -q

4pe0

y

x -1

2d

Ê Ë

ˆ ¯

2

+ y2 + z 2Ê Ë Á

ˆ ¯ ˜

3/ 2

- 22 -

E z x,y,z( ) = -∂V

∂z=

q

4pe0

z

x +1

2d

Ê Ë

ˆ ¯

2

+ y2 + z 2Ê Ë Á

ˆ ¯ ˜

3/ 2 -q

4pe0

z

x -1

2d

Ê Ë

ˆ ¯

2

+ y2 + z 2Ê Ë Á

ˆ ¯ ˜

3/ 2

The components of the electric field at P = (0, 0, z) can now be calculated easily:

E x 0,0,z( ) =q

4pe0

d

14

d2 + z 2Ê Ë

ˆ ¯

3/ 2

E y 0,0,z( ) = 0

E z 0,0,z( ) = 0

b) Consider a small segment of the rod, centered at position x and with length dx. The charge

on this segment is equal to ldx. The potential generated by this segment at P is equal to

dV =1

4pe0

ldx

x2 + z 2

The total potential generated by the rod at P can be obtained by integrating dV between x = - L

and x = L

V =1

4pe0

ldx

x2 + z 2-L

L

Ú =l

4pe0

ln L2 + z 2 + L( ) - ln L2 + z 2 - L( )[ ]The z component of the electric field at P can be obtained from the potential V by calculating the

z component of the gradient of V. We obtain

Ez x,y,z( ) = -l

4pe0

∂∂z

ln L2 + z 2 + L( ) - ln L2 + z 2 - L( )[ ] =

= -l

4pe0

z

L2 + z 2

L2 + z 2 + L-

z

L2 + z 2

L2 + z 2 - L

È

Î

Í Í Í Í

˘

˚

˙ ˙ ˙ ˙

=l

4pe0

2L

z L2 + z 2

c) Consider a ring of radius r and width dr. The charge on this ring is equal to

dq = s p r + dr( )2 - pr 2[ ] = 2p ◊ s ◊ rdr

- 23 -

The electric potential dV at P generated by this ring is equal to

dV =1

4pe0

dq

r 2 + z 2=

s2e0

rdr

r 2 + z 2

The total electric potential at P can be obtained by integrating dV between r = 0 and r = R:

V =s2e0

rdr

r 2 + z 20

R

Ú =s2e0

R2 + z 2 - z[ ]The z component of the electric field generated by this charge distribution can be obtained by

taking the gradient of V:

Ez = -s2e0

∂∂z

R2 + z 2 - z[ ] = -s2e0

z

R2 + z 2-1

È

Î Í Í

˘

˚ ˙ ˙


Find the electric field a distance z above the center of a circular loop of radius r, which

carries a uniform line charge l.

The total charge Q on the ring is equal to

Q = 2pRl

The total electric potential V at P is equal to

V =1

4pe0

2pRlR2 + z 2

=l

2e0

R

R2 + z 2

The z component of the electric field at P can be obtained by calculating the gradient of V:

Ez = -l2e0

∂∂z

R

R2 + z 2=

l2e0

Rz

R2 + z 2( )3/ 2

This is the same answer we obtained in the beginning of this Chapter by taking the vector sum of

the segments of the ring.

We have seen so far that there are three fundamental quantities of electrostatics:

- 24 -

1. The charge density r

2. The electric field E

3. The electric potential V

If one of these quantities is known, the others can be calculated:

Known Ø r E V

r E =1

4pe0

ˆ r

r 2 rdtÚ V =1

4pe0

rr

dtÚE r = e0— ∑ E V = - E ∑ dl ÚV r = -e0 — 2V E = -— V

In general the charge density r and the electric field E do not have to be continuous. Consider

for example an infinitesimal thin charge sheet with surface charge s. The relation between the

electric field above and below the sheet can be obtained using Gauss's law. Consider a

rectangular box of height e and area A (see Figure 2.9). The electric flux through the surface of

the box, in the limit e Æ 0, is equal to


Ú = E^above - E^below( )A

Eabove

Ebelows

A e

Figure 2.9. Electric field near a charge sheet.

- 25 -

where E^,above and E^,below are the perpendicular components of the electric field above and

below the charge sheet. Using Gauss's law and the rectangular box shown in Figure 2.9 as

integration volume we obtain

Qencl

e0

=sA

e0

= E^, above - E^, below( )A

This equation shows that the electric field perpendicular to the charge sheet is discontinuous at

the boundary. The difference between the perpendicular component of the electric field above

and below the charge sheet is equal to

E^,above - E^, below =se0

The tangential component of the electric field is always continuous at any boundary. This

can be demonstrated by calculating the line integral of E around a rectangular loop of length L

and height e (see Figure 2.10). The line integral of E , in the limit e Æ 0, is equal to

E ∑ dl Ú = E ||,above ∑ dl Ú + E ||,below ∑ dl Ú = E||,above - E||,below( )L

Eabove

Ebelows

e

L

Figure 2.10. Parallel field close to charge sheet.

Since the line integral of E around any closed loop is zero we conclude that

E||,above - E||,below( )L = 0

or

E||,above = E||,below

These boundary conditions for E can be combined into a single formula:

- 26 -

E above - E below =se0

ˆ n

where ˆ n is a unit vector perpendicular to the surface and pointing towards the above region.

The electric potential is continuous across any boundary. This is a direct results of the

definition of V in terms of the line integral of E :

Vabove - Vbelow = E ∑ dl above

below

Ú

If the path shrinks the line integral will approach zero, independent of whether E is continuous

or discontinuous. Thus

Vabove = Vbelow


a) Check that the results of examples 4 and 5 of Griffiths are consistent with the boundary

conditions for E .

b) Use Gauss's law to find the field inside and outside a long hollow cylindrical tube which

carries a uniform surface charge s. Check that your results are consistent with the boundary

conditions for E .

c) Check that the result of example 7 of Griffiths is consistent with the boundary conditions for

V.

a) Example 4 (Griffiths): The electric field generated by an infinite plane carrying a uniform

surface charge s is directed perpendicular to the sheet and has a magnitude equal to

E above =s2e0

ˆ k

E below = -s2e0

ˆ k

Therefore,

E above - E below =se0

ˆ k =s sheet

e0

ˆ k

which is in agreement with the boundary conditions for E .

- 27 -

Example 5 (Griffiths): The electric field generated by the two charge sheets is directed

perpendicular to the sheets and has a magnitude equal to

E I = 0

E II =se0

ˆ i

E III = 0

The change in the strength of the electric field at the left sheet is equal to

E II - E I =se0

ˆ i =s left

e0

ˆ i

The change in the strength of the electric field at the right sheet is equal to

E III - E II = -se0

ˆ i =s right

e0

ˆ i

These relations show agreement with the boundary conditions for E .

b) Consider a Gaussian surface of length L and radius r. As a result of the symmetry of the

system, the electric field will be directed radially. The electric flux through this Gaussian

surface is therefore equal to the electric flux through its curved surface which is equal to

F E = E ∑ da Ú = 2p r L E r( )

The charge enclosed by the Gaussian surface is equal to zero when r < R. Therefore

E r ( ) = 0

when r < R. When r > R the charge enclosed by the Gaussian surface is equal to

Qenclosed = 2p R L s

The electric field for r > R, obtained using Gauss' law, is equal to

E r ( ) =1

2p r L

Qenclosed

e0

ˆ r =se0

R

rˆ r

The magnitude of the electric field just outside the cylinder, directed radially, is equal to

- 28 -

E outside = E R ( ) =se0

ˆ r

The magnitude of the electric field just inside the cylinder is equal to

E inside = 0

Therefore,

E outside - E inside =se0

ˆ r

which is consistent with the boundary conditions for E.

c) Example 7 (Griffiths): the electric potential just outside the charged spherical shell is equal

to

Voutside = Voutside z = R( ) =R2se0 R

=s R

e0

The electric potential just inside the charged spherical shell is equal to

Vinside = Vinside z = R( ) =s R

e0

These two equations show that the electric potential is continuous at the boundary.

2.4. Work and Energy in Electrostatics

Consider a point charge q1 located at the origin. A point charge q2 is moved from infinity to

a point a distance r2 from the origin. We will assume that the point charge q1 remains fixed at

the origin when point charge q2 is moved. The force exerted by q1 on q2 is equal to

F 12 = q2E 1

where E 1 is the electric field generated by q1. In order to move charge q2 we will have to exert a

force opposite to F 12 . Therefore, the total work that must be done to move q2 from infinity to r2

is equal to

- 29 -

W = - F 12 ∑ dl •

r2

Ú = - q2E 1 ∑ dl •

r2

Ú = q2V1 r 2( )

where V1 r 2( ) is the electric potential generated by q1 at position r2. Using the equation of V for a

point charge, the work W can be rewritten as

W =1

4pe0

q1 q2

r2

This work W is the work necessary to assemble the system of two point charges and is also called

the electrostatic potential energy of the system. The energy of a system of more than two point

charges can be found in a similar manner using the superposition principle. For example, for a

system consisting of three point charges (see Figure 2.11) the electrostatic potential energy is

equal to

W =1

4pe0

q1 q2

r12

+q1 q3

r13

+q2 q3

r23

Ï Ì Ó

¸ ˝ ˛

q1

q3

q2r23

r12r13

Figure 2.11. System of three point charges.

In this equation we have added the electrostatic energies of each pair of point charges. The

general expression of the electrostatic potential energy for n point charges is

W =1

4pe0 i =1

n

Âqi qj

rijj =i +1

n

Â

The lower limit of j (= i + 1) insures that each pair of point charges is only counted once. The

electrostatic potential energy can also be written as

- 30 -

W =1

2

1

4pe0 i =1

n

Âqi qj

rijj =1j πi

n

ÂÈ

Î

Í Í Í

˘

˚

˙ ˙ ˙

=1

2qi

i =1

n

Â 1

4pe0

qj

rij

=1

2qiVi

i =1

n

Âj =1j πi

n

Â

where Vi is the electrostatic potential at the location of qi due to all other point charges.

When the charge of the system is not distributed as point charges, but rather as a continuous

charge distribution r, then the electrostatic potential energy of the system must be rewritten as

W =1

2rVdt

VolumeÚ

For continuous surface and line charges the electrostatic potential energy is equal to

W =1

2sVda

SurfaceÚ

and

W =1

2lVdl

LineÚ

However, we have already seen in this Chapter that r, V , and E carry the same equivalent

information. The charge density r, for example, is related to the electric field E :

r = e0 — ∑ E ( )

Using this relation we can rewrite the electrostatic potential energy as

W =e0

2— ∑ E ( )Vdt

VolumeÚ =

e0

2— ∑ VE ( )( )dt

VolumeÚ -

e0

2E ∑ — V( )( )dt

VolumeÚ =

=e0

2VE ∑ da

SurfaceÚ +

e0

2E ∑ E ( )dt

VolumeÚ

where we have used one of the product rules of vector derivatives and the definition of E in

terms of V. In deriving this expression we have not made any assumptions about the volume

considered. This expression is therefore valid for any volume. If we consider all space, then the

contribution of the surface integral approaches zero since VE will approach zero faster than 1/r2.

Thus the total electrostatic potential energy of the system is equal to

- 31 -

W =e0

2E2dt

VolumeÚ


A sphere of radius R carries a charge density r r ( ) = k r (where k is a constant). Find the

energy of the configuration. Check your answer by calculating it in at least two different ways.

Method 1:

The first method we will use to calculate the electrostatic potential energy of the charged

sphere uses the volume integral of E2 to calculate W. The electric field generated by the charged

sphere can be obtained using Gauss's law. We will use a concentric sphere of radius r as the

Gaussian surface. First consider the case in which r < R. The charge enclosed by the Gaussian

surface can be obtained by volume integration of the charge distribution:

QEnclosed = r r ( )dtSphere

Ú = 4p k r r 2dr0

r

Ú = p k r4

The electric flux through the Gaussian surface is equal to

F E = 4p r 2E r( )

Applying Gauss's law we find for the electric field inside the sphere (r < R):

E r( ) =1

4p r 2

QEnclosed

e0

=1

4p r 2

p k r4

e0

=k r2

4e0

The electric field outside the sphere (r > R) can also be obtained using Gauss's law:

E r( ) =1

4p r 2

QEnclosed

e0

=1

4p r 2

p k R4

e0

=k R4

4e0 r2

The total electrostatic energy can be obtained from the electric field:

W =12

e0 E2 r( )dtAll Space

Ú = 2pe0

k r2

4e0

È

Î Í

˘

˚ ˙

2

r 2dr0

R

Ú + 2pe0

k R4

4e0 r2

È

Î Í

˘

˚ ˙

2

r 2dr0

R

Ú =

=pk2

8e0

r 6dr0

R

Ú +pk2

8e0

R8

r 2 drR

•

Ú =pk2

8e0

1

7R7 + R7È

Î Í ˘ ˚ ˙ =

pk2R7

7e0

- 32 -

Method 2:

An alternative way calculate the electrostatic potential energy is to use the following relation:

W =1

2rVdt

VolumeÚ

The electrostatic potential V can be obtained immediately from the electric field E . To evaluate

the volume integral of rV we only need to know the electrostatic potential V inside the charged

sphere:

V r ( ) = - E ∑ dl •

r

Ú = -1

4e0

k R4

r 2 dr•

R

Ú + kr2drR

r

ÚÈ

Î Í

˘

˚ ˙ =

k

12e0

4R3 - r 3( )

The electrostatic potential energy of the system is thus equal to

W =1

2r r ( )V r ( )dt

SphereÚ =

4p2

k rk

12e0

4R3 - r 3( )r 2dr0

R

Ú =4p2

k2

12e0

R7 -1

7R7Ê

Ë ˆ ¯ =

p k2

7e0

R7

which is equal to the energy calculated using method 1.

2.5. Metallic Conductors

In a metallic conductor one or more electrons per atom are free to move around through the

material. Metallic conductors have the following electrostatic properties:

1. The electric field inside the conductor is equal to zero.

If there would be an electric field inside the conductor, the free charges would move and

produce an electric field of their own opposite to the initial electric field. Free charges will

continue to flow until the cancellation of the initial field is complete.

2. The charge density inside a conductor is equal to zero.

This property is a direct result of property 1. If the electric field inside a conductor is equal

to zero, then the electric flux through any arbitrary closed surface inside the conductor is

equal to zero. This immediately implies that the charge density inside the conductor is equal

to zero everywhere (Gauss's law).

3. Any net charge of a conductor resides on the surface.

- 33 -

Since the charge density inside a conductor is equal to zero, any net charge can only reside

on the surface.

4. The electrostatic potential V is constant throughout the conductor.

Consider two arbitrary points a and b inside a conductor (see Figure 2.12). The potential

difference between a and b is equal to

V b( ) - V a( ) = - E ∑ dl a

b

Ú

a

b

Figure 2.12. Potential inside metallic conductor.

Since the electric field inside a conductor is equal to zero, the line integral of E between a

and b is equal to zero. Thus

V b( ) - V a( ) = 0

or

V b( ) = V a( )

5. The electric field is perpendicular to the surface, just outside the conductor.

If there would be a tangential component of the electric field at the surface, then the surface

charge would immediately flow around the surface until it cancels this tangential component.

Example: A spherical conducting shell

a) Suppose we place a point charge q at the center of a neutral spherical conducting shell (see

Figure 2.13). It will attract negative charge to the inner surface of the conductor. How much

induced charge will accumulate here?

b) Find E and V as function of r in the three regions r < a, a < r < b, and r > b.

- 34 -

a

bq

Figure 2.13. A spherical conducting shell.

a) The electric field inside the conducting shell is equal to zero (property 1 of conductors).

Therefore, the electric flux through any concentric spherical Gaussian surface of radius r (a<r<b)

is equal to zero. However, according to Gauss's law this implies that the charge enclosed by this

surface is equal to zero. This can only be achieved if the charge accumulated on the inside of the

conducting shell is equal to -q. Since the conducting shell is neutral and any net charge must

reside on the surface, the charge on the outside of the conducting shell must be equal to +q.

b) The electric field generated by this system can be calculated using Gauss's law. In the three

different regions the electric field is equal to

E r( ) =1

4pe0

q

r 2 for b < r

E r( ) = 0 for a < r < b

E r( ) =1

4pe0

q

r 2 for r < a

The electrostatic potential V(r) can be obtained by calculating the line integral of E from infinity

to a point a distance r from the origin. Taking the reference point at infinity and setting the value

of the electrostatic potential to zero there we can calculate the electrostatic potential. The line

integral of E has to be evaluated for each of the three regions separately.

For b < r:

V r( ) = - E r( ) ∑ dl •

r

Ú = -1

4pe0

q

r '2dr '

•

r

Ú =1

4pe0

q

r

- 35 -

For a < r < b:

V r( ) = - E r( ) ∑ dl •

r

Ú = -1

4pe0

q

r '2dr '

•

b

Ú =1

4pe0

q

b

For r < a:

V r( ) = - E r( ) ∑ dl •

r

Ú = -1

4pe0

q

r '2dr '-

1

4pe0

q

r '2dr '

a

r

Ú•

b

Ú =1

4pe0

q

b-

q

a+

q

rÈ Î Í

˘ ˚ ˙

q

Figure 2.14. Arbitrarily shaped conductor.

In this example we have looked at a symmetric system but the general conclusions are also valid

for an arbitrarily shaped conductor. For example, consider the conductor with a cavity shown in

Figure 2.14. Consider also a Gaussian surface that completely surrounds the cavity (see for

example the dashed line in Figure 2.14). Since the electric field inside the conductor is equal to

zero, the electric flux through the Gaussian surface is equal to zero. Gauss's law immediately

implies that the charge enclosed by the surface is equal to zero. Therefore, if there is a charge q

inside the cavity there will be an induced charge equal to -q on the walls of the cavity. On the

other hand, if there is no charge inside the cavity then there will be no charge on the walls of the

cavity. In this case, the electric field inside the cavity will be equal to zero. This can be

demonstrated by assuming that the electric field inside the cavity is not equal to zero. In this

case, there must be at least one field line inside the cavity. Since field lines originate on a

positive charge and terminate on a negative charge, and since there is no charge inside the cavity,

this field line must start and end on the cavity walls (see for example Figure 2.15). Now

consider a closed loop, which follows the field line inside the cavity and has an arbitrary shape

inside the conductor (see Figure 2.15). The line integral of E inside the cavity is definitely not

equal to zero since the magnitude of E is not equal to zero and since the path is defined such that

E and dl are parallel. Since the electric field inside the conductor is equal to zero, the path

- 36 -

integral of E inside the conductor is equal to zero. Therefore, the path integral of E along the

path indicated in Figure 2.15 is not equal to zero if the magnitude of E is not equal to zero inside

the cavity. However, the line integral of E along any closed path must be equal to zero and

consequently the electric field inside the cavity must be equal to zero.

Figure 2.15. Field line in cavity.


A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell

(inner radius a, outer radius b, see Figure 2.16). The shell carries no net charge.

a) Find the surface charge density s at R, at a, and at b.

b) Find the potential at the center of the sphere, using infinity as reference.

c) Now the outer surface is touched to a grounding wire, which lowers its potential to zero

(same as at infinity). How do your answers to a) and b) change?

a) Since the net charge of a conductor resides on its surface, the charge q of the metal sphere

will reside its surface. The charge density on this surface will therefore be equal to

sR =q

4p R2

As a result of the charge on the metal sphere there will be a charge equal to -q induced on the

inner surface of the metal shell. Its surface charge density will therefore be equal to

sa =-q

4p a2

Since the metal shell is neutral there will be a charge equal to +q on the outside of the shell. The

surface charge density on the outside of the shell will therefore be equal to

- 37 -

sb =q

4p b2

R

ab


b) The potential at the center of the metal sphere can be found by calculating the line integral of

E between infinity and the center. The electric field in the regions outside the sphere and shell

can be found using Gauss's law. The electric field inside the shell and sphere is equal to zero.

Therefore,

Vcenter = - E ∑ dl •

0

Ú = -q

4pe0

1

r 2 dr•

b

Ú -q

4pe0

1

r 2 dra

R

Ú =q

4pe0

1

b-

1

a+

1

RÈ Î Í

˘ ˚ ˙

c) When the outside of the shell is grounded, the charge density on the outside will become

zero. The charge density on the inside of the shell and on the metal sphere will remain the same.

The electric potential at the center of the system will also change as a result of grounding the

outer shell. Since the electric potential of the outer shell is zero, we do not need to consider the

line integral of E in the region outside the shell to determine the potential at the center of the

sphere. Thus

Vcenter = - E ∑ dl b

0

Ú = -q

4pe0

1

r 2 dra

R

Ú =q

4pe0

1

R-

1

aÈ Î Í

˘ ˚ ˙

Consider a conductor with surface charge s, placed in an external electric field. Because the

electric field inside the conductor is zero, the boundary conditions for the electric field require

that the field immediately above the conductor is equal to

- 38 -

E above =se0

ˆ n

dA

Ep,above

Eother

Ep,below

Figure 2.17. Patch of surface of conductor.

This electric field will exert a force on the surface charge.

Consider a small, infinitely thin, patch of the surface with surface area dA (see Figure 2.17).

The electric field directly above and below the patch is equal to the vector sum of the electric

field generated by the patch, the electric field generated by the rest of the conductor and the

external electric field. The electric field generated by the patch is equal to

E p, above =se0

ˆ n

E p, below = -se0

ˆ n

The remaining field, E other , is continuous across the patch, and consequently the total electric

field above and below the patch is equal to

E above = E other +se0

ˆ n

E below = E other -se0

ˆ n

These two equations show that E other is equal to

- 39 -

E other =1

2E above + E below( )

In this case the electric field below the surface is equal to zero and the electric field above the

surface is directly determined by the boundary condition for the electric field at the surface.

Thus

E other =1

2

se0

ˆ n + 0Ê Ë Á

ˆ ¯ ˜ =

s2e0

ˆ n

Since the patch cannot exert a force on itself, the electric force exerted on it is entirely due to the

electric field E other . The charge on the patch is equal to s dA. Therefore, the force exerted on

the patch is equal to

dF = s dA E other =s 2

2e0

dA ˆ n

The force per unit area of the conductor is equal to

f =dF

dA=

s 2

2e0

ˆ n

This equation can be rewritten in terms of the electric field just outside the conductor as

f =1

2e0

e0E( )2ˆ n =

e0

2E2 ˆ n

This force is directed outwards. It is called the radiation pressure.

2.6. Capacitors

Consider two conductors (see Figure 2.18), one with a charge equal to +Q and one with a

charge equal to -Q. The potential difference between the two conductors is equal to

DV = V+ - V- = - E ∑ dl -

+

Ú

- 40 -

Since the electric field E is proportional to the charge Q, the potential difference DV will also be

proportional to Q. The constant of proportionality is called the capacitance C of the system and

is defined as

C =Q

DV

-Q

+Q

Figure 2.18. Two conductors.

The capacitance C is determined by the size, the shape, and the separation distance of the two

conductors. The unit of capacitance is the farad (F). The capacitance of a system of conductors

can in general be calculated by carrying out the following steps:

1. Place a charge +Q on one of the conductors. Place a charge of -Q on the other conductor (for

a two conductor system).

2. Calculate the electric field in the region between the two conductors.

3. Use the electric field calculated in step 2 to calculate the potential difference between the two

conductors.

4. Apply the result of part 3 to calculate the capacitance:

C = Q

DV

We will now discuss two examples in which we follow these steps to calculate the capacitance.

Example: Example 2.11 (Griffiths)

Find the capacitance of two concentric shells, with radii a and b.

Place a charge +Q on the inner shell and a charge -Q on the outer shell. The electric field

between the shells can be found using Gauss's law and is equal to

- 41 -

E r ( ) =1

4pe0

Q

r 2 ˆ r a < r < b

The potential difference between the outer shell and the inner shell is equal to

V a( ) - V b( ) = - E r ( ) ∑ dl a

b

Ú = -1

4pe0

Q

r 2 dra

b

Ú =Q

4pe0

1

a-

1

bÊ Ë

ˆ ¯

The capacitance of this system is equal to

C =Q

DV =

QQ

4pe0

1a

-1b

Ê Ë

ˆ ¯

= 4pe0

ab

b - a

ab

+Q-Q

Figure 2.19. Example 2.11.

A system does not have to have two conductors in order to have a capacitance. Consider for

example a single spherical shell of radius R. The capacitance of this system of conductors can be

calculated by following the same steps as in Example 12. First of all, put a charge Q on the

conductor. Gauss's law can be used to calculate the electric field generated by this system with

the following result:

E r ( ) =1

4pe0

Q

r 2 ˆ r

Taking infinity as the reference point we can calculate the electrostatic potential on the surface of

the shell:

V R( ) = - E r ( ) ∑ dl •

R

Ú = -1

4pe0

Q

r 2 dr•

R

Ú =1

4pe0

Q

R

- 42 -

Therefore, the capacitance of the shell is equal to

C =Q

DV =

Q1

4pe0

QR

= 4pe0R

Let us now consider a parallel-plate capacitor. The work required to charge up the parallel-

plate capacitor can be calculated in various ways:

Method 1: Since we are free to chose the reference point and reference value of the potential we

will chose it such that the potential of the positively charges plate is +DV / 2 and the potential of

the negatively charged plate is -DV / 2. The energy of this charge distribution is then equal to

W =1

2rVdtÚ =

1

2Q

DV

2+ -Q( ) -DV

2Ê Ë

ˆ ¯

È Î Í

˘ ˚ ˙ =

1

2Q DV( ) =

1

2C DV( )2

Method 2: Let us look at the charging process in detail. Initially both conductors are uncharged

and DV = 0. At some intermediate step in the charging process the charge on the positively

charged conductor will be equal to q. The potential difference between the conductors will then

be equal to

DV =q

C

To increase the charge on the positively charged conductor by dq we have to move this charge

dq across this potential difference DV . The work required to do this is equal to

dW = DV dq =q

Cdq

Therefore, the total work required to charge up the capacitor from q = 0 to q = Q is equal to

W =q

Cdq

0

Q

Ú =1

2

Q 2

C=

1

2C DV( )2

Example: Problem 2.40.

Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal

distance e, as a result of their mutual attraction.

- 43 -

a) Use equation (2.45) of Griffiths to express the amount of work done by electrostatic forces,

in terms of the field E and the area of the plates A.

b) Use equation (2.40) of Griffiths to express the energy lost by the field in this process.

a) We will assume that the parallel-plate capacitor is an ideal capacitor with a homogeneous

electric field E between the plates and no electric field outside the plates. The electrostatic force

per unit surface area is equal to

f =e0

2E2 ˆ n

The total force exerted on each plate is therefore equal to

F = Af =e0

2AE2 ˆ n

As a result of this force, the plates of the parallel-plate capacitor move closer together by an

infinitesimal distance e. The work done by the electrostatic forces during this movement is equal

to

W = F ∑ dl Ú =e0

2A e E2

b) The total energy stored in the electric field is equal to

W =e0

2E2dtÚ

In an ideal capacitor the electric field is constant between the plates and consequently we can

easily evaluate the volume integral of E2:

W =e0

2A d E2

where d is the distance between the plates. If the distance between the plates is reduced, then the

energy stored in the field will also be reduced. A reduction in d of e will reduce the energy

stored by an amount DW equal to

DW =e0

2A d E2 -

e0

2A d -e( ) E2 =

e0

2A e E2

which is equal to the work done by the electrostatic forces on the capacitor plates (see part a).

