Potenital of a cylindrical, linear dialectric in a E-Field

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Homework Statement


A very long cylinder of linear dialectric material is placed in an otherwise uniform electric field ##E_0##. Find the resulting field within the cylinder. (The radius is ##a##, the susceptibility ##X_e##, and the axis is perpendicular to ##E_0##)

Homework Equations


Boundary conditions
##V_i = V_o## at ##s=a##

##\epsilon \frac{\partial V_i}{\partial n} = \epsilon_0 \frac {\partial V_o}{\partial n}## because there is no free charge

##V = -E_0*s*cos(\theta)## for ##r>>a##

The Attempt at a Solution


I've gotten this far

##\epsilon_r[-na^{-n-1}* (Acos(n*\theta) + Bsin(n*\theta))] = na^{n-1} * (Ccos(n*\theta) + Dsin(n*\theta)) -E_0cos(\theta)##

where n is what I'm summing over. I'm unsure of how to restrict the constants of the trig functions to reduce the problem.
 
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This equation isn't correct, as written. There should be a summation over ##n## on each side (except for the term ##-E_0 \cos\theta ~##, which is not summed over ##n##).

For each value of ##n##, the coefficient of ##\cos(n \theta)## on the left side must match the coefficient of ##\cos(n \theta)## on the right side. Similarly for ##\sin(n \theta)##.

So, what condition on the coefficients do you get when matching ##\cos(n \theta)## for the special case of ##n = 1##? What condition on the coefficients do you get when matching ##\cos(n \theta)## for ##n \neq 1##?

Similarly, get equations from matching ##\sin (n\theta)##.

There is also another boundary condition on ##V## (which you stated in the relevant equations section) that you need to invoke where you can again match coefficients on each side.
 
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