Jackson Bourjaily Sol

8/10/2019 Jackson Bourjaily Sol

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2 JACOB LEWIS BOURJAILY

Therefore, we see that G(x, x) will automatically be a symmetric function if

F(x) F(x) = 1S

(G(x, y)G(x, y)) da.

In particular, this will be true if we define F(x) by

F(x) =1

S

G(x, y)da.

We have shown that for any arbitrary Greens function, G(x, x), satisfying Neumannboundary conditions there exists a Greens function G(x, x) G(x, x) 1

S

G(x, y)da,which satisfies the same boundary conditions as G(x, x) and has the property that

G(x, y)

nG(x, y) G(x, y)

nG(x, y)

da= 0.

From our derivation in part (a), this implies that the Greens function G(x, x) issymmetric.

o

2. Capacitance I

a) We are to determine the capacitance of two large, flat, parallel conducting sheets of area Aseparated by a distance d.

If we chose a Gaussian region that completely encloses one of the plates such that theedges are arbitrarily small, then the surface integral of the electric field will give EAwhereEis the magnitude of the electric field. Notice that we have used the fact thatthe electric field will be non-vanishing only between the plates.

Using Gauss law, was see that the surface integral is equal to the total charge containedwithin the region divided by 0. Specifically, we have that

E A= Q0

= E= 0

,

where is the charge density on the surface of one of the plates.

The magnitude of the voltage difference between the two plates is equal to the lineintegral of the electric field from one plate to the other. Because we know that in theregion between the two plates the electric field is independent of position, this willbe simply V = d

0.

Therefore, using the definition of capacitance, we see that

C= A0

d .

b) We are to determine the capacitance of two concentric conducting spheres with radii a and bwhere b > a.

If we chose a Gaussian region that completely encloses the inner sphere, then the surfaceintegral of the electric field will give E A where Eis the magnitude of the electricfield and A= 4r

2

, the area of the boundary of the region.Using Gauss law, was see that the surface integral is equal to the total charge contained

within the region divided by 0. Specifically, we have that

E 4r2 = Q0

= E= Q40r2

,

where a < r < b and Q is the charge on one of the spheres.The magnitude of the voltage difference between the two spheres is equal to the line

integral of the electric field from one to the other. Specifically,

V =

b

a

Ed = Q

40

b

a

1

r2dr=

Q

40

b aab

.


C= 40abb a .


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PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 1 3

c) We are to determine the capacitance of two concentric conducting cylinders of length L withradii a and b where b > a.

If we chose a Gaussian region that completely encloses one the inner cylinder, then thesurface integral of the electric field will give E A where E is the magnitude of theelectric field andA= 2rL, the area of the boundary of the region.


E 2rL = Q0

= E= Q20Lr

,

where a < r < b and Q is the charge on one of the cylinders.The magnitude of the voltage difference between the two cylinders is equal to the line


V =

ba

Ed = Q

20L

ba

dr

r =

Q

40log

b

a

.


C= 20L

logb

a

.

3. Capacitance II

We are to approximate the capacitance per unit length of two parallel, cylindrical conductorswith radii a1 and a2 which are separated by a distance d.

Let us work within the coordinate system such that the center of the first cylinder, withradius a1, is located at r = 0 and the second cylinder, with radius a2, is located atr= d.

Because the electric field is linear, we can consider the field caused by each of theconductors separately. Specifically, for points between the two cylinders, we can addthe electric fields produced by each cylinder separately. We can determine the electricfield per unit length induced by each cylinder by imagining a Gaussian region thatcompletely encloses a unit length of either cylinder. Therefore, for a point collinearwith the centers of each cylinder, we have that

E= Q

20r+

Q

20(d r).

The magnitude of the voltage difference between the two cylinders is equal to the lineintegral of the electric field from one to the other. Specifically,

V = Q

20

da2a1

1

r+

1

d r

dr,

= Q

20

log

d a2

a1

log

a2

d a1

,

= Q

20 log(d a2)(d a1)

a1a2

,

Q20

log

d2

a1a2

,

= Q

0log

d

a1a2

.

Therefore, using the definition of capacitance per unit length, we see that

C 0log

da1a2

.


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c) We are to determine the capacitance of two concentric conducting cylinders of length L withradii a and b where b > a.

If we chose a Gaussian region that completely encloses one the inner cylinder, then thesurface integral of the electric field will give E A where E is the magnitude of theelectric field andA= 2rL, the area of the boundary of the region.


E 2rL = Q0

= E= Q20Lr

,

where a < r < b and Q is the charge on one of the cylinders.The magnitude of the voltage difference between the two cylinders is equal to the line


V =

b

a

Ed = Q

20L

b

a

dr

r =

Q

40log

b

a

.


C= 20L

logb

a

.

3. Capacitance II

We are to approximate the capacitance per unit length of two parallel, cylindrical conductorswith radii a1 and a2 which are separated by a distance d.

Let us work within the coordinate system such that the center of the first cylinder, withradius a1, is located at r = 0 and the second cylinder, with radius a2, is located atr= d.

Because the electric field is linear, we can consider the field caused by each of theconductors separately. Specifically, for points between the two cylinders, we can addthe electric fields produced by each cylinder separately. We can determine the electricfield per unit length induced by each cylinder by imagining a Gaussian region thatcompletely encloses a unit length of either cylinder. Therefore, for a point collinearwith the centers of each cylinder, we have that

E= Q

20r+

Q

20(d r).

The magnitude of the voltage difference between the two cylinders is equal to the lineintegral of the electric field from one to the other. Specifically,

V = Q

20

da2

a1

1

r+

1

d r

dr,

= Q

20

log

d a2

a1

log

a2

d a1

,

= Q

20 log(d a2)(d a1)

a1a2

,

Q20

log

d2

a1a2

,

= Q

0log

d

a1a2

.

Therefore, using the definition of capacitance per unit length, we see that

C 0log

da1a2

.


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4. The Force Between Conductors

a) We are to find the attractive force between the conductors of the parallel plate capacitor describedin problem (2.a) and the parallel cylinders of problem (3) for fixed charges on each conductor.

We know that the energy stored between an arbitrary capacitor with fixed charges is

given by Q2

/2Cwhere Q is the charge on each conductor and C is the capacitanceof the system.

Let us fist consider the parallel plate capacitor. Using our results from problem (2.a), we

can easily determine that the energy of the capacitor is given by W = Q2d

2A0. Because

the force F = Wd

, we have that

F = Q2

2A0.

Similarly, we can use our results from problem (3) for the capacitance of the parallel

cylinder system to arrive at the energy stored per unit length, W = Q2

20log

da1a2

.

Therefore, we see that

F =

Q2

20d .

b) We are to find the attractive force between the conductors of the parallel plate capacitor describedin problem (2.a) and the parallel cylinders of problem (3) for fixed potential difference of theconductors.

We know that the energy stored between an arbitrary capacitor with fixed potentials oneach conductor is given by 1

2CV 2 whereV is the voltage difference between the two

conductors and Cis the capacitance of the system.Let us fist consider the parallel plate capacitor. Using our results from problem (2.a),

we can determine that the energy of the capacitor is given by W = 0A2d

V2. Because

the force F = Wd

, we have that

F = 0A2d2V2.

Similarly, we can use our results from problem (3) for the capacitance of the parallelcylinder system to arrive at the energy stored per unit length, W = 0

2log

da1a2

V2.


F = 0

2log

da1a2

2d

V2.

5. Thomsons Theorem

If an empty region is bounded by a number of equipotential surfaces, then the electrostaticenergy inside the region is absolutely minimized.

proof: Let us consider the energy within the bounded, compact region with boundary which is composed of equipotential surfaces. We will show that the electrostaticenergyWof the region is absolutely minimized.

Recall from the discussion in section (1.11) of Jacksons text that, in general, the elec-trostatic energy of a region is given by

W =0

2

2 d3x,

where(x) is the scalar potential. Integrating this expression by parts, we see that

W =0

2

da

||2 d3x

.

Using the definition of the scalar potential , E= , we see that this implies that

W =0

2

E da +0

2

||2 d3x.


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Physics 505, Classical ElectrodynamicsHomework 1

Due Thursday, 16th September 2004

Jacob Lewis Bourjaily

1. Symmetric Greens Functions

a) Any Greens function, G(x, x

), which satisfies Dirichlet boundary conditions is automaticallysymmetric: G(x, x) = G(x, x).

proof: Let us say that the Greens functionG(x, x) satisfies Dirichlet boundary conditions.That is, for a compact, bounded region with boundary ,G(x, x) = 0 x .

BecauseG(x, x) is a scalar field, Greens theorem implies that

G(x, y)2G(x, y) G(x, y)2G(x, y)

d3y =

G(x, y)

nG(x, y) G(x, y)

nG(x, y)

da.

Now, becauseG(x, x) satisfies Dirichlet boundary conditions, the surface integral onthe right hand side of the expression above vanishes identically so that

G(x, y)2G(x, y) G(x, y)2G(x, y)

d3y= 0.

