Finding the Potential Between Two Coaxial Cylinders Using Laplace's Equation

[cscott]

Homework Statement

Two coaxial cylinders, radii {a,b} where b>a. Find the potential between the two cylinder surfaces.

Boundary conditions:
V(a,\phi) = 2 \cos \phi
V(b,\phi) = 12 \sin \phi

Homework Equations

Solution by separation of variables:
V(r,\phi) = a_0 + b_0 \ln s + \sum_k \left[ r^k(a_k \cos k\phi + b_k \sin k\phi)+r^{-k}(c_k\cos k\phi + d_k \sin k\phi)\right]

The Attempt at a Solution

I don't think I can eliminate the r^{-k} term because the origin isn't between the two cylinders.

I think k=1 is the only term in the summation that is required for the solution.

V(r,\phi) = r(a_1 \cos \phi + b_1 \sin \phi)+\frac{1}{r}(c_1\cos \phi + d_1 \sin \phi)

I don't see how to have the cosines vanish for V(b) and sines vanish for V(a) because of the common k in both.

 

[fzero]

You should actually write down the boundary conditions for k=1. There are nontrivial solutions.

 

[cscott]

I made a typo in my boundary conditions

Boundary conditions (in volts):
V(a,\phi) = 2 \cos \phi
V(b,\phi) = 12 \sin \phiTaking V(r,\phi)_{k=1} gives,
V(r,\phi) = r(a_1 \cos \phi + b_1 \sin \phi)+\frac{1}{r}(c_1\cos \phi + d_1 \sin \phi)

I will take a look at this...