PHYS463 Electricity& Magnetism III (2003-2004)Problems Solutions (assignment #3)

1. (Problem 3.38, p.126)

Solution:

Use equation (3.95)

Φ =1

4π²0

∞Xn=0

1

rn+1

Z ³r

0´nPn

³cosϑ

0´ρ³−→r 0

´dτ

0

Nowλ = Q/2a − a < z < +a

for 0 < z < a, r0= z

0, ϑ

0= ϑ.

for −a < z < 0, r0= −z0

=¯̄z

0 ¯̄, ϑ

0= π − ϑ and

cosϑ0= cos (π − ϑ) = − cosϑ

So

Φ =λ

4π²0

∞Xn=0

1

rn+1

Z a

−a

³¯̄̄z

0¯̄̄´n

Pn

³cosϑ

0´dz

0

Phys. 463 E & M 2003-04 1 / Solution3

evaluateZ a

−a

³¯̄̄z

0¯̄̄´n

Pn³cosϑ

0´dz

0=

Z 0

−a

³¯̄̄z

0¯̄̄´n

Pn³cosϑ

0´dz

0+

Z a

0

³¯̄̄z

0¯̄̄´n

Pn³cosϑ

0´dz

0

=

Z 0

−a

³¯̄̄z

0¯̄̄´n

Pn (cosϑ) dz0+

Z a

0

³¯̄̄z

0¯̄̄´n

Pn (− cosϑ) dz0

=

Z 0

−a

³¯̄̄z

0¯̄̄´n

Pn (cosϑ) dz0+

Z a

0

³¯̄̄z

0¯̄̄´n

(−1)n Pn (cos ϑ) dz0

for odd nZ a

−a

³¯̄̄z

0¯̄̄´n

Pn³cosϑ

0´dz

0= Pn (cosϑ)

·Z 0

−a

³−z0

´ndz

0+

Z a

0

³z

0´n(−1)n dz0

¸= (−1)n Pn (cos ϑ)

·Z 0

−a

³z

0´ndz

0+

Z a

0

³z

0´ndz

0¸

= (−1)n Pn (cos ϑ)·Z a

−a

³z

0´ndz

0¸

= (−1)n Pn (cos ϑ)"¡z

0¢n+1

n+ 1

#a−a

=

2Pn (cosϑ)an+1

n+1n = 0, 2, 4, · · ·

0 n = 1, 3, 5, · · ·

Φ =λ

4π²0

∞Xn=0

1

rn+1

Z a

−a

³¯̄̄z

0¯̄̄´n

Pn³cosϑ

0´dz

0

=1

4π²0

Q

2a

∞Xn=even

1

rn+12Pn (cosϑ)

an+1

n+ 1

=Q

4π²0

1

r

∞Xn=even

1

n+ 1

³ar

´nPn (cosϑ)

=Q

4π²0

1

r

µ1 +

³ar

´2

P2 (cosϑ) +³ar

´4

P4 (cosϑ) + · · ·¶

2.

−→S =

1

µ0

−→E ×−→B =

1

µ0ErBθ

−→e z

S =1

µ0ErBθ =

1

µ0

V

ln ba

1

r

µ0I

2πr=

V I

2π ln ba

1

r2

Phys. 463 E & M 2003-04 2 / Solution3

Power through the cross-section a < r < b

P =

Z b

a

−→S · d−→A =

Z b

a

S2πrdr

=

Z b

a

V I

2π ln ba

1

r22πrdr =

V I

ln ba

Z b

a

1

rdr

=V I

ln ba

lnb

a= V I

Exactly the power delivered by the power supply.

3. Problem 8.2 (ref. to Problem 7.31, p. 324)

A fat wire, radius a,carries a constant current I, uniformly distributed over its crosssection. A narrow gap in the wire, of width w¿ a, forms a parallel-plate capacitor.

(a) Find the electric and magnetic field in the gap, as functions of the distance s fromthe axis and the time t . (Assume the charge is zero at t = 0.)

(b) Find the energy density uem and the Poynting vector−→S in the gap. Note especially

the direction of−→S . Check that equation 8.14 is satisfied.

Determine the total energy in the gap, as a function of time. Calculate the total powerflowing into the gap, by integrating the Poynting vector over the appropriatesurface. Check that the power input is equal to the rate of increase of energy inthe gap (eq. 8.9 — in this case W = 0, Because there is no charge in the gap).[If you worried about the fringing fields, do it for a volume of radius b < a wellinside the gap.]

