PHY 5346
HW Set 5 Solutions – Kimel
1. 2.13 The system is pictured in the following figure:
a) Notice from the figure, Φρ,−φ = Φρ,φ; thus from Eq. (2.71) in the text,
Φρ,φ = a0 +∑
n=1
∞
anρn cosnφ
∫
−π/2
3π/2
Φb,φ = 2πa0 = πV1 + πV2 → a0 =
V1 + V2
2
Using
∫
−π/2
3π/2
cosmφ cosnφdφ = δnmπ
Applying this to Φ, only odd terms m contribute in the sum and
am =
2V1 − V2
πmbm
−1
m−1
2
Thus
Φρ,φ = V1 + V2
2
+
2V1 − V2
π
Im ∑
m odd
imρme imφ
mbm
Using
2 ∑
m odd
xm
m = ln
1 + x
1 − x
and
Im lnA + iB = tan−1B/A
we get
Φρ,φ = V1 + V2
2
+
V1 − V2
π
tan−1
2 ρ
b
cosφ
1 − ρ
2
b2
as desired.
b)
σ = −0 ∂∂ρ
Φρ,φ|ρ=b
σ = −0
V1 − V2
π
∂
∂ρ
tan−1
2 ρ
b
cosφ
1 − ρ
2
b2
ρ=b
σ = −20
V1 − V2
π
bcosφ
b2 + ρ2
b4 − 2b2ρ2 + ρ4 + 4ρ2b2 cos2φ
|ρ=b = −0
V1 − V2
πbcosφ
: