Problem Set 4 Solution
Phys 182 - Fall 2010
Assigned: Sunday, Sept. 26 Due: Friday, Oct. 1
1
Griffiths 3.23
Correction!
∂V
1 ∂ 2V
1 ∂
s
+ 2
= 0 , take solutions of the form V (s, φ) = S(s)Φ(φ).
s ∂s
∂s
s ∂φ2
1 ∂
∂S
1 ∂2Φ
s
+ 2 S 2 = 0 , multiply by s2 and divide by V = Sφ.
Φ
s ∂s
∂s
s ∂φ
s ∂
∂S
1 ∂2Φ
s
+
= 0 , so we may write:
S ∂s
∂s
Φ ∂φ2
∂S
1 ∂2Φ
s ∂
s
= k2 ,
= −k 2 .
S ∂s
∂s
Φ ∂φ2
For k 6= 0
∂2Φ
= −k 2 Φ ⇒ Φ(φ) = A cos(kφ) + B sin(kφ)
∂φ2
∂
∂S
s
= k 2 S ⇒ S(s) = Csk + Ds−k
∂s
∂s
For k = 0
∂2Φ
= 0 ⇒ Φ(φ) = E + F φ, but F = 0 because Φ(φ + 2π) = Φ(φ)
∂φ2
∂
∂S
∂S
s
=0⇒s
= const. ⇒ S(s) = G ln(s) + H
∂s
∂s
∂s
The general solution is
V (s, φ) = a0 + b0 ln(s) +
∞
X
[(ak sk + bk s−k )(ck cos(kφ) + dk sin(kφ))]
k=1
2
Griffiths 3.24
Pick V = 0 on the yz plane, with E0 in the x direction, we have:
(
(i)V = 0,
when s = R,
(ii)V → −E0 x = −E0 s cos φ when s ≫ R.
1
So a0 = b0 = bk = dk = 0, and ak = ck = 0 except for k = 1.
c1 cos φ
V (s, φ) = a1 s +
s
(
(i) ⇒ c1 = −a1 R2
(ii) ⇒ a1 = −E0
Therefore
#
" 2
E0 R2
R
V (s, φ) = −E0 s +
− 1 cos φ
cos φ = −E0 s
s
s
∂V R2
σ = −ǫ0
= 2ǫ0 E0 cos φ
= −ǫ0 E0 − 2 − 1 cos φ
∂s s=R
s
s=R
3
Griffiths 3.33
p = (p · r̂)r̂ + (p · θ̂)θ̂ = p cos(θ)r̂ − p sin(θ)θ̂.
So 3(p · r̂)r̂ − p = 3p cos(θ)r̂ − p cos(θ)r̂ + p sin(θ)θ̂ = 2p cos(θ)r̂ + p sin(θ)θ̂
4
Griffiths 4.1
E = V /x = 500/10−3 = 5 × 105 . From Table 4.1 α/4πǫ0 = 0.66 × 10−33 , so α = 7.34 × 10−41 .
d
p = αE = ed ⇒ d = αE/e = 2.29 × 10−16 m.
= 4.6 × 10−6 .
R
To ionize, let d = R. Then R = αE/e = αV /ex ⇒ V = Rex/α = 108 V.
5
Griffiths 4.4
1 q
αq
r̂. Induced dipole moment of atom: p = αE =
r̂.
2
4πǫ0 r2
4πǫ
0r
2αq
1 1
(to the right).
Field of this dipole, at the location of q: E =
4πǫ0 r3 4πǫ0 r2
2
q
1
Force on q due to this field: F = 2α
r 5 (attractive).
4πǫ0
Field of q:
2
6
Griffiths 4.7
If the potential is zero at infinity, the energy of a point charge Q is W = QV (r). For a physical dipole, with
−q at r and +q at r + d,
#
" Z
r+d
E · dl .
U = qV (r + d) − qV (r) = q −
r
For an ideal dipole the integral reduces to E · d
U = −qE · d = −p · E, since p = qd.
If you do not (or cannot) use infinity as the reference point, the result still holds as long as you bring the
two charges in from the same point.
7
Griffiths 4.8
U = −p1 · E2 , but E2 =
8
1 1
1 1
[3(p2 · r̂)r̂ − p2 ]. So U =
[p1 · p2 − 3(p1 · r̂)(p2 · r̂)].
4πǫ0 r3
4πǫ0 r3
Griffiths 4.10
(a) σb = P · n̂ = kR, ρb = −∇ · P = −
(b) For r < R, E =
1 ∂ 2
1
(r kr) = − 2 3kr2 = −3k.
2
r ∂r
r
1
ρrr̂, so E = −(k/ǫ0 )r.
3ǫ0
4
For r > R, same as if all charge at center, but Qtot = (kR)(4πR2 ) + (−3k)( πR3 ) = 0, so E = 0.
3
9
Griffiths 4.13
Think of it as two cylinders of opposite uniform charge density ±ρ.
Inside, the field at a distance s from the axis of a uniformly charged cylinder is given by Gauss’s Law:
E2πsl = ǫ10 ρπs2 l ⇒ E = (ρ/2ǫ0 )s. For two such cylinders, one positively charged and the other negatively
charged, the net field inside is E = E+ + E− = (ρ/2ǫ0 )(s+ − s− ). But S+ − s− = −d, so E = −ρd/(2ǫ0 ),
where d is the vector from the negative axis to the positive axis. In this case the total dipole moment of a
chunk of length l is P(πa2 l) = (ρπa2 l)d. So ρd = P, and E = −P/(2ǫ0 ) for s < a.
Outside, Gauss’s law gives E2πsl =
1
2
ǫ0 ρπa l
⇒ E =
3
ρa2 ŝ
2ǫ0 s ,
for one cylinder. For the combination, E =
E+ + E− =
ŝ+
s+
ρa2
2ǫ0
ŝ+
s+
−
ŝ−
s−
, where
d
s± = s ∓
2
−1
−1
d2
s·d
s·d
d
d
s±
d
1
1
2
∼
∼
s +
1∓ 2
1± 2
= s∓
∓s·d
= 2 s∓
= 2 s∓
s2±
2
4
s
2
s
s
2
s
(s · d) d
1
(keeping only 1st order terms in d).
= 2 s±s 2 ∓
s
s
2
(s · d) s
1
(s · d) d
s(s · d)
1
ŝ−
− s−s 2 +
= 2 2
s+s 2 −
= 2
−
−d .
s−
s
s
2
s
2
s
s2
E(s) =
10
a2 1
[2(P · ŝ)ŝ − P], for s > a.
2ǫ0 s2
Griffiths 4.15
1 ∂
(a) ρb = −∇ · P = − 2
r ∂r
Gauss’s law ⇒ E =
(
+P · r̂ = k/b
r=b
k
2k
r
= − 2 . σb = P · n̂ =
r
r
−P · r̂ = −k/a r = a
1 Qenc
4πǫ0 r 2 r̂.
For r < a, Qenc = 0,so E = 0. For r > b, Qenc = 0,so E = 0.
Rr
′2 ′
For a < r < b, Qenc = −k
(4πa2 ) + a −k
a
r 2 4πr dr = −4πka − 4πk(r − a) = −4πkr, so E = −(k/ǫ0 r)r̂.
I
(b)
D · da = Qfenc = 0 ⇒ D = 0 everywhere. D = ǫ0 E + P = 0 ⇒ E = (−1/ǫ0 )P, so
E = 0 (for r < a and r > b); E = −(k/ǫ0 r)r̂ (for a < r < b).
4