ejercicios jackson electrodinamica

8/9/2019 Ejercicios Jackson Electrodinamica

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4.9

a) The system is described by

Since there is azimuthal symmetry, choosing the z-axis through q,

out 140

∑l

B lr −l−1Pl

q

| x − x′ |

out 140

∑l

B lr −l−1Pl

qr ∑

l

r r

lPl

in 140

∑l

A lr lPl

qr ∑

l

r r

lPl

Boundary conditions: At the surface, r ′ d r , r a r .

1) out in |r a, or

B l A la2l1

2) ∂

∂r in ∂

∂r out |r a, or letting k

0

k ∑l

lA lal−1Pl

q

d l a

l−1

d l Pl ∑

l

−l 1 Bla−l−2Pl q

d l a

l−1

d l Pl

∑l

−l 1 A lal−1Pl q

d l a

l−1

d l Pl

or

Al a1 − k l1 k l 1d l1

Bl a1 − k la2l1

1 k l 1d l1

Remember that P l 421 Y l0, and substitute the above coefficients into the expansion to get the

answer requested by the problem.

4.10

The system is described by

a) Since there is azimutal symmetry,

r , ∑l

Alr l B lr −l−1 Plcos

Also

D x E − x∇r ,

Dr − x∑l

lAlr l−1 − l 1 B lr −l−2Plcos

between the spheres,

Dr d r 2 Q , and is independent of r .Thus

Al 0, B l 0, l 0 → D r x B0

r 2

Dr d r 2 2 B0 0 −1

0

d cos 0

1

d cos 2 B00 Q

B0 Q

201 0

E Q2 01 0 r

2 r ̂

b)

Dr dA D r A f A → f D r x E r

f Q

201 0 r 2 , cos ≥ 0


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f Q

21 0 r 2 , cos 0

c)

poldV pol A −∇ PdV −PA → pol −P −0e E

pol − x/0 − 1 Q

21 0 r 2

Notice

pol f tot Q

21 0 r 2

0 E , as expected.

More Problems for Chapter 4

r/a

-Φ1 q

4πε0 a

1 2 3 4 5


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5.1


We want to show

m − 0 I 4

Suppose the observation point is moved by a displacement x, or equivlently that the loop isdisplaced by − x.

If we are to have B −∇m, then

m − x B

Using the law of Biot and Savart,

m − 0 I 4 x

dl′ r

r 3 − 0 I

4 r x dl′

r 3 − 0 I

4 r ̂ x dl′

r 2

m − 0 I 4

r ̂ dAr 2

− 0 I 4

Or,

m − 0 I 4

5.2

a) The system is described by

First consider a point at the axis of the solenoid at point z 0. Using the results of problem 5.1,

d m 04

NIdz

From the figure,

r ̂ dAr 2

dA cosr 2

2 z 0

R d

2 z 23/2 2 − z

R2 z 2 1

m 0

2 NI

z0

z − 1

R2 z 2 1 z dz

02

NI − z0 R2 z 02

Br − 02

NI ∂∂ z0

− z0 R2 z 02 02 NI − z0 R

2

z 02

R2 z 0

2

In the limit z0 → 0

Br 0

2 NI

By symmetry, thej loops to the left of z0 give the same contribution, so

B B l B r 0 NI

H NI

By symmetry, B is directed along the z axis, so

m − B 0

if is directed to the z axis. Thus for a given z, m is independent of , and consequently

H NI

everywhere within the solenoid.

If you are on the outside of the solenoid at position z 0, by symmetry the magnetic field must be in

the z direction. Thus using the above argument, m must not depend on . Set us take far awayfrom the axis of the solenoid, so that we can replace the loops by elementary dipoles m directed alongthe z axis. Thus for any point z0 we will have a contributions

m m r 1

r 13

m r 2

r 23

where m r 1 −m r 2 and r 1 r 2. Thus

H 0


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5.8

Using the same arguments that lead to Eq. (5.35), we can write

A 04

d 3 x

′

cos′ J r ′,′| x − x′ |

Choose x in the x − z plane. Then we use the expansion

1| x − x′ |

l,m

42l 1

r l

r l1

Y lm∗ ′,′Y l

m,0

The cos ′ factor leads to only an m 1 contribution in the expansion. Using

Y lm,0 2l 1

4l − m!l m!

P lmcos

and l−1!

l1! 1

ll1 , we have on the inside

A 04

l

1ll 1

r lPl1cos d 3 x′ P l

1cos′ J r ′, ′r ′l1

which can be written

A − 04

l

m lr lPl

1cos

with

m l − 1ll 1

d 3 x′ P l1cos ′ J r ′, ′

r ′l1

A similar expression can be written on the outside by redefining r and r .