- 1 -

Chapter 3. Special Techniques for Calculating Potentials

Given a stationary charge distribution r r ( ) we can, in principle, calculate the electric field:

E r ( ) =1

4pe0

Dˆ r

Dr( )2Ú r r '( )dt '

where Dr = r '-r . This integral involves a vector as an integrand and is, in general, difficult to

calculate. In most cases it is easier to evaluate first the electrostatic potential V which is defined

as

V r ( ) =1

4pe0

1

DrÚ r r '( ) dt '

since the integrand of the integral is a scalar. The corresponding electric field E can then be

obtained from the gradient of V since

E = -— V

The electrostatic potential V can only be evaluated analytically for the simplest charge

configurations. In addition, in many electrostatic problems, conductors are involved and the

charge distribution r is not known in advance (only the total charge on each conductor is

known).

A better approach to determine the electrostatic potential is to start with Poisson's equation

— 2V = -re0

Very often we only want to determine the potential in a region where r = 0. In this region

Poisson's equation reduces to Laplace's equation

— 2V = 0

There are an infinite number of functions that satisfy Laplace's equation and the appropriate

solution is selected by specifying the appropriate boundary conditions. This Chapter will

concentrate on the various techniques that can be used to calculate the solutions of Laplace's

equation and on the boundary conditions required to uniquely determine a solution.

- 2 -

3.1. Solutions of Laplace's Equation in One-, Two, and Three Dimensions

3.1.1. Laplace's Equation in One Dimension

In one dimension the electrostatic potential V depends on only one variable x . The

electrostatic potential V(x) is a solution of the one-dimensional Laplace equation

d2V

dx2 = 0

The general solution of this equation is

V x( ) = sx + b

where s and b are arbitrary constants. These constants are fixed when the value of the potential

is specified at two different positions.

Example

Consider a one-dimensional world with two point conductors located at x = 0 m and at x = 10

m. The conductor at x = 0 m is grounded (V = 0 V) and the conductor at x = 10 m is kept at a

constant potential of 200 V. Determine V.

The boundary conditions for V are

V 0( ) = b = 0V

and

V 10( ) =10s + b = 200V

The first boundary condition shows that b = 0 V. The second boundary condition shows that s =

20 V/m. The electrostatic potential for this system of conductors is thus

V x( ) = 20x

The corresponding electric field can be obtained from the gradient of V

E x( ) = -dV x( )

dx= -20 V / m

- 3 -

The boundary conditions used here, can be used to specify the electrostatic potential between x =

0 m and x = 10 m but not in the region x < 0 m and x > 10 m. If the solution obtained here was

the general solution for all x, then V would approach infinity when x approaches infinity and V

would approach minus infinity when x approaches minus infinity. The boundary conditions

therefore provide the information necessary to uniquely define a solution to Laplace's equation,

but they also define the boundary of the region where this solution is valid (in this example 0 m

< x < 10 m).

The following properties are true for any solution of the one-dimensional Laplace equation:

Property 1:

V(x) is the average of V(x + R) and V(x - R) for any R as long as x + R and x - R are located in

the region between the boundary points. This property is easy to proof:

V x + r( ) + V x - R( )2

=s x + R( ) + b + s x - R( ) + b

2= sx + b = V x( )

This property immediately suggests a powerful analytical method to determine the solution

of Laplace's equation. If the boundary values of V are

V x = a( ) = Va

and

V x = b( ) = Vb

then property 1 can be used to determine the value of the potential at (a + b)/2:

V x =a + b

2Ê Ë

ˆ ¯ =

1

2Va + Vb[ ]

Next we can determine the value of the potential at x = (3 a + b)/4 and at x = (a + 3 b)/4 :

V x =3a + b

2Ê Ë

ˆ ¯ =

1

2V x = a( ) + V x =

a + b

2Ê Ë

ˆ ¯

È Î Í

˘ ˚ ˙ =

1

2

3

2Va +

1

2Vb

È Î Í

˘ ˚ ˙

V x =a + 3b

2Ê Ë

ˆ ¯ =

1

2V x =

a + b

2Ê Ë

ˆ ¯ + V x = b( )È

Î Í

˘ ˚ ˙ =

1

2

1

2Va +

3

2Vb

È Î Í

˘ ˚ ˙

- 4 -

This process can be repeated and V can be calculated in this manner at any point between x =

a and x = b (but not in the region x > b and x < a).

Property 2:

The solution of Laplace's equation can not have local maxima or minima. Extreme values

must occur at the end points (the boundaries). This is a direct consequence of property 1.

Property 2 has an important consequence: a charged particle can not be held in stable

equilibrium by electrostatic forces alone (Earnshaw's Theorem). A particle is in a stable

equilibrium if it is located at a position where the potential has a minimum value. A small

displacement away from the equilibrium position will increase the electrostatic potential of

the particle, and a restoring force will try to move the particle back to its equilibrium

position. However, since there can be no local maxima or minima in the electrostatic

potential, the particle can not be held in stable equilibrium by just electrostatic forces.

3.1.2. Laplace's Equation in Two Dimensions

In two dimensions the electrostatic potential depends on two variables x and y. Laplace's

equation now becomes

∂ 2V

∂x2 +∂ 2V

∂y2 = 0

This equation does not have a simple analytical solution as the one-dimensional Laplace

equation does. However, the properties of solutions of the one-dimensional Laplace equation are

also valid for solutions of the two-dimensional Laplace equation:

Property 1:

The value of V at a point (x, y) is equal to the average value of V around this point

V x,y( ) =1

2pRVRdf

CircleÚ

where the path integral is along a circle of arbitrary radius, centered at (x, y) and with radius

R.

Property 2:

V has no local maxima or minima; all extremes occur at the boundaries.

- 5 -

3.1.3. Laplace's Equation in Three Dimensions

In three dimensions the electrostatic potential depends on three variables x , y , and z.

Laplace's equation now becomes

∂ 2V

∂x2 +∂ 2V

∂y2 +∂ 2V

∂z 2 = 0

This equation does not have a simple analytical solution as the one-dimensional Laplace

equation does. However, the properties of solutions of the one-dimensional Laplace equation are

also valid for solutions of the three-dimensional Laplace equation:

Property 1:

The value of V at a point (x, y, z) is equal to the average value of V around this point

V x,y,z( ) =1

4pR2 VR2 sinq dq dfSphere

Ú

where the surface integral is across the surface of a sphere of arbitrary radius, centered at

(x,y,z) and with radius R.

x

y

z

r

R d

P

qq

Figure 3.1. Proof of property 1.

- 6 -

To proof this property of V consider the electrostatic potential generated by a point charge q

located on the z axis, a distance r away from the center of a sphere of radius R (see Figure

3.1). The potential at P, generated by charge q, is equal to

VP =1

4pe0

q

d

where d is the distance between P and q. Using the cosine rule we can express d in terms of

r, R and q

d2 = r 2 + R2 - 2rR cosq

The potential at P due to charge q is therefore equal to

VP =1

4pe0

q

r 2 + R2 - 2rR cosq

The average potential on the surface of the sphere can be obtained by integrating VP across

the surface of the sphere. The average potential is equal to

Vaverage =1

4pR2 VPR2 sinq dq dfSphere

Ú =1

4p1

4pe0

q

r 2 + R2 - 2rR cosq2p sinq dqÚ =

=q

8pe0

r 2 + R2 - 2rR cosqrR

0

p

=q

8pe0

r + R

rR-

r - R

rRÊ Ë

ˆ ¯ =

1

4pe0

q

r

which is equal to the potential due to q at the center of the sphere. Applying the principle of

superposition it is easy to show that the average potential generated by a collection of point

charges is equal to the net potential they produce at the center of the sphere.

Property 2:

The electrostatic potential V has no local maxima or minima; all extremes occur at the

boundaries.


Find the general solution to Laplace's equation in spherical coordinates, for the case where V

depends only on r. Then do the same for cylindrical coordinates.

Laplace's equation in spherical coordinates is given by

- 7 -

1

r 2

∂∂r

r 2 ∂V

∂rÊ Ë

ˆ ¯ +

1

r 2 sinq∂

∂qsinq

∂V

∂qÊ Ë

ˆ ¯ +

1

r 2 sin2q∂ 2V

∂f 2 = 0

If V is only a function of r then

∂V

∂q= 0

and

∂V

∂f= 0

Therefore, Laplace's equation can be rewritten as

1

r 2

∂∂r

r 2 ∂V

∂rÊ Ë

ˆ ¯ = 0

The solution V of this second-order differential equation must satisfy the following first-order

differential equation:

r 2 ∂V

∂r= a = constant

This differential equation can be rewritten as

∂V

∂r=

a

r 2

The general solution of this first-order differential equation is

V r( ) = -a

r+ b

where b is a constant. If V = 0 at infinity then b must be equal to zero, and consequently

V r( ) = -a

r

Laplace's equation in cylindrical coordinates is

- 8 -

1

r

∂∂r

r∂V

∂rÊ Ë

ˆ ¯ +

1

r 2

∂ 2V

∂f 2 +∂ 2V

∂z 2 = 0

If V is only a function of r then

∂V

∂f= 0

and

∂V

∂z= 0

Therefore, Laplace's equation can be rewritten as

1

r

∂∂r

r∂V

∂rÊ Ë

ˆ ¯ = 0

The solution V of this second-order differential equation must satisfy the following first-order

differential equation:

r∂V

∂r= a = constant

This differential equation can be rewritten as

∂V

∂r=

a

r

The general solution of this first-order differential equation is

V r( ) = a ln r( ) + b

where b is a constant. The constants a and b are determined by the boundary conditions.

3.1.4. Uniqueness Theorems

Consider a volume (see Figure 3.2) within which the charge density is equal to zero.

Suppose that the value of the electrostatic potential is specified at every point on the surface of

this volume. The first uniqueness theorem states that in this case the solution of Laplace's

equation is uniquely defined.

- 9 -

Volume

Boundary

Figure 3.2. First Uniqueness Theorem

To proof the first uniqueness theorem we will consider what happens when there are two

solutions V1 and V2 of Laplace's equation in the volume shown in Figure 3.2. Since V1 and V2

are solutions of Laplace's equation we know that

— 2V1 = 0

and

— 2V2 = 0

Since both V1 and V2 are solutions, they must have the same value on the boundary. Thus V1 =

V2 on the boundary of the volume. Now consider a third function V3, which is the difference

between V1 and V2

V3 = V1 - V2

The function V3 is also a solution of Laplace's equation. This can be demonstrated easily:

— 2V3 = — 2V1 - — 2V2 = 0

The value of the function V3 is equal to zero on the boundary of the volume since V1 = V2 there.

However, property 2 of any solution of Laplace's equation states that it can have no local

maxima or minima and that the extreme values of the solution must occur at the boundaries.

Since V3 is a solution of Laplace's equation and its value is zero everywhere on the boundary of

the volume, the maximum and minimum value of V3 must be equal to zero. Therefore, V3 must

be equal to zero everywhere. This immediately implies that

- 10 -

V1 = V2

everywhere. This proves that there can be no two different functions V1 and V2 that are solutions

of Laplace's equation and satisfy the same boundary conditions. Therefore, the solution of

Laplace's equation is uniquely determined if its value is a specified function on all boundaries of

the region. This also indicates that it does not matter how you come by your solution: if (a) it is

a solution of Laplace's equation, and (b) it has the correct value on the boundaries, then it is the

right and only solution.

Boundary

q1

q2q3

q4

Figure 3.3. System with conductors.

The first uniqueness theorem can only be applied in those regions that are free of charge and

surrounded by a boundary with a known potential (not necessarily constant). In the laboratory

the boundaries are usually conductors connected to batteries to keep them at a fixed potential. In

many other electrostatic problems we do not know the potential at the boundaries of the system.

Instead we might know the total charge on the various conductors that make up the system (note:

knowing the total charge on a conductor does not imply a knowledge of the charge distribution rsince it is influenced by the presence of the other conductors). In addition to the conductors that

make up the system, there might be a charge distribution r filling the regions between the

conductors (see Figure 3.3). For this type of system the first uniqueness theorem does not apply.

The second uniqueness theorem states that the electric field is uniquely determined if the total

charge on each conductor is given and the charge distribution in the regions between the

conductors is known.

- 11 -

The proof of the second uniqueness theorem is similar to the proof of the first uniqueness

theorem. Suppose that there are two fields E 1 and E 2 that are solutions of Poisson's equation in

the region between the conductors. Thus

— ∑ E 1 =re0

and

— ∑ E 2 =re0

where r is the charge density at the point where the electric field is evaluated. The surface

integrals of E 1 and E 2, evaluated using a surface that is just outside one of the conductors with

charge Qi, are equal to Qi / e0 . Thus

E 1 ∑ da Surfaceconductor i

Ú =Qi

e0


Ú =Qi

e0

The difference between E 1 and E 2, E 3 = E 1 - E 2, satisfies the following equations:

— ∑ E 3 = — ∑ E 1 - — ∑ E 2 =re0

-r

e0

= 0


Ú = E 1 ∑ da Surfaceconductor i

Ú - E 2 ∑ da Surfaceconductor i

Ú =Qi

e0

-Qi

e0

= 0

Consider the surface integral of E 3, integrated over all surfaces (the surface of all conductors and

the outer surface). Since the potential on the surface of any conductor is constant, the

electrostatic potential associated with E 1 and E 2 must also be constant on the surface of each

conductor. Therefore, V3 = V1 - V2 will also be constant on the surface of each conductor. The

surface integral of V3E 3 over the surface of conductor i can be written as

V3E 3 ∑ da Surfaceconductor i

Ú = V3 E 3 ∑ da Surfaceconductor i

Ú = 0

- 12 -

Since the surface integral of V3E 3 over the surface of conductor i is equal to zero, the surface

integral of V3E 3 over all conductor surfaces will also be equal to zero. The surface integral of

V3E 3 over the outer surface will also be equal to zero since V3 = 0 on this surface. Thus

V3E 3 ∑ da All surfaces

Ú = 0

The surface integral of V3E 3 can be rewritten using Green's identity as

0 = V3E 3 ∑ da All surfaces

Ú = - V3 — V3 ∑ da All surfaces

Ú = - V3— 2V3 + — V3( ) ∑ — V3( )( )dtVolumebetweenconductors

Ú =

= - -V3 — ∑ E 3( ) + E 3 ∑ E 3( )dtVolumebetweenconductors

Ú = - E32dt

Volumebetweenconductors

Ú = 0

where the volume integration is over all space between the conductors and the outer surface.

Since E32 is always positive, the volume integral of E3

2 can only be equal to zero if E32 = 0

everywhere. This implies immediately that E1 = E2 everywhere, and proves the second

uniqueness theorem.

3.2. Method of Images

Consider a point charge q held as a distance d above an infinite grounded conducting plane

(see Figure 3.4). The electrostatic potential of this system must satisfy the following two

boundary conditions:

V x,y,0( ) = 0

V x,y,z( ) Æ 0 when

x Æ •y Æ •z Æ •

Ï

Ì Ô

Ó Ô

A direct calculation of the electrostatic potential can not be carried out since the charge

distribution on the grounded conductor is unknown. Note: the charge distribution on the surface

of a grounded conductor does not need to be zero.

- 13 -

z axis

q

d

Figure 3.4. Method of images.

Consider a second system, consisting of two point charges with charges +q and -q, located at

z = d and z = -d, respectively (see Figure 3.5). The electrostatic potential generated by these two

charges can be calculated directly at any point in space. At a point P = (x, y, 0) on the xy plane

the electrostatic potential is equal to

V x,y,0( ) =1

4pe0

q

x2 + y2 + d2+

-q

x2 + y2 + d2

È

Î Í Í

˘

˚ ˙ ˙

= 0

q

z axis

d

-q

d

P

Figure 3.5. Charge and image charge.

The potential of this system at infinity will approach zero since the potential generated by each

charge will decrease as 1/r with increasing distance r. Therefore, the electrostatic potential

- 14 -

generated by the two charges shown in Figure 3.5 satisfies the same boundary conditions as the

system shown in Figure 3.4. Since the charge distribution in the region z > 0 (bounded by the xy

plane boundary and the boundary at infinity) for the two systems is identical, the corollary of the

first uniqueness theorem states that the electrostatic potential in this region is uniquely defined.

Therefore, if we find any function that satisfies the boundary conditions and Poisson's equation,

it will be the right answer. Consider a point (x, y, z) with z > 0. The electrostatic potential at this

point can be calculated easily for the charge distribution shown in Figure 3.5. It is equal to

V x,y,z( ) =1

4pe0

q

x2 + y2 + z - d( )2+

-q

x2 + y2 + z + d( )2

È

Î Í Í

˘

˚ ˙ ˙

Since this solution satisfies the boundary conditions, it must be the correct solution in the region

z > 0 for the system shown in Figure 3.4. This technique of using image charges to obtain the

electrostatic potential in some region of space is called the method of images.

The electrostatic potential can be used to calculate the charge distribution on the grounded

conductor. Since the electric field inside the conductor is equal to zero, the boundary condition

for E (see Chapter 2) shows that the electric field right outside the conductor is equal to

E Outside =se0

ˆ n =se0

ˆ k

where s is the surface charge density and ˆ n is the unit vector normal to the surface of the

conductor. Expressing the electric field in terms of the electrostatic potential V we can rewrite

this equation as

s = e0Ez = -e0

∂V

∂z z=0

Substituting the solution for V in this equation we find

s = -q

4p- z - d( )

x2 + y2 + z - d( )2( )3/ 2 +z + d( )

x2 + y2 + z + d( )2( )3/ 2

È

Î

Í Í Í

˘

˚

˙ ˙ ˙

z=0

= -q

2pd

x2 + y2 + d2( )3/ 2

Only in the last step of this calculation have we substituted z = 0. The induced charge

distribution is negative and the charge density is greatest at (x = 0, y = 0, z = 0). The total charge

on the conductor can be calculated by surface integrating of s:

- 15 -

Qtotal = s daSurface

Ú =0

2p

Ú s r( )rdrdq0

•

Ú

where r 2 = x2 + y2. Substituting the expression for s in the integral we obtain

Qtotal = -qd1

r 2 + d2( )3/ 2 rdr0

•

Ú =qd

r 2 + d20

•

= qd 0 -1

dÈ Î Í

˘ ˚ ˙ = -q

As a result of the induced surface charge on the conductor, the point charge q will be

attracted towards the conductor. Since the electrostatic potential generated by the charge image-

charge system is the same as the charge-conductor system in the region where z > 0, the

associated electric field (and consequently the force on point charge q) will also be the same.

The force exerted on point charge q can be obtained immediately by calculating the force exerted

on the point charge by the image charge. This force is equal to

F = -1

4pe0

q2

2d( )2ˆ k

There is however one important difference between the image-charge system and the real

system. This difference is the total electrostatic energy of the system. The electric field in the

image-charge system is present everywhere, and the magnitude of the electric field at (x, y, z)

will be the same as the magnitude of the electric field at (x, y, -z). On the other hand, in the real

system the electric field will only be non zero in the region with z > 0. Since the electrostatic

energy of a system is proportional to the volume integral of E2 the electrostatic energy of the

real system will be 1/2 of the electrostatic energy of the image-charge system (only 1/2 of the

total volume has a non-zero electric field in the real system). The electrostatic energy of the

image-charge system is equal to

Wimage =1

4pe0

q2

2d

The electrostatic energy of the real system is therefore equal to

Wreal =1

2Wimage = -

1

4pe0

q2

4d

The electrostatic energy of the real system can also be obtained by calculating the work required

to be done to assemble the system. In order to move the charge q to its final position we will

- 16 -

have to exert a force opposite to the force exerted on it by the grounded conductor. The work

done to move the charge from infinity along the z axis to z = d is equal to

Wreal =1

4pe0

q2

4z 2 dz•

d

Ú =1

4pe0

-q2

4z•

d

= -1

4pe0

q2

4d

which is identical to the result obtained using the electrostatic potential energy of the image-

charge system.

Example: Example 3.2 + Problem 3.7

A point charge q is situated a distance s from the center of a grounded conducting sphere of

radius R (see Figure 3.6).

a) Find the potential everywhere.

b) Find the induced surface charge on the sphere, as function of q. Integrate this to get the total

induced charge.

c) Calculate the electrostatic energy of this configuration.

R

s

q

V = 0

Figure 3.6. Example 3.2 + Problem 3.7.

a) Consider a system consisting of two charges q and q', located on the z axis at z = s and z = z',

respectively. If the potential produced by this system is identical everywhere to the potential

produced by the system shown in Figure 3.6 then the position of point charge q' must be chosen

such that the potential on the surface of a sphere of radius R, centered at the origin, is equal to

zero (in this case the boundary conditions for the potential generated by both systems are

identical).

We will start with determining the correct position of point charge q'. The electrostatic

potential at P (see Figure 3.7) is equal to

- 17 -

VP = 0 =1

4pe0

q

s - R+

1

4pe0

q'

R - z '


q'= -q

s - RR - z '( )

R

s

qV = 0

q'

Q

z'

d'd

P'

P

Figure 3.7. Image-charge system.

The electrostatic potential at Q is equal to

VQ = 0 =1

4pe0

q

s + R+

1

4pe0

q'

R + z '


q'= -q

s + RR + z '( )

Combining the two expression for q' we obtain

q

s - RR - z '( ) =

q

s + RR + z '( )

or

s + R( ) R - z '( ) = s - R( ) R + z '( )


z' s - R( ) + z ' s + R( ) = 2sz '= R s + R( ) - R s - R( ) = 2R2

- 18 -

The position of the image charge is equal to

z'=R2

s

The value of the image charge is equal to

q'= -q

s + RR + z '( ) = -

q

s + RR +

R2

s

Ê Ë Á

ˆ ¯ ˜ = -q

R

s

Now consider an arbitrary point P' on the circle. The distance between P' and charge q is d and

the distance between P' and charge q' is equal to d'. Using the cosine rule (see Figure 3.7) we

can express d and d' in terms of R, s, and q:

d = R2 + s2 - 2Rscosq

d'= R2 + z '2 -2Rz 'cosq = R2 +R4

s2 - 2RR2

scosq

The electrostatic potential at P' is equal to

VP ' =1

4pe0

q

d+

q'

d 'È Î Í

˘ ˚ ˙ =

1

4pe0

q

R2 + s2 - 2Rscosq+

-qRs

R2 +R4

s2 - 2RR2

scosq

È

Î

Í Í Í Í

˘

˚

˙ ˙ ˙ ˙

=

=1

4pe0

q

R2 + s2 - 2Rscosq-

q

R2 + s2 - 2Rscosq

È

Î Í

˘

˚ ˙ = 0

Thus we conclude that the configuration of charge and image charge produces an electrostatic

potential that is zero at any point on a sphere with radius R and centered at the origin. Therefore,

this charge configuration produces an electrostatic potential that satisfies exactly the same

boundary conditions as the potential produced by the charge-sphere system. In the region

outside the sphere, the electrostatic potential is therefore equal to the electrostatic potential

produced by the charge and image charge. Consider an arbitrary point r ,q ,f( ) . The distance

between this point and charge q is d and the distance between this point and charge q' is equal to

d'. These distances can be expressed in terms of r, s, and q using the cosine rule:

d = r 2 + s2 - 2rs cosq

- 19 -

d'= r 2 + z '2 -2rz 'cosq = r 2 +R4

s2 - 2rR2

scosq

The electrostatic potential at r ,q ,f( ) will therefore be equal to

V r ,q ,f( ) =1

4pe0

q

d+

q'

d 'È Î Í

˘ ˚ ˙ =

1

4pe0

{q

r 2 + s2 - 2rs cosq+

-qRs

r 2 +R4

s2 - 2rR2

scosq

} =

=1

4pe0

{q

r 2 + s2 - 2rs cosq-

q

rs

RÊ Ë

ˆ ¯

2

+ R2 - 2rs cosq

}

b) The surface charge density s on the sphere can be obtained from the boundary conditions of

E

E outside - E inside = E outside =se0

ˆ n =se0

ˆ r

where we have used the fact that the electric field inside the sphere is zero. This equation can be

rewritten as

s = e0Er = -e0

∂V

∂r

Substituting the general expression for V into this equation we obtain

s = -q

4p{

-r + scosqr 2 + s2 - 2rs cosq( )3/ 2 -

-rs2

R2 + scosq

rs

RÊ Ë

ˆ ¯

2

+ R2 - 2rs cosqÊ

Ë Á ˆ

¯ ˜

3/ 2 }

r =R

=

= -q

4p{

-R + scosqR2 + s2 - 2Rscosq( )3/ 2 -

-s2

R+ scosq

s2 + R2 - 2Rscosq( )3/ 2 } =

= -q

4pRs2 - R2

R2 + s2 - 2Rscosq( )3/ 2

- 20 -

The total charge on the sphere can be obtained by integrating s over the surface of the sphere.

The result is

Q = sR2 sinq dq dfÚ = -q

2R s2 - R2( ) sinq dq

R2 + s2 - 2Rscosq( )3/ 20

p

Ú =

= -q2

R s2 - R2( )-

1

sRR2 + s2 - 2Rscosq

È

Î

Í Í Í

˘

˚

˙ ˙ ˙

0

p

=q2

s2 - R2

s1

R + s -

1 R - s

È Î Í

˘ ˚ ˙ = -q

Rs

c) To obtain the electrostatic energy of the system we can determine the work it takes to

assemble the system by calculating the path integral of the force that we need to exert in charge q

in order to move it from infinity to its final position (z = s). Charge q will feel an attractive force

exerted by the induced charge on the sphere. The strength of this force is equal to the force on

charge q exerted by the image charge q'. This force is equal to

F qq ' =1

4pe0

qq'

s - z '( )2ˆ k =

1

4pe0

q -R

sq

Ê Ë

ˆ ¯

s -R2

s

Ê Ë Á

ˆ ¯ ˜

2ˆ k = -

1

4pe0

sRq2

s2 - R2( )2ˆ k

The force that we must exert on q to move it from infinity to its current position is opposite to

F qq ' . The total work required to move the charge is therefore equal to

W = -F qq ' ∑ dl •

s

Ú =1

4pe0

zRq2

z 2 - R2( )2 dz•

s

Ú =1

4pe0

-Rq2

2 z 2 - R2( )•

s

= -1

8pe0

Rq2

s2 - R2


Two semi-infinite grounded conducting planes meet at right angles. In the region between

them, there is a point charge q, situated as shown in Figure 3.8. Set up the image configuration,

and calculate the potential in this region. What charges do you need, and where should they be

located? What is the force on q? How much work did it take to bring q in from infinity?

