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The solution to problem 1 of homework #5 in physics 6346 (fall 2008), which involves finding the electrostatic potential around a long, hollow conducting cylinder using separation of variables and green's functions in cylindrical coordinates. The solution includes the general form of the potential, series expansions for the potential inside and outside the cylinder, and the surface charge density on the cylinder.
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PHY 6346 Fall 2008 Homework #5, Due Wednesday, October 1
(a) Use separation of variables to show that the general two-dimensional solution to ∇^2 Φ = 0 in cylindrical coordinates (ρ, φ) can be written as Φ = R(ρ)Q(φ), with R(ρ) = ρ±m^ and Q(φ) = e±imφ, where m is zero or a positive integer. (For m = 0, “ ρ−^0 ” means ln ρ).
(b) Fix the coefficients to match the given potential at ρ = b and write a series for Φ(ρ, φ) inside and outside the cylinder (two different series).
(c) Sum the series and show that inside the cylinder (ρ < b)
Φ(ρ, φ) =
π
tan−^1
2 bρ cos φ b^2 − ρ^2
What is the solution for ρ > b?
(d) Calculate and sketch the surface charge density on the cylinder.
(a) Show that
δ(φ − φ′) =
2 π
m=−∞
eim(φ−φ
2 π
m=
cos m(φ − φ′)
(b) From the differential equation
∇′^2 G = − 4 π
δ(ρ − ρ′) ρ
δ(φ − φ′)
we know that ∇′^2 G vanishes almost everywhere. Expand G in one series G< for ρ′^ < ρ and another G> for ρ′^ > ρ. Apply boundary conditions: that G = 0 at ρ′^ = b; that G is nonsingular at ρ′^ → 0; and that G is smooth at ρ′^ = ρ,
G = C 0 (ln ρ> − ln b) +
m=
Cm
ρ (^) <m ρ (^) >m
ρ (^) <m ρ (^) >m b^2 m
cos m(φ − φ′).
(c) Integrate the differential equation over the δ-function in ρ and show that
G = −2 ln ρ> + 2 ln b +
m=
m
ρm< ρm>
ρm< ρm> b^2 m
cos m(φ − φ′).
(d) Sum the series to obtain
G = ln
ρ^2 ρ′^2 b−^2 + b^2 − 2 ρρ′^ cos(φ − φ′) ρ^2 + ρ′^2 − 2 ρρ′^ cos(φ − φ′)