Do not hand in optional parts!

Problem 1. A point charge in a rectangular tube

Consider a point charge placed in an infinitely long grounded rectangular tube as shownbelow. The sides of the square cross sectional area of the tube have length a.

b

a

(a) (Optional) Show that the solutions to the homogeneous Laplace equation (i.e. withoutthe extra point charge) are linear combinations of functions of the form

�(kx

x)�(ky

y) e±zz where �(u) =ncos(u) or sin(u) (1)

for specific values of kx

, ky

and

z

. Determine the allowed the values of kx

, ky

and

z

and their associated functions.

(b) Now consider a point charge displaced from the center of the tube by a distance b inthe x direction, i.e. the coordinatess of the charge are r

o

= (x, y, z) = (b, 0, 0). Usethe method of images to determine the potential. You will need an infinite number ofimage charges of both sign.

(c) As an alternative to the method of images, use a series expansion in terms of thehomogeneous solutions of part (a) to determine the potential from the point chargedescribed in part (b). You should find the form

�(r; ro

) =4q

a

2

X

n

X

m

X

n

(x)Xn

(b)Ym

(y)Ym

(0)e

�n,m|z|

2n,m

(2)

where X

n

(x) = �n

(kn

x) and Y

m

(y) = �m

(km

y) and

n,m

=pk

2n

+ k

2m

.

(d) Determine the asymptotic form of the surface charge density, and the force per areaon the walls of the rectangular tube far from the point charge, i.e. z � a. You shouldfind that the force per area on the bottom plate (far from the charge) is

F

y

A

' q

2

a

4cos2(⇡x/a) cos2(⇡b/a) e�2

p2⇡|z|/a

. (3)

1

Solution

1. The Laplace equation is�r2' = 0 (2)

Separating variables with ' = X(x)Y (y)Z(z) we must have

�d2X

dx2

=k2

xX (3a)

�d2Y

dy2

=k2

yY (3b)

�d2Z

dz2

=k2

zZ (3c)

In order to satisfy Eq. (2), the separation constants satisfy

k2

x + k2

y + k2

z = 0 (4)

and thusd2Z

dz2

= 2Z with =

q

k2

x + k2

y (5)

The solutions to Eq. (3a) may be either sines or cosines

X(x) = �(kxx) , (6)

with kx at this point still arbitrary. In order to satisfy the boundary conditionsX(±a/2) = 0, we require for the cosine functions that

kxa/2 = (n +

1

2

)⇡ . (7)

Similarly, for the sin functionskxa/2 = n⇡ . (8)

Thus, the general form is

Xn(x) = �n(knx) n = 0, 1, . . . , (9)

with kn = (n + 1)⇡/a and

�n(u) =

(

cos(u) n evensin(u) n odd

. (10)

The Y (y) direction follows by analogy

Ym(y) = �m(kmx) m = 0, 1, . . . , (11)

with km = (m + 1)⇡/a. The solutions to the z direction are

Z(z) = e±z =

p

k2

n + k2

m . (12)

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Figure 1: Arrangement of image charges. The black circles idicate plus charges, while thewhite circles indicate negative charges. The origin of the coordinate system is indicated bythe dashed lines. The real charge is displaced by a distance b from the origin.

2. The image charges may be placed in a rectangular lattice as shown below. Their arefour types of charges with coordinates

r1

(n, m) =(b + 2na)

ˆx + 2maˆy (13)r

2

(n, m) =((2n + 1)a � b)ˆx + 2maˆy (14)r

3

(n, m) =(b + 2na)

ˆx + (2m + 1)aˆy (15)r

4

(n, m) =((2n + 1)a � b)ˆx + (2m + 1)aˆy (16)

where n, m are integers. Then the potential is

�(r) =

q

4⇡

1X

n,m=0

1

|r � r1

(n, m)|�1

|r � r2

(n, m)|�1

|r � r3

(n, m)|+1

|r � r4

(n, m)| (17)