Consider the system of four charges shown in Figure 3.9. The electrostatic potential

generated by this charge distribution is zero at every point on the yz plane and at every point on

the xz plane. Therefore, the electrostatic potential generated by this image charge distribution

satisfies the same boundary conditions as the electrostatic potential of the original system. The

- 21 -

potential generated by the image charge distribution in the region where x > 0 and y > 0 will be

identical to the potential of the original system. The potential at a point P = (x, y, z) is equal to

VP =1

4pe0

q

x - a( )2 + y - b( )2+ z 2

+1

4pe0

-q

x - a( )2 + y + b( )2+ z 2

+

+1

4pe0

-q

x + a( )2 + y - b( )2+ z 2

+1

4pe0

q

x + a( )2 + y + b( )2+ z 2

y

x

b

aq

V=0

V=0


y

x

bb

b b

a

a

q

q -q

-q

a

a

Figure 3.9. Image charges for problem 3.10.

The force exerted on q can be obtained by calculating the force exerted on q by the image

charges. The total force is equal to the vector sum of the forces exerted by each of the three

- 22 -

image charges. The force exerted by the image charge located at (-a, b, 0) is directed along the

negative x axis and is equal to

F 1 = -1

4pe0

q2

4a2ˆ i

The force exerted by the image charge located at (a, -b, 0) is directed along the negative y axis

and is equal to

F 2 = -1

4pe0

q2

4b2ˆ j

The force exerted by the image charge located at (-a, -b, 0) is directed along the vector

connecting (-a, -b, 0) and (a, b, 0) and is equal to

F 3 =1

4pe0

q2

4a2 + 4b2

aˆ i + b j

a2 + b2=

1

16pe0

q2

a2 + b2( )3/ 2 aˆ i + b j ( )

The total force on charge q is the vector sum of F 1 , F 2 and F 3 :

F tot = F 1 + F 2 + F 3 = -q2

16pe0

1

a2 -a

a2 + b2( )3/ 2

Ê

Ë Á

ˆ

¯ ˜ ˆ i +

1

b2 -b

a2 + b2( )3/ 2

Ê

Ë Á

ˆ

¯ ˜ ˆ j

Ê

Ë Á Á

ˆ

¯ ˜ ˜

The electrostatic potential energy of the system can, in principle, be obtained by calculating the

path integral of -F tot between infinity and (a, b, 0). However, this is not trivial since the force

-F tot is a rather complex function of a and b. An easier technique is to calculate the electrostatic

potential energy of the system with charge and image charges. The potential energy of this

system is equal to

Wimage =1

4pe0

-q2

2a+

-q2

2b+

q2

4a2 + 4b2

Ê Ë Á

ˆ ¯ ˜ =

q2

8pe0

1

a2 + b2-

1

a-

1

b

Ê Ë Á

ˆ ¯ ˜

However, in the real system the electric field is only non-zero in the region where x > 0 and y >

0. Therefore, the total electrostatic potential energy of the real system is only 1/4 of the total

electrostatic potential energy of the image charge system. Thus

Wreal =1

4Wimage =

q2

32pe0

1

a2 + b2-

1

a-

1

b

Ê Ë Á

ˆ ¯ ˜

- 23 -

3.3. Separation of Variables

3.3.1. Separation of variables: Cartesian coordinates

A powerful technique very frequently used to solve partial differential equations is

separation of variables. In this section we will demonstrate the power of this technique by

discussing several examples.

Example: Example 3.3 (Griffiths)

Two infinite, grounded, metal plates lie parallel to the xz plane, one at y = 0, the other at y = p(see Figure 3.10). The left end, at x = 0, is closed off with an infinite strip insulated from the two

plates and maintained at a specified potential V0 y( ) . Find the potential inside this "slot".

y

z

x

V=0

V=0

V(y)

p

Figure 3.10. Example 3.3 (Griffiths).

The electrostatic potential in the slot must satisfy the three-dimensional Laplace equation.

However, since V does not have a z dependence, the three-dimensional Laplace equation reduces

to the two-dimensional Laplace equation:

∂ 2V

∂x2 +∂ 2V

∂y2 = 0

The boundary conditions for the solution of Laplace's equation are:

1. V(x, y = 0) = 0 (grounded bottom plate).

- 24 -

2. V(x, y = p) = 0 (grounded top plate).

3. V(x = 0, y) = V0(y) (plate at x = 0).

4. V Æ 0 when x Æ •.

These four boundary conditions specify the value of the potential on all boundaries surrounding

the slot and are therefore sufficient to uniquely determine the solution of Laplace's equation

inside the slot. Therefore, if we find one solution of Laplace's equation satisfying these

boundary conditions than it must be the correct one. Consider solutions of the following form:

V x,y( ) = X x( )Y y( )

If this is a solution of the two-dimensional Laplace equation than we must require that

∂ 2

∂x2 X x( )Y y( )[ ] +∂ 2

∂y2 X x( )Y y( )[ ] = Y y( ) ∂ 2X x( )∂x2 + X x( ) ∂ 2Y y( )

∂y2 = 0


1

X x( )∂ 2X x( )

∂x2 +1

Y y( )∂ 2Y y( )

∂y2 = 0

The first term of the left-hand side of this equation depends only on x while the second term

depends only on y. Therefore, if this equation must hold for all x and y in the slot we must

require that

1

X x( )∂ 2X x( )

∂x2 = C1 = constant

and

1

Y y( )∂ 2Y y( )

∂y2 = -C1

The differential equation for X can be rewritten as

∂ 2X x( )∂x2 - C1X x( ) = 0

- 25 -

If C1 is a negative number than this equation can be rewritten as

∂ 2X x( )∂x2 + k2X x( ) = 0

where k2 = -C1 . The most general solution of this equation is

X x( ) = A cos kx( ) + B sin kx( )

However, this function is an oscillatory function and does not satisfy boundary condition # 4,

which requires that V approaches zero when x approaches infinity. We therefore conclude that

C1 can not be a negative number. If C1 is a positive number then the differential equation for X

can be written as

∂ 2X x( )∂x2 - k2X x( ) = 0

The most general solution of this equation is

X x( ) = Aekx + Be-kx

This solution will approach zero when x approaches infinity if A = 0. Thus

X x( ) = Be-kx

The solution for Y can be obtained by solving the following differential equation:

∂ 2Y y( )∂y2 + k2Y y( ) = 0

The most general solution of this equation is

Y y( ) = C cos ky( ) + D sin ky( )

Therefore, the general solution for the electrostatic potential V(x,y) is equal to

V x,y( ) = e-kx C cos ky( ) + D sin ky( )( )

- 26 -

where we have absorbed the constant B into the constants C and D. The constants C and D must

be chosen such that the remaining three boundary conditions (1, 2, and 3) are satisfied. The first

boundary condition requires that V(x, y = 0) = 0:

V x,y = 0( ) = e-kx C cos 0( ) + D sin 0( )( ) = Ce-kx = 0

which requires that C = 0. The second boundary condition requires that V(x, y = p) = 0:

V x,y = p( ) = De -kx sin kp( ) = 0

which requires that sin kp = 0. This condition limits the possible values of k to positive

integers:

k = 1, 2, 3, 4, .....

Note: negative values of k are not allowed since exp(-kx) approaches zero at infinity only if k >

0. To satisfy boundary condition # 3 we must require that

V x = 0,y( ) = D sin ky( ) = V0 y( )

This last expression suggests that the only time at which we can find a solution of Laplace's

equation that satisfies all four boundary conditions has the form exp – kx sin ky is when

V0 y( ) happens to have the form sin ky . However, since k can take on an infinite number of

values, there will be an infinite number of solutions of Laplace's equation satisfying boundary

conditions # 1, # 2 and # 4. The most general form of the solution of Laplace's equation will be a

linear superposition of all possible solutions. Thus

V x,y( ) = Dke-kx sin ky( )

k =1

•

Â

Boundary condition # 3 can now be written as

V x = 0,y( ) = Dk sin ky( )k =1

•

Â = V0 y( )

Multiplying both sides by sin(ny) and integrating each side between y = 0 and y = p we obtain

Dk sin ny( )sin ky( )dy0

p

Úk =1

•

Â = sin ny( )V0 y( )dy0

p

Ú

- 27 -

The integral on the left-hand side of this equation is equal to zero for all values of k except k = n.

Thus

Dk sin ny( )sin ky( )dy0

p

Úk =1

•

Â = Dk

p2

d kn =k =1

•

Â p2

Dn

The coefficients Dk can thus be calculated easily:

Dk =2

psin ky( )V0 y( )dy

0

p

Ú

The coefficients Dk are called the Fourier coefficients of V0 y( ) . The solution of Laplace's

equation in the slot is therefore equal to


k =1

•

Â

where

Dk =2

psin ky( )V0 y( )dy

0

p

Ú

Now consider the special case in which V0 y( ) = constant = V0 . In this case the coefficients Dk

are equal to

Dk =2

psin ky( )V0 y( )dy

0

p

Ú = -2

pV0

cos ky( )k

0

p

=2

pV0

k1 - cos kp( )( ) =

0 if k is even

4

pV0

kif k is odd

Ï

Ì Ô Ô

Ó Ô Ô

The solution of Laplace's equation is thus equal to

V x,y( ) =4V0

p1

ke-kx sin ky( )

k =1, 3,5, ..

•

Â


Find the potential in the infinite slot of Example 3.3 (Griffiths) if the boundary at x = 0

consists to two metal stripes: one, from y = 0 to y = p/2, is held at constant potential V0 , and the

other, from y = p/2 to y = p is at potential - V0 .

- 28 -

The boundary condition at x = 0 is

V 0,y( ) =V0 for 0 < y < p / 2

-V0 for p / 2 < y < p

Ï

Ì Ô

Ó Ô

The Fourier coefficients of the function V0 y( ) are equal to

Dk =2

psin ky( )V0 y( )dy

0

p

Ú =2

pV0 sin ky( )dy

0

p / 2

Ú -2

pV0 sin ky( )dy

p / 2

p

Ú =

=2

pV0

k1 + cos kp( ) - 2cos

1

2kp

Ê Ë

ˆ ¯

Ê Ë Á

ˆ ¯ ˜ =

2

pV0

kCk

The values for the first four C coefficients are

C1 = 0 C3 = 0

C2 = 4 C4 = 0

It is easy to see that Ck + 4 = Ck and therefore we conclude that

Ck =4 for k = 2,6,10,...

0 otherwise

Ï

Ì Ô

Ó Ô

The Fourier coefficients Ck are thus equal to

Dk =

8V0

kpfor k = 2,6,10,...

0 otherwise

Ï

Ì Ô Ô

Ó Ô Ô

The electrostatic potential is thus equal to

V x,y( ) =8V0

p1

ke-kx sin ky( )

k =2,6,10, ...

•

Â


For the infinite slot (Example 3.3 Griffiths) determine the charge density s y( ) on the strip at

x=0, assuming it is a conductor at constant potential V0 .

- 29 -

The electrostatic potential in the slot is equal to

V x,y( ) =4V0

p1

ke-kx sin ky( )

k =1, 3,5, ..

•

Â

The charge density at the plate at x = 0 can be obtained using the boundary condition for the

electric field at a boundary:

E x = +0 - E x =-0 = E x = +0 =se0

ˆ n

where ˆ n is directed along the positive x axis. Since E = -— V this boundary condition can be

rewritten as

∂V

∂x x = +0

= -se0

Differentiating V(x,y) with respect to x we obtain

∂V

∂x= -

4V0

pe-kx sin ky( )

k =1, 3,5, ...

•

Â

At the x = 0 boundary we obtain

∂V

∂x x = +0

= -4V0

psin ky( )

k =1, 3,5, ...

•

Â

The charge density s on the x = 0 strip is therefore equal to

s = -e0

∂V

∂x x = +0

=4V0e0

psin ky( )

k =1, 3,5, ...

•

Â

Example: Double infinite slots

The slot of example 3.3 in Griffiths and its mirror image at negative x are separated by an

insulating strip at x = 0. If the charge density s(y) on the dividing strip is given, determine the

potential in the slot.

The boundary condition at x = 0 requires that

- 30 -

E x = +0 - E x =-0 = 2E x = +0 =se0

ˆ n

where ˆ n is directed along the positive x axis. Here we have used the symmetry of the

configuration which requires that the electric field in the region x < 0 is the mirror image of the

field in the region x > 0. Since E = -— V this boundary condition can be rewritten as

∂V

∂x x = +0

= -1

2

s y( )e0

We will first determine the potential in the x > 0 region. Following the same procedure as in

Example 3 we obtain for the electrostatic potential


k =1

•

Â

where the constants Dk must be chosen such that the boundary condition at x = 0 is satisfied.

This requires that

∂V

∂x x = +0

= - kDke-kx sin ky( )

k =1

•

Âx = +0

= - kDk sin ky( ) = -1

2

s y( )e0k =1

•

Â

Thus

kDk sin ky( )k =1

•

Â =1

2

s y( )e0

The constants Dk can be determined by multiplying both sides of this equation with sin my( ) and

integrating both sides with respect to y between y = 0 and y = p. The result is

1

2e0

s y( )sin my( )dy0

p

Ú = kDk sin ky( )sin my( )dy0

p

Ú = mDm

p2k =1

•

Â

The constants Ck are thus equal to

Dk =1

pe0ks y( )sin ky( )dy

0

p

Ú

The electrostatic potential is thus equal to

- 31 -

V x,y( ) =1

pe0

1

ks y( )sin ky( )dy

0

p

Ú{ }e-kx sin ky( )È Î Í

˘ ˚ ˙

k =1

•

Â

3.3.2. Separation of variables: spherical coordinates

Consider a spherical symmetric system. If we want to solve Laplace's equation it is natural to

use spherical coordinates. Assuming that the system has azimuthal symmetry ( ∂V / ∂f = 0)

Laplace's equation reads

1

r 2

∂∂r

r 2 ∂V

∂rÊ Ë

ˆ ¯ +

1

r 2 sinq∂

∂qsinq

∂V

∂qÊ Ë

ˆ ¯ = 0

Multiplying both sides by r2 we obtain

∂∂r

r 2 ∂V

∂rÊ Ë

ˆ ¯ +

1

sinq∂

∂qsinq

∂V

∂qÊ Ë

ˆ ¯ = 0

Consider the possibility that the general solution of this equation is the product of a function

R r( ) , which depends only on the distance r, and a function a q( ) , which depends only on the

angle q:

V r ,q( ) = R r( )a q( )

Substituting this "solution" into Laplace's equation we obtain

a q( ) ∂∂r

r 2 ∂R r( )∂r

Ê Ë Á

ˆ ¯ ˜ +

R r( )sinq

∂∂q

sinq∂a q( )

∂qÊ Ë Á

ˆ ¯ ˜ = 0

Dividing each term of this equation by R r a q we obtain

1

R r( )∂

∂rr 2 ∂R r( )

∂r

Ê Ë Á

ˆ ¯ ˜ +

1

a q( )1

sinq∂

∂qsinq

∂a q( )∂q

Ê Ë Á

ˆ ¯ ˜ = 0

The first term in this expression depends only on the distance r while the second term depends

only on the angle q. This equation can only be true for all r and q if

1

R r( )∂

∂rr 2 ∂R r( )

∂r

Ê Ë Á

ˆ ¯ ˜ = m m + 1( ) = constant

and

- 32 -

1

a q( )1

sinq∂

∂qsinq

∂a q( )∂q

Ê Ë Á

ˆ ¯ ˜ = -m m + 1( )

Consider a solution for R of the following form:

R r( ) = Ark

where A and k are arbitrary constants. Substituting this expression in the differential equation for

R(r) we obtain

1

Ark

∂∂r

kr2Ark -1( ) =k

r k k + 1( )r k = k k + 1( ) = m m + 1( )

Therefore, the constant k must satisfy the following relation:

k k + 1( ) = k2 + k = m m + 1( )

This equation gives us the following expression for k

k =-1 ± 1 + 4m m + 1( )

2=

-1 ± 2 m +1

2Ê Ë

ˆ ¯

2=

m

or

- m + 1( )

Ï

Ì Ô

Ó Ô

The general solution for R r( ) is thus given by

R r( ) = Arm +B

r m +1

where A and B are arbitrary constants.

The angle dependent part of the solution of Laplace's equation must satisfy the following

equation

∂∂q

sinq∂a q( )

∂qÊ Ë Á

ˆ ¯ ˜ + m m + 1( )a q( )sinq = 0

The solutions of this equation are known as the Legendre polynomial Pm cosq( ) . The Legendre

polynomials have the following properties:

1. if m is even: Pm x( ) = Pm -x( )

2. if m is odd: Pm x( ) = -Pm -x( )

- 33 -

3. Pm 1( ) = 1 for all m

4. Pn x( )Pm x( )dx-1

1

Ú =2

2m + 1d nm or Pn cosq( )Pm cosq( )sinqdq

-1

1

Ú =2

2m + 1d nm

Combining the solutions for R r( ) and a q( ) we obtain the most general solution of Laplace's

equation in a spherical symmetric system with azimuthal symmetry:

V r ,q( ) = Amr m +Bm

r m +1

Ê Ë

ˆ ¯ Pm cosq( )

m =0

•

Â


The potential at the surface of a sphere is given by

V0 q( ) = k cos 3q( )

where k is some constant. Find the potential inside and outside the sphere, as well as the surface

charge density s q( ) on the sphere. (Assume that there is no charge inside or outside of the

sphere.)

The most general solution of Laplace's equation in spherical coordinates is


r m +1

Ê Ë

ˆ ¯ Pm cosq( )

m =0

•

Â

First consider the region inside the sphere (r < R). In this region Bm = 0 since otherwise V r ,q( )would blow up at r = 0. Thus

V r ,q( ) = Amr mPm cosq( )m =0

•

Â

The potential at r = R is therefore equal to

V R,q( ) = AmRmPm cosq( )m =0

•

Â = k cos 3q( )

Using trigonometric relations we can rewrite cos 3q( ) as

cos 3q( ) = 4 cos3q - 3cosq =8

5P3 cosq( ) -

3

5P1 cosq( )

- 34 -

Substituting this expression in the equation for V R,q( ) we obtain


•

Â =8k

5P3 cosq( ) -

3k

5P1 cosq( )

This equation immediately shows that Am = 0 unless m = 1 or m = 3. If m = 1 or m = 3 then

A1 = -3

5

k

R

A3 =8

5

k

R3

The electrostatic potential inside the sphere is therefore equal to

V r ,q( ) =8k

5

r 3

R3 P3 cosq( ) -3k

5

r

RP1 cosq( )

Now consider the region outsider the sphere (r > R). In this region Am = 0 since otherwise

V r ,q( ) would blow up at infinity. The solution of Laplace's equation in this region is therefore

equal to

V r ,q( ) =Bm

r m +1 Pm cosq( )m =0

•

Â

The potential at r = R is therefore equal to

V R,q( ) =Bm

Rm +1 Pm cosq( )m =0

•

Â =8k

5P3 cosq( ) -

3k

5P1 cosq( )

The equation immediately shows that Bm = 0 except when m = 1 or m = 3. If m = 1 or m = 3

then

B1 = -3

5kR2

B3 =8

5kR4

The electrostatic potential outside the sphere is thus equal to

V r ,q( ) = -3k

5

R2

r 2 P1 cosq( ) +8k

5

R4

r 4 P3 cosq( )

- 35 -

The charge density on the sphere can be obtained using the boundary conditions for the electric

field at a boundary:

E r =R + - E r =R - =s q( )

e0

ˆ r

Since E = -— V this boundary condition can be rewritten as

∂V

∂r r =R +-

∂V

∂r r =R -= -

s q( )e0

The first term on the left-hand side of this equation can be calculated using the electrostatic

potential just obtained:

∂V

∂r r =R +=

6k

5

R2

r 3 P1 cosq( ) -32k

5

R4

r 5 P3 cosq( )Ê Ë Á

ˆ ¯ ˜

r =R +

=k

5R6P1 cosq( ) - 32P3 cosq( )( )

In the same manner we obtain

∂V

∂r r =R -= -

3k

5

1

RP1 cosq( ) +

24k

5

r 2

R3 P3 cosq( )Ê Ë Á

ˆ ¯ ˜

r =R -

=k

5R-3P1 cosq( ) + 24P3 cosq( )( )

Therefore,

∂V

∂r r =R +-

∂V

∂r r =R -=

k

5R9P1 cosq( ) - 56P3 cosq( )( ) = -

s q( )e0

The charge density on the sphere is thus equal to

s q( ) =ke0

5R-9P1 cosq( ) + 56P3 cosq( )( )


Suppose the potential V0 q( ) at the surface of a sphere is specified, and there is no charge

inside or outside the sphere. Show that the charge density on the sphere is given by

s q( ) =e0

2R2m + 1( )2

CmPm cosq( )m =0

•

Â

where

- 36 -

Cm = V0 q( )Pm cosq( )sinqdq0

p

Ú

Most of the solution of this problem is very similar to the solution of Problem 3.18. First

consider the electrostatic potential inside the sphere. The electrostatic potential in this region is

given by


•

Â

and the boundary condition is


•

Â = V0 q( )

The coefficients Am can be determined by multiplying both sides of this equation by

Pn cosq( )sinq and integrating with respect to q between q = 0 and q = p:

V0 q( )Pn cosq( )sinqdq0

p

Ú = AmRm

m =0

•

Â Pm cosq( )Pn cosq( )sinqdq0

p

Ú =2

2n + 1AnR

n

Thus

Am =2m + 1

2

1

Rm V0 q( )Pm cosq( )sinqdq0

p

Ú

In the region outside the sphere the electrostatic potential is given by

V r ,q( ) =Bm


•

Â

and the boundary condition is

V R,q( ) =Bm

Rm +1 Pm cosq( )m =0

•

Â = V0 q( )

The coefficients Bm are given by

Bm =2m + 1

2Rm +1 V0 q( )Pm cosq( )sinqdq

0

p

Ú

The charge density s q( ) on the surface of the sphere is equal to

- 37 -

s q( ) = -e0

∂V

∂r r =R +-

∂V

∂r r =R -

Ï Ì Ô

Ó Ô ¸ ˝ Ô

˛ Ô

Differentiating V r ,q( ) with respect to r in the region r > R we obtain

∂V

∂r r =R += - m + 1( ) Bm

Rm + 2 Pm cosq( )m =0

•

Â

Differentiating V r ,q( ) with respect to r in the region r < R we obtain

∂V

∂r r =R -= mAmRm -1Pm cosq( )

m =0

•

Â

The charge density is therefore equal to

s q( ) = -e0 - m + 1( ) Bm

Rm + 2 Pm cosq( )È Î Í

˘ ˚ ˙

m =0

•

Â - mAmRm -1Pm cosq( )[ ]m =0

•

ÂÏ Ì Ó

¸ ˝ ˛

=

= e0 m + 1( ) Bm

Rm + 2 + mAmRm -1Ê Ë

ˆ ¯ Pm cosq( )È

Î Í

˘ ˚ ˙

m =0

•

ÂÏ Ì Ó

¸ ˝ ˛

Substituting the expressions for Am and Bm into this equation we obtain

s q( ) = e0 m + 1( ) 2m + 1( )Rm +1

2Rm + 2 + m2m + 1( )2Rm Rm -1Ê

Ë Á ˆ ¯ ˜ CmPm cosq( )È

Î Í Í

˘

˚ ˙ ˙ m =0

•

ÂÏ Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô =

=e0

2R2m + 1( )2

CmPm cosq( )[ ]m =0

•

ÂÏ Ì Ó

¸ ˝ ˛

where

Cm = V0 q( )Pm cosq( )sinqdq0

p

Ú


Solve Laplace's equation by separation of variables in cylindrical coordinates, assuming there

is no dependence on z (cylindrical symmetry). Make sure that you find all solutions to the radial

equation. Does your result accommodate the case of an infinite line charge?

- 38 -

For a system with cylindrical symmetry the electrostatic potential does not depend on z. This

immediately implies that ∂V / ∂z = 0. Under this assumption Laplace's equation reads

1

r

∂∂r

r∂V

∂rÊ Ë

ˆ ¯ +

1

r 2

∂ 2V

∂f 2 = 0

Consider as a possible solution of V:

V r ,f( ) = R r( )a f( )

Substituting this solution into Laplace's equation we obtain

a f( )r

∂∂r

r∂R r( )

∂r

Ê Ë Á

ˆ ¯ ˜ +

R r( )r 2

∂ 2a f( )∂f 2 = 0

Multiplying each term in this equation by r2 and dividing by R r( )a f( ) we obtain

r

R r( )∂

∂rr

∂R r( )∂r

Ê Ë Á

ˆ ¯ ˜ +

1

a f( )∂ 2a f( )

∂f 2 = 0

The first term in this equation depends only on r while the second term in this equation depends

only on f. This equation can therefore be only valid for every r and every f if each term is equal

to a constant. Thus we require that

1

R r( )∂

∂rr

∂R r( )∂r

Ê Ë Á

ˆ ¯ ˜ = constant = g

and

1

a f( )∂ 2a f( )

∂f 2 = -g

First consider the case in which g = -m2 < 0. The differential equation for a f( ) can be

rewritten as

∂ 2a f( )∂f 2 - m2a f( ) = 0

The most general solution of this differential solution is

- 39 -

am f( ) = Cmemf + Dme-mf

However, in cylindrical coordinates we require that any solution for a given f is equal to the

solution for f + 2p. Obviously this condition is not satisfied for this solution, and we conclude

that g = m2 ≥ 0. The differential equation for a f( ) can be rewritten as

∂ 2a f( )∂f 2 + m2a f( ) = 0

The most general solution of this differential solution is

am f( ) = Cm cos mf( ) + Dm sin mf( )

The condition that am f( ) =a m f + 2p( ) requires that m is an integer. Now consider the radial

function R r( ) . We will first consider the case in which g = m2 > 0. Consider the following

solution for R r( ) :

R r( ) = Ark

Substituting this solution into the previous differential equation we obtain

r

Ark

∂∂r

r∂

∂rArk( )Ê

Ë ˆ ¯ =

1

Ark -1

∂∂r

r kArk -1( )( ) =1

Ark -1

∂∂r

kArk( ) =1

Ark -1 k2Ark -1= k2 = m2

Therefore, the constant k can take on the following two values:

k+ = m

k- = -m

The most general solution for R r( ) under the assumption that m2 > 0 is therefore

R r( ) = Amr m +Bm

r m

Now consider the solutions for R r( ) when m2 = 0. In this case we require that

∂∂r

r∂R r( )

∂r

Ê Ë Á

ˆ ¯ ˜ = 0

or

- 40 -

r∂R r( )

∂r= constant = a0


∂R r( )∂r

=a0

r

If a0 = 0 then the solution of this differential equation is

R r( ) = b0 = constant

If a0 π 0 then the solution of this differential equation is

R r( ) = a0 ln r( ) + b0

Combining the solutions obtained for m2 = 0 with the solutions obtained for m2 > 0 we conclude

that the most general solution for R r( ) is given by

R r( ) = a0 ln r( ) + b0 + Amr m +Bm

r m

È Î Í

˘ ˚ ˙

m =1

•

Â

Therefore, the most general solution of Laplace's equation for a system with cylindrical

symmetry is

V r ,f( ) = a0 ln r( ) + b0 + Amr m +Bm

r m

Ê Ë

ˆ ¯ Cm cos mf( ) + Dm sin mf( )( )È

Î Í

˘ ˚ ˙

m =1

•

Â


A charge density

s = a sin 5f( )

is glued over the surface of an infinite cylinder of radius R. Find the potential inside and outside

the cylinder.

