Physics 322 Homework 5
Assigned: Wed. 10/5/16
Due: Wed. 10/12/16
Note: First midterm is 10/14/16. You may want to photocopy your hw5 solutions for your midterm studies.
1. Griffiths 3.30
ans
a) The dipole moment formula from lecture is
Z
~p
=
dτ 0~r0 ρ(~r0 )
Z 1
=
−1
Z 1
=
−1
d cos θ 0
Z 2π
dφ 0 R3 r̂k cos θ 0
0
d cos θ 0
Z 2π
dφ 0 R3 (sin θ 0 cos φ 0 x̂ + sin θ 0 sin φ 0 ŷ + cos θ 0 ẑ)k cos θ 0
0
Since φ 0 integral is trivially zero for x̂ and ŷ components, we have
~p
3
Z 1
= 2πR ẑk
dww2
−1
=
k
4π 3
R ẑ
3
b) Dipole potential from lecture is
Vdip =
1 ~p · r̂
4πε0 r2
Vdip =
kR3 cos θ
3ε0 r2
which gives
Since this matches the exact result, we conclude that all higher multipoles are zero.
2. A potential for a conducting sphere of radius R in a uniform field E0 was derived in one of the lectures:
(
0 r≤R
V (r) =
R3
E0 r2 − r cos θ r > R
a) Find the electric field, everywhere.
1
b) We computed the induced charge σ (θ ) in class. Compute the potential due to σ (θ ) alone for r < R.
c) Find the electric field due to σ (θ ).
ans
For r < R, compute using V or part b)
~E
= −~∇V
1
1
= −(∂rV r̂ + ∂θ V θ̂ +
∂φ V φ̂ )
r
r sin θ
=
−E0 cos θ r̂ + E0 sin θ θ̂ r < R
For r > R, we noted in lecture
V = E0
R3
cos θ .
r2
Hence, we find from part a)
3
3
~E = E0 2R cos θ r̂ + R sin θ θ̂
r>R
r3
r3
3. Griffiths 3.46
2
ans
a) monopole moment:
Z a
k
−a
dz0 cos(
πz0
2a
π
−π
4ak
) = k [sin − sin
]=
2a
π
2
2
π
1 4ak
4πε0 r π
≈
V
ak
π 2 ε0 r
=
b) The monopole moment for this case is zero since cos π = cos −π. The dipole term numerator is
Z a
Z a
Z a
πz0
πz0
0 0
0 0
0 0
0
k
dz r cos α sin(
) = k cos θ
dz z sin
=
k cos θ
dz z sin(bz ) |b= πa
a
a
−a
−a
−a
Z
d a 0
0
=
−k cos θ
dz cos(bz ) |b= πa
db −a
d 2
=
−k cos θ
sin(ba) |b= πa
db b
2
2
=
−k cos θ − 2 sin(ba) + a cos(ba) |b= πa
b
b
=
2ka2
cos θ
π
Hence,
V
≈
1 2ka2
cos θ
4πε0 πr2
=
ka2
cos θ
2ε0 π 2 r2
c) Monopole is zero since sin (±π) = 0. Dipole is zero since z cos πz
a is an odd function of z. Note that The quadrupole
term numerator is
Z a
k
−a
dz0 (r0 )2 P2 (cos α) cos(
πz0
) = kP2 (cos θ )
a
Z a
−a
dz0 (z0 )2 cos(
πz0
)
a
because P2 (cos α) is an even function of cos α (i.e. the fact that as z0 crosses zero, cos α goes from cos θ to
− cos θ does not change P2 (cos θ )). Hence, we find
2 Za
Z a
πz0
d
0
0
0 0 2
) = −kP2 (cos θ )
dz cos(bz ) |b= πa
kP2 (cos θ )
dz (z ) cos(
a
db2 −a
−a
2
d 2
= −kP2 (cos θ )
[ sin(ba)] |b= πa
db2 b
d
2
2a
[− sin(ba) + cos(ba)] |b= πa
= −kP2 (cos θ )
db b2
b
4
4a
2a2
= −kP2 (cos θ ) 3 sin(ba) − 2 cos(ba) −
sin(ba) |b= πa
b
b
b
= −kP2 (cos θ )
3
4a3
π2
which gives
V
≈ −
=
4a3
1
kP2 (cos θ ) 2
3
4πε0 r
π
−
ka3 P2 (cos θ )
π 3 ε0
r3
4. Griffiths 4.8
ans
5. Griffiths 4.16
4
(Hint: In part b), neglect the asymmetry of the needle at the top and at the bottom. Long needle means that if A is the
cross-sectional area of the needle, A|~P|/(4πε0 [L/2]2 ) |~E0 |.)
ans
a) Since
~E0 = ~Eext + ~E pol
before the object is hollowed out. Hence, after taking the spherical cavity out, the field at the center of the cavity
is
~Ecavity |center = ~E0 − ~E pol |center .
In lecture 13, we worked out
~
~E pol |center = − P
3ε0
Hence, we conclude
~
~Ecavity |center = ~E0 + P .
3ε0
The ~D at the center is
~Dcavity |center
= ε0 ~Ecavity |center + ~Pcavity |center
#
"
~P
~
= ε0 E0 +
3ε0
=
~
~D0 − 2P
3
b) As in part a), we have
~Ecavity |center = ~E0 − ~E pol |center .
There is no volume polarization charge for a needle shaped polarized material with constant ~P since ~∇ · ~P = 0.
Since the effective surface charge is ~P · n̂, we have this being non-zero only at the two ends: there is a positive
charge of magnitude A|~P| (if ~P points upward) at the upper end of the needle and a negative charge at the bottom
end of the needle. The hint tells us
A|~P|/(4πε0 [L/2]2 ) |~E0 |
which means that we can neglect the ~E pol |center from this end-point charge at the center of the needle. Hence,
we have
~Ecavity |center ≈ ~E0 .
5
~Dcavity |center
= ε0 ~Ecavity |center + ~Pcavity |center
= ε0 ~E0
=
~D0 − ~P
c) As in part a), we have
~Ecavity |center = ~E0 − ~E pol |center .
Repeating the arguments of b), we have the upper end of the cylinder having a unform surface charge distribution:
σb ≈ |~P|
on the upper end and σb ≈ −|~P| on the lower end of ~P is upwards. This situation is more like a parallel plate
capacitor. Hence, we conclude
~
~E pol |center = − σb P̂ = − P
ε0
ε0
which means
~
~Ecavity |center = ~E0 + P .
ε0
The displacement is
~Dcavity |center
= ε0 ~Ecavity |center + ~Pcavity |center
= ε0 ~E0 + ~P
=
~D0
6. Griffiths 4.21
ans
Adding a test charge Q to the inner conductor, we can compute the electric field is computed using Gauss’ law over a
pill box of length L and radius r:
1
2πLrEr = Q
ε0
Er =
1 Q/L
.
2πε0 r
6
for a < r < b. Becuase ~E is radial, we know ~P is radial between b < r < c. Hence, the electric field for b < r < c is
Er =
1 Q/L
.
2πε r
We conclude the potential difference between the two conductors is
Q/L 1 b 1 1
dr +
[
2π ε0 a
r ε
b 1 c
Q/L
[ln + ln ]
2πε0 a εr b
Z
V
=
=
Z c
b
1
dr ]
r
Hence, we conclude
C/L =
2πε0
ln + ε1r ln bc
b
a
7. A capacitor of area A = w2 and separation d is partially filled by a dielectric slab with dielectric constant K as
shown.
The potenial across the capacitor is V . Find the horizontal force on the slab.
ans
7
8