5.10

a) From Eq. (5.35)

Ar , 04

I a

r ′2dr ′d ′ sin ′ cos′cos ′r ′ − a| x − x′ |

Using the expansion of 1/| x − x′ | given by Eq. (3.149),

1| x − x′ |

4 0

dk cosk z − z ′ 1

2 I 0k K 0k ∑

m1

cosm − ′ I mk K mk

We orient the coordinate system so 0, and because of the cos′ factor, m 1. Thus,

Ar , 04

I a

4

0

dk r ′2dr ′d cos ′ sin′cos′r ′ − acoskz I 1k K 1k

Ar , 0 aI

0

dk coskz I 1k K 1k

where is the smaller (larger) of a and .

b) From problem 3.16 b),

1| x − x′ |

∑m−

0

dkeim−

′ J mk J mk ′e−k | z|

Note z ′ 0, and 0, so

A 0 Ia

2

0

dke−k | z| J 1k J 1ka


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5.16

a) The system is shown in the figure

I shall use the magnetic potential approach and will call inside the sphere region 1 and outside thesphere region 2.

1 loop ∑l

A lr lPl

2 loop ∑l

B lr −l−1Pl

where H −∇, and we have the boundary conditions,

H 1|| H 2|| → 1r b 2r b

0 ∂∂r 1r b ∂∂r

2r b

We are given that b a , so

loop 14

m cosr 2

with m a2 I . (From the form of loop, only the l 1 term contributes.) The boundaryconditions give

A1b1 B 1b−1−1

− 20m4b3

0 A1 − 2m4b3

− 2 B1b−3

So

A1 − 24

mb3

− 02 0

On the inside, at the center of the loop

H −∇ loop − ∇ A1r cos

From Eq. (5.40), we are given −∇ loop at the center of the loop, which is directed in the z direction.

H z 10 − B − A1

If 0

A1 → − 14

mb3

and from (5.40), at r 0

H z I 2a

14

mb3

I 2a

I 4

a2

b3 I

2a 1 a

3

2b3

5.18

a) From the results of Problem 5.17, we can replace the problem stated by the system

where I∗ is equidistant from the interface and is equal to I ∗ r −1r 1 I . The radius of each currentloop is a. Now from Eq. (5.7)

F on I I dl Br

dl B dl B r dl B dlB r −̂ dlB r ̂

By symmetry, only the z − component survives, so, from the figure

dl B ẑ

dlB r a

4d 2 a 2 dlB 2d

4d 2 a 2

So

F z 2aI 4d 2 a 2

aBr 2dB

with B r and B given by Eqs. (5.48) and (5.49) and cos 2d 4d 2a2

, r 4d 2 a 2 , and I → I ∗.

c) To determine the limiting term, simply let r → 2d and take the lowest non-vanishing term in theexpansion of the magnetic flux density.

F z aI d aBr 2dB

F z aI d

a 0 I ∗a

4d a

2d 2 2d − 0 I

∗a2

41

2d 3 − a

2d

F z → − 30

32a4 I I ∗

d 4

The minus sign shows the force is attractive if I and I ∗ are in the same direction. This same resultcan be gotten more directly, using

F z ∇ zmB z

with m a2 I , and (from Eq. (5.64))

B z 04

2m∗

z3

with m∗ a2 I ∗ , and z 2d

F z 04

2a2 I ∗a2 I − 32d 4

− 30

32a4 I I ∗

d 4

with agrees with out previous result.


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5,19


The effective volume magnetic charge density is zero, since M is constant within the cylinder. The

effective surface charge density (n̂ M from Eq. (5.99)) is M 0, on the top surface and − M 0 on the bottom surface. From the bottom surface the potential is (for z 0

b 14

− M 0 2 0

a d

2 z 2 1/2 − M 0

2 a2 z2 − z

By symmetry, the potential from the top surface is (on the inside)

t M 0

2 a2 L − z 2 − L − z

The total magnetic potential is

b t − M 0

2 a2 z2 − z M 0

2 a2 L − z 2 − L − z

So, on the inside of the cylinder,

H z − ∂∂ z − M 0

2 a2 z 2 − z M 0

2 a2 L − z2 − L − z

H z − M 0

2 2 − z

a2 z2 − L − z

a2 L − z2

while above the cylinder,

H z − M 0

2 − z

a2 z 2 z − L

a2 L − z2

with a similar expression below the cylinder.