The electrostatic potential can be obtained using the general solution of Laplace's equation

for a system with cylindrical symmetry obtained in Problem 3.24. In the region inside the

cylinder the coefficient Bm must be equal to zero since otherwise V r ,f( ) would blow up at r = 0.

For the same reason a0 = 0. Thus

- 41 -

V r ,f( ) = bin, 0 + r m Cin,m cos mf( ) + Din,m sin mf( )( )[ ]m =1

•

Â

In the region outside the cylinder the coefficients Am must be equal to zero since otherwise

V r ,f( ) would blow up at infinity. For the same reason a0 = 0. Thus

V r ,f( ) = bout , 0 +1

r m Cout , m cos mf( ) + Dout , m sin mf( )( )È Î Í

˘ ˚ ˙

m =1

•

Â

Since V r ,f( ) must approach 0 when r approaches infinity, we must also require that bout, 0 is

equal to 0. The charge density on the surface of the cylinder is equal to

s f( ) = -e0

∂V

∂r r =R +-

∂V

∂r r =R -

È

Î Í Í

˘

˚ ˙ ˙

Differentiating V r ,f( ) in the region r > R and setting r = R we obtain

∂V

∂r r =R += -

m

Rm +1 Cout , m cos mf( ) + Dout , m sin mf( )( )È Î Í

˘ ˚ ˙

m =1

•

Â

Differentiating V r ,f( ) in the region r < R and setting r = R we obtain

∂V

∂r r =R -= mRm -1 Cin,m cos mf( ) + Din, m sin mf( )( )[ ]

m =1

•

Â

The charge density on the surface of the cylinder is therefore equal to

s f( ) = e0 mRm -1Cin,m +m

Rm +1 Cout , m

Ê Ë

ˆ ¯ cos mf( ) + mRm -1Din,m +

m

Rm +1 Dout, m

Ê Ë

ˆ ¯ sin mf( )È

Î Í

˘ ˚ ˙

m =1

•

Â

Since the charge density is proportional to sin 5f( ) we can conclude immediately that

Cin, m = Cout , m = 0 for all m and that Din, m = Dout , m = 0 for all m except m = 5. Therefore

s f( ) = e0 5R4Din, 5 +5

R6 Dout , 5

Ê Ë

ˆ ¯ sin 5f( ) = a sin 5f( )

This requires that

5R4Din, 5 +5

R6 Dout , 5 =a

e0

- 42 -

A second relation between Din, 5 and Dout, 5 can be obtained using the condition that the

electrostatic potential is continuous at any boundary. This requires that

Vin R,f( ) = bin, 0 + R5Din, 5 sin 5f( ) = Vout R,f( ) =Dout , 5

R5 sin 5f( )

Thus

bin, 0 = 0

and

Dout, 5 = R10Din, 5

We now have two equations with two unknown, Din, 5 and Dout, 5 , which can be solved with the

following result:

Din, 5 =a

10e0

1

R4

and

Dout, 5 =a

10e0

R6

The electrostatic potential inside the cylinder is thus equal to

Vin r ,f( ) = r 5Din, 5 sin 5f( ) =a

10e0

r 5

R4 sin 5f( )

The electrostatic potential outside the cylinder is thus equal to

Vout r ,f( ) =Dout , 5

r 5 sin 5f( ) =a

10e0

R6

r 5 sin 5f( )


A conducting sphere of radius a, at potential V0 , is surrounded by a thin concentric spherical

shell of radius b, over which someone has glued a surface charge

s q( ) = s 0 cosq

- 43 -

where s0 is a constant.

a) Find the electrostatic potential in each region:

i) r > b

ii) a < r < b

b) Find the induced surface charge s q( ) on the conductor.

c) What is the total charge of the system? Check that your answer is consistent with the

behavior of V at large r.

a) The system has spherical symmetry and we can therefore use the most general solution of

Laplace's equation in spherical coordinates:


r m +1

Ê Ë

ˆ ¯ Pm cosq( )

m =0

•

Â

In the region inside the sphere Bm = 0 since otherwise V r ,q( ) would blow up at r = 0. Therefore


•

Â

The boundary condition for V r ,q( ) is that it is equal to V0 at r = a:

V a ,q( ) = AmamPm cosq( )m =0

•

Â = V0 = V0P0 cosq( )

This immediately shows that Am = 0 for all m except m = 0:

A0 = V0

The electrostatic potential inside the sphere is thus given by

Vr <a r ,q( ) = V0

which should not come as a surprise.

In the region outside the shell Am = 0 since otherwise V r ,q( ) would blow up at infinity.

Thus

Vr >b r ,q( ) =Bout , m


•

Â

In the region between the sphere and the shell the most general solution for V r, q is given by

- 44 -

Va<r <b r ,q( ) = Ain,mr m +Bin,m

r m +1

Ê Ë Á

ˆ ¯ ˜ Pm cosq( )

m =0

•

Â

The boundary condition for Va<r <b r ,q( ) at r = a is

Va<r <b a ,q( ) = Ain,mam +Bin,m

am +1

Ê Ë Á

ˆ ¯ ˜ Pm cosq( )

m =0

•

Â = V0 = V0P0 cosq( )

This equation can only be satisfied if

Ain, mam +Bin,m

am +1 = 0 if m > 0

Ain, 0 +Bin,0

a= V0 if m = 0

The requirement that the electrostatic potential is continuous at r = b requires that

Ain,mbm +Bin,m

bm +1

Ê Ë Á

ˆ ¯ ˜ Pm cosq( )

m =0

•

Â =Bout , m

bm +1 Pm cosq( )m =0

•

Â

or

Ain, mbm +Bin,m

bm +1 =Bout , m

bm +1

This condition can be rewritten as

Bout, m - Bin,m = Ain,mb2m +1

The other boundary condition for the electrostatic potential at r = b is that it must produce the

charge distribution given in the problem. This requires that

s f( ) = -e0

∂V

∂r r =b +

-∂V

∂r r =b-

È

Î Í Í

˘

˚ ˙ ˙

= e0

m + 1

bm + 2 Bout, m - Bin,m( ) + mAin,mbm -1Ê Ë

ˆ ¯ Pm cosq( )

m =0

•

Â =

= s 0 cosq = s 0P1 cosq( )

This condition is satisfied if

m + 1

bm + 2 Bout , m - Bin,m( ) + mAin,mbm -1 = 0 if m π1

- 45 -

2

b3 Bout ,1 - Bin,1( ) + Ain,1 =s 0

e0

if m =1

Substituting the relation between the various coefficients obtained by applying the continuity

condition we obtain

m + 1

bm + 2 Ain, mb2m +1 + mAin,mbm -1 = 2m + 1( )Ain,mbm -1 = 0 if m π1

2

b3 Ain,1b3 + Ain,1 = 3Ain,1 =

s 0

e0

if m =1

These equations show that

Ain, m = 0 if m π1

Ain,1 =s 0

3e0

if m =1

Using these values for Ain, m we can show that

Bout, m - Bin,m = 0 if m π1

Bout,1 - Bin,1 =s 0

3e0

b3 if m =1

The boundary condition for V at r = a shows that

Bin, m = -Ain,ma2m +1 = 0 if m ≥ 2

Bin,1 = -Ain,1a3 = -

s 0

3e0

a3 if m =1

Bin, 0 = a V0 - Ain,0( ) = aV0 if m = 0

These values for Bin, m immediately fix the values for Bout, m :

Bout, m = Bin,m = 0 if m ≥ 2

Bout,1 =s 0

3e0

b3 + Bin,1 =s 0

3e0

b3 - a3( ) if m = 1

Bout, 0 = Bin,0 = aV0 if m = 0

- 46 -

The potential in the region outside the shell is therefore equal to

Vr >b r ,q( ) =aV0

rP0 cosq( ) +

s 0

3e0

1

r 2 b3 - a3( )P1 cosq( )

The potential in the region between the sphere and the shell is equal to

Va<r <b r ,q( ) =aV0

rP0 cosq( ) +

s 0

3e0

r -a3

r 2

Ê Ë Á

ˆ ¯ ˜ P1 cosq( )

b) The charge density on the surface of the sphere can be found by calculating the slope of the

electrostatic potential at this surface:

s q( ) = -e0

∂V

∂r r =a +-

∂V

∂r r =a-

È

Î Í Í

˘

˚ ˙ ˙

= -e0 -V0

a+

s 0

e0

cosqÈ

Î Í

˘

˚ ˙ =

e0V0

a- s 0 cosq

c) The total charge on the sphere is equal to

Qa =0

p

Ú s q( ) a2 sinq dq df0

2p

Ú = 2pa2 2e0V0

a- s 0 cosq sinq dq

0

p

ÚÏ Ì Ó

¸ ˝ ˛

= 4p ae0V

The total charge on the shell is equal to zero. Therefore the total charge of the system is equal to

Qtotal = 4p ae0V0

The electrostatic potential at large distances will therefore be approximately equal to

V =1

4pe0

Qtotal

r=

1

4pe0

4p ae0V

r=

aV0

r

This is equal to limit of the exact electrostatic potential when r Æ •.

3.4. Multipole Expansions

Consider a given charge distribution r. The potential at a point P (see Figure 3.11) is equal

to

V P( ) =1

4pe0

rd

dtVolume

Ú

- 47 -

where d is the distance between P and a infinitesimal segment of the charge distribution. Figure

3.11 shows that d can be written as a function of r, r' and q:

d2 = r 2 + r'2 -2rr 'cosq = r 2 1 +r '

rÊ Ë

ˆ ¯

2

- 2r '

rÊ Ë

ˆ ¯ cosq

Ê

Ë Á ˆ

¯ ˜

y

xr'

r

dPq

Figure 3.11. Charge distribution rrrr.


1

d=

1

r

1

1 +r'

rÊ Ë

ˆ ¯

2

- 2r '

rÊ Ë

ˆ ¯ cosq

Ê

Ë Á ˆ

¯ ˜

At large distances from the charge distribution r >> r' and consequently r' /r <<1. Using the

following expansion for 1 / 1 + x( ) :

1

1 + x( )= 1 -

1

2x +

3

8x2 -

5

16x3 + ....

we can rewrite 1/d as

1

dª

1

r1 -

1

2

r '

rÊ Ë

ˆ ¯

r '

rÊ Ë

ˆ ¯ - 2cosq

Ê Ë Á

ˆ ¯ ˜ +

3

8

r '

rÊ Ë

ˆ ¯

2 r'

rÊ Ë

ˆ ¯ - 2cosq

Ê Ë Á

ˆ ¯ ˜

2

- .....Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô =

=1

r1 +

r '

rÊ Ë

ˆ ¯ cosq +

r '

rÊ Ë

ˆ ¯

2 3

2cos2q -

1

2Ê Ë

ˆ ¯ + .....

Ï Ì Ô

Ó Ô ¸ ˝ Ô

˛ Ô =

1

r

r '

rÊ Ë

ˆ ¯

n

Pn cosq( )n=0

•

Â

Using this expansion of 1/d we can rewrite the electrostatic potential at P as

- 48 -

V P( ) =1

4pe0

1

r n +1 r r '( )r 'n Pn cosq( )dtVolume

Ún=0

•

Â

This expression is valid for all r (not only r >> r'). However, if r >> r' then the potential at P

will be dominated by the first non-zero term in this expansion. This expansion is known as the

multipole expansion. In the limit of r >> r' only the first terms in the expansion need to be

considered:

V P( ) =1

4pe0

1

rrdt

VolumeÚ +

1

r 2 rr 'cosqdtVolume

Ú +1

r 3 rr '23

2cos2q -

1

2Ê Ë

ˆ ¯ dt

VolumeÚ + ....

Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô

The first term in this expression, proportional to 1/r, is called the monopole term. The second

term in this expression, proportional to 1/r2, is called the dipole term. The third term in this

expression, proportional to 1/r3, is called the quadrupole term.

3.4.1. The monopole term.

If the total charge of the system is non zero then the electrostatic potential at large distances

is dominated by the monopole term:

V P( ) =1

4pe0

1

rrdt

VolumeÚ =

1

4pe0

Q

r

where Q is the total charge of the charge distribution.

The electric field associated with the monopole term can be obtained by calculating the

gradient of V P :

E P( ) = - — V P( ) = -Q

4pe0

— 1

rÊ Ë

ˆ ¯ =

1

4pe0

Q

r 2 ˆ r

3.4.2. The dipole term.

If the total charge of the charge distribution is equal to zero (Q = 0) then the monopole term

in the multipole expansion will be equal to zero. In this case the dipole term will dominate the

electrostatic potential at large distances

V P( ) =1

4pe0

1

r 2 rr 'cosq dtVolume

Ú

- 49 -

Since q is the angle between r and r ' we can rewrite r'cosq as

r'cosq = ˆ r ∑ r '

The electrostatic potential at P can therefore be rewritten as

V P( ) =1

4pe0

ˆ r

r 2 ∑ rr ' dtVolume

Ú =1

4pe0

p ∑ ˆ r

r 2

In this expression p is the dipole moment of the charge distribution which is defined as

p = rr ' dtVolume

Ú

The electric field associated with the dipole term can be obtained by calculating the gradient of

V P( ) :

Er P( ) = -∂V P( )

∂r=

2p cosq4pe0

1

r 3

Eq P( ) = -1

r

∂V P( )∂q

=p sinq4pe0

1

r 3

Ef P( ) = -1

r sinq∂V P( )

∂f= 0

Example

Consider a system of two point charges shown in Figure 3.12. The total charge of this

system is zero, and therefore the monopole term is equal to zero. The dipole moment of this

system is equal to

p = -q( )r - + + q( )r + = q r + - r -( ) = qs

where s is the vector pointing from -q to +q.

The dipole moment of a charge distribution depends on the origin of the coordinate system

chosen. Consider a coordinate system S and a charge distribution r. The dipole moment of this

charge distribution is equal to

- 50 -

p S = r r S dtVolume

Ú

A second coordinate system S' is displaced by d with respect to S:

r S ' = r S + d

The dipole moment of the charge distribution in S' is equal to

p S ' = r r S ' dtVolume

Ú = r r S dtVolume

Ú + d r dtVolume

Ú = p S + d Q

This equation shows that if the total charge of the system is zero (Q = 0) then the dipole moment

of the charge distribution is independent of the choice of the origin of the coordinate system.

y

x

r-

r+

s

+q

-q

Figure 3.12. Electric dipole moment.


A thin insulating rod, running from z = -a to z = +a, carries the following line charges:

a) l = l0 cosp z

2aÊ Ë

ˆ ¯

b) l = l0 sinp z

aÊ Ë

ˆ ¯

c) l = l0 cosp z

aÊ Ë

ˆ ¯

In each case, find the leading term in the multipole expansion of the potential.

a) The total charge on the rod is equal to

Qtot = ldz-a

+a

Ú = l0 cosp z

2aÊ Ë

ˆ ¯ dz

-a

+ a

Ú =4a

pl0

- 51 -

Since Qtot π 0, the monopole term will dominate the electrostatic potential at large distances.

Thus

VP =1

4pe0

4a

pl0

1

r

b) The total charge on the rod is equal to zero. Therefore, the electrostatic potential at large

distances will be dominated by the dipole term (if non-zero). The dipole moment of the rod is

equal to

p = zldz-a

+a

Ú = zl0 sinp z

aÊ Ë

ˆ ¯ dz

-a

+ a

Ú =2a2

pl0

Since the dipole moment of the rod is not equal to zero, the dipole term will dominate the

electrostatic potential at large distances. Therefore

VP =1

4pe0

2a2

pl0

1

r 2 cosq

c) For this charge distribution the total charge is equal to zero and the dipole moment is equal to

zero. The electrostatic potential of this charge distribution is dominated by the quadrupole term.

I2 = z 2ldz-a

+a

Ú = z 2l0 cosp z

aÊ Ë

ˆ ¯ dz

-a

+ a

Ú =4a3

p 2 l0

The electrostatic potential at large distance from the rod will be equal to

VP = 14 p e 0

– 4 a3

p2l 0

1r 3

12

3 cos2 q – 1


Four particles (one of charge q, one of charge 3q, and two of charge -2q) are placed as shown

in Figure 3.12, each a distance d from the origin. Find a simple approximate formula for the

electrostatic potential, valid at a point P far from the origin.

The total charge of the system is equal to zero and therefore the monopole term in the

multipole expansion is equal to zero. The dipole moment of this charge distribution is equal to

- 52 -

p = qir ii

Â = -2q( )d j + q( ) -d( ) ˆ k + -2q( ) -d( ) j + 3q( )d ˆ k = 2qd ˆ k

The Cartesian coordinates of P are

x = r sinq cosf

y = r sinq sinf

z = r cosq

The scalar product between p and ˆ r is therefore

p ∑ ˆ r = 2qd cosq

The electrostatic potential at P is therefore equal to

VP =1

4pe0

p ∑ ˆ r

r 2 =1

4pe0

2qd cosqr 2

z

y

-2q

3q

-2q

+q



A charge Q is distributed uniformly along the z axis from z = -a to z =a. Show that the

electric potential at a point r ,q( ) is given by

V r ,q( ) =Q

4pe0

1

r1 +

1

3

a

rÊ Ë

ˆ ¯

2

P2 cosq( ) +1

5

a

rÊ Ë

ˆ ¯

4

P4 cosq( ) + ....Ê

Ë Á ˆ

¯ ˜

- 53 -

for r > a.

The charge density along this segment of the z axis is equal to

r =Q

2a

Therefore, the nth moment of the charge distribution is equal to

In = z nrdz-a

a

Ú =Q

2az ndz

-a

a

Ú =Q

2a

z n +1

n + 1-a

a

=Q

2a

an +1

n + 11 - -1( )n +1{ } =

Q

2

an

n + 11 - -1( )n +1{ }

This equation immediately shows that

In =an

n + 1Q if n is even

In = 0 if n is odd

The electrostatic potential at P is therefore equal to

V r ,q( ) =1

4pe0

1

r n +1 InPn cosq( )n=0

•

Â =Q

4pe0

1

r1 +

1

3

a

rÊ Ë

ˆ ¯

2

P2 cosq( ) +1

5

a

rÊ Ë

ˆ ¯

4

P4 cosq( ) + ....Ê

Ë Á ˆ

¯ ˜

- 1 -

Chapter 4. Electrostatic Fields in Matter

4.1. Polarization

A neutral atom, placed in an external electric field, will experience no net force. However,

even though the atom as a whole is neutral, the positive charge is concentrated in the nucleus

(radius = 10-14 m) while the negative charge forms an electron cloud (radius = 10-10 m)

surrounding the nucleus (see Figure 4.1). The nucleus of the atom will experience a force

pointing in the same direction as the external electric field (to the right in Figure 4.1) and of

magnitude qEext. The negatively charged electron cloud will experience a force of the same

magnitude, but pointed in a direction opposite to the direction of the electric field. As a result of

the external force, the nucleus will move in the direction of the electric field until the external

force on it is canceled by the force exerted on the nucleus by the electron cloud.

F-

+

F+

Eext

Figure 4.1. Atom in external electric field.

Consider an electron cloud with a constant volume charge density r and a radius a. If the

total charge of the electron cloud is -q then the corresponding charge density r is equal to

r =-q

43

pa3= -

3q

4pa3

The electric field inside the uniformly charged cloud is equal to

- 2 -

E r( ) = -1

4pe0

qr

a3

where r is the distance from the center of the cloud. Suppose that as a result of the external

electric field the nucleus moves by a distance d with respect to the center of the electron cloud.

The electric force exerted on the nucleus by the electron cloud is equal to

Fcloud = qE d( ) = -1

4pe0

q2d

a3

The equilibrium position of the nucleus is that position where the external force is canceled by

the force exerted on it by the electron cloud:

F cloud + F ext = 0

This expression can be rewritten as

qEext -1

4pe0

q2d

a3 = 0

The equilibrium distance d is thus equal to

d = 4pe0a3 Eext

q

The induced dipole moment p of the atom is defined as

p = qd = 4pe0a3E ext

Therefore, the magnitude of the induced dipole moment is proportional to the magnitude of the

external electric field, and its direction is equal to the direction of the external electric field. The

constant of proportionality is called the atomic polarizability a and is defined as

a =p

E ext

= 4pe0a3

Although this model of the atom is extremely crude, it produces results that are in reasonable

agreement with direct measurements of the atomic polarizability.


- 3 -

According to quantum mechanics, the electron cloud for a hydrogen atom in its ground state

has a charge density equal to

r r( ) =q

pa3 e-2r / a

where q is the charge of the electron and a is the Bohr radius. Find the atomic polarizability of

such an atom.

As a result of an external electric field the nucleus of the atom will be displaced by a distance

d with respect to the center of the electron cloud. The force exerted on the nucleus by the

electron cloud is equal to

F cloud d( ) = qE cloud d( )

where E cloud is the electric field generated by the electron cloud. The electric field generated by

the electron cloud can be calculated using Gauss's law:

E cloud d( ) =1

4pd2

Qencl

e0

=1

e0d2

q

pa3 e-2r / ar 2dr0

d

Ú =1

4pe0

q

d2 1 - e-2d / a 1 + 2d

a+ 2

d2

a2

Ê Ë Á

ˆ ¯ ˜

È

Î Í Í

˘

˚ ˙ ˙

The displacement of the nucleus will be very small compared to the size of the electron cloud

(d«a). Therefore, we can expand exp(-2d/a) in terms of d/a:

E cloud d( ) =1

4pe0

q

d2 1 - 1 - 2d

a+ 2

d

aÊ Ë

ˆ ¯

2

-4

3

d

aÊ Ë

ˆ ¯

3

...Ê

Ë Á ˆ

¯ ˜ 1 + 2d

a+ 2

d2

a2

Ê Ë Á

ˆ ¯ ˜

È

Î Í Í

˘

˚ ˙ ˙

@

@1

4pe0

q

d2

4

3

d

aÊ Ë

ˆ ¯

3

=1

3pe0

qd

a3 =1

3pe0

p

a3

The nucleus will be in an equilibrium position when the electric force exerted on it by the

external field is equal to the electric force exerted on it by the electron cloud. This occurs when

the electric field at the position of the nucleus, generated by he electron cloud, is equal in

magnitude to the externally applied electric field, but pointing in the opposite direction. The

dipole moment of the dipole can therefore be expressed in terms of the external field:

p = 3pe0a3E ext

The electric polarizability of the material is therefore equal to

- 4 -

a =p

E ext

= 3pe0a3

which is close to the result obtained using the classical model of the atom.

qE

-qE

+q

-q

sE

Figure 4.2. Torque on dipole in an electric field.

Besides polarizing the atoms of a material, the external electric field can align its molecules.

Some molecules, like water, have a permanent dipole moment. Normally, the dipole moments of

the water molecules will be directed randomly, and the average dipole moment is zero. When

the water is exposed to an external electric field, a torque is exerted on the water molecule, and it

will try to align its dipole moment with the external electric field. This is schematically

illustrated in Figure 4.2. Figure 4.2 shows a dipole p = qs placed in an electric field, directed

along the x axis. The net force on the dipole is zero since the net charge is equal to zero. The

torque on the dipole with respect to its center is equal to

N = r + ¥ F +( ) + r - ¥ F -( ) =1

2s

Ê Ë

ˆ ¯ ¥ qE ( )Ê

Ë Á ˆ ¯ ˜ + -

1

2s

Ê Ë

ˆ ¯ ¥ -qE ( )Ê

Ë Á ˆ ¯ ˜ = qs ¥ E = p ¥ E

As a result of this torque, the dipole will try to align itself with the electric field. When the

dipole moment is pointing in the same direction as the electric field the torque on the dipole will

be equal to zero.


A dipole with dipole moment p is situated a distance d above an infinite grounded

conducting plane (see Figure 4.3). The dipole makes and angle q with the perpendicular to the

plane. Find the torque on p . If the dipole is free to rotate, in what direction will it come to rest?

- 5 -

q p

d


q p

2d

p

q

p

2dp

a) b)

Figure 4.4. Method of images (Problem 4.6).

This problem can be solved using the method of images (see Figure 4.4a). Note that the

method of images, when applied to a dipole, does not produce an exact mirror image of the

dipole. After defining the image dipole, we chose a new coordinate system such that the image

- 6 -

dipole is located at the origin, and pointing upwards (along the positive z axis, see Figure 4.4b).

The electric field at the position of the real dipole due to the image dipole is equal to

E image =1

4pe0

1

2d( )3 3 p image ∑ ˆ r ( ) ˆ r - p image{ } =1

32pe0

1

d3 3p cosq ˆ r - p ˆ k { }

The torque on the real dipole is equal to

N = p ¥ E image =1

32pe0

1

d3 3p cosq p ¥ ˆ r ( ) - p p ¥ ˆ k ( ){ } =

=1

32pe0

1

d3 3p2 cosq sinq ˆ j - p2 sin 2q j { } =1

64pe0

1

d3 p2 sin 2q j

The torque on the dipole is positive when 0 < q < p/2 and as a consequence the dipole will rotate

counter clockwise towards the stable orientation of q = 0. The torque on the dipole is negative

when p/2 < q < p and as a consequence the dipole will rotate clockwise towards the stable

orientation of q = p.


Show that the energy of a dipole in an electric field is given by

U = -p ∑ E

x axis

z axis

pq


Consider the dipole located at the origin of a coordinate system. The z axis of the coordinate

system coincides with the direction of the electric field and the angle between the dipole and the

z axis is equal to q (see Figure 4.5). The energy of the system can be determined by calculating

- 7 -

the work to be done to move the dipole from infinity to its present location. Assume the dipole is

initially oriented parallel to the x axis and is first moved from infinity along the x axis to r = 0.

The force exerted on the dipole by the electric field is directed perpendicular to the displacement

and therefore the work done by this force is equal to zero. The dipole is then rotated to its final

position (from p/2 to q). The torque exerted by the electric field on the dipole is equal to

N = p ¥ E = pE sinq ˆ k

In order to rotate the dipole I must supply a torque opposite to N :

N supplied = -pE sinq ˆ k

Therefore, the work done by me is equal to

W = - N supplied dqp / 2

q

Ú = pE sinqdqp / 2

q

Ú = - pE cosq = - p ∑ E

The potential energy of the dipole is therefore equal to

U = -p ∑ E

and reaches a minimum when p is parallel to E (the dipole is aligned with the electric field).

4.2. The Field of a Polarized Object

Consider a piece of polarized material with a dipole moment per unit volume equal to P .