By the definition of a Greens function, 2G(x, y) = 4(3)(x y). Therefore, we seethat

G(x, y)2G(x, y) G(x, y)2G(x, y)

d3y = 4

G(x, y)(3)(x y) G(x, y)(3)(x y)

d3y,

= 4 (G(x, x) G(x, x)) = 0,

G(x, x) =G(x, x).o

b) For any Greens function, G(x, x), which satisfies Neumann boundary conditions, there exists a

symmetric Greens function G(x, x) which satisfies the same boundary conditions.

proof: Let us say that the Greens function G(x, x) satisfies Neumann boundary condi-

tions. That is, for a compact, bounded region with boundary , we have that

nG(x, x) = 4/S x whereSis the surface area of the boundary .

Recall that for any Greens function G(x, x), the function G(x, x) = G(x, x) + F(x, x)is another Greens function if F(x, x) satisfies the Laplace equation. In general,however, G(x, x) will not satisfy the same boundary conditions ofG(x, x).

IfG(x, x) is to satisfy the same Neumann boundary conditions as G(x, x), then notonly must the Laplacian ofF(x, x) vanish, but alsoF(x, x) must satisfy the relation

nF(x, x) = 0 x .

A wide class of functions satisfy this condition. In particular, any function F(x, x) whichis independent ofx automatically satisfies

nF(x, x) = 0it also has a vanishing

Laplacian. Therefore, let us investigate if any functionF(x) F(x, x) can be chosen

so that G(x, x) =G(x, x) +F(x) is symmetric.

From our work in part (a) above, we know that G(x, x) will be symmetric if

G(x, y)

nG(x, y) G(x, y)

nG(x, y)

da= 0.

Using the form ofG(x, x) and the fact that G(x, x) satisfies Neumann boundaryconditions, we have that

G(x, y)

nG(x, y) G(x, y)

nG(x, y)

da=

4

S

G(x, y) G(x, y)

da,

=4

S

(G(x, y) +F(x) G(x, y) F(x)) da,

=4

S

(G(x, y) G(x, y)) da 4

S (F(x) F(x))

da,

=4

S

(G(x, y) G(x, y)) da 4 (F(x) F(x)) .

1


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Therefore, we see that G(x, x) will automatically be a symmetric function if

F(x) F(x) = 1

S

(G(x, y) G(x, y)) da.

In particular, this will be true if we define F(x) by

F(x) =1

S

G(x, y)da.

We have shown that for any arbitrary Greens function, G(x, x), satisfying Neumann

boundary conditions there exists a Greens function G(x, x) G(x, x) 1S

G(x, y)da,which satisfies the same boundary conditions as G(x, x) and has the property that

G(x, y)

nG(x, y) G(x, y)

nG(x, y)

da= 0.

From our derivation in part (a), this implies that the Greens function G(x, x) issymmetric. o

2. Capacitance I

a) We are to determine the capacitance of two large, flat, parallel conducting sheets of area Aseparated by a distanced.

If we chose a Gaussian region that completely encloses one of the plates such that theedges are arbitrarily small, then the surface integral of the electric field will give EAwhereEis the magnitude of the electric field. Notice that we have used the fact thatthe electric field will be non-vanishing only between the plates.


E A= Q

0= E=

0,

where is the charge density on the surface of one of the plates.

The magnitude of the voltage difference between the two plates is equal to the lineintegral of the electric field from one plate to the other. Because we know that in theregion between the two plates the electric field is independent of position, this willbe simply V = d

0.


C=A0

d .

b) We are to determine the capacitance of two concentric conducting spheres with radii a and bwhereb > a.

If we chose a Gaussian region that completely encloses the inner sphere, then the surfaceintegral of the electric field will give E A where Eis the magnitude of the electric

field and A= 4r2

, the area of the boundary of the region.Using Gauss law, was see that the surface integral is equal to the total charge contained

within the region divided by 0. Specifically, we have that

E 4r2 = Q

0= E=

Q

40r2,

wherea < r < b andQ is the charge on one of the spheres.The magnitude of the voltage difference between the two spheres is equal to the line


V =

b

a

Ed = Q

40

b

a

1

r2dr=

Q

40

b a

ab .


C= 40abb a

.


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4. The Force Between Conductors

a) We are to find the attractive force between the conductors of the parallel plate capacitor describedin problem (2.a) and the parallel cylinders of problem (3) for fixed charges on each conductor.

We know that the energy stored between an arbitrary capacitor with fixed charges is

given by Q2

/2Cwhere Q is the charge on each conductor and C is the capacitanceof the system.

Let us fist consider the parallel plate capacitor. Using our results from problem (2.a), we

can easily determine that the energy of the capacitor is given by W = Q2d

2A0. Because

the force F = Wd

, we have that

F = Q2

2A0.

Similarly, we can use our results from problem (3) for the capacitance of the parallel

cylinder system to arrive at the energy stored per unit length, W = Q2

20log

da1a2

.


F =

Q2

20d .

b) We are to find the attractive force between the conductors of the parallel plate capacitor describedin problem (2.a) and the parallel cylinders of problem (3) for fixed potential difference of theconductors.

We know that the energy stored between an arbitrary capacitor with fixed potentials oneach conductor is given by 1

2CV2 whereV is the voltage difference between the two

conductors and Cis the capacitance of the system.Let us fist consider the parallel plate capacitor. Using our results from problem (2.a),

we can determine that the energy of the capacitor is given by W = 0A2d

V2 . Because

the force F = Wd

, we have that

F = 0A2d2V2 .

Similarly, we can use our results from problem (3) for the capacitance of the parallelcylinder system to arrive at the energy stored per unit length, W = 0

2log

da1a2

V2 .


F = 0

2log

da1a2

2d

V2 .

5. Thomsons Theorem

If an empty region is bounded by a number of equipotential surfaces, then the electrostaticenergy inside the region is absolutely minimized.

proof: Let us consider the energy within the bounded, compact region with boundary which is composed of equipotential surfaces. We will show that the electrostaticenergyWof the region is absolutely minimized.

Recall from the discussion in section (1.11) of Jacksons text that, in general, the elec-trostatic energy of a region is given by

W =0

2

2 d3x,

where(x) is the scalar potential. Integrating this expression by parts, we see that

W =0

2

da

||2 d3x

.

Using the definition of the scalar potential , E= , we see that this implies that

W =0

2

E da+0

2

||2 d3x.


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Because the boundary is composed of equipotential pieces, (x) must be constanton the boundary. Therefore, the first integral can be reduced using Gauss law.Specifically, we have that

W =0

2

E da+0

2

||2 d3x,

=Qint

2 |+

02

||2 d3x,

=0

2

||2 d3x,

where we have used the fact that is empty and therefore Qint = 0. It should benoted that each term is positive definite and, in particular, the first term is absolutelyminimized by an equipotential boundaryindeed, it vanishes entirely. However, itremains for us to show that the remaining source of electrostatic energy is minimized.

We will briefly digress to discover what conditionsmust satisfy so thatWis extremized.Let us imagine that the potential function extremizes the energy W. In particular,this means that a first-order variation in the potential function should not vary

the energy W (keeping = 0 on the boundary).Therefore, the condition thatWis extremized by is equivalent to the requirement that

W = 0 for any first-order variation , fixed on the boundary . In particular,using integration by parts and the fact that vanishes on the boundary, we havethat

W = 0 =0

2

2 ()d3x,

=0

2

2 ()d3x,

=0 | 0

2 d3x,

= 0

2

d3

x= 0,

2= 0.

In the last step we have used the fact that because the variation was arbitrary,2 must precisely vanish everywhere in .

Therefore, we have shown that the energy is extremized by a potential which satisfiesLaplaces equation. In particular, this implies that the bulk term of the electrostaticenergy is extremized by a vacuum interior.

The Laplace equation is satisfied by the region because it free of charge. Therefore,the second source of electrostatic energy is extremizedhopefully, minimized. Weshowed above that because the equipotential distribution of charges on the surfacecomponents minimizes the first term. Therefore, if a region is bounded by an equipo-tential surface, the electrostatic energy of that region is minimized.

o


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Due Thursday, 23rd September 2004


2.1 A point charge q is brought a distanced away from an infinite, conducting, grounded plane. Weare to solve the following using the method of images.

a) Let us find the surface-charge density induced on the plane and plot it.

From the symmetry of the problem, it is clear that the potential (x) is equivalent tothat produced by the charge q together with an image charge q = q located adistanced on the opposite side of the plane. Specifically, the potential is given by

(x) = q

40

1

(x21

+x22

+ (x3d)2)1/2

1

(x21

+x22

+ (x3+d)2)1/2

,

in cartesian coordinates where the point charge is located at the point z (0, 0, d).

The charge density is then computed,

(x) = 0

x3

x3=0

,

= q

4

x3d

(x21

+x22

+ (x3d)2)3/2+

x3+d

(x21

+x22

+ (x3+d)2)3/2

x3=0

,

= q

4

d

(x21

+x22

+d2)3/2+

d

(x21

+x22

+d2)3/2

,

(x) = qd

2(x21

+x22

+d2)3/2

In Figure 1 below, we have plotted the charge density of the plane.

Figure 1. Problem (2.1): Induced charge density as a function of position on the plane.Thez -axis labels the density in arbitrary units. Specifically,{qd/(2), d} 1

b) We are to compute the force between the plane and the charge using Coulombs law for the forcebetween the charge and its image.

Quite directly we see that

F = q2

401

(2d)2z.