3. For the potential at r due to a point charge at ro = (b, 0, 0), we expand the potentialas

�(r; ro) =

✓

2

a

◆

2

1,1X

n,m=0

Xn(x)Xn(b) Ym(y)Ym(0) gn,m(z) (18)

and substitute into the Poisson equation

�r2'(r; ro) = q�(x � b)�(y)�(z) . (19)

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The leading factors (2/a)

2 arise from the fact that we have not normalized the eigen-functions X and Y

Z a/2

�a/2

dx Xn(x)Xn0(x) =

a

2

�n,n0 (20)Z a/2

�a/2

dy Ym(y)Ym0(y) =

a

2

�m,m0 (21)

If gn,m(z) satisfies✓

k2

n + k2

m � @2

@z2

◆

gn,m(z) = q�(z) , (22)

then using the completeness relation

2

a

X

n

Xn(x)Xn(xo) =�(x � xo) (23)

2

a

X

m

Ym(x)Ym(xo) =�(y � yo) (24)

it is not difficult to show that Eq. (19) is satisfied.The solution to Eq. (22) is

gn,m(z) =

(

Ae�n,m

z z > 0

Aen,m

z z < 0

(25)

Integrating across the �-fcn in Eq. (22) we have

� dg

dz

�

�

�

�

z=0

+

+

dg

dz

�

�

�

�

0

�= q (26)

With this requirement A =

q2

n,m

and

�(r; ro) =

4q

a2

1,1X

n,m=0

Xn(x)Xn(b) Ym(y)Ym(0)

e�n,m

|z|

2n,m

(27)

4. At asymptotic distances the terms with the smallest n,m dominate the sum. We thenhave only the contribution from n = m = 0 mode, and

0,0 =

p2⇡/a . (28)

The potential reads

�(r; ro) '4q

a2

cos(⇡x/a) cos(⇡b/a) cos(⇡y/a)

e�0,0|z|

20,0

(29)

27

or

�(r; ro) 'p

2q

⇡acos(⇡x/a) cos(⇡b/a) cos(⇡y/a)e�

p2⇡|z|/a (30)

Let us calculate the charge density on the bottom plate

� = n ·E = �@y�|y=�a/2

, (31)

= �p

2q

a2

cos(⇡x/a) cos(⇡b/a) e�p

2⇡|z|/a . (32)

Finally, the force per area on the bottom plate is

F y

A=

�2

2

, (33)

=

q2

a4

cos

2

(⇡x/a) cos

2

(⇡b/a) e�2

p2⇡|z|/a . (34)

The direction of the force is into the tube. The other walls of the tube have the sameforce per area.

28

ϕ = Voϕ = 0 ϕ = 0

x

y

D

Problem 1. Potential from a strip.

An infinite conducting strip of width D (between 0 < x < D) is maintained at potential Vo,while on either side of the strip are grounded conducting planes. The strip and the planesare separated by a tiny gap as shown below.

(a) Following a similar example given in class, determine the potential everywhere in theupper half plane y > 0.

(b) Determine the surface charge density on the strip and on the grounded planes, andmake a graph.

1

Problem 2. The electric stress tensor

Recall that the stress tensor is the force per area. The force per volume f j is (minus) thedivergence of the stress tensor (see class notes)

f j = −∂iT ij (1)

This follows from the conservation law

∂tgj + ∂iT

ij = 0 (2)

where gj is the momentum per volume, and the basic notion that the force is the timederivative of the momentum.

The force per volume in electrostatics is

f j = ρEj (3)

This form must be the divergence of something. As you will show in this excercise

ρEj = −∂iT ijE (4)

whereT ijE ≡ −E

iEj + 12E2δij (5)

(a) (Optional) First write the electrostatic Maxwell equations ∇ ·E = ρ and ∇×E = 0using tensor notation, and explain why ∂iEj = ∂jEi.

(b) Within the limits of electrostatics, show that the electric force on a charged body isrelated to a surface integral of the (electric) stress tensor:

F j =

∫V

d3r ρ(r)Ej = −∫S

dS niTijE (6)

where T ijE = −EiEj + 1

2E2δij, i.e. show that ρEj = −∂iT ij

E

2

Problem 3. A stress tensor tutorial

Do not turn in the optional parts.