B 0 H M

Thus inside the cylinder,

B z 0 − M 0

2 2 − z

a2 z 2 − L − z

a2 L − z2 M 0

B z

0 M 02

z

a2 z 2 L − za2 L − z 2

while above the cylinder,

B z 0 M 0

2 z

a2 z 2 − z − L

a2 L − z 2

First we plot B z in units of a for L 5a

g z

12

z

1 z2 5− z

15− z2if z 5

12

z

1 z2− z−5

15− z2if 5 z

g z

0

0.2

0.4

0.6

0.8

1

2 4 6 8 10z

And similarly, H z in units of a for L 5a.

f z

− 12

2 − z1 z2

− 5− z15− z2

if z 5

− 12

− z1 z2

z−515− z2

if 5 z

f z

-0.4

-0.2

0

0.2

0.4

2 4 6 8 1 0z

5.26


Since the wires are nonpermeable, 0. The system is made of parts with cylindricalsymmetry, so we can determine B using Ampere’s law.

∇ B 0 J , or B dl 0 J da

On the outside of each wire,

B dl B2 0 I → B out 0 I 2

On the inside of each wire

B dl B 2 0 I 2

R2 , B in

0 I 2

R2

with R a , b

From the right-hand rule, the B from each wire is in the ̂ direction. From the above figure, usingthe general expression for the vector potential, we see A is in the ẑ direction. Since ∇ A B,

B z − ∂∂ A z → A z − B zd

Thus

A z − 0 I

2 ln R C − 0 I

4 ln 2

R2 1 on the outside

− 0 I 4

2

R 2 , on the inside

where I’ve determined C 1/2, from the requirement that A z be continuous at R . Let l be the


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length of the wire. Then we know the total potential energy is given by

W 12 J Ad 3 x l

2 J a Adaa J b Adab

Consider the second term l2 J b Adab. The system is pictured as

From the figure

a d b, a2 d 2 b2 − 2d b cos

so, since J b I

b2

l2 J b Adab l

2 I b2

Aout a A inb bd bd

l2

I b2

0 I 4 ln

a2

a2 1 −

b2

b2 bd bd

≃ l2

I b2

0 I 4

2 0

b

ln d 2

a2 1 −

b2

b2 bd b

l2

I b2

0 I 4

2 14

b 2 1 2ln d 2

a2 l

204

12 2ln d a I

2

The first term l2 J a Adaa is equal to

l2 J a Adaa l

204

12 2ln d

b I 2

Thus

W l2

04

1 2ln d 2

ab I 2 l

2 Ll

I 2

or

L

l

0

4

1 2ln d 2

ab

5.27


Using Ampere’s law in integral form

B dl 0 I enclosed

we get

B 0 I

2b2

, b

B 0 I

21 , b a

B 0, a

Now the energy in the magnetic field is given by ( l is the length of the wires)

W 12 B H d 3 x 1

20 B2d 3 x

120

0 I 2

2

l 2 0

b b2

2

d 2 b

a1

2d

120

0 I 2

2

l 12

2ln ab

l2

Ll

I 2

→ Ll

04

12

2ln ab

If the inner wire is hollow, B 0, b , so

Ll

02

ln ab

5.29


This problem is very much like 5.26, except the wires are superconducting. We know from section5.13 that the magnetic field within a superconductor is zero. We will be using

W 12 J Ad 3 x l

2 J a Adaa J b Adab

Using the same arguments as applied in problem 5.26,

A z − I

2 ln R

C − I 4 ln

2

R 2 0 on the outside

0, on the inside

Thus if we consider the second term l2 J b Adab,

l2 J b Adab l

2 I b2

Aout a A inb bd bd

≃ l

2

I

b2 I

4 2 0b

ln

d 2

a2 bd b

l

2

4 2ln d

a I 2

The first term l2 J a Adaa is equal to

l2 J a Adaa l

24

2ln d b

I 2

Thus

W l2

4

2ln d 2

ab I 2 l

2 Ll

I 2

so

Ll

4 2ln d

2

ab

Now using the methods of problem 1.6, assuming the left wire has charge Q, and the right wire charge−Q, we find

12 b

d −a Edr

Q

l

2 bd −a 1

r 1d − r

dr ≃Q

l

2 ln d

2

ab

C l

Q

l

12 2ln d

2

ab

Thus

Ll

C l

4 2ln d

2

ab 2

ln d 2

ab


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Solutions to Problems in Jackson,

Classical Electrodynamics , Third Edition

Homer Reid

October 8, 2000

Chapter 4: Problems 8-13

Problem 4.8

A very long, right circular, cylindrical shell of dielectric constant /0 and inner andouter radii a and b, respectively, is placed in a previously uniform electric field E0with its axis perpendicular to the field. The medium inside and outside the cylinderhas a dielectric constant of unity.

