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):
$$V(a,\phi) = 2 \cos \phi$$
$$V(b,\phi) = 12 \sin \phi$$Taking $V(r,\phi)_{k=1}$ 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...