(a) (Optional) Consider a plane of charge with surface charge density σ, use the boundaryconditions (i.e. Gauss Law) to show that the electric field on either side is σ/2

(b) (Optional) Consider an ideal infinite parallel plate capacitor with surface charge densi-ties σ and −σ respectively. Without using the stress tensor machinery, show that theforce per area on each of the plates is σ2/2

(c) (Optional) Consider a charged perfectly conducting solid object of any shape. Explainphysically why the electric field is: (i) normal to the surface, (ii) zero on the inside,(iii) and equal to

E = σn or Ei = σ ni (7)

(d) Without using the stress tensor machinery, show that the force per area on the wallsof any metal surface is σ2/2. (Hint: how large is the self field? Use part (a).)

The physics of the stress tensor is easy illustrated by knowing that the stress tensor ofideal gas is T ij

gas = p δij, where p is the pressure (force per area). Thus, if one considers awall separating two gasses of left and right pressures pL and pR (i.e. the normal vector is1,nj = δjx), then the net force per area on the wall is

niTijL − niT

ijR = (pL − pR)nj (8)

Note: that it is only the differences in the stress tensor which are physically important.

(e) (Optional) Recall that the net force on any object

F j = −∮dS niT

ij , (9)

which we derived from the conservation law

∂tgj + ∂iT

ij = 0 . (10)

Deduce from this that the net force per area on a wall separating two regions is

ni(Tijout − T

ijin ) . (11)

(f) Using the electric stress tensor T ijE = −EiEj + 1

2E2δij, show that the force per area on

the surface of a charged metal object is

force-per-area =σ2

2nj (12)

where n points from inside the metal to out.

1The notation is to confuse/educate you – I could have written n = (1, 0, 0) or n = x.

3

(g) Now consider a charged and isolated parallel plate capacitor with charge per area −σand +σ on the left and right plates (so that the normal is nj = δjx). A plane of chargewith charge per area σ/2 lies halfway between the plates.

(i) Compute all non-zero components of the stress tensor in the regions to the leftand right of the plane of charge.

(ii) Use the stress tensor to compute the force per area on the plane of charge, andshow that it agrees with a simple minded approach.

4

Problem 4. Practice with the stress tensor

(a) Calculate the force between two (solid and insulating) uniformly charged hemisphereseach with total charge Q and radius R that are separated by a small gap as shownbelow. You should find

F =3Q2

16πR2(13)

5

Problem 5. Green function of a sphere

Consider a grounded, metallic, hollow spherical shell of radius R. A point charge of chargeq is placed at a distance, a, from the center of the sphere along the z-axis. For simplicitytake a > R.

(a) Start by momentarily setting R = 1, and therefore measure all lengths in units of R.a is then shorthand for a/R in this system of units. With these units, show that thedistance from the point r = a z to any point, n, on the surface of the sphere is equal(up to a constant factor of a) to the distance from a point at r = (1/a) z to the samepoint n on the sphere.

(b) Use the result of part (a) to construct the Green function of the grounded sphere ofradius R using images, i.e. find the potential due to a point charge at r = az in thepresence of a grounded sphere.

(c) Compute the surface charge density, and show that it is correct by directly integratingto find the total induced charge on the sphere of part (b). You should find that thetotal induced charge is equal to the enclosed image charge (why?). Please do not useMathematica to do integrals.

(d) Now consider a point charge of charge q at a point r = zz above a metallic hemisphereof radius R in contact with a grounded plane (see below). Determine the force on thecharge as a function of z. You should find that at a distance z = 2R the force is

F z = − Q2

4πR2

(737

3600

)(14)

(e) Show that at large distances, z, the Taylor series expansion for F z is

F z ' Q2

4πR2

[−1

4u2− 4

u5+ . . .

]where u = z/R. Explicitly explain the coefficients of the series expansion (i.e. the−1/4 and −4) in terms of the multiple moments of the image solution.

6