(a) Determine the potential and electric fields in the three regions, neglecting endeffects.

(b) Sketch the lines of force for a typical case of b ≈ 2a.(c) Discuss the limiting forms of your solution appropriate for a solid dielectric

cylinder in a uniform field, and a cylindrical cavity in a uniform dielectric.

We will take the axis of the cylinder to be the z axis and the electric field tobe aligned with the x axis: E0 = E 0̂i. Since the cylinder is very long and we’retold to neglect end effects, we can ignore the z direction altogether and treatthis as a two-dimensional problem.

(a) The general solution of the Laplace equation in two dimensional polar co-ordinates is

Φ(r,ϕ ) =

[Anrn + Bnr

−n][C n sin(nϕ) + Dn cos(nϕ)]

For the region inside the shell (r < a), the B coefficients must vanish to keepthe potential from blowing up at the origin. Also, in the region outside the shell

1

Homer Reid’s Solutions to Jackson Problems: Chapter 4 2

(r > b), the only positive power of r in the sum must be that which gives riseto the external electric field, i.e. −E 0r cos ϕ with An = 0 for n > 1. With theseobservations we may write expressions for the potential in the three regions:

Φ(r,ϕ ) =

rn[An sin nϕ + Bn cos nϕ], r < a

rn[C n sin nϕ + Dn cos nϕ] + r−n[E n sin nϕ + F n cos nϕ], a < r < b

−E 0r cos ϕ +

r−n[Gn sin nϕ + H n cos ϕ], r > b

The normal boundary condition at r = a is

0∂ Φ

∂r

x=a−

= ∂ Φ

∂r

x=a+

or

0

nan−1[An sin nϕ + Bn cos nϕ] =

nan−1[C n sin nϕ + Dn cos nϕ] − na−(n+1)[E n sin nϕ + F n cos nϕ]

From this we obtain two equations:

0

An = C n − E na−2n (1)0

Bn = Dn − F na−2n (2)

Next, the tangential boundary condition at r = a is

∂ Φ

∂ϕ

x=a+

= ∂ Φ

∂ϕ

x=a−

ornan[An cos nϕ − Bn sin nϕ] =

nan[C n cos nϕ − Dn sin nϕ] + na−n[E n cos nϕ − F n sin nϕ]

from which we obtain two more equations:

An = C n + E na−2n (3)

Bn = Dn + F na−2n (4)

Similarly, from the normal boundary condition at r = b we obtain

−0

E 0 cos ϕ − 0

nb−(n+1)[Gn sin nϕ + H n cos ϕ] =

nbn−1[C n sin nϕ + Dn cos nϕ] − nb−(n+1)[E n sin nϕ + F n cos ϕ]


which leads to

−0

Gn = C nb2n − E n (5)

−0

b2E 0δ n1 − 0

H n = Dnb2n − F n (6)

Finally, we have the tangential boundary condition at r = b:

bE 0 sin ϕ +

nb−n[Gn cos nϕ − H n sin nϕ] =nbn[C n cos nϕ − Dn sin nϕ] + nb−n[E n cos nϕ − F n sin nϕ]

giving

Gn = C nb2n + E n (7)

−b2E 0δ n1 + H n = Dnb2n + F n. (8)

The four equations (1), (3), (5), and (7) specify a degenerate system of linearequations, which can only be satisfied by taking An = C n = E n = Gn = 0 forall n. Next, for n = 1, the system of equations (2), (4), (6), and (8) specify thesame degenerate system of equations, so Bn = Dn = F n = Gn = 0 for n = 0.However, for n = 1, we have

0

B1 = D1 − F 1a−2 D1 = 12

1 +

0

B1

⇒B1 = D1 + F 1a

−2 F 1 = 1

2a2

1 − 0

B1.

and

−H 1 = b

2E 0 +

0

D1b2

−

0

F 1

H 1 = b2E 0 + D1b

2 + F 1

→ 0 = 2b2E 0 + b2

1 +

0

D1 +

1 −

0

F 1

Substituting from above,

−4b2E 0 = 10

b2( + 0)

2 − a2( − 0)2

B1

or

B1 = −40b2

b2( + 0)2 − a2( − 0)2E 0.


