Electric potential in a sector between two planes

[phibonacci]

Homework Statement

Two conducting planes meet at the origin at an angle $$\beta$$. Solve Laplace's equation in polar coordinates under the boundary condition that the potential is $$V_0$$ on the planes. Calculate the components of the electrical field near the origin. Explain how the charge density behaves at the following values of $$\beta$$: $$\pi/4,\pi/2,3\pi/2$$.

Homework Equations

The Attempt at a Solution

I placed the x-axis so that the angles between the x-axis and the planes are $$\beta/2$$ and $$-\beta/2$$ since this appears to give rise to the most symmetry. From what I know, the solution to the laplace equation should have the form

$$\begin{equation} V(r,\theta) = \sum (A_nr^n+B_nr^{-n})(C_nsin(n\theta)+D_ncos(n\theta)+A_0ln(\frac{r}{r_0})(C_0\theta +D_0) \end{equation}.$$

I changed the last term by writing $$ln(\frac{r}{r_0})=ln(r)-ln(r_0)$$ and renaming some constants, so I get

$$\begin{equation} V(r,\theta)=\sum_{n=1}^{\infty} (A_nr^n+B_nr^{-n})(C_nsin(n\theta)+D_ncos(n\theta)+ln(r)(C\theta+D)+k\theta+d \end{equation}.$$
For symmetry reasons we should have $$V(r,-\theta)=V(r,\theta)$$ which seems to imply $$C_n=C=k=0$$.
Now the solution should be
$$\begin{equation} V(r,\theta)=\sum_{n=1}^{\infty} (K_n r^n + L_n r^{-n})cos(n\theta)+Dln(r)+d \end{equation}.$$
Because the potential should be continuous near the origin, we must have $$L_n=0=D$$ so the solution should be
$$V(r,\theta)=\sum_{n=1}^{\infty} K_n r^n cos(n\theta)+d$$.
But this solution seems very strange. I don't see how it could be constant on the planes, unless the potential is constant. But then there should be no electric field, so the other questions in the problem do not make sense.

 

[AvroArrow]

Did I make a mistake in my solution? If so, what did I do wrong? It would be much appreciated if anyone could help me out.