1


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c) We are to compute the total force acting on the plane by directly integrating over 2/20.

Let us proceed to compute the force directly

F =

da

2

20=

q2d2

820

rdrd

(r2 +d2)3/2,

= q2d2

820

dudu3/2

,

= q2d2

160

1

u2

d2,

= q2d2

160d4,

F = q2

40

1

(2d)2z.

d) We are to compute the work required to remove the charge from its position to infinity.

By direct computation, we see that the work required is

W =

d

F d= q2

160

d

dz

z2 =

q2

160

1

z

d

= q2

160d.

e) Let us compute the potential energy of the charge-image-charge system and compare this resultwith that of part (d) above.

From the definition of potential energy, we see that

U= q2

80d.

This is precisely twice the energy required to pull the charge q to infinity. This isexpected: the energy is twice that of the original system because we also have energyassociated with the electric field of the image charge. In the original system, there

is only the electric field of a single charge. Furthermore, notice that the positionand hence energy of the image charge depends on the charge and position ofq. Thisimplies that it is negligent to consider qmoving in a static field produced by q.

2.2 Let us consider the problem of a point charge q located inside a hollow, grounded, conductingsphere of inner radius a.

a) We are to find the potential function describing the inside of the sphere.

This problem is superficially similar to that discussed in Jacksons text regarding acharge located outside a conducting, grounded sphere. It should be clear that if thecharge is located at the point r, then, by axial symmetry, the image charge mustbe located along the direction ofr at a distance ofr > aif the method of imagecharges applies. Therefore, the potential at a point x will be given in the form

(x) = 1

40

q

|xr|+

q

|xr|

,

Because the sphere is grounded, we must have (x) = 0 x| x = a. This situation isidentical to that required for the situation described in Jackson and therefore wemust have that

q

a=

q

rand

r

a=

a

r.

In particular, this implies that

(x) = q

40

1

(x2 +r2 2xr cos )1/2

a

r

x2 + a4

r2 2xa2

r cos 1/2

.


11/89


c) We are to compute the total force acting on the plane by directly integrating over 2/20.

Let us proceed to compute the force directly

F =

da

2

20=

q2d2

820

rdrd

(r2 +d2)3/2,

= q2d2

820

dudu3/2

,

= q2d2

160

1

u2

d2,

= q2d2

160d4,

F = q2

40

1

(2d)2z.

d) We are to compute the work required to remove the charge from its position to infinity.

By direct computation, we see that the work required is

W =

d

F d= q2

160

d

dz

z2 =

q2

160

1

z

d

= q2

160d.

e) Let us compute the potential energy of the charge-image-charge system and compare this resultwith that of part (d) above.

From the definition of potential energy, we see that

U= q2

80d.

This is precisely twice the energy required to pull the charge q to infinity. This isexpected: the energy is twice that of the original system because we also have energyassociated with the electric field of the image charge. In the original system, there

is only the electric field of a single charge. Furthermore, notice that the positionand hence energy of the image charge depends on the charge and position ofq. Thisimplies that it is negligent to consider qmoving in a static field produced by q.

2.2 Let us consider the problem of a point charge q located inside a hollow, grounded, conductingsphere of inner radius a.

a) We are to find the potential function describing the inside of the sphere.

This problem is superficially similar to that discussed in Jacksons text regarding acharge located outside a conducting, grounded sphere. It should be clear that if thecharge is located at the point r, then, by axial symmetry, the image charge mustbe located along the direction ofr at a distance ofr > aif the method of imagecharges applies. Therefore, the potential at a point x will be given in the form

(x) = 1

40

q|x r|

+ q

|x r|

,

Because the sphere is grounded, we must have (x) = 0 x| x = a. This situation isidentical to that required for the situation described in Jackson and therefore wemust have that

q

a=

q

r and

r

a=

a

r.

In particular, this implies that

(x) = q

40

1

(x2 +r2 2xr cos )1/2

a

r

x2 + a

4

r2 2xa2

r cos 1/2

.


12/89


b) Let us compute the charge distribution induced on the sphere.

We call on the simple symmetry of the charge-image-charge system noting that this isidentical to the charge distribution induced by the image charge on the outside of thesphere. Because we have already derived this expression in classand in Jacksonup

to a redefinition ofr and r

, we simply have that

(x) = 0

x

|x|=a

= q

4ar

1 a2

r21 + a

2

r2 2ar cos

3/2 ,where is the angle between r and x.

c) Let us compute the force acting on the charge q.

Using our results above, we see that

F = q2

40

ar

(a2 r2)2r.

d) We are to discuss how the work above is altered, if at all, if the sphere were kept at fixed potentialVor if there were total charge Q on its inner and outer surfaces.

Neither of the two situations alters the work above because neither would effect theinterior of the sphereonly the outside. By Gau law, we know that the electric fieldinsidea charged or fixed-potential sphere is identically zero. Because electrostaticsis linear, the field inside the sphere will be the linear sum of that described aboveand that caused by the spherewhich is vanishing. Hence, there is no alteration.

2.7 Let us consider the space R3 satisfying Dirichlet boundary conditions on the plane R3.

a) We are to find the appropriate Greens function describing this system.

In many ways, this problem is similar to that describing a point charge and an infinite

conducting plane. Specifically, we see that the Greens function given by

G(x, x) = 1

((x1 x1)2 + (x2 x2)

2 + (x3 x3)2)

1/2

1

((x1 x1)2 + (x2 x2)

2 + (x3+x3)2)

1/2,

is of the correct form and satisfies the Dirichlet boundary conditions. In particular,we manifestly have that G(x, x) = 0 x|x3 = 0. Hence, this is our required Greensfunction.

b) Let us say that the potential on the plane x3 = 0 is specified to be = V inside a circle ofradiusa and vanish outside the circle. We are to find an integral expression for the potential incylindrical coordinates.

In general, we know that the potential function for a problem with a Greens function

satisfying Dirichlet boundary conditions is given by

(x) = 1

40

R3

(x)G(x, x)d3x 1

4

R3

(x)G(x, x)

n da.

Because the space is empty of charges, the first integral identically vanishes and wemust only consider the boundary integral.

Up to a sign which we will set a posteriori, we see that

G(x, x)

x3

x3

=0

=

x3 x3

((x1 x1)2 + (x2 x2)

2 + (x3 x3)2)

3/2+

x3+x3

((x1 x1)2 + (x2 x2)

2 + (x3+x3)2)

3/2

x3

=0

,

= 2x3

((x1 x1)2 + (x2 x2)

2 +x23

)3/2

.

Later, we will see that the n-direction should coincide withx3so that the potentialat the surface is positive; this is identical to inserting the seemingly spurious minussign in the above calculation.


13/89


b) Let us compute the charge distribution induced on the sphere.

We call on the simple symmetry of the charge-image-charge system noting that this isidentical to the charge distribution induced by the image charge on the outside of thesphere. Because we have already derived this expression in classand in Jacksonup

to a redefinition ofr and r

, we simply have that

(x) = 0 x

|x|=a

= q

4ar

1 a2r21 + a

2

r2 2ar cos 3/2 ,

where is the angle between r and x.

c) Let us compute the force acting on the charge q.

Using our results above, we see that

F = q2

40

ar

(a2 r2)2 r.

d) We are to discuss how the work above is altered, if at all, if the sphere were kept at fixed potentialVor if there were total charge Q on its inner and outer surfaces.

Neither of the two situations alters the work above because neither would effect theinterior of the sphereonly the outside. By Gau law, we know that the electric fieldinsidea charged or fixed-potential sphere is identically zero. Because electrostaticsis linear, the field inside the sphere will be the linear sum of that described aboveand that caused by the spherewhich is vanishing. Hence, there is no alteration.

2.7 Let us consider the space R3 satisfying Dirichlet boundary conditions on the plane R3.

a) We are to find the appropriate Greens function describing this system.

In many ways, this problem is similar to that describing a point charge and an infinite

conducting plane. Specifically, we see that the Greens function given by

G(x, x) = 1

((x1 x1)2 + (x2 x2)2 + (x3 x3)2)1/2 1

((x1 x1)2 + (x2 x2)2 + (x3+x3)2)1/2,

is of the correct form and satisfies the Dirichlet boundary conditions. In particular,we manifestly have that G(x, x) = 0x|x3 = 0. Hence, this is our required Greensfunction.

b) Let us say that the potential on the plane x3 = 0 is specified to be = V inside a circle ofradiusa and vanish outside the circle. We are to find an integral expression for the potential incylindrical coordinates.

In general, we know that the potential function for a problem with a Greens function

satisfying Dirichlet boundary conditions is given by

(x) = 1

40

R3

(x)G(x, x)d3x 14

R3

(x)G(x, x)

n da.

Because the space is empty of charges, the first integral identically vanishes and wemust only consider the boundary integral.

Up to a sign which we will set a posteriori, we see that

G(x, x)

x3

x3

=0

=

x3 x3((x1 x1)2 + (x2 x2)2 + (x3 x3)2)3/2

+ x3+x

3

((x1 x1)2 + (x2 x2)2 + (x3+x3)2)3/2

x3

=0

,

= 2x3((x1 x1)2 + (x2 x2)2 +x23)3/2

.