Solution

(a) Electric field:

from the definition of the capacitance

C =Q

V=It

Ew=It

Ew

Phys. 463 E & M 2003-04 3 / Solution3

on the other hand, the capacitance of parallel plate is

C = ²0A

w= ²0

πa2

w

SoIt

Ew= ²0

πa2

wor

E =1

π²0

I

a2t

−→E =

1

π²0

I

a2t−→e z

independent of s

Magnetic fields:

within the gap,−→j = 0. So

∇×−→B = µ0

Ã−→j + ²0

∂−→E

∂t

!= µ0²0

∂−→E

∂t= µ0

1

π

I

a2−→e z

integrate over a circular cross section of radius sZ ³∇×−→B

´· d−→A =

Zµ01

π

I

a2−→e z · d−→A

LHS: Z ³∇×−→B

´· d−→A =

I −→B · d−→l = 2πsBθ

RHS: Zµ01

π

I

a2−→e z · d−→A = µ0

1

π

I

a2πs2

so

2πsBθ = µ01

π

I

a2πs2

or

Bθ = µ01

2π

I

a2s

or −→B = µ0

1

2π

I

a2s−→e θ

Phys. 463 E & M 2003-04 4 / Solution3

(b)

uem =1

2²0E

2 +1

2µ0B2

=1

2²0

µ1

π²0

I

a2t

¶2

+1

2µ0

µµ01

2π

I

a2s

¶2

−→S =

1

µ0

−→E ×−→B =

µ1

µ0

1

π²0

I

a2t

¶µµ01

2π

I

a2s

¶−→e z ×−→e θ

=

µ1

2π²0

I2

πa4st

¶(−−→e r) =

µ− 1

2π²0

I2

πa4st

¶−→e r

radially inwards. Now check

∂

∂t(umech + uem) = −∇ ·−→S

In this case, umech = 0 and

∂uem∂t

= ²0

µ1

π²0

I

a2

¶2

t =1

π²0

I2

πa4t

−∇ ·−→S = −1s

∂

∂s(sSs) = −1

s

∂

∂s

µ− 1

2π²0

I2

πa4s2t

¶=

1

π²0

I2

πa4t

therefore∂

∂t(umech + uem) = −∇ ·−→S

(c) Total energy in the gap

Uem =

Zuemdτ = 2π

Z s

0

"1

2²0

µ1

π²0

I

a2t

¶2

+1

2µ0

µµ01

2π

I

a2s

¶2#wsds

= 2π

"1

4²0

µ1

π²0

I

a2t

¶2

ws2 +w

8µ0

µµ01

2π

I

a2

¶2

s4

#

=1

2π²0

µI

a2t

¶2

ws2 +µ0

16π

µI

a2

¶2

ws4

∂Uem∂t

=1

π²0

µI

a2

¶2

ws2t

Phys. 463 E & M 2003-04 5 / Solution3

I −→S · d−→A =

Z s

0

SsdA = Ss (2πsw)

=

µ− 1

2π²0

I2

πa4st

¶(2πsw)

= − 1

π²0

I2

a4ws2t

AlsodW

dt= 0

−∂Uem∂t

−I −→S · d−→A = − 1

π²0

µI

a2

¶2

ws2t−µ− 1

π²0

I2

a4ws2t

¶= 0

4. c−→r = sinϑ cosφ−→e x + sinϑ sinφ−→e y + cosϑ−→e zOn the bowl

d−→A = dAc−→r = R2 sinϑdϑdφ [sinϑ cosφ−→e x + sinϑ sinφ−→e y + cosϑ−→e z]