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Then

D1 = −20( + 0)b2

b2( + 0)2 − a2( − 0)2 E 0

F 1 = −20( − 0)a2b2

b2( + 0)2 − a2( − 0)2E 0

H 1 = −b2(b2 − a2)(20 − 2)b2( + 0)2 − a2( − 0)2 E 0.

The potential is

Φ(r,ϕ ) =

−40b2b2( + 0)2 − a2( − 0)2

· E 0rcos ϕ, r < a

−20b2b2( + 0)2 − a2( − 0)2

( + 0)r + ( − 0) a

2

r

E 0cos ϕ, a < r < b

−(b2 − a2)(20 − 2)

b2

( + 0)2

− a2

( − 0)2

·

b2

r

E 0cos ϕ

−E 0rcos ϕ, b < r.

As → 0, Φ → −E 0r cos ϕ in all three regions, which is reassuring.The electric field is

E(r,ϕ) =

40b2

b2( + 0)2 − a2( − 0)2E 0 [cos ϕr̂− sin ϕϕ̂] , r < a

20b2

b2( + 0)2 − a2( − 0)2

( + 0) − ( − 0)a2

r2

E 0 cos ϕr̂

−

( + 0) + ( − 0)a2

r2

E 0 sin ϕϕ̂

, a < r < b

− (b2 − a2)(20 − 2)

b2( + 0)2 − a2( − 0)2 ·

b

r

2E 0 [cos ϕr̂ + sin ϕϕ̂]

+E 0 [cos ϕr̂− sin ϕϕ̂] , b < r.

(b) In Figure 4.1 I’ve plotted the field lines for b = 2a, = 50. Also, as anappendix to this document I’ve included the C program I wrote to generate thisplot.

(c) For a solid dielectric cylinder in a uniform field, we would have a → 0. Inthat case the field would look like

E(r,ϕ ) =

20 + 0

E 0̂i, r < b

E 0̂i− (20 − 2)

( + 0)2

b

r

2E 0[cos ϕr̂ + sin ϕϕ̂], r > b

On the other hand, a cylindrical cavity in a uniform dielectric corresponds to


Figure 1: Field lines in Problem 4.8 for b = 2 a, = 5 0.

b → ∞, in which case the field becomes

E(r,ϕ ) =

40( + 0)2

E 0̂i, r < a

20( + 0)

E 0̂i− 20( − 0)( + 0)2

ar

2E 0[cos ϕr̂ + sin ϕϕ̂], r > a.


Problem 4.9

A point charge q is located in free space a distance d away from the center of adielectric sphere of radius a (a < d) and dielectric constant /0.

(a) Find the potential at all points in space as an expansion in spherical harmonics.

(b) Calculate the rectangular components of the electric field near the center of the sphere.

(c) Verify that, in the limit /0 → ∞, your result is the same as that for theconducting sphere.

We will take the origin of coordinates at the center of the sphere, and putthe point charge on the z axis at z = +h. Then the problem has azimuthalsymmetry.

(a) Since there is no free charge within the sphere, ∇·D = 0 there. But since thepermittivity is uniform within the sphere, we may also write ∇·(D/) = ∇·E = 0there. This means that polarization charge only exists on the surface of thesphere, so within the sphere the potential satisfies the normal Laplace equation,whence

Φ(r,θ ) =l

AlrlP l(cos θ) (r < a).

Now, in the region r > a, the potential may be written as the sum of twocomponents Φ1 and Φ2, where Φ1 comes from the polarization charge on thesurface of the sphere, while Φ2 comes from the external point charge. SinceΦ1 satisfies the Laplace equation for r > a, we may expand it in Legendrepolynomials:

Φ1(r,θ ) = l

Blr−(l+1)P l(cos θ) (r > a).

On the other hand, Φ2 is just the potential due to a point charge at z = d:

Φ2(r,θ ) =

q

4π0

rldl+1

P l(cos θ), r < d

q

4π0

dlrl+1

P l(cos θ), r > d.

(9)

Putting this all together we may write the potential in the three regions as

Φ(r,θ ) =

Alr

lP l(cos θ), r < aBlr

−(l+1) + q

4π0

rl

dl+1

P l(cos θ), a < r < d

Bl +

qdl

4π0

r−(l+1)P l(cos θ), r > d.