Later, we will see that the n-direction should coincide with x3so that the potentialat the surface is positive; this is identical to inserting the seemingly spurious minussign in the above calculation.


14/89


We can now compute the potential. Using our work above, it is clear that

(,,z) = 14

R3

(x)G(x, x)

n da,

=

1

4 2

0

a0 V

2zdd

(( cos cos )2 + ( sin sin )2 +z2)3/2 ,

(,,z) = zV

2

20

a0

dd

(2 +2 2 cos( ) +z2)3/2

c) Let us compute the potential along the line = 0.

Using our work above, we may compute the potential directly.

(0, 0, z) = zV

2

20

a0

dd

(2 +z2)3/2

,

= zV

2 a2+z2

z

2

du

u3/2,

= zV 1u

a2+z2

z2,

= zV

1a2 +z2

1z

,

(0, 0, z) =V

1 z

a2 +z2

.

d) We are to explicitly compute the potential by expanding its expression in the limit where 2 +z2 >> a2.

Let us define the variable 2

2 + z2. In general, we can rewrite the potential derived

above as

(,,z) = zV

23

20

a0

dd

(1 +2(2 2 cos( )))3/2.

Let us Taylor expand the integrand toO(4). We see that

(1 +2(2 2 cos( )))3/2 =

1 3

22

2 2 cos( )+15

84

2 2 cos( )2 + O(6) .

When we integrate over the angle , all terms independent of will be multipliedby a factor of 2, those directly proportional to cos( ) will integrate to zero,and the term proportional to cos2( ) will obtain a factor of. Therefore,

(,,z) = zV

23 2

0

a0

dd

(1 +2(2 2 cos( )))3/2 ,

= zV

23

20

a0

dd

1 32

2

2 2 cos( )+158

4

2 2 cos( )2 + O(6),=

zV

3

a0

d

1 32

22 +15

442 +

15

844 + O(6)

,

= zV

3

2

2 3

4

82 +

1542

164 +

56

164+ O(6)

a

0

,

= zV a2

23

1 3a

2

42 +

5(32a2 +a4)

84 + O(6)

.

Substituting the definition of , we arrive at the desired expression:

(,,z) = zV a2

2(2 +z2)3/2

1 3a

2

4(2 +z2)+

5(32a2 +a4)

8(2 +z2)2 + O((2 +z2)6)

.


15/89


2.8 Let us consider two parallel, straight line charges separated by a distance R and with equal andopposite linear charge densities .

a) Let us find the surfaces of constant potential. We will show that these are circular cylinders.

With the anticipation only available in hindsight, let us work with polar coordinates

suppressing the longitudinal directiondefined such that the origin is located at thecenter of one of the circles of constant potential. Specifically, let us say that the firstline charge is located at (, 0) and the second at (R+, 0). Although we do not yetknow the displacement from the first line charge, we know that if is properlyspecified, the potential should be independent of the angular coordinate.

In general, the potential will be the linear superposition the potentials of each of thetwo line-charges. Having calculated this in homework one, we see

(, ) =

40

log(2 2 cos +2) log((R+)2 2(R+) cos +2)

.

The requirement that be constant on a cylinder centered at the origin is equivalent tothe condition that = 0. Using this identity, we compute

= 0 = 20

sin 2 2 cos +2 (R+) sin (R+)2 2(R+) cos +2

,

= sin

20

2 2 cos +2

(R+)

(R+)2 2(R+) cos +2

,

2 2 cos +2 =

(R+)

(R+)2 2(R+) cos +2.

Expanding and collecting terms, we arrive at the constraint

2 =(+R).

We must now determine the radius such that (, ) = V . To do this, we willinsert the expression derived above for 2 into our expression for the potential withthe condition that = V. Because this will be independent of the angle, let us

choose = /2 to simplify our expressions. We will need to remember this choicelater.

Computing directly, we see that

(, ) =V =

40log

2 2 cos +2

(R+)2 2(R+) cos +2

,

=

40log

2 +2 +R

(R+)2 +2 +R

,

=

40log

22 +R

22 + 3R+R2

.

By rearranging terms, exponentiating, and simplifying, we obtain the quadratic ex-pression

22

1 e40V/

+R

1 3e40V/R2e40V/ = 0.

Simply using the quadratic formula and a bit of algebra, we see that

= R

2(1 e40V/)

1 3e40V/ 1 e40V/

,

= R e40V/

1 e40V/ =

R

e40V/ 1 =

Re20V/

2 sinh(20V /) or = R/2.

Notice that the solution of the quadratic, = R/2, simply demonstrates that be-tween the two wires the potential is constant along the line (,/2). We could haveanticipated this solution because we simplified our work to determine using the

condition that at , (,/2) =(,/2) =V . The fact that this cannotbe a so-lution is evident from the last line of our expression for (, ) above: the numeratorin the logarithm vanishes if = R/2 and hence this is not a physical solution.


16/89


Using our expression for , we have

2 =R2 e80V/

(1 e40V/)2+R2

e40V/

(1 e40V/) =R2

e40V/

(1 e40V/)2,

= R

e20V/

1 e40V/ =

Re20V/

e40V/ 1 =

R

2 sinh(20V /) .

Therefore, we have shown that if

= R

2 sinh(20V /) and =

Re20V/

2 sinh(20V /),

the potential along a right circular cylinder in the longitudinal direction of radius positioned precisely to the leftof the first line-charge is constant and equal to V.

It is important to notice that the sign of is somewhat important. In particular, ifV/ < 0 then is a positivedisplacement to the left of and encloses the first line-charge. However, ifV / >0, then is a negativedisplacement to the left of the firstline-charge; in particular, > Rand so the cylinder is displaced to the right of andencloses the second line-charge.

b) We are to demonstrate that the capacitance per unit length of two cylindrical conductors withradiia and b separated by a distance d is given by

C= 20

arccoshd2a2b2

2ab

.proof: We know from the definition of capacitance that it is given byC/L = V+V . We

must show that the above equation is consistent with this fact.First, we know that the required problem is equivalent to one in which there are only

two line charges, separated by a distance R = ++ with radii such that thetwo cylinders are at constant voltages V+ and V, respectively. Specifically, we see,reorganizing the expressions derived above,

d= += R(e40V/ 1)

R(e40V+/ 1)

,

and,

a= Re20V+/

(e40V+/ 1) and b=

Re20V/

(e40V/ 1).

Looking at the equation we must verify, it may be helpful to compute d2a2b2

2ab and seeif it points us toward the solution. Let us first compute the numerator. We find that

d2 a2 b2 =

R2

e40V

12 2R

2e40V

1

e40V+

1 + R2

e40V+

12 R

2e40V+

e40V+

1)2 R

2e40V

e40V

12 ,

=R2 1 e

40V+

e40V+

12 + 1 e

40V

e40V

12 + 2

e40V

1

e40V+

1 ,

= R2

e40V

1

e40V+

1 e40V+ e40V .

Now, because

2ab= 2R2e20(V+V)/

e40V

1

e40V+

1 ,

we have that

d2 a2 b2

2ab =

1

2e20(V+V)/ e40V+/ +e40V/ ,

= 1

2

e20(V+V)/ +e20(V+V)/

,


17/89


d2 a2 b2

2ab = cosh

20(V+ V)

.

Therefore, we see that the capacitance per unit length agrees with the desired formula:

C= 20

arccoshd2a

2b2

2ab = 20

20(V+V)

=

V+ V.

o

c) Let us verify that the expression above for the capacitance agrees with that derived in homeworkone for a similar problem in the limit where d2 >> a2 +b2.

proof: Similar to our previous results, we can make extraordinary progress by simply guess-ing the form of the answer. First, it should be true that

arccosh

d2 a2 b2

2ab

= log ,

for some. In fact, because we are simply verifying a result, we could presume it tobe true and then it would be obvious what must be. However, let us simply see ifsuch an equation makes sense.

By taking the hyperbolic cosine of each side, we effectively exponentiate the logarithm.Therefore,must satisfy

d2 a2 b2

2ab =

1

2

elog +e log

=

1

2

+

1

=

2 + 1

2 .

This reduces to the following quadratic equation,

2

d2 a2 b2

ab

+ 1 = 0.

It is clear that in the limit where d2a2b2

ab >>1, the solutions approach the trivialsolution and that given by

d2 a2 b2

ab

.

Furthermore, in the limit ofd2 >> a2 +b2, this approaches d2

ab .Therefore, we see that

C= 0

arccoshd2a2b2

2ab

0log

d2a2b2

ab

0log

d2

ab

02log

dab

.

C 20

log

dab

.o


18/89


2.10 Let us consider a large parallel plate capacitor which is made up of two plane conducting sheetsseparated by a distance D. One of the sheets has a hemispherical boss of radiusa on its innersurface and D >> a. The conductor and the boss are grounded and the electric field betweenthe plates is E0.

a) We are to calculate and sketch the surface-charge density on the sheet with the boss.This problem appears to be identical to that of a conducting sphere in a uniform electric

field which was worked out in class and in Jackson. For that problem, the potentialwas constant over a sphere in a uniform field. By the symmetry of that problem, theplane orthogonal to the field would automatically be at zero potential. Therefore,we suspect that the problem is precisely identical.