Tzx = ε0EzEx = ε0

µ1

4πε0

Q

R2

¶2

cosϑ sinϑ cosφ

Tzy = ε0EzEy = ε0

µ1

4πε0

Q

R2

¶2

cosϑ sinϑ sinφ

Tzz =1

2ε0

¡E2z −E2

x − E2y

¢=1

2ε0

µ1

4πε0

Q

R2

¶2 ¡cos2 ϑ− sin2 ϑ

¢µ←→

T · d−→A¶z

= TzxdAx + TzydAy + TzzdAz

= (Tzx sinϑ cosφ+ Tzy sinϑ sinφ+ Tzz cosϑ)R2 sinϑdϑdφ

=

"ε0

µ1

4πε0

Q

R2

¶2

cosϑ sinϑ cosφ sinϑ cosφ

+ε0

µ1

4πε0

Q

R2

¶2

cosϑ sinϑ sinφ sinϑ sinφ

+1

2ε0

µ1

4πε0

Q

R2

¶2 ¡cos2 ϑ− sin2 ϑ

¢cosϑ

#R2 sinϑdϑdφ

Phys. 463 E & M 2003-04 6 / Solution3

= ε0

µ1

4πε0

Q

R2

¶2 £cosϑ sin2 ϑ cos2 φ+ cosϑ sin2 ϑ sin2 φ

+1

2

¡cos2 ϑ− sin2 ϑ

¢cosϑ

¸R2 sinϑdϑdφ

= ε0

µ1

4πε0

Q

R2

¶2 ·cosϑ sin2 ϑ+

1

2

¡cos2 ϑ− sin2 ϑ

¢cos ϑ

¸R2 sinϑdϑdφ

= ε0

µ1

4πε0

Q

R2

¶2 ·1

2

¡cos2 ϑ+ sin2 ϑ

¢cosϑ

¸R2 sinϑdϑdφ

=1

2ε0

µ1

4πε0

Q

R2

¶2

R2 sinϑ cosϑdϑdφ

5. Problem 8.4 (hints: read the example 8.2 for integral over the disk first. You need to findthe electric field in the mid plane first)

• Solution:Create a semi hemisphere consisting of the mid-plane and the half spherical surface onthe right hand side with radius →∞ The force on the charge on the left hand side is

−→F =

I ←→T · d−→A =

Zmid−plane

←→T · d−→A +

Zspher. surface

←→T · d−→A

Calculate the field on the mid-plane now. consider a point on the x axis at (r, 0, 0).Due to left-right symmetry, the field is on the x axis direction. Field due to one charge

E =1

4π²0

q

R2

Field due to both charges

Ex = 2 · 1

4π²0

q

R2· cos θ = 1

2π²0

q

R2· rR=

1

2π²0

qr

(r2 + a2)3/2

Due to symmetry about z axis. the field on the x-y plane in a cylindrical coordinatesystem is radially outwards..

Er(r) =1

2π²0

q

R2· rR=

1

2π²0

qr

(r2 + a2)3/2

Phys. 463 E & M 2003-04 7 / Solution3

Sinced−→A = dA−→e z

←→T · d−→A = TxzdA

−→e x + TyzdA−→e y + TzzdA−→e zbut since Ez = 0

Txz = ²0ExEz = 0, Tyz = ²0EyEz = 0

and

Tzz =²02

¡E2z − E2

x −E2y

¢= −²0

2

¡E2x + E

2y

¢= −²0

2E2

= −²02

Ã1

2π²0

qr

(r2 + a2)3/2

!2

= − q2

8π2²0

r2

(r2 + a2)3

The surface integral on the mid-plane isZmid−plane

←→T · d−→A =

Zmid−plane

TzzdA−→e z

= −→e zZmid−plane

·− q2

8π2²0

r2

(r2 + a2)3

¸dA

= − q2

8π2²0−→e z

Z ∞

0

r2

(r2 + a2)3(2πrdr)

= − q2

8π²0−→e z

Z ∞

0

r2

(r2 + a2)3dr2

I =

Z ∞

0

r2

(r2 + a2)3dr2 =

Z ∞

0

t

(t+ a2)3dt = −1

2

Z ∞

0

td1

(t+ a2)2

Phys. 463 E & M 2003-04 8 / Solution3

= −12

t

(t+ a2)2

¯̄̄̄∞0

+1

2

Z ∞

0

1

(t+ a2)2dt = 0 +

1

2

Z ∞

0

1

(t+ a2)2dt

= −12

1

(t+ a2)

¯̄̄̄∞0

=1

2a2

So Zmid−plane

←→T · d−→A = − q2

8π²0−→e z

µ1

2a2

¶= − q2

16π²0a2−→e z

On the half spherical surface

|T | ∝ 1

r4, dA ∝ r2

So Zspher. surface

←→T · d−→A = 0

Total force

−→F =

I ←→T · d−→A =

Zmid−plane

←→T · d−→A = − q2

16π²0a2−→e z

Direct calculation reveals the force on the charge on the left hand side

F =1

4π²0

q2

(2a)2=

q2

16π²0a2in the −−→e z direction

6. Problem 8.6

Solution

1. A charged parallel plate capacitor (with uniform electric field−→E = E−→e z) is placed in

a uniform magnetic field−→B = B−→e x as shown.

