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PHY 6346 Fall 2008 HW #5: Electrostatic Potential of Hollow Conducting Cylinder, Assignments of Physics

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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Pre 2010

Uploaded on 03/11/2009

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PHY 6346 Fall 2008
Homework #5, Due Wednesday, October 1
1. A long, hollow conducting cylinder of radius bis centered on the z-axis. The right (x>0)
and left (x<0) halves of the cylinder are separated by small insulating gaps and are kept
at potentials ±V0respectively.
(a) Use separation of variables to show that the general two-dimensional solution to 2Φ=0
in cylindrical coordinates (ρ, φ) can be written as Φ = R(ρ)Q(φ), with R(ρ)=ρ±mand
Q(φ)=e±imφ,wheremis zero or a positive integer. (For m=0,“ρ0”meanslnρ).
(b) Fix the coefficients to match the given potential at ρ=band 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)=2V0
πtan
12 cos φ
b2ρ2.
What is the solution for ρ>b?
(d) Calculate and sketch the surface charge density on the cylinder.
2. Construct the 2-d Green’s function (∂/∂z = 0) for the interior of a cylinder of radius b:
(a) Show that
δ(φφ0)= 1
2π
X
m=−∞
eim(φφ0)=1
2π"1+2
X
m=1
cos m(φφ0)#.
(b) From the differential equation
02G=4πδ(ρρ0)
ρδ(φφ0)
we know that 02Gvanishes almost everywhere. Expand Gin one series G<for ρ0
and another G>for ρ0. Apply boundary conditions: that G=0atρ0=b;thatGis
nonsingular at ρ00; and that Gis smooth at ρ0=ρ,
G=C0(ln ρ>ln b)+
X
m=1
Cmρm
<
ρm
>
ρm
<ρm
>
b2mcos m(φφ0).
(c) Integrate the differential equation over the δ-function in ρand show that
G=2lnρ>+2lnb+
X
m=1
2
mρm
<
ρm
>
ρm
<ρm
>
b2mcos m(φφ0).
(d) Sum the series to obtain
G=lnρ2ρ02b2+b22ρρ0cos(φφ0)
ρ2+ρ022ρρ0cos(φφ0).

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PHY 6346 Fall 2008 Homework #5, Due Wednesday, October 1

  1. A long, hollow conducting cylinder of radius b is centered on the z-axis. The right (x > 0) and left (x < 0) halves of the cylinder are separated by small insulating gaps and are kept at potentials ±V 0 respectively.

(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) =

2 V 0

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.

  1. Construct the 2-d Green’s function (∂/∂z = 0) for the interior of a cylinder of radius b:

(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′)

]