The electrostatic potential generated by this material is equal to

V r ( ) =1

4pe0

Dˆ r ∑ P

Dr( )2 dtVolume

Ú =1

4pe0

P ∑ — 1

DrÊ Ë

ˆ ¯ dt

VolumeÚ

where Dr = r - r '. Using the following relation (one of the product rules of the vector operator)

— ∑1

DrP

Ê Ë

ˆ ¯ =

1

Dr— ∑ P ( ) + P ∑ —

1

DrÊ Ë

ˆ ¯

we can rewrite the expression for the electric potential as

- 8 -

V r ( ) =1

4pe0

— ∑1

DrP

Ê Ë

ˆ ¯ dt

VolumeÚ -

1

4pe0

1

Dr— ∑ P ( )dt

VolumeÚ =

=1

4pe0

1

DrP ∑ da

SurfaceÚ -

1

4pe0

1

Dr— ∑ P ( )dt

VolumeÚ =

=1

4pe0

1Dr

s bdaSurface

Ú +1

4pe0

1Dr

rbdtVolume

Ú

where

sb = P ∑ d ˆ n (bound surface charge)

and

rb = - — ∑ P ( ) (bound volume charge)

Here the unit vector ˆ n is perpendicular to the integration surface (and pointing outwards). The

equation for the electrostatic potential shows that the potential (and therefore also the electric

field) generated by a polarized object is equal to the potential generated by an object with surface

charge density sb and volume charge density rb .


A sphere of radius R carries a polarization

P r ( ) = kr

where k is a constant and r is the vector from the center.

a) Calculate the bound charges sb and rb .

b) Find the field inside and outside the sphere.

a) The unit vector ˆ n on the surface of the sphere is equal to the radial unit vector. The bound

surface charge is equal to

sb = P ∑ ˆ n r =R

= kr ∑ ˆ r r =R = kR

The bound volume charge is equal to

rb = - — ∑ P ( ) = -1

r 2

∂∂r

r 2kr( ) = -3k

- 9 -

b) First consider the region outside the sphere. The electric field in this region due to the

surface charge is equal to

E surface r( ) =1

4pe0

4pR2s b

r 2 ˆ r =kR3

e0r2 ˆ r

The electric field in this region due to the volume charge is equal to

E volume r( ) =1

4pe0

43

pR3rb

r 2 ˆ r = -kR3

e0r2 ˆ r

Therefore, the total electric field outside the sphere is equal to zero.

Now consider the region inside the sphere. The electric field in this region due to the surface

charge is equal to zero. The electric field due to the volume charge is equal to

E volume r( ) =1

4pe0

43

pr 3rb

r 2 ˆ r = -kr

e0

ˆ r

The bound charges introduced in this Section are not just mathematical artifacts, but are real

charges, bound to the individual dipoles of the material. Consider for example the three dipoles

shown in Figure 4.6a. When they are aligned (lengthwise) the center charges cancel, and the

system looks like a single dipole with dipole moment 3dq (see Figure 4.6b).

d

-q q

d

-q q

d

-q q q

3d

-q

a) b)

Figure 4.6. Aligned dipoles.

In a uniformly polarized material of thickness s and polarization P all dipoles are perfectly

aligned (see Figure 4.7). The net result of the alignment of the individual dipoles is a positive

surface charge on one side of the material and negative surface charge on the opposite side.

Consider a cylinder with surface area A whose axis is aligned with the direction of polarization

of the polarized material. The total dipole moment of this cylinder is equal to

- 10 -

p cylinder = AsP

-q q

-q q

-q q

-q q

-q q

-q q

-q qP

-s +s

Figure. 4.7. Uniform polarization.

Since the only charge of the system resides on the end caps of the cylinder (volume charges

cancel in a uniformly polarized material: see Figure 4.6), the net charge there must be equal to

qend =pcylinder

s= AP

The charge density on the surface is therefore equal to

s =qend

A= P

If the surface of the material is not perpendicular to the direction of polarization then surface

charge density will be less than P (surface charge distributed over a larger area) and equal to

s = P ∑ ˆ n

where ˆ n is the unit vector perpendicular to the surface of the material, pointing outwards. For

the material shown in Figure 4.7 this equation immediately shows that a positive surface charge

resides on the right surface ( P parallel to ˆ n ) and a negative surface charge resides on the left

surface ( P anti parallel to ˆ n ). Since these charges reside on the surface and are bound to the

dipoles they are called the bound surface charge or sb .

If the material is uniformly polarized then the volume charge density is equal to zero (see

Figure 4.6). However, if the polarization is not uniform then there will be a net volume charge

inside the material. Consider a system of three aligned dipoles (see Figure 4.8). If the

- 11 -

polarization is not uniform then the strength of the individual dipoles will vary. Assuming that

the physical size (length) of the dipoles shown in Figure 4.8 is the same, then the varying dipole

strength is a result of variations in the charge on the ends of the dipoles. Since the net charge on

the polarized material must be equal to zero, the sum of the volume charges and surface charges

must be equal to zero. Thus

sb daSurface

Ú + rb dtVolume

Ú = 0

This equation can be rewritten by substituting the expression for the surface charge density and

applying the fundamental theorem of divergences:

rb dtVolume

Ú = - s b daSurface

Ú = - P ∑ da Surface

Ú = - — ∑ P ( )dtVolume

Ú

Since this relation holds for any volume we can conclude that

rb = - — ∑ P ( )

d

-q q

d

-1.2q 1.2q

d

-0.8q 0.8q

a)

3d

-q 0.8q

b)

-0.2q 0.4q

Figure 4.8. Non-uniform polarization.


A dielectric cube of side s, centered at the origin, carries a "frozen-in" polarization P = kr ,

where k is a constant. Find all the bound charges, and check that they add up to zero.

The bound volume charge density rb is equal to

rb = - — ∑ P [ ] = -1

r 2

∂∂r

r 2kr( ) = -3k

Since the bound volume charge density is constant, the total bound volume charge in the cube is

equal to product of the charge density and the volume:

- 12 -

qvolume = -3ks3

The surface charge density sb is equal to

sb = P ∑ ˆ n = kr ∑ ˆ n

The scalar product between r and ˆ n can be evaluate easily (see Figure 4.9) and is equal to

r ∑ ˆ n = r cosq =1

2s

s

s/2 r

P

n

q

q


Therefore the surface charge density is equal to

sb = kr ∑ ˆ n =1

2ks

The surface charge density is constant across the surface of the cube and consequently the total

surface charge on the cube is equal to the product of the surface charge density and the total

surface area of the cube:

qsurface =1

2ks

Ê Ë

ˆ ¯ 6s2( ) = 3ks3

The total bound charge on the cube is equal to

- 13 -

qtotal = qvolume + qsurface = -3ks3 + 3ks3 = 0

4.3. The Electric Displacement

The electric field generated by a polarized material is equal to the electric field produced by

its bound charges. If free charges are also present then the total electric field produced by this

system is equal to the vector sum of the electric fields produced by the bound charges and by the

free charges. Gauss's law can also be used for this type of systems to calculate the electric field

as long as we include both free and bound charges:

— ∑ E =re0

=rbound + rfree

e0

=1

e0

-— ∑ P + rfree( )

where P is the polarization of the material. This expression can be rewritten as

— ∑ e0E + P ( ) = rfree

The expression in parenthesis is called the electric displacement D which is defined as

D = e0E + P

In terms of D , Gauss's law can be rewritten as

— ∑ D = rfree (Gauss' s law in differential form)

and

D ∑ da Surface

Ú = Qfree (Gauss' s law in integral form)

These two versions of Gauss's law are particularly useful since they make reference only to free

charges, which are the charges we can control.

Although it seems that the displacement D has properties similar to the electric field E there

are some very significant differences. For example, the curl of D is equal to

— ¥ D = e0 — ¥ E + — ¥ P = — ¥ P

and is in general not equal to zero. Since the curl of D is not necessarily equal to zero, there is

in general no potential that generates D .

- 14 -

The Helmholtz theorem tell us that if we know the curl and the divergence of a vector

function v then this is sufficient information to uniquely define the vector function v .

Therefore, the electric field E is uniquely defined by Gauss's law since we know that he curl of

E is zero, everywhere. The displacement current D on the other hand is not uniquely

determined by the free charge distribution, but requires additional information (like for example

P ).


Suppose the field inside a large piece of dielectric is E 0, so that the electric displacement is

equal to D 0 = e0E 0 + P .

a) Now, a small spherical cavity is hollowed out of the material. Find the field at the center of

the cavity in terms of E 0 and P . Also find the displacement at the center of the cavity in

terms of D 0 and P .

b) Do the same for a long needle-shaped cavity running parallel to P .

c) Do the same for a thin wafer-shaped cavity perpendicular to P .

a) Consider a large piece of dielectric material with polarization P and a small sphere with

polarization -P superimposed on it. The field generated by this system is equal to the field

generated by the dielectric material with a small spherical cavity hollowed out (principle of

superposition). The electric field inside a sphere with polarization -P is uniform and equal to

E sphere = -1

3e0

-P ( ) =1

3e0

P

(see Example 2 of Griffiths). The field at the center of the cavity is therefore equal to

E center = E 0 + E sphere = E 0 +1

3e0

P

The corresponding electric displacement at the center of the cavity is equal to

D center = e0E center = e0E 0 +1

3P = D 0 -

2

3P

b) Consider a large piece of dielectric material with polarization P and a small long needle-

shaped piece with polarization -P superimposed on it. The field generated by this system is

equal to the field generated by the dielectric material with a small long needle-shaped cavity

hollowed out (principle of superposition). The electric field of a polarized needle of length s is

- 15 -

equal to that of two point charges (+q and -q) located a distance s apart. The charge on top of the

needle will be negative, while the charge on the bottom of the needle will be positive. The

charge density on the end caps of the needle is equal to P. Therefore,

q = sbA = PA

where A is the surface area of the end caps of the needle. The electric field generated by the

needle at its center is equal to

E needle =1

4pe0

+ PA( )14

s2

ˆ k -1

4pe0

-PA( )14

s2

ˆ k =2

pe0

PA

s2ˆ k

In the needle limit A Æ 0 and therefore E needle Æ 0. Thus at the center of the needle cavity

E center = E 0

The electric displacement at this point is equal to

D center = e0E center = e0E 0 = D 0 - P

c) Consider a large piece of dielectric material with polarization P and a thin wafer-shaped

piece of dielectric material with polarization -P superimposed on it. The field generated by this

system is equal to the field generated by the dielectric material with a thin wafer-shaped cavity

hollowed out (principle of superposition). The electric field inside the wafer will be that of two

parallel plates with charge densities equal to -s on the top and +s on the bottom. For a thin

wafer-shaped cavity the electric field between the plates will be equal to the field of a parallel-

plate capacitor with infinitely large plates. Thus

E wafer =se0

ˆ k =1

e0

P

The net electric field in the center of the cavity is therefore equal to

E center = E 0 + E wafer = E 0 +1

e0

P

The electric displacement at the center of the cavity is equal to

D center = e0E center = e0E 0 + P = D

- 16 -

4.4. Linear Dielectrics

Most dielectric materials become polarized when they are placed in an external electric field.

In many materials the polarization is proportional to the electric field:

P = e0 c eE

where E is the total electric field (external + internal). The constant of proportionality, ce , is

called the electric susceptibility. Materials in which the induced polarization is proportional to

the electric field are called linear dielectrics.

The electric displacement in a linear dielectric is also proportional to the total electric field:

D = e0E + P = e0 1 + c e( )E = eE

where e is called the permittivity of the material which is equal to

e = e0 1 + c e( )

The constant 1 + ce is called the dielectric constant K of the material.

Consider a volume V entirely filled with linear dielectric material with dielectric constant K.

The polarization P of this material is equal to

P = e0 c eE

and is therefore proportional to E everywhere. Therefore

— ¥ P = e0 c e — ¥ E ( ) = 0

and consequently

— ¥ D = e0 — ¥ E ( ) + — ¥ P = 0

The electric displacement D therefore satisfies the following two conditions:

— ¥ D = 0

and

— ∑ D = rfree

- 17 -

The electric field generated by the free charges when the dielectric is not present satisfies the

following two equations:

— ¥ E free = 0

and

— ∑ E free =rfree

e0

Comparing the two sets of differential equations for D and E free we conclude that

D = e0E free

The displacement D can also be expressed in terms of the total field inside the dielectric:

D = e0 1 + c e( )E = eE

These two equations show that

E =e0

eE free =

1

KE free

The presence of the dielectric material therefore reduces the electric field by a factor K.


The space between the plates of a parallel-plate capacitor (see Figure 4.10) is filled with two

slabs of linear dielectric material. Each slab has thickness s, so that the total distance between

the plates is 2s. Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric constant of 1.5.

The free charge density on the top plate is s and on the bottom plate is -s.

a) Find the electric displacement D in each slab.

b) Find the electric field E in each slab.

c) Find the polarization P in each slab.

d) Find the potential difference between the plates.

e) Find the location and amount of all bound charge.

f) Now that you know all charges (free and bound), recalculate the field in each slab, and

compare with your answers to b).

- 18 -

+s

-s

Slab 1

Slab 2


a) The electric displacement D 1 in slab 1 can be calculated using "Gauss's law". Consider a

cylinder with cross sectional area A and axis parallel to the z axis, being used as a Gaussian

surface. The top of he cylinder is located inside the top metal plate (where the electric

displacement is zero) and the bottom of the cylinder is located inside the dielectric of slab 1. The

electric displacement is directed parallel to the z axis and pointed downwards. Therefore, the

displacement flux through this surface is equal to

F D = D1A

The free charge enclosed by this surface is equal to

Qfree,encl = sA

Combining these two equations we obtain

D1 =F D

A=

Qfree,encl

A= s

In vector notation

D 1 = -s ˆ k

In the same way we obtained for slab 2

D 2 = -s ˆ k

b) The electric field E 1 in slab 1 is equal to

E 1 =1

K1e0

D 1 = -s

K1e0

ˆ k = -s2e0

ˆ k

- 19 -

The electric field E 2 in slab 2 is equal to

E 2 =1

K2e0

D 2 = -s

K2e0

ˆ k = -2s3e0

ˆ k

c) Once D and E are known, the polarization P can be calculated:

P = D - e0E

Therefore, the polarization of slab 1 is equal to

P 1 = D 1 - e0E 1 = -s ˆ k +s2

ˆ k = -s2

ˆ k

The polarization of slab 2 is equal to

P 2 = D 2 - e0E 2 = -s ˆ k +2s3

ˆ k = -s3

ˆ k

d) The potential difference between the top plate and the bottom plate is equal to

DV = Vtop - Vbottom = - E ∑ dl bottom

top

Ú = E1s + E2s =s2e0

+2s3e0

Ê Ë Á

ˆ ¯ ˜ s =

7s6e0

e) There are no bound volume charges (constant polarization). The bound surface charge

density on the surface of a dielectric with polarization P is equal to P ∑ ˆ n . For slab 1 the

polarization is equal to

P 1 = -s2

ˆ k

The surface charge density on the top of slab 1 is equal to

s top,1 = P 1 ∑ ˆ n = -s2

ˆ k Ê Ë

ˆ ¯ ∑ ˆ k = -

s2

The surface charge density on the bottom of slab 1 is equal to

sbottom ,1 = P 1 ∑ ˆ n = -s2

ˆ k Ê Ë

ˆ ¯ ∑ - ˆ k ( ) =

s2

For slab 2 the polarization is equal to

- 20 -

P 2 = -s3

ˆ k

The surface charge density on the top of slab 2 is equal to

s top, 2 = P 2 ∑ ˆ n = -s3

ˆ k Ê Ë

ˆ ¯ ∑ ˆ k ( ) = -

s3

The surface charge density on the bottom of slab 2 is equal to

sbottom , 2 = P 2 ∑ ˆ n = -s3

ˆ k Ê Ë

ˆ ¯ ∑ - ˆ k ( ) =

s3

f) The total charge above slab 1 is equal to s - s/2 = s/2. This charge will produce an electric

field in slab 1 equal to

E above,1 = -s4e0

ˆ k

The total charge below slab 1 is equal to s/2 - s/3 + s/3 - s = - s/2. This charge will produce an

electric field in slab 1 equal to

E below,1 = -s4e0

ˆ k

The total electric field in slab 1 is the vector sum of these two fields and is equal to

E 1 = E above,1 + E below,1 = -s2e0

ˆ k

The total charge above slab 2 is equal to s - s/2 + s/2 - s/3 = 2s/3. This charge will produce an

electric field in slab 2 equal to

E above, 2 = -s3e0

ˆ k

The total charge below slab 1 is equal to s/3 - s = - 2s/3. This charge will produce an electric

field in slab 1 equal to

E below,2 = -s3e0

ˆ k

- 21 -

The total electric field in slab 1 is the vector sum of these two fields and is equal to

E 2 = E above, 2 + E below, 2 = -s3e0

ˆ k -s3e0

ˆ k = -2s3e0

ˆ k

These answers are in agreement with the results obtained in part b).


A sphere of linear dielectric material has embedded in it a uniform free charge density r.

Find the potential at the center of the sphere, if its radius is R and its dielectric constant is K.

The system has spherical symmetry and therefore the electric displacement D is easy to

calculate since — ∑ D = rfree and — ¥ D = 0. The calculation of D is very similar to the

calculation of E using Gauss's law:

D r( ) =1

4pr 2 Qfree, encl =rR3

3r 2 r > R

D r( ) =1

4pr 2 Qfree, encl =1

3rr r < R

The corresponding electric field is equal to

E r( ) =D r( )Ke0

=rR3

3e0r2 r > R

E r( ) =D r( )Ke0

=1

3Ke0

rr r < R

Here we have used the fact that K = 1 in the region outside the sphere (r > R). The potential at

the center of the sphere can be calculated using this electric field:

V r( ) = - E r( ) ∑ dl •

0

Ú = -rR3

3e0r2 dr

•

R

Ú -1

3Ke0

rrdrR

0

Ú =

=rR3

3e0r •

R

-1

6Ke0

rr 2

R

0

=rR2

3e0

1 +1

2KÏ Ì Ó

¸ ˝ ˛

- 22 -

The examples of calculations involving polarized material that have been discussed so far are

either artificial, in the sense that the polarization is specified at the start, or highly symmetric, so

that the electric displacement can be obtained directly from the free charge. In the next couple of

examples we will encounter systems where these special conditions do not apply.

Example: Example 7 (Griffiths) and Problem 4.23

A sphere of linear dielectric material (dielectric constant K) is placed in an originally uniform

electric field E 0 (note: we will assume that this electric field is directed along the positive z axis).

a) Find the new field inside the sphere.

b) Solve for the field inside the sphere by the method of separation of variables. Note that: (1)

V is continuous at R; (2) the discontinuity in the normal derivative of V at the surface is equal to

- sb / e0 ; (3) because the dielectric is linear sb = e0 c e E ∑ ˆ n ( ) = e0 c e E ∑ ˆ r ( ) .

a) Suppose the electric field inside the sphere is equal to E 0. Since the material is a linear

dielectric the polarization is proportional to the total electric field:

P 0 = e0 c eE 0

However, a uniformly polarized sphere with polarization P produces an internal electric field

equal to

E = -1

3e0

P

The electric field produced by the polarization of the sphere will therefore reduce the electric

field inside the sphere by

E 1 = -1

3ceE 0

This change in the electric field will change the polarization of the sphere by

P 1 = e0 c eE 1 = -1

3e0 c e

2E 0

This change in the polarization of the sphere will again change the electric field inside the

sphere. This change of the electric field strength is equal to

E 2 = -1

3e0

P 1 = -c e

3Ê Ë

ˆ ¯

2

E 0

- 23 -

This iterative process will continue indefinitely, and the final electric field will be equal to

E final = E 0 + E 1 + E 2 + .... = 1 + -c e

3Ê Ë

ˆ ¯ + -

c e

3Ê Ë

ˆ ¯

2

+ ....Ê

Ë Á ˆ

¯ ˜ E 0 =

= -c e

3Ê Ë

ˆ ¯

n

E 0n=0

n

Â =1

1 +c e

3

E 0 =3

2 + KE 0

The final polarization of the sphere is therefore equal to

P final = e0 c eE final =3e0 K -1( )

2 + KE 0

b) Since the dielectric will be uniformly polarized, all the bound charge will reside on the

surface of the sphere:

sb = e0 c e E ∑ ˆ n ( ) = e0 c e E ∑ ˆ r ( )

Therefore, the charge density is zero everywhere except on the surface of the sphere. The

electrostatic potential of this system must therefore satisfy Laplace's equation (see Chapter 3).

The most general solution of Laplace's equation for this system is

V r ,q( ) = AnrnPn cosq( )

n=0

•

Â r < R

V r ,q( ) = -E0r cosq +Bn

r n +1 Pn cosq( )n=0

•

Â r > R

Note that the potential does not approach zero when r approaches infinity since the electric field

at infinity is equal to E 0. The electrostatic potential has to be continuous at r = R. Thus

AnRnPn cosq( )

n=0

•

Â = -E0R cosq +Bn

Rn +1 Pn cosq( )n=0

•

Â

This relation requires that

AnRn =

Bn

Rn +1 n π 1

A1R = -E0R +B1

R2 n = 1

- 24 -

These two equations can be rewritten as

Bn = AnR2n +1 n π1

B1 = A1R3 + E0R

3 n =1

The normal derivative of V at the surface of the sphere must satisfy the following boundary

condition:

∂V

∂r r =R +-

∂V

∂r r =R -= -

s b

e0

Note that since the sphere is neutral, there is no free charge present. Therefore, the total surface

charge on the sphere is equal to the bound surface charge. Substituting the general solution for V

in this equation we obtain

-E0 cosq - n + 1( ) Bn

Rn + 2 Pn cosq( )n=0

•

Â - nAnRn-1Pn cosq( )

n=0

•

Â = -s b

e0

This equation can be rewritten by using the expressions for Bn in terms of An with the following

result:

-sb

e0

= -E0 cosq - n + 1( )AnRn-1Pn cosq( )

n=0

•

Â - 2E0 cosq - nAnRn-1Pn cosq( )

n=0

•

Â =

= -3E0 cosq - 2n + 1( )AnRn-1Pn cosq( )

n=0

•

Â

The bound charge is determined by the electric field, and therefore by the gradient of the

potential:

sb = e0 c e E r =R - ∑ ˆ r ( ) = -e0 c e

∂V

∂r r =R -= -e0 c e nAnR

n-1Pn cosq( )n=0

•

Â

Combining the last two equations we obtain

ce nAnRn-1Pn cosq( )

n=0

•

Â = -3E0 cosq - 2n + 1( )AnRn-1Pn cosq( )

n=0

•

Â

This equation shows that for n = 1:

- 25 -

ceA1 = -3E0 - 3A1

and for n π 1:

nceAnRn-1 = - 2n + 1( )AnR

n-1

These two equations can be rewritten as

A1 = -3E0

3 + ce

and for n π 1

An =0

cen + 2n + 1( ) = 0

The electrostatic potential since the sphere is thus equal to

V r ,q( ) = -3E0

3 + c e

r cosq = -3E0

2 + Kz

We conclude that the electrostatic potential inside the sphere only depends on the z coordinate.

The electric field inside the sphere can be obtained from the gradient of the electrostatic

potential:

E = -— V = -∂V

∂zˆ k =

3E0

2 + Kˆ k

which is identical to the result we obtained in a).


Prove the following uniqueness theorem: A region S contains a specified free charge

distribution rf and various pieces of linear dielectric material, with the susceptibility of each one

given. If the potential is specified on the boundary of S (and V = 0 at infinity) then the potential

throughout S is uniquely defined.

Suppose that there are two different solutions V1 and V2 . The corresponding electric fields

are E 1 = -— V1 and E 2 = -— V2 , respectively. The corresponding electric displacements are

D 1 = eE 1 and D 2 = eE 2. Consider a third function V3 = V2 - V1. Since V1 and V2 must have the

- 26 -

same value on the border, V3 = 0 there. Now consider the volume integral (over volume S) of

— ∑ V3D 3( ) :

— ∑ V3D 3( )dtVolume S

Ú = V3D 3 ∑ da Surface S

Ú = 0

since V3 = 0 on the surface of volume S. The left-hand side of this equation can be rewritten as

— ∑ V3D 3( )dtVolume S

Ú = — V3 ∑ D 3( )dtVolume S

Ú + V3 — ∑ D 3( )dtVolume S

Ú = 0

But the divergence of D 3 is equal to zero since

— ∑ D 3( ) = — ∑ D 2( ) - — ∑ D 1( ) = rfree - rfree = 0

Therefore

— V3 ∑ D 3( )dtVolume S

Ú + V3 — ∑ D 3( )dtVolume S

Ú = — V3 ∑ D 3( )dtVolume S

Ú = 0

The integral on the right-hand side of this equation can be rewritten in terms of E 3 using the

following relations:

— V3 = — V2 - — V1 = -E 2 + E 1 = -E 3

and

D 3 = D 2 - D 1 = eE 2 - eE 1 = eE 3

Therefore,

— V3 ∑ D 3( )dtVolume S

Ú = -E 3( ) ∑ eE 3( )( )dtVolume S

Ú = -e E32dt

Volume SÚ = 0

Since e > 0 this equation can only be satisfied if E3 = 0. This requires that

E 1 = E 2

and

V1 = V2

- 27 -

everywhere. We therefore conclude that there are no two different electrostatic potentials that

satisfy the same boundary conditions. The electrostatic potential is therefore uniquely defined if

its value is specified on the surface of the volume S.


A conducting sphere at potential V0 is half embedded in linear dielectric material of

susceptibility ce , which occupies the region z < 0 (see Figure 4.11). Claim: the potential

everywhere is exactly the same as it would have been in the absence of the dielectric! Check this

claim as follows:

a) Write down the formula for the suggested potential V r( ) , in terms of V0 , R, and r. Use it to

determine the field, the polarization, the bound charge, and the free charge distribution on the

sphere.

b) Show that the total charge configuration would indeed produce the potential V r( ) .

c) Appeal to the uniqueness theorem in Problem 4.35 to complete the argument.

Rz = 0


a) In the absence of the dielectric, the electrostatic potential of this system is constant inside the

sphere and is given by

V r( ) = V0

R

r

in the region outside the sphere. The electric field in the region outside the sphere is equal to the

gradient of V and is therefore given by

E r( ) = - — V r( ) = V0

R

r 2 ˆ r

- 28 -

If this solution satisfies the boundary conditions on the surface of the sphere when the dielectric

is present then it is the only solution (uniqueness theorem of problem 4.35). The boundary

conditions for the electrostatic potential are:

1. V is continuous on the surface of the sphere. This boundary condition is satisfied by the

proposed solution.

2. The difference in the normal derivative of V on the surface of the sphere is equal to

∂V

∂r r =R +-

∂V

∂r r =R -= -

s total

e0

For the proposed solution this requires that

s total = -e0

∂V

∂r r =R +-

∂V

∂r r =R -

Ê

Ë Á ˆ

¯ ˜ =e0V0

R

which shows that the total charge is uniformly distributed across the surface of the sphere.

The polarization of the dielectric material in the region z < 0 (and r > R) can be obtained from

the electric field:

P = e0 c eE = e0 c eV0

R

r 2 ˆ r

In the region z > 0 the polarization is equal to zero since no dielectric material is present there.

The bound surface charge on the surface of the dielectric is equal to

sb = P ∑ ˆ n

where ˆ n is the surface vector (perpendicular to the surface and pointing out of the dielectric).