Following this, we suppose that the potential is given by

(r) =E0

r a3

r2

cos ,

wherea is the radius of the sphere, r is the magnitude of the vector r, and is theangle relative to the electric field E.

Notice that whenr = a, the potential vanishes, thereby satisfying the boundary condition

on the boss. Furthermore, the potential identically vanishes everywhere on the planeorthogonal to the electric field, because cos = 0.

Therefore, is the correct electric potential.We can now trivially solve for the surface-charge density. We can determine the density

on the boss or plane by computing the derivative of in the direction normal to theboss or plane. On the boss, this is given by

(r) =0 r

|r|=a

=E0

1 + 2

a3

a3

cos = 30E0cos .

For points on the plane, the normal direction is dz =rd and so we have

(r) =0 z

=/2

= 0

r

=/2

=E0

1 a3

r3

( sin /2) = 0E0

1 a

3

r3

.

In figure 2 below, we have plotted the surface charge density of the sheet-boss system.The shading indicates the density of charge.

Figure 2. Problem (2.10): Induced charge density as a function of position on theconductor. The shading labels the charge density and is in arbitrary units. Specifically,

{0E0, a

} 1.


19/89


b) Let us compute the magnitude of the charge on the boss.

Quite directly, we have that

Q=

da = 2a2

1

0

30E0cos d cos = 30E0a2.

c) Let us now consider the charge induced on the boss given a charge located a distance d from itscenter. We are to compute the charge induced on the boss.

proof: This situation is nearly identical to that worked out in class and in Jackson wherewe considered the surface charge density induced by a point charge a distance dfrom a conducting, grounded sphere. By the symmetry here, it is quite clear thatthis description appliesat least for our analysis of the boss. To make the potentialvanish on the sheet, however, we must do a little more work. However, it is not tohard to guess what need: if we add two image charges to this system, symmetrizedalong the direction of the point charge then the potential will precisely vanish alongthe conducting plane. Specifically, if the charge q is a distance d from the sphere,

then in the direction q, there should be image charges q =adq atd = a2

d, q = ad

placed atd =a2

d and q =qplaced atd.When this system is set up, it is clear that the fixed, vanishing potential on the boss

is achieved because each pair of charges separately leave = 0 on the boss. Thepotential on the plane vanishes by the planar symmetry of the problem. Therefore,this system satisfies our boundary conditions and can be used to compute the density.

The calculation of the density follows precisely that of Jacksons for each charge-pariseparately. Therefore, we may conclude that the charge density on the sphere isgiven by

Therefore, we can call upon our previous work to notice that surface charge densityshould be given by

() = q4ad 1

a2

d2

1

1 +

a2

d22a

dcos 3/2

1

1 +

a2

d2 + 2a

dcos 3/2

.

Let us compute the total induced charge directly.

q =

da = q

4ad

1 a

2

d2

2a2

10

1

1 + a2

d22 adcos 3/2 1

1 + a2

d2 + 2adcos

3/2 d cos ,

= q

4

1 a

2

d2

(1ad)21+a

2

d2

du

u3/2 +

(1+ ad )21+ a

2

d2

du

u3/2

,

=q2

1 a

2

d2

1u1/2

(1ad)2

1+ a2

d2

+ 1

u1/2

(1+ ad )2

1+a2

d2

,

=q2

1 + a

d

1 a

d

11 ad

+ 11 + ad

11 + a

2

d2

11 + a

2

d2

,

=q2

1 + a

d+ 1 a

d2

1 a2d2

1 a2d2

,

q =q

1 d2 a2

d

d2 a2

.

o


20/89


Due Thursday, 30th September 2004


2.12 We are to obtain the potential insidea cylinder of radius b in the form of Poissons integral bysolving the general solution of the two-dimensional potential.

Because we are considering the interior of the cylinder, there should no 1 terms in theseries expansion. Therefore, the potential inside will be given by

(, ) = 0/2 +n=1

nn cos(n) +n

n sin(n),

where the coefficientsm andm are to be determined. Notice that we have chosento work both sine and cosine terms in the series to avoid worrying about the phaseterms.

Proceeding almost canonically, we see that

n = 1

bn

20

(b, ) cos(n)d, and n= 1

bn

20

(b, )sin(n)d.

Notice that the above expression for n includes the case where n = 0.Using these expressions above in our general series, we can proceed by direct computation

(, ) = 1

2

20

(b, )d +n=1

1

bnn cos(n)

20

(b, ) cos(n)d + 1

bnn sin(n)

20

(b, )sin(n)d

,

= 1

20

(b, )

1

2+

n=1

b

n(sin(n)sin(n) + cos(n)cos(n))

d,

= 1

2

0

(b, )12

+

n=1

bn cos(n( )) d,

= 1

2

20

(b, )

1 +

n=1

b

n ein(

) +ein()

d,

= 1

2

20

(b, )

1

1 b

ei() +

1

1 b

ei() 1

d,

= 1

2

20

(b, )

1 b

ei() + 1

bei(

)

1

bei(

)

1 b

ei()

1

bei()

1

bei()

d,

= 1

2

2

0

(b, )2

b

ei(

) +ei() 1

2

b2 +

b

ei(

) +ei()

1 + 2

b2

b

ei()

+ei() d,

= 1

2

20

(b, )

1

2

b2

1 + 2

b2 2

bcos( )

d,

= 1

2

20

(b, )

b2 2

b2 +2 2b cos( )

d,

(, ) = 1

2

20

(b, )

b2 2

b2 +2 2b cos( )

d

o

1


21/89


2.13 a) We ware to compute the interior potential of a cylinder composed of two separated, conductinghalves which are kept at potentials V1 and V2.

From our experience with this problem, we should expect that the potential can beexpanded as a cosine series over the azimuthal coordinate. Specifically, we expect

(, ) = 0+

m=1

mm cos(m),

where the coefficients m are determined using the orthogonality conditions. We seethat

0 = 1

2

3/2

/2

(b, )d= 1

2 (V1+V2) =

V1+V22

,

and, similarly,

m= 1

bm

/2/2

V1cos(m)d+V2

3/2

/2

V2cos(m)d

=

(V1 V2) (1)m

mbm |m (2Z + 1).

Therefore, we can use some of Jacksons tricks in elementary complex analysis to see

(, ) = V1+V22

+ V1 V2

modd

(1)m

mm

bmcos(m),

= V1+V2

2 +

V1 V2

Im

modd

immeim

mbm

,

= V1+V2

2 +

V1 V2

Im

log

1 +i(/b)ei

1 i(/b)ei

,

= V1+V2

2 +

V1 V2

arctan

2/b cos

1 2/b2

,

(, ) = V1+V2

2 +

V1 V2

arctan

2b cos

b2 2 .

o

b) Let us calculate the surface charge density on each half of the cylinder.

We recall that the surface charge density is given by

() = 0(, )

=b

=0V1 V2

arctan

2b cos

b2 2

.

Using a computer algebra package to evaluate the derivativetime becomes tooprecious to evaluate by handwe see that

() =0(V1 V2)

2b(2b2)cos

2b4 + 2b4 cos(2),

= 20(V1 V2)

cos

b(1 + cos(2)),

=20(V1 V2)

cos

2b cos2 ,

() = 0(V1 V2)

b cos .

o


22/89


2.17 a) We are to construct a the general, free-space Greens function for two dimensional electrostaticsby integrating 1/Rwith respect to (z z) between the limits Zwhere|Z| 1. We mayneglect constant terms that do not depend on the planar coordinates.

Let us define the variable

(x x)2 + (y y)2

1/2

. We are to determine the

general, electrostatic Greens function by performing the integration

G(, )

ZZ

d

(2 +2)1/2,

in the limit where Zis large.The above integral is easily evaluated using the method of trigonometric substitution, Z

Z

d

(2 +2)1/2= 2 arcsinh

Z

.

Notice that this implies that arcsinh(Z/) = /2. If we take the hyperbolic sine ofboth sides, we obtain

1

2

e/2 e/2

= Z/.

Because we are assuming that Z , we may set e/2 0, at least to leadingorder in /Z. Therefore, we see

e/2 2Z/,

which implies that

2 log 2 + 2 log Z 2log log 2.

Therefore, up to constants independent on the planar coordinates, we have shown that

G(, ) = log 2 = log

(x x)2 + (y y)2

= log

2 +2 2 cos( )

.

o

b) We are to show explicitly that the Greens function can be expressed as a Fourier series in theazimuthal coordinate,

G(, ) = 1

2

m=

eim()gm(,

),

where the radial Greens functions satisfy1

m2

2

gm =

2gm = 4( )

.

Let us show that the Greens functionG(, ) given above is in fact a Greens function.This can be demonstrated by the following, mindless calculation

2G(, ) = 1

2

2

m=

eim()gm(,

),

= 1

2

m=

eim()2gm(,

),

= 2( )

m=

eim(),

= 4( )( )

.

This is precisely the form desired for a general Greens function in polar coordinatesbecause

2G(, )dd = 4

( )( )

dd = 4.

Therefore, the Greens function can be expressed as a Fourier series in the azimuthalcoordinate.


23/89


c) We are to complete the expression above to determine an explicit form of the free-space Greensfunction in polar coordinates.

We will use the notation +() to be the larger (smaller) of and .