Phys. 463 E & M 2003-04 9 / Solution3

(a) Find the electromagnetic momentum in the space between the plates.

−→p momentum = µ0²0−→S = ²0

−→E ×−→B

= ²0EB−→e y

Assume the distance between the plates is d, surface area of A, the total momen-tum is −→p total = ²0EBAd−→e y

(b) Now a resistive wire is connected between the two plates, along the z-axis, so thatthe capacitor slowly discharges. The current through the wire will experience a

magnetic force; What is the total impulse (R −→F dt) delivered to the system, during

the discharge?

Assume the current on the wire is I(t). It should flow from the bottom plate tothe top plate. The force on the wire is

−→F =

Z h

0

I(t)d−→l ×−→B = I(t)dB−→e y

But

I(t) =dq

dt

AlsoE =

σ

²0=

q

A²0

So

I(t) = A²0dE

dt

and −→F = AdB²0

dE

dt−→e y

The impulse

∆p =

Z −→F dt = −→e y

ZAdB²0

dE

dt

= ²0BEAd−→e y

This is exactly the initial EM momentum stored between the plates.

(c) Determine all nine elements of the stress tensor, in the region between the plates.Display your answer as a 3× 3 matrix.

Tij = ²0

µEiEj − 1

2δijE

¶+1

µ0

µBiBj − 1

2δijB

¶Phys. 463 E & M 2003-04 10 / Solution3

For i 6= j

Tij = ²0EiEj +1

µ0

BiBj

Note−→E = E−→e z, −→B = B−→e xTij = 0, when i 6= j

When i = j

Tii = ²0

µE2i −

1

2E2

¶+1

µ0

µB2i −

1

2B2

¶

T11 = ²0

µE2x −

1

2E2

¶+1

µ0

µB2x −

1

2B2

¶= −²0

2E2 +

1

2µ0B2

T22 = ²0

µE2y −

1

2E2

¶+1

µ0

µB2y −

1

2B2

¶= −²0

2E2 − 1

2µ0B2

T33 = ²0

µE2z −

1

2E2

¶+1

µ0

µB2z −

1

2B2

¶=²02E2 − 1

2µ0B2

So

T =1

2

−²0E2 + 1µ0B2 0 0

0 −²0E2 − 1µ0B2 0

0 0 ²0E2 − 1

µ0B2

7. Problem 8.15

SolutiondW

dt=

Z ³−→E ·−→J

´dτ

Now

∇×−→H =−→J f +

∂−→D

∂t

−→E ·−→J =

−→E ·

Ã∇×−→H − ∂

−→D

∂t

!

=−→E ·

³∇×−→H

´−−→E · ∂

−→D

∂t

=h−→H ·

³∇×−→E

´−∇ ·

³−→E ×−→H

´i−−→E · ∂

−→D

∂t

Phys. 463 E & M 2003-04 11 / Solution3

= −∇ ·³−→E ×−→H

´+−→H ·

Ã−∂−→B

∂t

!−−→E · ∂

−→D

∂t

= −∇ ·³−→E ×−→H

´−Ã−→E · ∂

−→D

∂t+−→H · ∂

−→B

∂t

!= −∇ ·

³−→S´− ∂uem

∂t

where −→S =

−→E ×−→H

and∂uem

∂t=−→E · ∂

−→D

∂t+−→H · ∂

−→B

∂t

For linear media, ² and µ are independent of−→E and

−→B respectively

−→E · ∂

−→D

∂t+−→H · ∂

−→B

∂t= ²

−→E · ∂

−→E

∂t+1

µ

−→B · ∂

−→B

∂t

= ²1

2

∂

∂t

³−→E ·−→E

´+1

2µ

∂

∂t

³−→B ·−→B

´=

1

2

∂

∂t

µ²−→E ·−→E +

1

µ

−→B ·−→B

¶=

1

2

∂

∂t

³−→E ·−→D +

−→H ·−→B

´So

∂uem

∂t=1

2

∂

∂t

³−→E ·−→D +

−→B ·−→H

´or

uem =1

2

³−→E ·−→D +

−→B ·−→H

´

Phys. 463 E & M 2003-04 12 / Solution3