For the spherical surface ˆ n = - ˆ r and thus

sb = P R( ) ∑ ˆ n = -e0 c eV0

R

R2 ˆ r ∑ ˆ r ( ) = -e0 ceV0

1

R

There is no bound charge on the flat surface (z = 0) of the dielectric since ˆ n ^ ˆ r is there. The

bound volume charge is zero everywhere since

- 29 -

rb = - — ∑ P = -1

r 2

∂∂r

r 2e0 c eV0

R

r 2

Ê Ë

ˆ ¯ = 0

The free charge on the surface of the sphere can be determined from the electric displacement

D . The electric displacement D can be obtained from the electric field. In the region above the

dielectric (z > 0) and outside the sphere (r > R) the electric displacement D is equal to

D r ( ) = e0E r ( ) + P r ( ) = e0 c eV0

R

r 2 ˆ r

In the region z < 0 and outside the sphere (r > R) the electric displacement D is equal to

D r ( ) = e0E r ( ) + P r ( ) = e0 1 + c e( )V0

R

r 2 ˆ r

The free charge on the bottom hemisphere and part of the z = 0 plane (see Figure 4.12a) is equal

to

Qfree = D r ( ) ∑ da Bottom Surface

Ú = 2pR2e0 1 + ce( )V0

1

R

There is no contribution to the surface charge from the z = 0 plane since D r ( )^da there. The

free charge density on the bottom hemisphere is therefore equal to

Qfree,z<0 =Qfree

2pR2 = e0 1 + c e( )V0

1

R

In the same manner we can calculate the free charge density on the top hemisphere:

Qfree,z>0 =Qfree

2pR2 = e0V0

1

R

The total charge density (bound charge + free charge) on the surface is therefore equal to

s total , z<0 = s free, z<0 + s bound , z<0 = e0 1 + c e( )V0

1

R- e0 c eV0

1

R= e0V0

1

R

s total , z>0 = s free, z>0 + s bound , z>0 = e0V0

1

R+ 0 = e0V0

1

R

Therefore, the total charge on the surface of the sphere is distributed uniformly, and has a value

consistent with the boundary condition for the normal derivative of V. Since the proposed

solution satisfies the boundary conditions for V it will be the only correct solution.

- 30 -

a) b)

Figure 4.12. Determination of free charge in Problem 4.36.

4.5. Energy in dielectric systems

Consider a capacitor with capacitance C and charged up to a potential V. The total energy

stored in the capacitor is equal to the work done during the charging process:

W =1

2CV 2

If the capacitor is filled with a linear dielectric (dielectric constant K) than the total capacitance

will increase by a factor K:

C = KCvac

and consequently the energy stored in the capacitor (when held at a constant potential) is

increased by a factor K. A general expression for the energy of a capacitor with dielectric

materials present can be found by studying the charging process in detail. Consider a free charge

rfree held at a potential V. During the charging process the free charge is increased by Drfree .

The work done on the extra free charge is equal to

DW = DrfreeVdtVolume

Ú

Since the divergence of the electric displacement D is equal to the free charge density rfree , the

divergence of DD is equal to Drfree . Therefore,

DW = — ∑ DD [ ]VdtVolume

Ú

- 31 -

Using the following relation

— ∑ VDD ( ) = — ∑ DD [ ]V + — V( ) ∑ DD ( )

we can rewrite the expression for DW as

DW = — ∑ VDD ( )[ ]dtVolume

Ú - — V( ) ∑ DD ( )[ ]dtVolume

Ú

The first term on the right-hand side of this equation can be rewritten as

— ∑ VDD ( )[ ]dtVolume

Ú = VDD ( ) ∑ da Surface

Ú = 0

since the product of potential and electric displacement approach zero faster than 1/r2 when r

approached infinity. Therefore,

DW = - — V( ) ∑ DD ( )[ ]dtVolume

Ú = E ∑ DD ( )[ ]dtVolume

Ú

Assuming that the materials present in the system are linear dielectrics then

D = eE

This relation can be used to rewrite E ∑ DD ( ) :

E ∑ DD ( ) = E ∑ eDE ( ) =1

2D eE ∑ E ( ) =

1

2D D ∑ E ( )

The expression for DW can thus be rewritten as

DW =1

2D D ∑ E ( )dt

VolumeÚ

The total work done during the charging process is therefore equal to

W =1

2D ∑ E ( )dt

VolumeÚ

Note: this equation can be used to calculate the energy for a system that contains linear

dielectrics. If some materials in the system are non-linear dielectrics than the derivation given

above is not correct ( E ∑ DD ( ) π 0.5D D ∑ E ( ) for non-linear dielectrics).

- 32 -


A spherical conductor, of radius a, carries a charge Q. It is surrounded by linear dielectric

material of susceptibility ce, out to a radius b. Find the energy of this configuration.

Since the system has spherical symmetry the electric displacement D is completely

determined by the free charge. It is equal to

D r ( ) =1

4p r 2 Qencl = 0 r < a

D r ( ) =1

4p r 2 Qencl =1

4pQ

r 2 r > a

Since we are dealing with linear dielectrics, the electric field E is equal to D / e0 1 + c e( )( ) .

Taking into account that the susceptibility of vacuum is zero and the susceptibility of a conductor

is infinite we obtain for E :

E r ( ) = 0 r < a

E r ( ) =D r ( )

e0 1 + c e( ) =1

4pe0 1 + c e( )Q

r 2 a < r < b

E r ( ) =D r ( )

e0

=1

4pe0

Q

r 2 b < r

The scalar product D ∑ E is equal to D E since E and D are parallel, everywhere. The

energy of the system is equal to

W D E d D E r dr

Q

rr dr

Q

rr dr

Q

a b b

Q

Volumea

ea

b

b

e e

= ∑( ) = =

=+( ) + =

=+( ) -Ê

Ëˆ¯ +

ÏÌÓ

¸˝˛

=+( )

Ú Ú

Ú Ú

•

•

12

2

21

16 12

116

81

11 1 1

8 1

2

20

2

42

20

2

42

2

0

2

0

t p

pp e c

pp e

pe c pe c

11a b

e+ÏÌÓ¸˛

c

- 33 -

4.6. Forces on dielectrics

A dielectric slab placed partly between the plates of a parallel-plate capacitor will be pulled

inside the capacitor. This force is a result of the fringing fields around the edges of the parallel-

plate capacitor (see Figure 4.13). Note: the field outside the capacitor can not be zero since

otherwise the line integral of the electric field around a closed loop, partly inside the capacitor

and partly outside the capacitor, would not be equal to zero.

- s

+ s

Figure 4.13. Fringing fields.

Inside the capacitor the electric field is uniform. The electric force exerted by the field on the

positive bound charge of the dielectric is directed upwards and is canceled by the electric force

on the negative bound charge (see Figure 4.14). Outside the capacitor the electric field is not

uniform and the electric force acting on the positive bound charge will not be canceled by the

electric force acting on the negative bound charge. For the system shown in Figure 4.14 the

vertical components of the two forces (outside the capacitor) will cancel, but the horizontal

components are pointing in the same direction and therefore do not cancel. The result is a net

force acting on the slab, directed towards the center of the capacitor.

- s

+ s

F+

F-

F+

F-

Figure 4.14. Forces on dielectric.

- 34 -

A direct calculation of this force requires a knowledge of the fringing fields of the capacitor

which are often not well known and difficult to calculate. An alternative method that can be

used is to determine this force is to calculate the change in the energy of the system when the

dielectric is displaced by a distance ds. The work to be done to pull the dielectric out by an

infinitesimal distance ds is equal to

dW = F usds

where F us is the force provided by us to pull the slab out of the capacitor. This force must just be

equal in magnitude but directed in a direction opposite to the force F field exerted by the electric

field on the slab. Thus

F field = -F us = -dW

ds

The work done by us to move the slab must be equal to the change in the energy of the capacitor

(conservation of energy). Consider the situation shown in Figure 4.15 where the slab of

dielectric is inserted to a depth s in the capacitor. The capacitance of this system is equal to

C = Cvac + Cdiel =e0 w - s( )a

d+ K

e0sa

d=

e0a

dw + c es( )

d

s

w

Figure 4.15. Calculation of Ffield .

If the total charge on the top plate is Q then the energy stored in the capacitor is equal to

W =1

2

Q 2

C=

1

2Q 2 d

e0a w + c es( )

The force on the dielectric can now be calculated and is equal to

- 35 -

Ffield = -dW

ds= -

1

2

Q 2

C2

dC

ds


Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a

tank of dielectric oil (susceptibility ce, mass density r). The inner one is maintained at potential

V, and the outer one is grounded. To what height h does the oil rise in the space between the

tubes?

The height of the oil is such that the electric force on the oil balances the gravitational force.

The capacitance of an empty cylindrical capacitor of height H is equal to

C =2pe0H

lnba

Ê Ë

ˆ ¯

If the oil rises to a height h then the capacitance of the capacitor is equal to

Ctot = Cvac + Coil =2pe0 H - h( )

lnba

Ê Ë

ˆ ¯

+ 1 + c e( ) 2pe0h

lnba

Ê Ë

ˆ ¯

=2pe0

lnba

Ê Ë

ˆ ¯

H + c eh( )

The electric force on the dielectric (the oil) is equal to

Fel =1

2V 2 dC

dh= c e

pe0

lnba

Ê Ë

ˆ ¯

V 2

and is directed upwards The gravitational force acting on the oil is equal to

Fgrav = p b2 - a2( )hrg

and is directed downwards. In the equilibrium position Fgrav = Fel . Thus

p b2 - a2( )hrg = c e

pe0

lnba

Ê Ë

ˆ ¯

V 2

or

- 36 -

h = ce

e0

rg b2 - a2( ) lnba

Ê Ë

ˆ ¯

V 2

For a linear dielectric, the polarization P is proportional to the total macroscopic field E total :

P = e0 c eE total

The polarization of the dielectric is equal to the vector sum of the polarization p of the

individual atoms or molecules:

P = N p

where N is the number of atoms or molecules per unit volume. The polarization of an individual

atom or molecule is proportional to the microscopic field at the position of the atom or molecule

due to everything except the particular atom or molecule under consideration:

p = a E else

The dipole moment of the atom or molecule will generate an electric field at its center equal to

E self = -1

4pe0

p

R3

where R is the radius of the atom or molecule. The total macroscopic field seen by the atom or

molecule is there for equal to

E total = E self + E else = -1

4pe0

aE else

R3 + E else = 1 -a

4pe0R3

Ê Ë Á

ˆ ¯ ˜ E else = 1 -

Na3e0

Ê Ë Á

ˆ ¯ ˜ E else

where N is the number of atoms per unit volume. The total polarization of the dielectric is thus

equal to

P = N p = Na E else =Na

1 -Na3e0

Ê Ë Á

ˆ ¯ ˜

E total = e0 ceE total

Therefore

- 37 -

ce =

Nae0

1 -Na3e0

=3Na

3e0 - Na

This equation can be rewritten in terms of the dielectric constants K as

K -1 =3Na

3e0 - Na

or

a =3e0

N

K -1

K + 2

This equation shows that a measurement of the macroscopic parameter K can be used to obtain

information about the microscopic parameter a. This equation is known as the Clausius-

Mossotti formula or the Lorentz-Lorenz equation.

- 1 -

Chapter 5. Magnetostatics

5.1. The Magnetic Field

Consider two parallel straight wires in which current is flowing. The wires are neutral andtherefore there is no net electric force between the wires. Nevertheless, if the current in bothwires is flowing in the same direction, the wires are found to attract each other. If the current inone of the wires is reversed, the wires are found to repel each other. The force responsible forthe attraction and repulsion is called the magnetic force. The magnetic force acting on a movingcharge q is defined in terms of the magnetic field:

F magnetic = q v ¥ B ( )

The vector product is required since observations show that the force acting on a moving chargeis perpendicular to the direction of the moving charge. In a region where there is an electric fieldand a magnetic field the total force on the moving force is equal to

F total = F electric + F magnetic = qE + q v ¥ B ( )

This equation is called the Lorentz force law and provides us with the total electromagneticforce acting on q. An important difference between the electric field and the magnetic field isthat the electric field does work on a charged particle (it produces acceleration or deceleration)while the magnetic field does not do any work on the moving charge. This is a directconsequence of the Lorentz force law:

dWmagnetic = F magnetic ∑ dl = q v ¥ B ( ) ∑ v [ ]dt = 0

We conclude that the magnetic force can alter the direction in which a particle moves, but cannot change its velocity.

Example: Problem 5.1A particle of charge q enters the region of uniform magnetic field B (pointing into the page).

The field deflects the particle a distance d above the original line of flight, as shown in Figure5.1. Is the charge positive or negative? In terms of a, d, B, and q, find the momentum of theparticle.

In order to produce the observed deflection, the force on q at the entrance of the field regionmust be directed upwards (see Figure 5.1). Since direction of motion of the particle and thedirection of the magnetic field are known, the Lorentz force law can be used to determine the

- 2 -

direction of the magnetic force acting on a positive charge and on a negative charge. The vectorproduct between v and B points upwards in Figure 5.1 (use the right-hand rule). This showsthat the charge of the particle is positive.

v x B

v

a

d q

Field Region

Figure 1. Problem 5.1.

The magnitude of the force acting on the moving charge is equal to

Fmagnetic = qvB

As a result of the magnetic force, the charged particle will follow a spherical trajectory. Theradius of the trajectory is determined by the requirement that the magnetic force provides thecentripetal force:

Fcent =mv2

r= Fmagnetic = qvB

In this equation r is the radius of the circle that describes the circular part of the trajectory ofcharge q. The equation can be used to calculate r:

r =mv

qB=

p

qB

where p is the momentum of the particle. Figure 5.2 shows the following relation between r, dand a:

r - d( )2 + a2 = r 2

This equation can be used to express r in terms of d and a:

- 3 -

r =d2 + a2

2d

The momentum of the charge q is therefore equal to

p = qBr =d2 + a2

2dqB

v

a

d q

q r

r

Figure 2. Problem 5.2.

The electric current in a wire is due to the motion of the electrons in the wire. The directionof current is defined to be the direction in which the positive charges move. Therefore, in aconductor the current is directed opposite to the direction of the electrons. The magnitude of thecurrent is defined as the total charge per unit time passing a given point of the wire (I = dq/dt). Ifthe current flows in a region with a non-zero magnetic field then each electron will experience amagnetic force. Consider a tiny segment of the wire of length dl. Assume that the electrondensity is -l C/m and that each electron is moving with a velocity v. The magnetic force exertedby the magnetic field on a single electron is equal to

dF 1e = -e v ¥ B ( )

A segment of the wire of length dl contains l dl/e electrons. Therefore the magnetic force actingin this segment is equal to

dF magnetic =ldl

edF 1e = -ldl v ¥ B ( ) = lv dl ¥ B ( ) = I dl ¥ B ( )

Here we have used the definition of the current I in terms of dq and dt:

- 4 -

I =dq

dt=

dq

dl

dl

dt= lv

In this derivation we have defined the direction of dl to be equal to the direction of the current(and therefore opposite to the direction of the velocity of the electrons). The total force on thewire is therefore equal to

F magnetic = dF magneticwireÚ = I dl ¥ B ( )

wireÚ

Here I have assumed that the current is constant throughout the wire. If the current is flowingover a surface, it is usually described by a surface current density K , which is the current perunit length-perpendicular-to-flow. The force on a surface current is equal to

F magnetic = K ¥ B ( )dasurface

Ú

If the current flows through a volume, is it is usually described in terms of a volume currentdensity J . The magnetic force on a volume current is equal to

F magnetic = J ¥ B ( )dtvolume

Ú

The surface integral of the current density J across the surface of a volume V is equal to thetotal charge leaving the volume per unit time (charge conservation):

J ∑ da Surface

Ú = -d

dtrdt

VolumeÚ

Using the divergence theorem we can rewrite this expression as

J ∑ da Surface

Ú = — ∑ J [ ]dtVolume

Ú = -d

dtrdt

VolumeÚ

Since this must hold for any volume V we must require that

— ∑ J = -drdt

This equation is known as the continuity equation.

- 5 -

5.2. The Biot-Savart Law

In this Section we will discuss the magnetic field produced by a steady current. A steadycurrent is a flow of charge that has been going on forever, and will be going on forever. Thesecurrents produce magnetic fields that are constant in time. The magnetic field produced by asteady line current is given by the Biot-Savart Law:

B P( ) =m0

4pI ¥ Dˆ r

Dr2 dlLineÚ =

m0I

4pdl ¥ Dˆ r

Dr2

LineÚ

where dl is an element of the wire, ˆ r is the vector connecting the element of the wire and P, andm0 is the permeability constant which is equal to

m0 = 4p 10-7 N / A2

The unit of the magnetic field is the Tesla (T). For surface and volume currents the Biot-Savartlaw can be rewritten as

B P( ) =m0

4pK ¥ Dˆ r

Dr2 daSurface

Ú

and

B P( ) =m0

4pJ ¥ Dˆ r

Dr2 dtVolume

Ú

Example: Problem 5.9Find the magnetic field at point P for each of the steady current configurations shown in

Figure 5.3.

a) The total magnetic field at P is the vector sum of the magnetic fields produced by the foursegments of the current loop. Along the two straight sections of the loop, ˆ r and dl are parallelor opposite, and thus dl ¥ ˆ r = 0. Therefore, the magnetic field produced by these two straightsegments is equal to zero. Along the two circular segments ˆ r and dl are perpendicular. Usingthe right-hand rule it is easy to show that

B b P( ) =m0I

4pdl ¥ ˆ b

b2

LineÚ = -

m0I

4p

12

pb

b2ˆ k = -

m0I

8bˆ k

and

- 6 -

B a P( ) =m0I

4pdl ¥ ˆ a

a2

LineÚ =

m0I

4p

12

pa

a2ˆ k =

m0I

8aˆ k

where ˆ z is pointing out of the paper. The total magnetic field at P is therefore equal to

B total P( ) =m0I

8

1

a-

1

bÊ Ë

ˆ ¯

ˆ k

a b

I

I

RI

I

b)a)

P

P


b) The magnetic field at P produced by the circular segment of the current loop is equal to

B R P( ) =m0I

4pdl ¥ ˆ R

R2

LineÚ = -

m0I

4ppR

R2ˆ k = -

m0I

4Rˆ k

where ˆ z is pointing out of the paper. The magnetic field produced at P by each of the two linearsegments will also be directed along the negative z axis. The magnitude of the magnetic fieldproduced by each linear segment is just half of the field produced by an infinitely long straightwire (see Example 5 in Griffiths):

B linear P( ) = -2m0I

4pRˆ k = -

m0I

2pRˆ k

The total field at P is therefore equal to

B total P( ) = -m0I

4Rˆ k -

m0I

2pRˆ k = -

m0I

R

1

4+

1

2pÊ Ë

ˆ ¯

ˆ k

- 7 -

Example: Problem 5.12Suppose you have two infinite straight-line charges l, a distance d apart, moving along at a

constant v (see Figure 5.4). How fast would v have to be in order for the magnetic attraction tobalance the electrical repulsion?

d

v

v

l

l


When a line charge moves it looks like a current of magnitude I = lv. The two parallelcurrents attract each other, and the attractive force per unit length is

fmagnetic =m0

2pI1I2

d=

m0

2pl2v2

d

and is attractive. The electric generated by one of the wires can be found using Gauss' law and isequal to

E r( ) =1

2pe0

lr

The electric force per unit length acting on the other wire is equal to

felectric = l E d( ) =1

2pe0

l2

d

and is repulsive (like charges). The electric and magnetic forces are balanced when

1

2pe0

l2

d=

m0

2pl2v2

d

or

m0v2 =

1

e0

- 8 -

This requires that

v =1

e0m0

= 3 108 m / s

This requires that the speed v is equal to the speed of light, and this can therefore never beachieved. Therefore, at all velocities the electric force will dominate.

5.3. The Divergence and Curl of B.

Using the Biot-Savart law for a volume current J we can calculate the divergence and curl ofB :

— ∑ B = 0

and

— ¥ B = m0J

This last equation is called Ampere's law in differential form. This equation can be rewritten,using Stokes' law, as

— ¥ B [ ] ∑ da Surface

Ú = B ∑ dl LineÚ = m0 J ∑ da

SurfaceÚ = m0Iencl

This equation is called Ampere's law in integral form. The direction of evaluation of the lineintegral and the direction of the surface element vector da must be consistent with the right-hand rule.

Ampere's law is always true, but is only a useful tool to evaluate the magnetic field if thesymmetry of the system enables you to pull B outside the line integral. The configurations thatcan be handled by Ampere's law are:

1. Infinite straight lines2. Infinite planes3. Infinite solenoids4. Toroids

Example: Problem 5.14A thick slab extending from z = -a to z = a carries a uniform volume current J = J ˆ i . Find

the magnetic field both inside and outside the slab.

- 9 -

z

z = +a

z = -a

y

L

B

z

J


Because of the symmetry of the problem the magnetic field will be directed parallel to the yaxis. The magnetic field in the region above the xy plane (z > 0) will be the mirror image of thefield in the region below the xy plane (z < 0). The magnetic field in the xy plane (z = 0) will beequal to zero. Consider the Amperian loop shown in Figure 5.5. The current is flowing out ofthe paper, and we choice the direction of da to be parallel to the direction of J . Therefore,

J ∑ da Surface

Ú = JzL 0 < z < a

J ∑ da Surface

Ú = JaL z > a

The direction of evaluation of the line integral of B must be consistent with our choice of thedirection of da (right-hand rule). This requires that the line integral of B must be evaluated in acounter-clockwise direction. The line integral of B is equal to

B ∑ dl LineÚ = BL

Applying Ampere's law we obtain for B :

B =m0

LJ ∑ da

SurfaceÚ = m0Jz 0 < z < a

B =m0

LJ ∑ da

SurfaceÚ = m0Ja z > a

Thus

- 10 -

B z( ) = -m0Jaˆ j a < z

B z( ) = -m0Jzˆ j -a < z < a

B z( ) = m0Jaˆ j z < -a

5.4. The Vector Potential

The magnetic field generated by a static current distribution is uniquely defined by the so-called Maxwell equations for magnetostatics:

— ∑ B = 0

— ¥ B = m0J

Similarly, the electric field generated by a static charge distribution is uniquely defined by theso-called Maxwell equations for electrostatics:

— ∑ E =re0

— ¥ E = 0

The fact that the divergence of B is equal to zero suggests that there are no point charges for B .Magnetic field lines therefore do not begin or end anywhere (in contrast to electric field lines thatstart on positive point charges and end on negative point charges). Since a magnetic field iscreated by moving charges, a magnetic field can never be present without an electric field beingpresent. In contrast, only an electric field will exist if the charges do not move.

Maxwell's equations for magnetostatics show that if the current density is known, both thedivergence and the curl of the magnetic field are known. The Helmholtz theorem indicates thatin that case there is a vector potential A such that

B = — ¥ A

However, the vector potential is not uniquely defined. We can add to it the gradient of any scalarfunction f without changing its curl:

— ¥ A + — f( ) = — ¥ A + — ¥ — f = — ¥ A

The divergence of A + — f is equal to

- 11 -

— ∑ A + — f( ) = — ∑ A + — ∑ — f = — ∑ A + — 2 f

It turns out that we can always find a scalar function f such that the vector potential A isdivergence-less. The main reason for imposing the requirement that — ∑ A = 0 is that itsimplifies many equations involving the vector potential. For example, Ampere's law rewrittenin terms of A is

— ¥ B = — ¥ — ¥ A ( ) = — — ∑ A ( ) - — 2A = - — 2A = m0J

or

— 2A = -m0J

This equation is similar to Poisson's equation for a charge distribution r:

— 2V = -re0

Therefore, the vector potential A can be calculated from the current J in a manner similar tohow we obtained V from r. Thus

A =m0

4pJ

Drdt

VolumeÚ for a volume current

A =m0

4pK

Drda

SurfaceÚ for a surface current

A =m0

4pI

Drdl

LineÚ =

m0I

4pdl

DrLineÚ for a line current

Note: these solutions require that the currents go to zero at infinity (similar to the requirementthat r goes to zero at infinity).

Example: Problem 5.22Find the magnetic vector potential of a finite segment of straight wire carrying a current I.

Check that your answer is consistent with eq. (5.35) of Griffiths.

The current at infinity is zero in this problem, and therefore we can use the expression for A in terms of the line integral of the current I. Consider the wire located along the z axis betweenz1 and z2 (see Figure 5.6) and use cylindrical coordinates. The vector potential at a point P isindependent of f (cylindrical symmetry) and equal to

- 12 -

A =m0

4pdl

DrLineÚ =

m0I

4pdz '

r 2 + z '2ˆ k

z1

z2

Ú =m0I

4pln

z2 + r 2 + z 22

z1 + r 2 + z12

È

Î Í Í

˘

˚ ˙ ˙

ˆ k

Here we have assumed that the origin of the coordinate system is chosen such that P has z = 0.The magnetic field at P can be obtained from the vector potential and is equal to

B = — ¥ A = -∂Az

∂rˆ f = -

m0I4p

r

r 2 + z22

1

z2 + r 2 + z22

-r

r 2 + z12

1

z1 + r 2 + z12

È

Î Í Í

˘

˚ ˙ ˙

ˆ f =

=m0I

4p r

z 2

r 2 + z22

-z1

r 2 + z12

È

Î Í Í

˘

˚ ˙ ˙

ˆ f =m0I

4p rsinq 2 - sinq1[ ] ˆ f

where q1 and q2 are defined in Figure 5.6. This result is identical to the result of Example 5 inGriffiths.

P

z

r

q1 q2

r

z'

z1

z2


Example: Problem 5.24If B is uniform, show that A = - r ¥ B ( ) / 2, where r is the vector from the origin to the

point in question. That is check that — ¥ A = B and — ∑ A = 0.

The curl of A = - r ¥ B ( ) / 2 is equal to

— ¥ A = -1

2— ¥ r ¥ B ( ) = -

1

2B ∑ — ( )r - r ∑ — ( )B + r — ∑ B ( ) - B — ∑ r ( )[ ]

- 13 -

Since B is uniform it is independent of r, q, and f and therefore the second and third term on theright-hand side of this equation are zero. The first term, expressed in Cartesian coordinates, isequal to

B ∑ — ( )r = Bx

∂∂x

+ By

∂∂y

+ Bz

∂∂z

Ê Ë Á

ˆ ¯ ˜ xˆ i + yˆ j + z ˆ k ( ) = Bx

ˆ i + Byˆ j + Bz

ˆ k = B

The fourth term, expressed in Cartesian coordinates, is equal to

B — ∑ r ( ) = B ∂

∂xx +

∂∂y

y +∂

∂zz

Ê Ë Á

ˆ ¯ ˜ = 3B

Therefore, the curl of A is equal to

— ¥ A = -1

2B - 3B ( ) = B

The divergence of A is equal to

— ∑ A = -1

2— ∑ r ¥ B ( ) = -

1

2B ∑ — ¥ r ( ) - r ∑ — ¥ B ( )[ ] = 0

Example: Problem 5.26Find the vector potential above and below the plane surface current of Example 5.8 in

Griffiths.