The normal boundary condition at r = a is

∂ Φ

∂rr=a−

= 0∂ Φ

∂rr=a+

→ 0

lAlal−1 = −(l + 1)Bla−(l+2) + lqa

l−1

4π0dl+1

→ Al = 0

−(l + 1)l

Bla−(2l+1) +

q

4π0dl+1

(10)

The tangential boundary condition at r = a is

∂ Φ

∂θ

r=a−

= ∂ Φ

∂θ

r=a+

→ Alal = Bla−(l+1) + q 4π0

al

d(l+1)

→ Bl = Ala2l+1 − q 4π0

a2l+1

dl+1 (11)

Combining (10) and (11), we obtain

Al = 10

+ l+1l

2l + 1

l

q

4π0dl+1

Bl = 10

+ l+1l

1 −

0

qa2l+1

4π0dl+1

In particular, as /0 → ∞ we have

Al → 0

as must happen, since the field within a conducting sphere vanishes; and

Bl → − qa2l+1

4π0dl+1. (12)

With the coefficients (12), the potential outside the sphere due to the polariza-tion charge at the sphere boundary is

Φ1(r,θ ) = 1

4π0

−qa

d

a2d

l1

rl+1P l(cos θ).

Comparing with (9) we see that this is just the potential of a charge −qa/d onthe z axis at z = a2/d. This is just the size and position of the image chargewe found in Chapter 2 for a point charge outside a conducting sphere.


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(b) Near the origin, we have

Φ(r,θ ) = A1rP 1(cos θ) + A2r2P 2(cos θ) + · ··

= q

4π0

30

d2( + 20)z +

1

2

50

d3(2 + 30)

(z 2 − x2 − y2) + ·· ·

so the field components are

E x = q

4π0d2 · 50

2 + 30

xd

+ · ··

E y = q

4π0d2 · 50

2 + 30

yd

+ ·· ·

E z = − q 4π0d2

30

+ 20+

502 + 30

z d

+ ·· ·

Problem 4.10

Two concentric conducting spheres of inner and outer radii a and b , respectively,carry charges ±Q. The empty space between the spheres is half-filled by a hemi-spherical shell of dielectric (of dielectric constant /0), as shown in the figure.

(a) Find the electric field everywhere between the spheres.

(b) Calculate the surface-charge distribution on the inner sphere.

(c) Calculate the polarization-chargedensity induced on the surface of the dielectricat r = a.

We’ll orient this problem such that the boundary between the dielectric-filled space and the empty space is the xy plane. Then the region occupiedby the dielectric is the region a < r < b , 0 < θ < π/2, and the problem hasazimuthal symmetry.

(a) Since the dielectric has uniform permittivity, all the polarization chargeexists on the boundary of the dielectric, so within its body we may take thepotential to be a solution of the normal Laplace equation. The potential in theregion between the spheres may then be written

Φ(r,θ ) =

[Alrl + Blr

−(l+1)]P l(cos θ), 0 < θ < π

2[C lr

l + Dlr−(l+1)]P l(cos θ),

π

2 < θ < π

First let’s apply the boundary conditions at the interface between the di-

electric and free space. That region is described by θ = π/2, a < r < b , and we


must have

∂ Φ

∂θ

θ=π/2+

= 0∂ Φ

∂θ

θ=π/2−

∂ Φ

∂r

θ=π/2+

= ∂ Φ

∂r

θ=π/2−

which leads to

0

Al − C l

P l (0)rl +

0Bl − Dl

P l (0)r

−l+1 = 0 (13)l [Al − C l] P (0)rl−1 − (l + 1)[Bl − Dl] P l(0)r−l+2 = 0. (14)

Since these equations must be satisfied for all r in the region a < r < b , thecoefficients of each power of r must vanish identically. In (13), this requirementis automatically satisfied for l even, since P l (0) vanishes for even l . Similarly,(14) is automatically satisfied for l odd. For other cases the vanishing of the

coefficients must be brought about by taking

0Al = C l

0Bl = Dl, l odd (15)

Al = C l Bl = Dl, l ev en . (1 6)

Next let’s consider the charge at the surface of the inner sphere. There areactually two components of this charge; one component comes from the surfacedistribution of the free charge +Q that exists on the sphere, and the othercomponent comes from the bound polarization charge on the inner surface of the dielectric

Problem 4.13

Two long, coaxial, cylindrical conducting surfaces of radii a a nd b are loweredvertically into a liquid dielectric. If the liquid rises an average height h between theelectrodes when a potential difference V is established between them, show that thesusceptibility of the liquid is

χe = (b2 − a2)ρgh ln(b/a)

0V 2

where ρ is the density of the liquid, g is the acceleration due to gravity, and thesusceptibility of air is neglected.