Noting that when =, gm is a solution to the Laplace equation, we immediately seethat it is given by

gm(, ) =

m

m < m

m > ,

where the coefficients m and m will be set by continuity and derivative conditions.Because when = the two functions must agree, we have

mm =m

m,

m= mm, and m = m

m.

Using our notation described above, we see that

gm(, ) = m

+

m.

To find the value ofm, we simply note that when = , there is a discontinuity in the

derivative ofgm which is characterized by the d-function. Specifically, we have that

4

=

dgmd

=

dgm

d

=

,

= 2mm

,

m= 2

m.

Let us briefly describe the g0 function. In Jackson, we have shown that this termwill be of the form 0 log(

) = 0 log(+). To determine the unknown coefficient,we note that the discontinuity in the derivative ofgm requires that

4

=

0

,

and so we see that 0= 4.Therefore, we can see that in general,

G(, ) = log(2+) +

m(=0)=

1

|m|

+

|m|eim(

),

= log(2+) +1

m=

1

m

+

meim(

) +

m=1

1

m

+

meim(

),

= log(2+) +

m=1

1

m

+

meim(

) +

m=1

1

m

+

meim(

),

= log(2+) + 2

m=1

1

m

+m

cos[m(

)] .

G(, ) = log(2+) + 2

m=1

1

m

+

mcos[m( )] .

o


24/89


18. a) Using the results obtained in problem (2.17) above, we are to find a closed-form expression forthe Greens function for the interior of a cylinder of radius b with Dirichlet boundary conditions.

From our general work with solutions to the Laplace equation with Greens functionsand from our work about in problem (2.17), we know that the Greens function for

the interior of the cylinder will be given by

G(, ) = 1

2

m=

eim()gm(,

),

where the functions gm(, ) satisfy the Laplace equation and will be given in the

form

gm=

m

m < mm +mm >

,

where the coefficientsm, m,andmwill be determined by the boundary conditionsand continuity requirements.

Specifically, because of Dirichlet boundary conditions are satisfied at = b, we musthave that gm= 0, we must demand

mbm +mb

m = 0,

which implies

m= mbm and m= mb

m,

for some set of coefficients m. Similarly, continuity requires that the two solutionsagree when = and so

mm =m

m +mm =m

b

m

b

m,

which implies that

m= m

m

bm

b

m

.Risking redundancy, we have shown that

gm=

m

m b

mb

m <

m

b

mb

m >

Let us now proceed to determine the coefficients m. These can be found by noting thediscontinuity in the derivative ofgm. Specifically, we have that

4

=

dgmd

=

dgm

d

=

,

=mmm1

b

m +mm

bm

m+1 mm

m1

m

bm

b

m

,m=

2

m

b

m.

Plugging this into our expression for gm, we see that

gm=

2m

b

m

m b

mb

m= 2

m

b2

m

m <

2m

b

m b

mb

m= 2

m

b2

m

m >

= 2m

+b2

m

+

m,

Let us now determine the m = 0 coefficients. We have demonstrated that, in general,g0= 0+ 0log

. Becauseg0(b) must vanish by the Dirichlet boundary conditions,we see that 0+0log b= 0] which immediately implies

0

= 0log(1/b).

Like we found in problem (2.17), we see that 0= 4. Therefore, we are now readyto write the explicit Greens function.


25/89


Putting this into our expression for the Greens function, we obtain

G(, ) = log

b2 log(2+) + 2

m=1

1

m

+

m

b2

mcos[m( )] ,

= log

b2 log(2+) + 2

m=1

1m

+

m

cos[m( )] 2m=1

1m

b2

m

cos[m( )] .

We will evaluate the two sums separately, but using the same technique. With someinspiration from Davis Fourier Series and Orthogonal Functions, we note that

1

mxm =

x0

m1d=

x0

1

md.

Therefore, substituting (/+)x in the above expressions, we can reexpress thefirst series as

2

m=1

1m

+

mcos[m( )] = 2

+0

1 m=1

m cos[m( )]

d,

Noting the similar expression used during the solution of problem (2.12) and recallingthat||< 1, we see that

m=1

m cos[m( )] = 1

2

m=1

m

eim() +eim(

)

,

= 1

2

1

1 ei()+

1

1 ei() 2

,

= 12

2 ei() +ei() 2 22 + 2ei() +ei()1 +2

ei() +ei()

,=

cos( ) 2

1 +2 2cos( ).

Inserting this relation into the integral above, we can determine the Greens functioncompletely. Calling trivial substitutional integration, we see that

2

m=1

1

m

+

mcos[m( )] = 2

+

0

1

m=1

m cos[m( )]

d,

= 2+

01 cos(

) 2

1 +2 2cos( ) d,

= 2

+

0

cos( )

1 +2 2cos( )d,

=

1+ +

2

2

+cos()

1

du

u,

= log

1 +

+

2 2

+

cos( )

.

Similarly, we see that

2

m=1

1

m

b2

m

cos[m( )] = log

1 +

b2

2

2

b2 cos( )

.


26/89



G(, ) = log

b2 log(2+) + 2

m=1

1

m

+

mcos[m( )] 2

m=1

1

m

+b2

mcos[m( )] ,

= log

b

2

2+

log

1 +

+

2 2 +

cos( )

+ log

1 +

+b22

2

b2 cos( )

,

= log

b2

2+

+ log

1 +

+b2

2 2+

b2 cos( )

1 ++

2 2

+cos( )

,

= log

b4

b22+

1 +

+b2

2 2+

b2 cos( )

1 ++

2 2

+cos( )

,

G(, ) = log

b4 +22 2b2 cos( )

b2 (2 +2 2 cos( )).

o

b) We are to show that the solution to the Laplace equation with the potential given as (b) onthe cylinder can be expressed as the Poissons integral of problem (2.12).

Because we consider a cylinder that is free of charges in the interior, the most generalsolution to Poissons equation with Greens functions is given by

(, ) = 1

4

(b, ) G(, )

=b

da.

Let us begin by computing the derivative of the Greens function in the directionnormal to the cylinder.

G(,

)

=b

= 22

2b2

cos(

)b4 +22 2b2 cos( )

2

2 cos(

)2 +2 2 cos( )

=b

,

= 22b 2b2 cos( )

b4 +2b2 2b3 cos( )

2b 2 cos( )

2 +b2 2b cos( ),

= 1

b

22 2b cos( )

b2 +2 2b cos( )

2b2 2b cos( )

2 +b2 2b cos( )

,

= 1

b

22 2b cos( ) 2b2 + 2b cos( )

b2 +2 2b cos( )

,

=2

b

b2 2

b2 +2 2b cos( )

.


(, ) = 1

4

20

(b, ) G(, )

=b

bd,

= 1

2

20

(b, )1

b

b2 2

b2 +2 2b cos( )

bd,

(, ) = 1

2

20

(b, )

b2 2

b2 +2 2b cos( )

d.

o


27/89


c) We are to determine what changes are necessary if the Greens function is desired to describetheexteriorproblem. Specifically, we are to show that the closed-form Greens function remainsthe same.

Following our now standard technique of generating the terms for the Fourier series,we see that the exterior Greens function can be expressed as a Fourier expansionover the azimuthal coordinate,

G(, ) = 1

2

m=

eim()gm(,

),


form

gm=

mm +mm < m

m > ,


If we continue mindlessly applying the continuity condition and the condition thatgm(b) = 0 m as before, we conclude that

m= mbm and m= mb

m and m= mm

b

m

b

m,

and we determinemby the now-tedious procedure of derivative-discontinuity check-

ing and obtain m = 2m

b

m. With these, we can reexpress the functions gm so

that they specifically refer to the current situation. We see that

gm=

m

b

m

b

m= 2

m

m

b2

m <

m

m b

mb

m= 2

m

m

b2

m >

= 2m

+

m

b2

m.

2.21 We are to use Cauchys theorem to derive the Poisson integral solution.

Let us consider the function (z) which is analytic over a disc,Cof radius b centered atz = 0. Then, as ever middle schooler knows from trivial application of the Residuetheorem,

(z) = 1

2i

C

(z) dz

z z z C,

because, by the analyticity of(z) in the region C, the only pole in Ccomes fromz =z which has a simple Residue of(z).

Furthermore, ifz C, then

1

2i

C

(z) dz

z b2/z = 0,

because z = b2/z / Cas trivially seen by the fact that |z b2/z| > b because|z|< b. Therefore, we have that for z C,

(z) = 12i

C

(z)

1z z

1z b2/z

dz.

Notice the trivial identity z b2/z = z /z

z z

, which can be seen by simple expan-sion. Noting that|z|= b on C, we have

(z) = 1

2i

C

(z)

1

z z

z

z

z z

dz,

= 1

2i

C

(z)

zz b2 zz+ |z|2

z

zz b2 |z|2 +zz

dz = 1

2i

C

(z (b, ))

b2 2

z (b2 +2 2b cos( ))

dz,

(, ) = 1

2

2

0

(b, ) b2 2

b2

+2

2b cos(

) d.

o


28/89


c) We are to determine what changes are necessary if the Greens function is desired to describethe exteriorproblem. Specifically, we are to show that the closed-form Greens function remainsthe same.