In Example 5.8 of Griffiths a uniform surface current is flowing in the xy plane, directedparallel to the x axis:

K = K ˆ i

However, since the surface current extends to infinity, we can not use the surface integral ofK / Dr to calculate A and an alternative method must be used to obtain A . Since Example 8showed that B is uniform above the plane of the surface current and B is uniform below theplane of the surface current, we can use the result of Problem 5.27 to calculate A :

A = -1

2r ¥ B ( )

In the region above the xy plane (z > 0) the magnetic field is equal to

- 14 -

B = -m0

2K ˆ j

Therefore,

A = -1

2r ¥ B ( ) = -

1

2

ˆ i ˆ j ˆ k

x y z

0 -m0

2K 0

= -m0

4K z ˆ i +

m0

4K x ˆ k

In the region below the xy plane (z < 0) the magnetic field is equal to

B =m0

2K ˆ j

Therefore,

A = -1

2r ¥ B ( ) = -

1

2

ˆ i ˆ j ˆ k

x y z

0m0

2K 0

=m0

4K z ˆ i -

m0

4K x ˆ k

We can verify that our solution for A is correct by calculating the curl of A (which must beequal to the magnetic field). For z > 0:

— ¥ A =

ˆ i ˆ j ˆ k

∂∂x

∂∂y

∂∂z

-m0

4K z 0

m0

4K z

= -m0

2K j = B

The vector potential A is however not uniquely defined. For example, A = - m0 / 2( )K z ˆ i andA = m0 / 2( )K x ˆ k are also possible solutions that generate the same magnetic field. Thesesolutions also satisfy the requirement that — ∑ A = 0.

- 15 -

5.5. The Three Fundamental Quantities of Magnetostatics

Our discussion of the magnetic fields produced by steady currents has shown that there arethree fundamental quantities of magnetostatics:

1. The current density J 2. The magnetic field B 3. The vector potential A

These three quantities are related and if one of them is known, the other two can be calculated.The following table summarizes the relations between J , B , and A :

Known Ø J = B = A =

J m0

4pJ ¥ Dˆ r

Dr2 dtÚm0

4pJ

DrdtÚ

B 1

m0

— ¥ B ( ) 1

4pB ¥ Dˆ r

Dr2 dtÚA -

1

m0

— 2A — ¥ A

5.6. The Boundary Conditions of B

In Chapter 2 we studied the boundary conditions of the electric field and concluded that theelectric field suffers a discontinuity at a surface charge. Similarly, the magnetic field suffers adiscontinuity at a surface current.

K

Figure 5.7. Boundary conditions for B .

Consider the surface current K (see Figure 5.7). The surface integral of B over a wafer thinpillbox is equal to

B ∑ da Surface

Ú = B^, above A - B^, below A

- 16 -

where A is the area of the top and bottom of the pill box. The surface integral of B can berewritten using the divergence theorem:

B ∑ da Surface

Ú = — ∑ B ( )dtVolume

Ú = 0

since — ∑ B = 0 for any magnetic field B . Therefore, the perpendicular component of themagnetic field is continuous at a surface current:

B^,above = B^, below

The line integral of B around the loop shown in Figure 5.8 (in the limit e Æ 0) is equal to

B ∑ dl Loop

Ú = B||,above L - B||,below L

According to Ampere's law the line integral of B around this loop is equal to

B ∑ dl Loop

Ú = m0Iencl = m0KL

L

K e

Figure 5.8. Boundary conditions for B .

Therefore, the boundary condition for the component of B , parallel to the surface andperpendicular to the current, is equal to

B||,above - B||,below = m0K

The boundary conditions for B can be combined into one equation:

B above - B below = m0 K ¥ ˆ n ( )

where ˆ n is a unit vector perpendicular to the surface and the surface current and pointing"upward". The vector potential A is continuous at a surface current, but its normal derivative isnot:

- 17 -

∂A above

∂n-

∂A below

∂n= -m0K

5.7. The Multipole Expansion of the Magnetic Field

To calculate the vector potential of a localized current distribution at large distances we canuse the multipole expansion. Consider a current loop with current I. The vector potential of thiscurrent loop can be written as

A =m0I

4pdl

DrLineÚ =

m0I

4p1

r n +1 r 'n Pn cosq( )dl LineÚ

È

Î Í Í

˘

˚ ˙ ˙ n=0

•

ÂÏ Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô

At large distance only the first couple of terms of the multipole expansion need to be considered:

A @m0I

4p1

rdl

LineÚ +

1

r 2 r 'cosq dl LineÚ + .....

Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô

The first term is called the monopole term and is equal to zero (since the line integral of dl isequal to zero for any closed loop). The second term, called the dipole term, is usually thedominant term. The vector potential generated by the dipole terms is equal to

A dipole =m0I

4p1

r 2 r 'cosq dl LineÚ =

m0I

4p1

r 2 r '∑ˆ r ( ) dl LineÚ


A dipole =m0I

4p1

r 2 -1

2ˆ r ¥ r '¥dl

LineÚ

Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô =

m0

4pm ¥ ˆ r

r 2

where m is called the magnetic dipole moment of the current loop. It is defined as

m =1

2I r '¥dl

LineÚ

If the current loop is a plane loop (current located on the surface of a plane) then r '¥dl ( ) / 2 isthe area of the triangle shown in Figure 5.9. Therefore,

1

2r '¥dl

LineÚ = a

- 18 -

where a is the area enclosed by the current loop. In this case, the dipole moment of the currentloop is equal to

m I a=

where the direction of a must be consistent with the direction of the current in the loop (right-hand rule).

I

r'

dl

Figure 5.9. Calculation of m .

Assuming that the magnetic dipole is located at the origin of our coordinate system and thatm is pointing along the positive z axis, we obtain for A :

A dipole =m0

4pm ¥ ˆ r

r 2 =m0

4pmsinq

r 2ˆ f

The corresponding magnetic field is equal to

B dipole = — ¥ A dipole =1

r sinq∂

∂qsinq

m0

4pmsinq

r 2

Ê Ë

ˆ ¯ ˆ r -

1r

∂∂r

rm0

4pmsinq

r 2

Ê Ë

ˆ ¯

ˆ q =

=m0

4pm

r 3 2cosq ˆ r + sinq ˆ q { }

The shape of the field generated by a magnetic dipole is identical to the shape of the fieldgenerated by an electric dipole.

Example: Problem 5.33Show that the magnetic field of a dipole can be written in the following coordinate free form:

B =m0

4p1

r 3 3 m ∑ ˆ r ( )ˆ r - m { }

- 19 -

P

r m q p/2 + q

z

r


Consider the configuration shown in Figure 5.10. The scalar product between ˆ r and m isequal to

m ∑ ˆ r = mcosq

The scalar product between ˆ q and m is equal to

m ∑ ˆ q = mcos1

2p + q

Ê Ë

ˆ ¯ = -msinq

Therefore,

B =m0

4p1

r 3 2mcosq ˆ r + msinq ˆ q { } =m0

4p1

r 3 2 m ∑ ˆ r ( ) ˆ r - m ∑ ˆ q ( ) ˆ q { } =

=m0

4p1

r 3 3 m ∑ ˆ r ( )ˆ r - m ∑ ˆ r ( )ˆ r - m ∑ ˆ q ( ) ˆ q { } =m0

4p1

r 3 3 m ∑ ˆ r ( )ˆ r - m { }

Example: Problem 5.34A circular loop of wire, with radius R, lies in the xy plane, centered at the origin, and carries a

current I running counterclockwise as viewed from the positive z axis.a) What is its magnetic dipole moment?b) What is its (approximate) magnetic field at points far from the origin?c) Show that, for points on the z axis, your answer is consistent with the exact field as calculatedin Example 6 of Griffiths.

- 20 -

a) Since the current loop is a plane loop, its dipole moment is easy to calculate. It is equal to

m = Ia = pR2I ˆ k

b) The magnetic field at large distances is approximately equal to

B =m0

4ppR2I

r 3 2cosq ˆ r + sinq ˆ q { }

c) For points on the positive z axis q = 0°. Therefore, for z>0

B =m0

4ppR2I

r 3 2 ˆ k =m0

2

R2I

r 3ˆ k

Fore points on the negative z axis q = 180°. Therefore, for z<0

B =m0

4ppR2I

r 3 -2 ˆ k ( ) = -m0

2

R2I

r 3ˆ k

The exact solution for B on the positive z axis is

B =m0I

2

R2

R2 + z 2( )3/ 2ˆ k

For z » R the field is approximately equal to

B @m0

2

R2I

z 32ˆ k

which is consistent with the dipole field of the current loop.

Example: Problem 5.35A phonograph record of radius R, carrying a uniform surface charge s, is rotating at constant

angular velocity w. Find its magnetic dipole moment.

The rotational period of the disk is equal to

T =2pw

Consider the disk to consist of a large number of thin rings. Consider a single ring of innerradius r and with dr. The charge on such a ring is equal to

dq = s p r + dr( )2 - pr 2( ) @ 2ps rdr

- 21 -

Since the charge is rotating, the moving charge corresponds to a current dI:

dI =dq

dt=

2ps rdr2pw

= sw rdr

The dipole moment of this ring is therefore equal to

dm = p r 2( )dI ˆ k = psw r3dr ˆ k

The total dipole moment of the disk is equal to

m = dm Ú = psw r 3dr0

R

Ú ˆ k =p4

swR4 ˆ k

- 1 -

Chapter 6. Magnetostatic Fields in Matter

6.1. Magnetization

Any macroscopic object consists of many atoms or molecules, each having electric chargesin motion. With each electron in an atom or molecule we can associate a tiny magnetic dipolemoment (due to its spin). Ordinarily, the individual dipoles cancel each other because of therandom orientation of their direction. However, when a magnetic field is applied, a netalignment of these magnetic dipoles occurs, and the material becomes magnetized. The state ofmagnetic polarization of a material is described by the parameter M which is called themagnetization of the material and is defined as

M = magnetic dipole moment per unit volume

In some material the magnetization is parallel to B . These materials are called paramagnetic. Inother materials the magnetization is opposite to B . These materials are called diamagnetic. Athird group of materials, also called Ferro magnetic materials, retain a substantial magnetizationindefinitely after the external field has been removed.

m

s1

q F

F

q

y

z

a)

z

x

Fr Fl

b)

I

s2

Figure 6.1. Force on a rectangular current loop.

- 2 -

6.1.1. Paramagnetism

Consider a rectangular current loop, with sides s1 and s2, located in a uniform magnetic field,pointing along the z axis. The magnetic dipole moment of the current loop makes an angle qwith the z axis (see Figure 6.1a). The magnetic forces on the left and right sides of the currentloop have the same magnitude but point in opposite directions (see Figure 6.1b). The net forceacting on the left and right side of the current loop is therefore equal to zero. The force on thetop and bottom part of the current loop (see Figure 6.1a) also have the same magnitude and pointin opposite directions. However since these forces are not collinear, the corresponding torque isnot equal to zero. The torque generated by magnetic forces acting on the top and the bottom ofthe current loop is equal to

N = r ¥ F Â =s1

2F sinq ˆ i +

s1

2F sinq ˆ i = s1Fsinq ˆ i

The magnitude of the force F is equal to

F = I dl ¥ B LineÚ = Is2B

Therefore, the torque on the current loop is equal to

N s s IB i mB i m B= = = ¥1 2 sin √ sin √q q

where m is the magnetic dipole moment of the current loop. As a result of the torque on thecurrent loop, it will rotate until its dipole moment is aligned with that of the external magneticfield.

In atoms we can associate a dipole moment with each electron (spin). An external magneticfield will line up the dipole moment of the individual electrons (where not excluded by the Pauliprinciple). The induced magnetization is therefore parallel to the direction of the externalmagnetic field. It is this mechanism that is responsible for paramagnetism.

In a uniform magnetic field the net force on any current loop is equal to zero:

F = I dl ¥ B [ ]Ú = I dl Ú[ ] ¥ B = 0

since the line integral of dl is equal to zero around any closed loop.If the magnetic field is non-uniform then, in general, there will be a net force on the current

loop. Consider an infinitesimal small current square of side e, located in the yz plane and with acurrent flowing in a counter-clockwise direction (see Figure 6.2). The force acting on the currentloop is the vector sum of the forces acting on each side:

- 3 -

F = I dl ¥ B Side 1

Ú + dl ¥ B Side 2

Ú + dl ¥ B Side 3

Ú + dl ¥ B Side 4

ÚÈ

Î Í Í

˘

˚ ˙ ˙

=

= I ˆ j ¥ B 0,y,0( )( )dy + ˆ k ¥ B 0,e ,z( )( )dz0

e

Ú + ˆ j ¥ B 0,y,e( )( )dy + ˆ k ¥ B 0,0,z( )( )dze

0

Úe

0

Ú0

e

ÚÈ Î Í

˘ ˚ ˙ =

= I ˆ j ¥ B 0,y,0( ) - B 0,y,e( )[ ]( )dy + ˆ k ¥ B 0,e ,z( ) - B 0,0,z( )[ ]( )dz0

e

Ú0

e

ÚÈ Î Í

˘ ˚ ˙ =

= I - ˆ j ¥ e∂B ∂z

(0,y, 0)

Ê

Ë Á

ˆ

¯ ˜ dy + ˆ k ¥ e

∂B ∂y

(0, 0,z)

Ê

Ë Á

ˆ

¯ ˜ dz

0

e

Ú0

e

ÚÈ

Î Í Í

˘

˚ ˙ ˙

3

y

z

x

I

e

e

1

24

Figure 6.2. Infinitesimal square current loop.

In this derivation we have used a first-order Taylor expansion of B :

B 0,e ,z( ) = B 0,0,z( ) + e∂B

∂y(0,0,z)

and

B 0,y,e( ) = B 0,y,0( ) + e∂B

∂z(0,y, 0)

Assuming that the current loop is so small that the derivatives of B are constant over theboundaries of the loop we can evaluate the integrals and obtain for the total force:

- 4 -

F = Ie 2 ˆ k ¥∂B

∂y- ˆ j ¥

∂B

∂z

È

Î Í

˘

˚ ˙ = m

∂Bx

∂yˆ j -

∂By

∂yˆ i

Ê Ë Á

ˆ ¯ ˜ -

∂Bz

∂zˆ i -

∂Bx

∂zˆ k

Ê Ë Á

ˆ ¯ ˜

È

Î Í Í

˘

˚ ˙ ˙

= m∂Bx

∂xˆ i +

∂Bx

∂yˆ j +

∂Bx

∂zˆ k

È

Î Í

˘

˚ ˙

where m is the magnetic dipole moment of the current loop. In this derivation we have used thefact that the divergence of B is equal to zero for any magnetic field and this requires that

∂Bx

∂x= -

∂By

∂y-

∂Bz

∂z

The magnetic dipole moment m of the current loop is equal to

m = mˆ i

Therefore, the equation for the force acting on the current loop can be rewritten in terms of m :

F =∂

∂xmBx( )ˆ i +

∂∂y

mBx( )ˆ j +∂

∂zmBx( )ˆ k = — m ∑ B ( )

Any current loop can be build up of infinitesimal current loops and therefore

F = — m ∑ B ( )

for any current loop.

Example: Problem 6.1.a) Calculate the torque exerted on the square loop shown in Figure 6.3 due to the circular loop(assume r is much larger than a or s).b) If the square loop is free to rotate, what will its equilibrium orientation be?

a) The dipole moment of the current loop is equal to

m = pa2I ˆ k

where we have defined the z axis to be the direction of the dipole. The magnetic field at theposition of the square loop, assuming that r»a, will be a dipole field with q = 90°:

B =m0

4pm

r 3ˆ q =

m0

4ppa2I

r 3ˆ q = -

m0

4

a2

r 3 I ˆ k

The dipole moment of the square loop is equal to

m square = s2I ˆ i

- 5 -

I

m

a

s I

r


Here we have assumed that the x axis coincides with the line connecting the center of the currentcircle and the center of the current square. The torque on the square loop is equal to

N = m square ¥ B =

ˆ i ˆ j ˆ k

s2I 0 0

0 0 -m0

4

a2

r 3 I

=m0

4

a2s2

r 3 I 2 ˆ j

b) Suppose the dipole moment of the square loop is equal to

m square = mxˆ i + my

ˆ j + mzˆ k

The torque on this dipole is equal to

N = m square ¥ B =

ˆ i ˆ j ˆ k

mx my mz

0 0 -m0

4

a2

r 3 I

= -m0

4

a2

r 3 I my ˆ i +m0

4

a2

r 3 I mx ˆ j

In the equilibrium position, the torque on the current loop must be equal to zero. This thereforerequires that

mx = my = 0

Thus, in the equilibrium position the dipole will have its dipole moment directed along the z axis.The energy of a magnetic dipole in a magnetic field is equal to

- 6 -

U = -m ∑ B

The system will minimize its energy if the dipole moment and the magnetic field are parallel.Since the magnetic field at the position of the square loop is pointing down, the equilibriumposition of the current loop will be with its magnetic dipole moment pointing down (along thenegative z axis).

6.1.2. Diamagnetism

Consider a very classical picture of a Hydrogen atom consisting of an electron revolving in acircular orbit of radius r around a nucleus (see Figure 6.4). Suppose that the velocity of theelectron is equal to v. Since the velocity of the electron is very high, the revolving electron lookslike a steady current of magnitude

I =e

T=

ev

2p r

B

r

I v

m

Figure 6.4. Electron in orbit.

The direction of the current is in a direction opposite to that of the electron. The dipole momentof this current is equal to

m = -1

2evr ˆ k

If the atom is placed in a magnetic field, it will be subject to a torque. However, it is verydifficult to tilt the entire orbit. Instead the electron will try to reduce its torque by changing its

- 7 -

velocity. With no magnetic field present, the velocity of the electron can be obtained byrequiring that the centripetal force is sustained by just the electric force:

1

4pe0

e2

r 2 = me

v2

r

In a magnetic field, the centripetal force will be sustained by both the electric and the magneticfield:

1

4pe0

e2

r 2 + ev'B = me

v'2

r

Here we have assumed that the magnetic field is pointing along the positive z axis (in a directionopposite to the direction of the magnetic dipole moment). We have also assumed that the size ofthe orbit (r) does not change when the magnetic field is applied. Combining the last twoequations we obtain

me

v2

r+ ev'B = me

v '2

r

or

ev'B =me

rv '2 -v2( ) =

me

rv '+ v( ) v '-v( )

Assuming that the change in the velocity is small we can use the following approximations:

v'@ v

and

v'-v @ Dv

Therefore,

evB = 2me

rvDv

This equation shows that the presence of the magnetic field will increase the speed of theelectron. An increase in the velocity of the electron will increase the magnitude of the dipolemoment of the revolving electron. The change in m is opposite to the direction of B . If the

- 8 -

electron would have been orbiting the other way, it would have been slowed down by themagnetic field. Again the change in the dipole moment is opposite to the direction of B .

In the presence of an external magnetic field the dipole moment of each orbit will be slightlymodified, and all these changes are anti-parallel to the external magnetic field. This is themagnetism that is responsible for diamagnetism. Diamagnetism is present in all materials, but isin general much weaker than paramagnetism. It can therefore only be observed in thosematerials where paramagnetism is not present.

6.2. The Field of a Magnetized Object

Consider a magnetized material with magnetization M . The associated vector potential A isequal to

A =m0

4pM ¥ Dˆ r

Dr2 dtÚ

Following the same procedure used in Chapter 4 to calculate the electrostatic potential of apolarized material, we obtain for A :

A =m0

4p— ¥ M

Drdt

SurfaceÚ +

m0

4pM ¥ da

Dr=

m0

4pJ bDr

dtSurface

Ú +m0

4pK bDr

daSurface

ÚSurface

Ú

where J b = — ¥ M is the bound volume current and K b = M ¥ ˆ n is the bound surface current. Ifthe material has a uniform magnetization then the bound volume current is zero. The fieldproduced by a magnetized object is equal to the field produced by the bound currents.

M

K

t

Figure 6.5. Bound surface current.

- 9 -

Consider a uniformly magnetized thin slab of material of thickness t. The material can be cutup into tiny current loops (see Figure 6.5). If each current loop has an area a then the dipolemoment due to a surface current I is equal to

m I = Iaˆ k

The volume of the current loop is at and therefore its dipole moment must be equal to

m I = Matˆ k

This requires that the surface current of the current loop is equal to

I = Mt

Since the magnetization is uniform, the current in each of the current loops will be constant andflowing in the same direction. Therefore, all volume currents cancel, and the only currentremaining will be a surface current, flowing on the surface of the material. The current flowingon the surface of the material will be equal to the current in each of the current loops. Therefore,the current density on the surface is equal to

K =I

t= M

In vector notation:

K = M ¥ ˆ n

This equation is also consistent with the fact that there is no current flowing on the top andbottom surfaces (where M ¥ ˆ n = 0).

Example: Problem 6.7An infinitely long circular cylinder carries a uniform magnetization M parallel to its axis.

Find the magnetic field (due to M ) inside and outside the cylinder.

Consider a coordinate system S in which the z axis coincides with the axis of the cylinder.The magnetization of the material is equal to

M = M ˆ k

Since the material is uniformly magnetized, its bound volume current is equal to zero. Thebound surface current is equal to

- 10 -

K b = M ¥ ˆ n = M ¥ ˆ r = M ˆ f

This current distribution is identical to the current distribution in an infinitely long solenoid. Themagnetic field outside an infinitely long solenoid is equal to zero (see Example 9, Chapter 5 ofGriffiths), and therefore also the field outside the magnetized cylinder will be equal to zero. Themagnetic field inside an infinitely long solenoid can be calculated easily using Ampere's law (seeExample 9, Chapter 5 of Griffiths). It is equal to

B = m0Kbˆ k = m0Mˆ k

6.3. The Auxiliary Field H

The magnetic field in a system containing magnetized materials and free currents can beobtained by calculating the field produced by the total current J where

J = J free + J bound

This approach is very similar to the approach taken in electrostatics where the total electric fieldproduced by a system containing dielectric materials is equal to the electric field produced by acharge distribution s where

s = s free + s bound

To calculate the magnetic field produced by a system containing magnetized materials we haveto use the following form of Ampere's law:

1

m0

— ¥ B ( ) = J free + J bound = J free + — ¥ M


— ¥1

m0

B - M Ê Ë Á

ˆ ¯ ˜ = J free

The quantity in parenthesis is called the H-field

H =1

m0

B - M

- 11 -

H plays a role in magnetostatics analogous to D in electrostatics. Ampere's law in terms of H reads

— ¥ H = J free

and

H ∑ dlÚ = I free,intercepted

Ampere's law for H tells us that the curl of H is equal to the free current density. However,a knowledge of the free current density is not sufficient to determine H . The Helmholtztheorem shows that besides knowing the curl of a vector function, we also need to know thedivergence of that vector function before it is uniquely defined. Although the divergence of B iszero for any magnetic field (and therefore Ampere's law for B defines B uniquely) thedivergence of H is not necessarily zero:

— ∑ H = — ∑1

m0

B - M Ê Ë Á

ˆ ¯ ˜ =

1

m0

— ∑ B - — ∑ M = -— ∑ M

Therefore, only for those systems where — ∑ M = 0 can we use Ampere's law for H directly tocalculate H . The divergence of H will be zero only for systems with cylindrical, plane,solenoidal, or toroidal symmetry.

The H field is a quantity that is used in the laboratory more often that the B field. This is aresult of the dependence of H on only the free currents (which are easy to control). The B fielddepends both on the free and on the bound currents, and thus requires a detailed knowledge ofthe magnetic properties of the materials used. In electrostatics, the electric field can be obtainedimmediately from the potential difference (which is easy to control). The electric displacementD depends only on the free charge distribution, but in most cases a direct measurement of thefree charge distribution is very difficult to carry out. Therefore, in electrostatics the electric fieldis in most cases a more useful parameter then the electric displacement D .

Example: Problem 6.12An infinitely long cylinder, of radius R, carries a "frozen-in" magnetization, parallel to the

axis,

M = kr ˆ k

where k is a constant and r is the distance from the axis (there is no free current anywhere). Findthe magnetic field inside and outside the cylinder by two different methods:a) Locate all the bound currents, and calculate the field they produce.

- 12 -

b) Use Ampere's law to find H , and then get B .

a) The magnetization of the material is directed along the z axis and is equal to

M = kr ˆ k

The bound volume current is equal to

J b = — ¥ M =1

r

∂Mz

∂fˆ r -

∂Mz

∂rˆ f = -k ˆ f

The bound surface current is equal to

K b = M ¥ ˆ n r =R

= kR k ¥ ˆ r = kR ˆ f

r L


The bound currents produce a solenoidal field. The field outside the cylinder will be equal tozero and the field inside the cylinder will be directed along the z axis. Its magnitude can be

- 13 -

obtained using Ampere's law. Consider the Amperian loop shown in Figure 6.6. The lineintegral of B along the Amperian loop is equal to

B ∑ dl Ú = -BL

The current intercepted by the Amperian loop is equal to

Iencl = -KbL + JbLdrr

R

Ú = -kLR + kLdrr

R

Ú = -kLR + kL R - r( ) = -kLr

Ampere's law can now be used to calculate the magnetic field:

B =m0Iencl

-Lˆ k = m0kr ˆ k

b) The divergence of M is equal to zero. Therefore, Ampere's law uniquely defines H . TheH field is pointing in the z direction. Using Ampere's law, in terms of the H field, weimmediately conclude that for the Amperian loop shown in Figure 6.6

H ∑ dl Ú = HL = I free,intercepted = 0

since there is no free current This can only be true if H = 0. This implies that

H =1

m0

B - M = 0

Therefore, the magnetic field B is equal to

B = m0M

In the region outside the cylinder the magnetization is equal to zero and therefore the magneticfield is equal to

B = 0

In the region inside the cylinder the magnetization is equal to

M = kr ˆ k

and therefore the magnetic field is equal to

B = m0kr ˆ k

- 14 -

which is identical to the result obtained in part a).