First let’s work out what happens when a battery of fixed voltage V is con-nected between two coaxial conducting cylinders with simple vacuum betweenthem. To begin, we can use Gauss’ law to determine the E field between the


cylinders. For our Gaussian pillbox we take a disk of thickness dz and radiusr, a < r < b centered on the axis of the cylinders. By symmetry there is nocomponent of E normal to the top or bottom boundary surfaces, and the com-

ponent normal to the side surfaces (the radial component) is uniform aroundthe disc. Hence E · dA = 2 π r d z E ρ = q

0=

1

0(2π a d z )σ

→ E ρ(ρ) = a σ

0r

where σ is the surface charge on the inner conductor. This must integrate togive the correct potential difference between the conductors:

V = − ba

E ρ(ρ)dρ = −aσ

0ln

b

a

which tells us that, to establish a potential difference V between the conductors,the battery has to flow enough charge to establish a surface charge of magnitude

σ = 0V

a ln(b/a) (17)

on the cylinder faces (the surface charges are of opposite sign on the two cylin-ders).

It is useful to figure out the energy per unit length stored in the electric fieldbetween the cylinder plates here. This is just

W v = 1

2

ba

2π0

E ·Dρdρdφ

= π0 b

a

E 2(ρ)ρdρ

= πa2σ2

0ln(b/a)

= π0V 2

ln(b/a) (18)

where the v subscript stands for ’vacuum’, since (18) is the energyper unit lengthstored in the field between the cylinders with just vacuum between them.

Now suppose we introduce a dielectric material between the cylinders. If thevoltage between the cylinders is kept at V , then the E field must be just thesame as it was in the no-dielectric case, because this field integrated from a tob must still give the same potential difference. However, in order to establishthis same E field in the presence of the retarding effects of the dielectric, thebattery now has to establish a surface charge that is greater that it was beforeby a factor (/0). With this greater charge on the electrodes, the D field willnow be bigger by a factor (/0) than it was in our above calculation. So the


energy per unit length stored in the field between the cylinders increases by afactor (/0 − 1) over the result (18):

∆W d = ( − 0) πV 2

ln(b/a) .

On the other hand, to get to this point the battery has had to flow enoughcharge to increase the surface charges to be of magnitude ( /0) times greaterthan (17). In doing this the internal energy of the battery decreases by anamount equal to the work it had to do to flow the excess charge, namely

∆W b = −V dQ = V (2π a d σ) = ( − 0) 2 πV 2

ln(b/a)

(per unit length). The energy lost by the battery is twice that gained by thedielectric, so the system with dielectric between the cylinders has lower overallenergy than the system with vacuum between the cylinders by a factor

∆W = ( − 0) πV 2

ln(b/a) (19)

(per unit length).Turning now to the situation in this problem, we’ll take the axis of the

cylinders as the z axis, so that the surface of the liquid is parallel to the xyplane. We’ll take the boundary between the liquid and the air above it to be atz = 0. With no potential between the cylinder plates, the liquid between thecylinders is at the same height as the liquid outside.

Now suppose a battery of fixed potential V is connected between the twocylinder plates. As we showed earlier, the combined system of battery and di-electric can lower its energy by having more of the dielectric rise up between thecylinders. However, at some point the energy win we get from this is balancedby the energy hit we take from the gravitational potential energy of havingthe excess liquid rise higher between the cylinders. The height at which we no

longer gain by having more liquid between the cylinders is the height to whichthe system will settle.

So suppose that, with a battery keeping a voltage V between the electrodes,the liquid between the electrodes rises to a height h above the surface of theliquid outside the electrodes. The decrease in electrostatic energy this affordsover the case with just vacuum filling that space is just (19) times the height,i.e.

E e = −h( − 0) πV 2

ln(b/a) (20)

This must be balanced by the gravitational potential energy E g of the excessliquid. E g is easily calculated by noting that the area between the cylinders isπ(b2 − a2), so the mass of liquid contained in a height dh between the cylindersis dm = ρπ(b2 − a2)dh, and if this mass is at a height h above the liquid surfaceits excess gravitational energy is

dE g = ( dm)gh = πgρ(b2 − a2)hdh.


17/46


Integrating over the excess height of liquid between the cylinders,

E g = πgρ(b2 − a2)

h0

h dh = 1

2πgρ(b2 − a2)h2. (21)

Comparing (20) to (21), we find that the gravitational penalty of the excessliquid just counterbalances the electrostatic energy reduction when

h = 2( − 0)V 2

ρg(b2 − a2)ln(b/a)=

2χe0V 2

ρg(b2 − a2)ln(b/a)

Solving for χe,

χe = ρgh (b2 − a2)ln(b/a)

20V 2 .