Following our now standard technique of generating the terms for the Fourier series,we see that the exterior Greens function can be expressed as a Fourier expansionover the azimuthal coordinate,

G(, ) = 1

2

m=

eim()gm(,

),


form

gm =

m

m +mm <

mm >

,


If we continue mindlessly applying the continuity condition and the condition thatgm(b) = 0 m as before, we conclude that

m= mbm and m = mb

m and m = mm

b

m

b

m,

and we determinemby the now-tedious procedure of derivative-discontinuity check-

ing and obtain m = 2m

b

m. With these, we can reexpress the functions gm so

that they specifically refer to the current situation. We see that

gm =

m

b

m

b

m= 2

m

m

b2

m <

m

m b

mb

m= 2

m

m

b2

m >

= 2m

+

m

b2

m.

2.21 We are to use Cauchys theorem to derive the Poisson integral solution.

Let us consider the function (z) which is analytic over a disc,Cof radius b centered atz = 0. Then, as ever middle schooler knows from trivial application of the Residuetheorem,

(z) = 1

2i

C

(z) dz

z z z C,

because, by the analyticity of(z) in the region C, the only pole in Ccomes fromz =z which has a simple Residue of(z).

Furthermore, ifz C, then

1

2i

C

(z) dz

z b2/z = 0,

because z = b2/z / Cas trivially seen by the fact that |z b2/z| > b because|z|< b. Therefore, we have that for z C,

(z) = 12i

C

(z)

1z z

1z b2/z

dz.

Notice the trivial identity z b2/z = z /z

z z

, which can be seen by simple expan-sion. Noting that|z|= b on C, we have

(z) = 1

2i

C

(z)

1

z z

z

z

z z

dz,

= 1

2i

C

(z)

zz b2 zz+ |z|2

z

zz b2 |z|2 +zz

dz = 1

2i

C

(z (b, ))

b2 2

z (b2 +2 2b cos( ))

dz,

(, ) = 1

2

2

0

(b, ) b2 2

b2

+2

2b cos(

) d.

o


29/89


2.23 a) We are to determine the potential inside as cuber with edge length a with two opposing facesheld at potentialV, all other faces having zero potential.

We can find the potential in analogy to the cube-potential problem worked in Jacksonstext. Let us use a cartesian system of coordinates chosen such that the faces atconstant potential are located at z =

a/2. We can later change this system to find

that desired by the more explicit problem.Similar to that of the cube describe in the text, we see that the potential inside should

be described by

(x) =

n,m=1

Anmsin(nx)sin(my) cosh(nmz),

where the terms n, m and nm are given by

n =n

a , m =

m

a , nm=

a

n2 +m2.

Notice that we have used the hyperbolic cosine because of the symmetry seen in thecoordinate system chosen. By the standard method of analysis, we see that

Anm = 16V

nm2 cosh

/2n2 +m2 |n, m (2Z+ 1).Therefore, we see that

(x) =

n,m=1,odd

16V sin(nx)sin(my) cosh(nmz)

nm2 cosh

/2

n2 +m2 .

In the system of coordinates used by the textbook question, one must set z (z+a/2).

b) We are to (numerically) find the potential at the center of the cube.

We are asked to calculate this numerically. Using a standard computer algebra package,one trivially demonstrates that (in the coordinate system used above)

(a/2, a/2, 0) = V

3.

c) We are to compute the surface charge density on one of the plates.

Proceeding as we have done many times before, we see that

(x, y) =0 z

z=a/2

= +016V

2

n,m=1,odd

sin(nx) sin(my)mnsinh(nmz)

nm cosh

/2

n2 +m2 ,

(x, y) = 016V

a

n,m=1,odd

m2 +n2

nm sin

nxa

sin

mya

tanh(/2

m2 +n2).


30/89


Due Thursday, 7th October 2004


Problem 3.1We are to determine the potential between two concentric spheres, having radii a, b(b > a), each

of which is divided into two hemispheres by the same horizontal plane. The upper hemisphere of theinner sphere and the lower hemisphere of the outer sphere are fixed at potential Vwhile the other twohemispheres are grounded. We are to calculate the potential as a series of Legendre polynomials up toorderl = 4.

By the azimuthal symmetry about the direction normal to the dividing plane, the potential willbe described by

(, ) ==0

+(+1)

P(cos ),

wherePare Legendre polynomials and the coefficientsandare determined by boundary conditions.Notice that by the orthogonality of the Legendre polynomials, we have 1

1

(, )P(cos )d cos = 22+ 1

+(+1)

.

Because we know the function (, ) when = a, b we can determine the coefficients and bysolving the following system of equations:

V

10

P(cos )d cos = 2

2+ 1

a

+a(+1)

,

V

01

P(cos )d cos = 2

2+ 1

b

+b(+1)

.

Solving for , we see that

= V(2+ 1)

2(a2+1

b2+1

)a+1

1

0

P(cos )d cos b+1

0

1

P(cos )d cos ,=

V(2+ 1)

2(a2+1 b2+1)

a+1 + (b)+1

10

P(cos )d cos .

Using, we see that

=V(2+ 1)

2 a+1

10

P(cos )d cos a2+1,

=V(2+ 1)

2 a+1

10

P(cos )d cos V(2+ 1)

2(a2+1 b2+1)

a+1 + (b)+1

a2+1

10

P(cos )d cos ,

=V(2+ 1)

2 a+1

10

P(cos )d cos

1 +

a

a+1 + (b)+1

a2+1 b2+1

.

Now, following the discussion near Jacksons equation (3.25), it is clear that 10

P(cos )d cos = (1/2)(1)/2 ( 2)!!

2((+ 1)/2)! | (2Z+ 1).

Specifically, this integral is nonzero only for odd values ofwith the exception of0. Evaluating theseexpressions,

0=V

2, 1=

3V

2(a3 b3)(a2 +b2)

1

2 =

3V

4

a2 +b2

a3 b3, 3=

7V

2

a4 +b4

a7 b7(1/2)

1

4 =

7V

16

a4 +b4

a7 b7;

1= 3V

4

a5 a2b3 a5 a3b2

a3 b3 =

3V

4

a2b2(a+b)

b3 a3 , 3=

7V

2 a4(1/2)

1

4+a7

7V

16

a4 +b4

a7 b7 =

7V

16

a4b4(a3 +b3)

a7 b7 .

Inserting this work into our general expression for the potential, we have

(, ) =V

2

1 +3

2

a2 +b2

a3 b3

a2b2

2

P1(cos ) 7

8

1

a7 b7

a4 +b4

3 a4b4(a3 +b3)

4

P3(cos ) +. . .

.

o 1


31/89


We should note that this more-or-less agrees with the expressions for the potential worked out inthe text for similar circumstances. In particular, Jacksons equation (3.36) describes a similar problem,where = V /2 on the inner sphere and the outer sphere is removed completely. Making thisredefinition of the potential on the inner two hemispheres and taking the limit where ab, (keeping onlytermsO(aibj) wherej i), we obtain Jacksons equation (3.36) as desired.

Problem 3.2Let us consider the potential induced by a uniform, spherical charge distribution of radius R, with

densityQ/4R2 except for a spherical cap centered on = 0 defined by the cone = , where there isno charge.

a) We are to show that the potential inside the spherical surface can be expressed as

(, ) = Q

80

=0

1

2+ 1

R+1[P+1(cos ) P1(cos )] P(cos ).

Quite generally, we can express the potential as a Poisson integral over the charge dis-tribution. Specifically,

(x) =

1

40(x)d3x

|x x| .From the discussion in Jackson near expression (3.70), we know that the function1/|x x| can be expressed in terms of spherical harmonics and so

(r,,) = 1

40

Q

4R2 4

=0

m=

1

2+ 1

r

r+1+Ym(, )Y

m(, )r2drd cos d,

wherer+ (r) is the larger (smaller) ofr, r and is the boundary of the spherical

charge distribution. If we first consider the potential inside the sphere, then r =

R= r+ and by azimuthal symmetry all m = 0. BecauseY0(, ) =

2+14

P(cos ),we see

(r, ) = Q

40R2

=0

r

R+1

1

4P(cos

)P(cos )R2d cos d,

= Q

80

=0

r

R+1P(cos )

cos1

P(cos )d cos .

Now, from Jacksons equation (3.28), we have that

(2+ 1)P(cos ) =dP+1(cos )

d cos

dP1(cos )

d cos ,

which clearly implies cos1

P(cos )d cos =

1

2+ 1 (P+1(cos

) P1(cos ))|

cos1 =

1

2+ 1(P+1(cos ) P1(cos )) .

Inserting this into our expression above, we have shown that

(r, ) = Q

80

=0

1

2+ 1

r


o

From our derivation, it is clear that if one desired to know the potential outside the

sphere, all one would need to do is substitute r

R+1 R

r+1 because of the expression

of 1/|x x| in terms of spherical harmonics.


32/89


We should note that this more-or-less agrees with the expressions for the potential worked out inthe text for similar circumstances. In particular, Jacksons equation (3.36) describes a similar problem,where = V /2 on the inner sphere and the outer sphere is removed completely. Making thisredefinition of the potential on the inner two hemispheres and taking the limit where ab, (keeping onlytermsO(aibj) wherej i), we obtain Jacksons equation (3.36) as desired.