Example: Problem 6.14Suppose the field inside a large piece of material is B 0, and the corresponding H field is

equal to

H 0 =1

m0

B 0 - M

a) A small spherical cavity is hollowed out of the material. Find the B field at the center of thecavity in terms of B 0 and M . Also find the H field at the center of the cavity in terms of H 0and M .b) Do the same for a long needle-shaped cavity running parallel to M .c) Do the same for a thin wafer-shaped cavity perpendicular to M .Assume the cavities are small enough so that M , B 0, and H 0 are essentially constant.

a) The field in the spherical cavity is the superposition of the field B 0 and the field produced bya sphere with magnetization -M . The bound volume current in the sphere is equal to zero(uniform magnetization). The bound surface current is equal to

K b = -M ( ) ¥ ˆ n = -M sinq ˆ f

Here we have assumed that the magnetization of the sphere is directed along the z axis. Nowconsider a uniformly charged sphere, rotating with an angular velocity w around the z axis. Thesystem carries a surface current equal to

K = sv = swRsinq ˆ f

Comparing these two equations for the surface current we conclude that

M = -swR

In Example 11 of Chapter 5 the magnetic field produced by a uniformly charged, rotating spherewas calculated. The magnetic field inside the sphere was found to be uniform and equal to

B =2

3m0swR ˆ k

But since M = -swR we can rewrite this expression as

B = -2

3m0M ˆ k = -

2

3m0M

- 15 -

The magnetic field inside the spherical cavity is therefore equal to

B cavity = B 0 + B sphere = B 0 -2

3m0M

The corresponding H field is equal to

H cavity =1

m0

B cavity - M cavity =1

m0

B cavity =1

m0

B 0 -2

3M = H 0 +

1

3M

Here we have used the fact that M cavity = 0 since no materials are present there.

b) The magnetic field inside the needle-shaped cavity is equal to the vector sum of the field B 0and the field produced by a needle-shaped cylindrical piece of material with magnetization -M .The field inside a needle-shaped cylinder of magnetization -M is approximately equal to thefield inside an infinitely long solenoid. This field was calculated in Problem 6.7, and for acylinder with a uniform magnetization -M it is equal to

B = -m0M

The magnetic field inside the needle-shaped cavity is thus equal to

B cavity = B 0 + B cylinder = B 0 - m0M


H cavity =1

m0


m0

B cavity =1

m0

B 0 - M = H 0

c) The magnetic field in the center of a thin wafer-shaped cavity is equal to the vector sum ofB 0 and the magnetic field inside a waver-shaped material with magnetization -M . Since thethickness of the wafer approaches zero, the total surface current on the material approaches zero,and consequently the magnetic field inside the waver approaches zero. Therefore, the magneticfield inside the cavity will be equal to

B cavity = B 0 + B wave = B 0


H cavity =1

m0


m0

B cavity =1

m0

B 0 = H 0 + M

- 16 -

6.4. Linear Media

In paramagnetic and diamagnetic materials, the magnetization is maintained by the externalmagnetic field. The magnetization disappears when the field is removed. Most paramagneticand diamagnetic materials are linear; that is their magnetization is proportional to the H field:

M = c mH

The constant of proportionality cm is called the magnetic susceptibility of the material. Invacuum the magnetic susceptibility is zero. In a linear medium, there is linear relation betweenthe magnetic field and the H field:

B = m0 H + M ( ) = m0 1 + c m( )H = mH

where m is called the permeability of the material. The permeability of free space is equal to m0.The linear relation between H and B does not automatically imply that the divergence of H

is zero. The divergence of H will only be equal to zero inside a linear material, but will be non-zero at the interface between two materials of different permeability. Consider for example theinterface between a linear material and vacuum (see Figure 6.7). The surface integral of M across the surface of the Gaussian pillbox shown in Figure 6.7 is definitely not equal to zero.According to the divergence theorem the surface integral of M is equal to the volume integral of— ∑ M :

M ∑ da Surface

Ú = — ∑ M ( )dtVolume

Ú

µ = 0

M = 0

mM

Figure 6.7. Interface of linear materials.

- 17 -

Therefore, if the surface integral of M is not equal to zero, the divergence of M can not be zeroeverywhere.

Example: Example 6.3An infinite solenoid (N turns per unit length, current I) is filled with linear material of

susceptibility cm. Find the magnetic field inside the solenoid.

I

H

L

Figure 6.8. Example 6.3.

Because of the symmetry of the problem, the divergence of H will be equal to zero,everywhere. Therefore, the H field can be obtained directly from Ampere's law. Consider theAmperian loop shown in Figure 6.8. The line integral of H around the loop is equal to

H ∑ dl LineÚ = HL

where the line integral is evaluate in the direction shown in Figure 6.8, and it is assumed that theH field is directed along the z axis. The free current intercepted by the Amperian loop is equalto

I free,intercepted = NIL

Ampere's law for the H field immediately shows that

H = NI ˆ k

The magnetic field inside the solenoid is equal to

B = m0 1 + c m( )H = m0 1 + c m( )NI ˆ k

The magnetization of the material is equal to

- 18 -

M = c mH = c m NI ˆ k

and is uniform. Therefore, there will be no bound volume currents in the material. The boundsurface current is equal to

K b = M ¥ ˆ n = c m H ¥ ˆ n ( ) = c m NI ˆ f

This last equation shows that the bound surface current flows in the same direction(paramagnetic materials) or in an opposite direction (diamagnetic materials) as the free current.

Example: Problem 6.18A sphere of linear magnetic material is placed in an originally uniform magnetic field B 0.

Find the new field inside the sphere.

This problem can be solved using a method similar to the method used in example 7 ofGriffiths (Chapter 4). The external field B 0 will magnetize the sphere:

M 0 = cmH 0 =c m

m0 1 + c m( ) B 0

This magnetization will produce a uniform magnetic field inside the sphere (see Example 6.1 ofGriffiths, Chapter 6):

B 1 =2

3m0M 0 =

2c m

3 1 + c m( ) B 0

This additional magnetic field magnetizes the sphere by an additional amount:

M 1 =c m

m0 1 + c m( ) B 1 =2c m

2

3m0 1 + c m( )2 B 0

This additional magnetization produces an additional magnetic field inside the sphere:

B 2 =2

3m0M 1 =

2c m

3 1 + c m( )Ê

Ë Á

ˆ

¯ ˜

2

B 0

The total magnetic field inside the sphere is therefore equal to

- 19 -

B =2c m

3 1 + c m( )Ê

Ë Á

ˆ

¯ ˜

n

B 0n=0

•

Â =1

1 -2c m

3 1 + c m( )B 0 =

3 1 + cm( )3 + cm

B 0

When can check the consistency of this answer by calculating the magnetization of the sphere:

M =c m

m0 1 + c m( ) B =c m

m0 1 + c m( )3 1 + c m( )3 + c m

B 0 =3c m

m0 3 + cm( ) B 0

The magnetic field inside the sphere due to the magnetization M is equal to

B M =2

3m0M =

2

3m0

3cm

m0 3 + cm( ) B 0 =2c m

3 + c m

B 0

The total magnetic field inside the sphere is therefore equal to

B total = B 0 + B M = 1 +2c m

3 + c m

Ê Ë Á

ˆ ¯ ˜ B 0 =

3 + 3c m

3 + c m

B 0

which is consistent with our assumption.

6.5. Nonlinear Media

The best known nonlinear media are the ferromagnetic materials. Ferromagnetic materialsdo not require external fields to sustain their magnetization (therefore, the magnetizationdefinitely depends in a nonlinear way on the field). The magnetization in ferromagneticmaterials involves the alignment of the dipole moments associated with the spin of unpairedelectrons. The difference between ferromagnetic materials and paramagnetic materials is that inferromagnetic materials the interaction between nearby dipoles makes them want to point in thesame direction, even when the magnetic field is removed. However, the alignment occurs inrelative small patches, called domains. When a ferromagnetic material is not located in amagnetic field, the dipole moments of the various domains are not aligned, and the material as awhole is not magnetized. When the ferromagnetic material is put into a magnetic field, theboundaries of the domain parallel to the field will increase at the expense of neighboringboundaries. If the field is strong enough, one domain takes over the entirely, and theferromagnetic material is said to be saturated (all unpaired electrons are aligned and therefore themagnetization reaches a maximum value). The magnetic susceptibility of ferromagnetic

- 20 -

materials is around 103 (roughly eighth orders of magnitude larger than the susceptibility ofparamagnetic materials). When the magnetic field is removed some magnetization remains (andwe have created a permanent magnet). For any ferromagnetic material, the magnetizationdepends not only on the applied magnetic field but also on the magnetization history. Thealignment of dipoles in a ferromagnet can be destroyed by random thermal motion. Thedestruction of the alignment occurs at a precise temperature (called the Curie point). When aferromagnetic material is heated above its Curie temperature it becomes paramagnetic.

- 1 -

Chapter 7. Electrodynamics

7.1. Electromotive Force

An electric current is flowing when the electric charges are in motion. In order to sustain an

electric current we have to apply a force on these charges. In most materials the current density

J is proportional to the force per unit charge:

J = s f

The constant of proportionality s is called the conductivity of the material. Instead of

specifying the conductivity, it is more common to specify the resistivity r:

r =1

s

For conductors the resistivity is typically 10-8 W-m; for semiconductor it varies between 0.01 W-

m and 1 W-m, and for insulators it varies between 105 W-m and 106 W-m. In most cases the

force on the charges is the electromagnetic force. In that case the current density is equal to:

J = s E + v ¥ B ( )

If the velocity of the charges is small the second term can be ignored, and the equation for J

reduces to Ohm's Law:

J = s E

Consider a wire of cross-sectional area A and length L. If a potential difference V is applied

between the ends of the wire, it will produce an electric field inside the wire of magnitude

E =V

L

The current density in the wire is therefore equal to

J = sV

L

The total current flowing through the wire is therefore equal to

- 2 -

I = JA = s AV

L

This equation shows that the current flowing from one electrode to the other electrode is

proportional to the potential difference between them. This is a rather surprising result since the

charge carriers are constantly accelerating. However, the proportionality between the current

and the potential difference has been found to be correct for most materials. This relation can be

written as

V = IR

The constant of proportionality R is called the resistance of the material. It is in general a

function of the geometry of the system and the conductivity of the materials between the

electrodes. The unit of resistance is the ohm (W). The resistance of the wire is equal to

R =V

I=

V

s AVL

=1

sL

A= r

L

A

To create a current we have to do work. The work required to move a unit of charge across a

potential difference V is equal to V. To establish a current I, we need to deliver a power P where

P = VI = I 2R

The unit of power is the Watt (1 W = 1 J/s). The work done by the electric force on the charge

carriers is converted into heat (Joule heating).


Two concentric metal spherical shells, of radius a and b, respectively, are separated by

weakly conducting material of conductivity s.

a) If they are maintained at a potential difference V, what current flows from one to the other?

b) What is the resistance between the shells?

a) Suppose a charge Q is placed on the inner shell. The electric field in the region between the

shells will be equal to

E =1

4pe0

Q

r 2 ˆ r

The corresponding potential difference between the spheres is equal to

- 3 -

Va - Vb = - E ∑ dr b

a

Ú =Q

4pe0

1

a-

1

bÊ Ë

ˆ ¯

Therefore, in order to maintain a potential difference V between the spheres, we must place a

charge Q equal to

Q =4pe0V1a

-1b

Ê Ë

ˆ ¯

on the center shell. The total current flowing between the two shells is equal to

I = J ∑ da Sphere

Ú = s E ∑ da Sphere

Ú = s1

4pe0

Q

r 2 4pr 2 = sQ

e0

= 4psV

1a

-1b

Ê Ë

ˆ ¯

b) The resistance between the shells can be obtained from Ohm's law:

R =V

I=

V

4psV

1

a-

1

bÊ Ë

ˆ ¯

=1

4ps1

a-

1

bÊ Ë

ˆ ¯


a) Two metal objects are embedded in weakly conducting material of conductivity s (see Figure

7.1). Show that the resistance between them is related to the capacitance of the arrangement

by

R =e0

s C

b) Suppose you connected a battery between 1 and 2 and charged them up to a potential

difference V0. If you then disconnect the battery, the charge will gradually leak off. Show

that V(t) = V0 exp(- t/t), and find the time constant t in terms of e0 and s.

a) Suppose a charge Q is placed on the positively charged conductor. The current flowing from

the positively charged conductor is equal to

- 4 -

I = J ∑ da Surface

Ú

where the surface integral is taken over a surface that encloses the positively charged conductor

(for example, the dashed surface in Figure 7.1). The expression for I can be rewritten in terms of

the electric field as

I = s E ∑ da Surface

Ú

1

2

s


Using Gauss's law to express the surface integral of E in terms of the total enclosed charge we

obtain

I = sQ

e0

The charge on the conductor is related to the capacitance of the arrangement and the potential

difference between the conductors:

Q = CV

The current I is therefore equal to

I =se0

CV

The resistance of the system can be calculated using Ohm's law:

- 5 -

R =V

I=

Vse0

CV=

e0

s C

b) The charge Q residing on the positively charged conductor is equal to

Q = CV = CRI = -CRdQ

dt


dQ

dt+

1

CRQ = 0

and has the following solution:

Q t( ) = Q0e-t / RC

The potential difference V is equal to

V t( ) =Q t( )

C=

Q0

Ce-t / RC = V0e

-t / RC

The decay constant t is equal to

t = RC =e0

s

In any electric circuit a current will only exist if a driving force is available. The most

common sources of the driving force are batteries and generators. When a circuit is hooked up to

a power source a current will start to flow. In a single-loop circuit the current will be the same

everywhere. Consider the situation in which the currents are not the same (see Figure 7.2). If

Iin > Iout then positive charge will accumulate in the middle. This accumulation of positive

charge will generate an electric field (see Figure 7.2) that slows down the incoming charges and

speeds up the outgoing charges. A reduction in the velocity of the incoming charges will reduce

the incoming current. An increase in the velocity of the outgoing charges will increase the

outgoing current. The current will change until Iin = Iout.

The total force f on the charge carriers (per unit charge) is equal to the sum of the source

force, fs, and the electric force:

f = f s + E

- 6 -

++

+

++

+

+

+

+

I in

I out

E E

E E

Figure 7.2. Current flow.

The work required to move one unit of charge once around the circuit is equal to

f ∑ dl Ú = f s ∑ dl Ú + E ∑ dl Ú = f s ∑ dl Ú = e

where e is called the electromotive force or emf. The emf determines the current flowing

through the circuit. This can be most easily seen bye rewriting the force f on the charge carriers

in terms of the current density J

e = f ∑ dl Ú =J

s∑ dl Ú =

I

asdlÚ = I

dl

asÚ = IR

Here, a is the cross-sectional area of the wire (perpendicular to the direction of the current).


a) Show that electrostatic force alone cannot be used to drive current around a circuit.

b) A rectangular loop of wire is situate so that one end is between the plates of a parallel-plate

capacitor (see Figure 7.3), oriented parallel to the field E = s/e0. The other end is way

outside, where the field is essentially zero. If the width of the loop is h and its total

resistance is R, what current flows? Explain.

- 7 -

h - s

+ s E

a

b c

d


a) If only electrostatic forces are present then the force per unit charge is equal to the

electrostatic force:

f = E

The associated emf is therefore equal to

e = f ∑ dl Ú = E ∑ dl Ú = 0

for any electrostatic field.

b) The only force on the charge carriers in the wire loop is the electric force. However, in part

a) we concluded that the emf associate with an electric force, generated by an electrostatic field,

is equal to zero. Therefore, the emf in the wire loop is equal to zero, and consequently the

current in the loop is also equal to zero. Note: at first sight it might appear that there is a net emf,

if we assume that the electric field generated by the capacitor is that of an ideal capacitor (that is

a homogeneous field inside and no field outside). Under that assumption, the emf is equal to

e = E ∑ dl Ú = E ∑ dl a

b

Ú =se0

h

The contribution of the path integral from c to d is equal to zero since the electric field is zero

there, and the contribution of the path integrals between b and c and between a and d is equal to

zero since the electric field and the displacement are perpendicular there. Clearly the calculated

emf is non-zero, and disagrees with the result of part a). The disagreement is a result of our

incorrect assumption that the electric field outside the capacitor is equal to zero (there are

fringing fields).

An important source of emf is the generator. In these devices the emf arises from the motion

of a conducting wire through a magnetic field. Consider the system shown in Figure 7.4 (note:

- 8 -

the magnetic field is only present in the region left of the dashed line). Consider the free charges

on the conductor. Since it is moving with a velocity v in a magnetic field it will experience a

magnetic force. The force on a positive charge q located ion segment ab of the wire loop is

equal to

Fq = qvB

d

cb

a

F

v

s Rh

Figure 7.4. The generator.

The magnetic force per unit charge is therefore equal to

fmag =Fq

q= vB

Since there are no other forces acting on the charges, the emf generated will be entirely due to

this magnetic force. The emf will be equal to

e = f mag ∑ dl = f mag ∑ dl a

b

Ú = vBhÚ

The magnetic flux intercepted by the wire loop is equal to

F = Bhs

The rate of change of the magnetic flux is equal to

dFdt

= Bhds

dt= -Bhv

Comparing the rate of change of enclosed magnetic flux and the induced emf we can conclude

that

- 9 -

e = -dFdt

This relation is called the flux rule for motional emf.

7.2. Faraday's Law

When a conducting wire moves in a constant magnetic field an emf is generated equal to

e = -dFdt

In this case, the magnetic force is responsible for the emf. However, the same emf is generated

when the wire is stationary and the magnetic field is moving. In this case, the magnetic force

does not play a role (since v = 0) and an electric field is responsible for the emf. This electric

field is not an electrostatic field (since electrostatic fields can not generate an emf; see Problem

7.5) but is induced by the changing magnetic field. The line integral of this electric field is equal

to

E ∑ dl LineÚ = e = -

dFdt

This equation can be rewritten by applying Stoke's theorem:

E ∑ dl LineÚ = — ¥ E ( ) ∑ da

SurfaceÚ = -

d

dtB ∑ da

SurfaceÚ = -

∂B

∂t∑ da

SurfaceÚ

Since we have not made any assumption about the surface, this equation can only be true if

— ¥ E = -∂B

∂t

This relation is called Faraday's law in differential form. The direction of the currents

generated by the changing magnetic field can be obtained most easily using Lenz's law which

states that

- 10 -

“ If a current flows, it will be in such a direction that the magnetic field it

produces tends to counteract the change in flux that induced the emf. “


A long solenoid of radius a, carrying N turns per unit length, is looped by a wire of resistance

R (see Figure 7.5).

a) If the current in the solenoid is increasing,

dI

dt= k = constant

what current flows in the loop, and which way (left or right) does it pass through the resistor.

b) If the current I in the solenoid is constant but the solenoid is pulled out of the loop and

reinserted in the opposite direction what total charge passes through the resistor?

R


a) Assume that the solenoid is an ideal solenoid; that is

B = m0NI ˆ k

If the current in the solenoid increases, the strength of the magnetic field also increases. The rate

of change in the strength of the magnetic field is equal to

dB

dt= m0N

dI

dtˆ k = m0 Nk ˆ k

The magnetic flux intercepted by the wire loop is equal to

- 11 -

F = p a2B

The corresponding rate of change of the magnetic flux is equal to

dFdt

= p a2 dB

dt= p a2m0Nk

The induced emf can be obtained from the flux law:

e = -dFdt

= -p a2m0 Nk

The current induced in the wire loop is equal to

I =eR

=p a2

Rm0Nk

The solenoidal magnetic field points from left to right. An increase in the strength of the

magnetic field will induce a current in the loop directed such that the magnetic field it produces

point from right to left (Lenz's law). Therefore, the current flows from left to right through the

resistor.

b) The change in the magnetic flux enclosed by the wire loop is equal to

DF = 2pa2m0 NI

The current flowing through the resistor is equal to

I =eR

=1

R

dFdt

=dQ

dt

This relation shows that

DQ =dQ

dtdt

-•

•

Ú = -1

R

dFdt

dt-•

•

Ú = -DFR

Substituting the expression for DF we obtain

DQ =1

R2pa2m0 NI( )

- 12 -

7.3. Inductance

Consider two loops: loop 1 and loop 2 (see Figure 7.6). A current I1 flowing through loop 1

will produce a magnetic field at the position of loop 2 equal to

B 1 =m0

4pI1

dl 1 ¥ D ˆ r

Dr2Ú

The magnetic flux through loop 2 is equal to

F 2 = B 1 ∑ da 2Ú = M21I1

Loop 2

Loop 1

Figure 7.6. Inductance.

Here, M21 is called the mutual inductance of the two loops. It is a purely geometrical quantity

that depends on the sizes, shapes and relative positions of the two loops. It does not change if we

switch the role of loop 1 and loop 2: the flux through loop 2 when we run a current I around

loop 1 is exactly the same as the flux through loop 1 when we send the same current I around

loop 2.

Besides inducing an emf in a nearby loop, the changing current in loop 1 also induces an emf

in loop 1. The flux through loop 1 generated by the current in loop 1 is equal to

F1 = LI1

The constant of proportionality is called the self inductance. The unit of inductance is the

Henrie (H).


- 13 -

A square loop of wire, of side s, lies midway between two long wires, 3s apart and in the

same place. (Actually, the long wires are sides of a large rectangular loop, but the short ends are

so far away that they can be neglected.) A clockwise current I in the square loop is gradually

increasing: dI/dt = k = constant. Find the emf induced in the big loop. Which way will the

induced current flow?

I

s

s

s


The system is schematically shown in Figure 7.7. To calculate the emf induces in the large

loop we need to determine the mutual inductance M. It is hard to calculate M in terms of a

current in the square loop since the magnetic field generated by this loop is rather complicated

(and therefore difficult to integrate). However, exploiting the equality of the mutual inductances,

we can also evaluate the inductance in terms of a current I in the large loop. The magnetic field

generated by the top wire of the large loop is that of an infinitely long straight wire carrying a

current I is equal to

B =m0

2pI

r

The flux associated with this magnetic field and intercepted by the square loop is equal to

F top =m0

2pI

rs dr

s

2s

Ú =m0

2pI s ln 2

The magnetic field generated by the bottom wire at the position of the square loop will be

pointing in the same direction as the magnetic field generated by the top wire at the position of

the square loop. Since the square loop is located at the same distance from the top wire as it is

from the bottom wire, the flux intercepted by the square loop is equal to

F = 2F top =m0

pI s ln 2

- 14 -

Therefore, the mutual inductance of the two loops is equal to

M =FI

=m0

ps ln 2

The induced emf in the large loop can now be calculated easily:

e = -dFdt

= -MdI

dt= -Mk = -

m0

pks ln 2

The direction of the flux through the large loop is pointing into the page. This can be seen most

easily by considering the magnetic field lines. Inside the square loop, the field lines point into

the page (right-hand rule). Since the field lines form closed loops, they must be pointing out of

the page anywhere outside the square loop. However, the large wire loop only covers a limited

fraction of space, and therefore definitely will not intercept all field lines outside the square loop.

Therefore, there will be more field lines pointing into the page then there are field lines pointing

out of the page. Consequently, the net magnetic flux will be pointing into the page. When the

current in the square loop increases the flux intercepted by the large loop will increase. The

induced emf will produce a magnetic field that counteracts this increase in flux, and therefore

produces a flux pointing out of the paper. The right-hand rule shows that the direction of the

current induced in the large loop must be flowing in a counter-clockwise direction.

7.4. The Maxwell Equations

The electric and magnetic fields in electrostatics and magnetostatics are described by the

following four equations:

— ∑ E =1

e0

r Gauss' Law

— ∑ B = 0

— ¥ E = -∂B

∂tFaraday's Law

— ¥ B = m0J Ampere's Law

In systems with non-steady currents not all of these equations are valid anymore. For example,

- 15 -

— ∑ — ¥ B ( ) = 0

for every vector function. However, according to Ampere's law

— ∑ — ¥ B ( ) = m0 — ∑ J ( )

which is only zero for steady currents (for which J is a constant, independent of position). For

non-steady currents

m0 — ∑ J ( ) π 0 = — ∑ — ¥ B ( )

We thus conclude that Ampere's law does not hold for non-steady currents. The failure of

Ampere's law can also be observed in a system in which a capacitor is being charged (see Figure

7.8). During the charging process a current I is flowing through the wire, and consequently there

will be a magnetic field present. The magnetic field generated by the charging current can be

calculated using Ampere's law. When we are far away from the capacitor the generated

magnetic field will be that of a line current. Consider an Amperian loop of radius r, centered on

the wire. The line integral of B around this loop is equal to

B ∑ dl Ú = 2p rB

According to Ampere's law the line integral of B around a closed loop is proportional to the

current intercepted by a surface spanned by this loop. For the system shown in Figure 7.8 the

intercepted current is ill defined. Consider first surface 1. The current intercepted by surface 1

is equal to I. Surface 2 is also spanned by the Amperian loop, but the current intercepted by this

loop is zero. We thus conclude that Ampere's law does not apply in systems where the current is

not continuous.

Surface 1

Surface 2

I

Figure 7.8. Charging a capacitor.

- 16 -

Maxwell modified Ampere's law in the following manner:

— ¥ B = m0 J + e0

∂E

∂t

Ê Ë Á

ˆ ¯ ˜

The term added by Maxwell is called the displacement current. It is defined as

J d = e0

∂E

∂t

Consider the region between the capacitor plates in Figure 7.8. The electric field in this region is

equal to

E =se0

ˆ k =Q

e0Aˆ k

where we have assumed that the field produced is that of an ideal capacitor with surface area A

and the z axis is in the direction of the current. The rate of change of the electric field is equal to

∂E

∂t=

1

e0

∂s∂t

ˆ k =1

e0A

∂Q

∂tˆ k =

I

e0Aˆ k

The surface integral of ∂E / ∂t across surface 2 is therefore equal to

∂E

∂t∑ da Ú =

I

e0

The surface integral of — ¥ B across surface 2 is equal to

— ¥ B ( ) ∑ da Surface 2

Ú = m0 J ∑ da Surface 2

Ú + m0e0

∂E

∂t∑ da

Surface 2Ú = m0I

The modification of Ampere's law by Maxwell insures that the surface integral of — ¥ B is

independent of the surface chosen. In electrostatics and magnetostatics the electric and magnetic

fields are constant in time, and therefore, the new form of Ampere's law reduces to the form of

Ampere's law we have been using so far.

In a region where there are no free charges or free currents Maxwell's equations become very

symmetric

- 17 -

— ∑ E = 0 — ∑ B = 0

— ¥ E = -∂B

∂t— ¥ B = m0e0

∂E

∂t

The symmetry is broken when electric charges are present, unless besides electric charges there

are magnetic monopoles. If the magnetic charge density is equal to h and the magnetic current

is equal to K then Maxwell's equation become

— ∑ E =re0

— ∑ B = m0h

— ¥ E = -m0K -∂B

∂t— ¥ B = m0J + m0e0

∂E

∂t

To obtain Maxwell's equations that describe the electric and magnetic fields in matter we

must take the bound charges and bound currents into account:

rb = - — ∑ P

J b = — ¥ M

In the non-static case, the polarization can be time dependent. Therefore, also the bound charge

density is time dependent, and a net current can be associated with the change in the bound

charge density. This current is called the polarization current J P and is equal to

J P =∂P

∂t

Maxwell's equations in matter are therefore equal to

— ∑ E =1e0

rf + rb( ) =1

e0

rf - — ∑ P ( ) Gauss' Law

— ∑ B = 0

— ¥ E = -∂B

∂tFaraday's Law

— ¥ B = m0 J f + J b + J P( ) + m0e0

∂E ∂t

= m0 J f + — ¥ M +∂P

∂tÊ Ë Á

ˆ ¯ ˜ + m0e0

∂E ∂t

Ampere's Law

- 18 -

It is common to rewrite Maxwell's equations in terms of the parameters we can control (the free

charge density and the free current density). Gauss's law can be rewritten as

— ∑ e0E + P ( ) = — ∑ D = rf

where D is called the electric displacement. Ampere's law can be rewritten as

— ¥B

m0

- M Ê Ë Á

ˆ ¯ ˜ = — ¥ H = J f +

∂∂t

e0E + P ( ) = J f +∂D

∂t

where H is called the H field. The most general form of Maxwell's equations, in terms of the

free charges and free currents, is given by

— ∑ D = rf — ∑ B = 0

— ¥ E = -∂B

∂t— ¥ H = J f +

∂D

∂t

Sem5 emphysic 217 electromagnet

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