So I seem to be off by a factor of 2 somewhere.Actually we should note one detail here. When the surface of the liquid

between the cylinders rises, the surface of the liquid outside the cylinders mustfall, since the total volume of the liquid is conserved. Hence there are reallytwo other contributions to the energy shift, namely, the change in gravitationaland electrostatic energies of the thin layer of liquid outside the cylinders thatfalls away when the liquid rises between the cylinders. But if the surface area of the vessel containing the liquid is sufficiently larger than the area between thecylinders, the difference layer will be thin and its energy shifts negligible.

Solutions to Problems in Jackson,

Classical Electrodynamics , Third Edition

Homer Reid

November 8, 2000


Problem 5.1

Starting with the differential expression

dB = µ0I

4π dl × x− x

|x− x|3

for the magnetic induction at the point P with coordinate x produced by an incre-ment of current I dl at x, show explicitly that for a closed loop carrying a currentI the magnetic induction at P is

B = µ0I

4π ∇Ω

where Ω is the solid angle subtended by the loop at the point P . This correspondsto a magnetic scalar potential, ΦM = −µ0I Ω/4π. The sign convention for thesolid angle is that Ω is positive if the point P views the “inner” side of the surfacespanning the loop, that is, if a unit normal n to the surface is defined by thedirection of current flow via the right-hand rule, Ω is positive if n points away fromthe point P , and negative otherwise. This is the same convention as in Section 1.6for the electric dipole layer.

I like to change the notation slightly: the observation point is r1, the coordi-nate of a point on the current loop is r2, and the displacement vector (pointingto the observation point) is r12 = r1 − r2.

The solid angle subtended by the current loop at r1 is given by a surfaceintegral over the loop:

Ω = S

cos γ dAr212

1

Solutions to Problems in Jackson,Classical Electrodynamics , Third Edition

Homer Reid

February 11, 2001


Problem 5.10

A circular current loop of radius a carrying a current I lies in the x − y plane withits center at the origin.

(a) Show that the only nonvanishing component of the vector potential is

Aφ(ρ,z ) = µ0Ia

π

∞0

dk cos kz I 1(kρ)

where ρ) is the smaller (larger) of a and ρ.

(b) Show that an alternative expression for Aφ is

Aφ(ρ,z ) = µ0Ia

2

∞0

dke−k|z|J 1(ka)J 1(kρ).

(c) Write down integral expressions for the components of magnetic induction,using the expressions of parts a and b. Evaluate explicitly the components of B on the z axis by performing the necessary integrations.

(a) Translating Jackson’s equation (5.33) into cylindrical coordinates, we have

J φ = Iδ (z )δ (ρ − a) (1)

Following Jackson, we take the observation point x on the x axis, so its coordi-nates are (ρ,φ = 0 , z ). Since there is no current in the z direction, and since the

1


current density is cylindrically symmetric, there is no vector potential in the ρor z directions. In the φ direction we have

Aφ = −Ax sin φ + Ay cos φ = Ay=

µ04π

J y(x

)

|x− x| dx

= µ04π

J φ(x)cos φ

|x− x| dx

= µ04π

Re

J φ(x)eiφ

|x− x| dx

= µ04π

Re

J φ(x

)eiφ

2

π

∞m=−∞

∞0

eim(φ−φ) cos[k(z − z )]I m(kρ) dk

dx

where we substituted in Jackson’s equation (3.148). Rearranging the order of integration and remembering that φ = 0, we have

Aφ = µ02π2

Re∞

m=−∞

∞0

J φ(x

)ei(1−m)φ

cos[k(z − z )]I m(kρ)dx

dk

If m = 1, the φ integral yields 2π; otherwise it vanishes. Thus

Aφ = µ0

π

∞0

∞0

∞−∞

J φ(r, z )cos[k(z − z )]I 1(kρ)ρ dz dr

dk

Substituting (1), we have

Aφ = I aµ0

π

∞0

cos kz I 1(kρ) dk.

(b) The procedure for obtaining this expression is identical to the one I justwent through, but with the expressionfrom Problem 3.16(b) used for the Green’sfunction instead of equation (3.148).

(c) Let’s suppose that the observation point is in the interior region of thecurrent loop, so ρ = a. Then

Bρ = [∇ ×A]ρ = − ∂Aφ∂z

= −Iaµ0π

∞0

k sin kz I 1(kρ)K 1(ka) dk

Bz = [ ∇ ×A]z = 1

ρAφ +

∂Aφ∂ρ

= I aµ0

π

ejercicios jackson electrodinamica

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