Problem 3.2Let us consider the potential induced by a uniform, spherical charge distribution of radius R, with

densityQ/4R2 except for a spherical cap centered on = 0 defined by the cone = , where there isno charge.

a) We are to show that the potential inside the spherical surface can be expressed as

(, ) = Q

80

=0

1

2+ 1


Quite generally, we can express the potential as a Poisson integral over the charge dis-tribution. Specifically,

(x) =

1

40(x)d3x

|x x| .From the discussion in Jackson near expression (3.70), we know that the function1/|x x| can be expressed in terms of spherical harmonics and so

(r,,) = 1

40

Q

4R2 4

=0

m=

1

2+ 1

r

r+1+Ym(, )Y

m(, )r2drd cos d,

wherer+ (r) is the larger (smaller) ofr, r and is the boundary of the spherical

charge distribution. If we first consider the potential inside the sphere, then r =

R= r+ and by azimuthal symmetry all m = 0. BecauseY0(, ) =

2+14 P(cos ),

we see

(r, ) = Q

40R2

=0

r

R+1

1

4P(cos

)P(cos )R2d cos d,

= Q

80

=0

r

R+1P(cos )

cos1

P(cos )d cos .

Now, from Jacksons equation (3.28), we have that

(2+ 1)P(cos ) =dP+1(cos )

d cos

dP1(cos )

d cos ,

which clearly implies cos1

P(cos )d cos =

1

2+ 1 (P+1(cos

) P1(cos ))|

cos1

= 1

2+ 1(P+1(cos ) P1(cos )) .

Inserting this into our expression above, we have shown that

(r, ) = Q

80

=0

1

2+ 1

r


o

From our derivation, it is clear that if one desired to know the potential outside the

sphere, all one would need to do is substitute r

R+1 R

r+1 because of the expression

of 1/|x x| in terms of spherical harmonics.


33/89


b) Let us find the magnitude and direction of the electric field at the origin.

By the azimuthal symmetry of the problem, the electric field must point in the directionz defined by = 0. Computing rather directly, we see

|E| = (r, )

rr=0

= Q

80

=1

2+ 1

r1

R+1 [P+1(cos ) P1(cos )] P(cos(0))

r=0

,

= Q

80

1

3R2P1(1)[P2(cos ) P0(cos )] ,

= Q

80

1

3R2

1

2

3cos2 1

1

,

= Q

160R2

cos2 1/3 2/3

,

E = Q sin2

160R2z.

o

c) Let us briefly discuss the limiting cases of the above results when 0, .

Because sin2() = 2 a4

3 + O(6), and is -periodic, it is clear that the electric field

approaches

E = Q2

1620z+ O(4),

for 0 and

E = Q( )2

1620z+ O(( )4),

for . This is expected. Notice that as the spherical charge distribution closes, 0, we approach the field inside a closed sphere, which vanishes by Gau law.The symmetric situation as also begins to vanish because the total charge onthe sphere decreases like ( )2 near . Therefore, the field will decrease like( )2.

Similarly, for 0, cos 1 and so P+1(1) P1(1) vanishes. Therefore, thepotential will vanish for 0 as expected for the interior of a charged sphere. For , the potential will decrease like ( )2 and will approach the potential of apoint charge of magnitude Q/8R2( )2.

Problem 3.5Let us consider the potential inside a sphere of radius a where the potential at the surface is specified.

We are to demonstrate that

(r,,) =a(a2 r2)

4

(, )d

(r2 +a2 2ar cos )3/2 =

=0

m=

Am

ra

Ym(, ),

where cos = cos cos + sin sin cos( ) and Am =

dY m(, )(, ).

From our work in Jacksons second chapter, we know that the first expression for the potential isthat obtained from the Greens function

G(r, r) = 1

|r r|

a

r|r a2

r2r|

.

Therefore, it is sufficient for us to show that the second expression for the potential is obtainable fromthe above Greens function to show that the two expressions for the potential are equivalent.


34/89


b) Let us find the magnitude and direction of the electric field at the origin.

By the azimuthal symmetry of the problem, the electric field must point in the directionz defined by = 0. Computing rather directly, we see

|E| = (r, )

rr=0

= Q

80

=1

2+ 1

r1

R+1 [P+1(cos ) P1(cos )] P(cos(0))

r=0

,

= Q

80

1

3R2P1(1)[P2(cos ) P0(cos )] ,

= Q

80

1

3R2

1

2

3cos2 1

1

,

= Q

160R2

cos2 1/3 2/3

,

E = Q sin2

160R2z.

o

c) Let us briefly discuss the limiting cases of the above results when 0, .

Because sin2() = 2 a4

3 + O(6), and is -periodic, it is clear that the electric field

approaches

E = Q2

1620z+ O(4),

for 0 and

E = Q( )2

1620z+ O(( )4),

for . This is expected. Notice that as the spherical charge distribution closes, 0, we approach the field inside a closed sphere, which vanishes by Gau law.The symmetric situation as also begins to vanish because the total charge onthe sphere decreases like ( )2 near . Therefore, the field will decrease like( )2.

Similarly, for 0, cos 1 and so P+1(1) P1(1) vanishes. Therefore, thepotential will vanish for 0 as expected for the interior of a charged sphere. For , the potential will decrease like ( )2 and will approach the potential of apoint charge of magnitude Q/8R2( )2.

Problem 3.5Let us consider the potential inside a sphere of radius a where the potential at the surface is specified.

We are to demonstrate that

(r,,) =a(a2 r2)

4

(, )d

(r2 +a2 2ar cos )3/2 =

=0

m=

Am

ra

Ym(, ),

where cos = cos cos + sin sin cos( ) and Am =

dY m(, )(, ).

From our work in Jacksons second chapter, we know that the first expression for the potential isthat obtained from the Greens function

G(r, r) = 1

|r r|

a

r|r a2

r2r|

.

Therefore, it is sufficient for us to show that the second expression for the potential is obtainable fromthe above Greens function to show that the two expressions for the potential are equivalent.


35/89


Let us begin by expressing G(x, x) in terms of spherical harmonics. Following the now standardprocedure used above, we see that

G(r, r) = 4

=0

m=

1

2+ 1Ym(

, )Ym(, ) r

r+1

a

rr

a2r2 r

+1 ,= 4

=0

m=

1

2+ 1Ym(

, )Ym(, )

r

r+1

rr

a2+1

.

To find the potential, we must compute the normal derivative of the Greens function in the directionofr , evaluated at r =a. Let us spend a moment and compute this.

G(r, r)

r

r=a

= 4=0

m=

1

2+ 1Y m(

, )Ym(, )

(+ 1)

r

r+2

r1r

a2+1

r=a

,

= 4

=0

m=1

2+ 1Y m(

, )Ym(, )(+ 1)

r

a+2

r

a2+2,

= 4=0

m=

1

2+ 1Ym(

, )Ym(, ) (2+ 1) r

a

1a2

,

= 4=0

m=

Ym(, )Ym(, )

ra

1a2

.

Now, we can directly compute the potential (r,,) using

(r,,) = 1

4

(, )

G(r, r)

r

r=a

a2d,

=

1

4

(

,

)4

=0

m= Y

m(

,

)Ym(, ) r

a 1

a2 a

2

d

,

=

(, )

=0

m=

Y m(, )Ym(, )

ra

d,

=

=0

m=

Ym(, ) r

a

(, )Ym(

, )d,

(r,,) ==0

m=

Am

ra

Ym(, ).

o

Problem 3.6Let us consider a system of two point charges of charge qlocated at z = a, respectively.

a) Let us find the electrostatic potential as an expansion in spherical harmonics and powers ofr.

First notice that the electrostatic potential is trivially given by

(x) = q

40

1

|xa|

1

|x+ a|

,

wherea (a, 0, 0) in spherical coordinates. As before, we can expand the function

1/|xa| following Jacksons equation (3.70). Furthermore, by azimuthal symmetry,it is clear that only m = 0 spherical harmonics contribute and so we can substi-tute Legendre polynomials in their place. Azimuthal symmetry also implies that


36/89


Let us begin by expressing G(x, x) in terms of spherical harmonics. Following the now standardprocedure used above, we see that

G(r, r) = 4

=0

m=

1

2+ 1Ym(

, )Ym(, ) r

r+1

a

rr

a2r2 r

+1 ,= 4

=0

m=

1

2+ 1Ym(

, )Ym(, )

r

r+1

rr

a2+1

.

To find the potential, we must compute the normal derivative of the Greens function in the directionofr , evaluated at r =a. Let us spend a moment and compute this.

G(r, r)

r

r=a

= 4=0

m=

1

2+ 1Ym(

, )Ym(, )

(+ 1)

r

r+2

r1r

a2+1

r=a

,

= 4

=0

m=1

2+ 1Ym(

, )Ym(, )(+ 1)

r

a+2

r

a2+2,

= 4=0

m=

1

2+ 1Ym(

, )Ym(, ) (2+ 1) r

a

1a2

,

= 4=0

m=

Ym(, )Ym(, )

ra

1a2

.

Now, we can directly compute the potential (r,,) using

(r,,) = 1

4

(, )

G(r, r)

r

r=a

a2d,

=

1

4

(

,

)4

=0

m= Y

m(

,

)Ym(, ) r

a 1

a2 a

2

d

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