# Potenial inside Wedge due to Line Charge

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## Homework Statement

We are given a wedge configuration with three sections. There are two conducting planes held at zero potential that make an angle of ##2\beta## with each other. They are connected by a portion of a conducting cylinder with radius ##a##. An infinite uniform line charge of ##\rho_l## is held at a distance of ##d## away from the center of the conducting cylinder. Use separation of variables to find an expression for the potential in the area between the conducting plates.

## Homework Equations

In the space between the plates, excluding the line charge, ##\nabla^2 \phi\left(r,\theta\right) = 0##. In cylindrical coordinates, the solution to Laplace's equation is:

\begin{align*}

\phi\left(r,\theta\right) &= \sum_{\alpha}R_{\alpha}\left(r\right)G_{\alpha}\left(\theta\right)

\end{align*}

given that ##\theta## is the angle measured from line connecting ##r=0## and ##r=d##. The full differential equation for this problem can be expressed as:

\begin{align*}

\left(\dfrac{\partial^2 }{\partial r^2} + \dfrac{1}{r}\dfrac{\partial }{\partial r} + \dfrac{1}{r^2}\dfrac{\partial^2 }{\partial \theta^2}\right)\phi\left(r,\theta\right) = -\dfrac{\rho_l}{\epsilon_0 r}\delta\left(r-d\right)\delta\left(\theta\right)

\end{align*}

The boundary conditions are:

\begin{align*}

&\phi\left(r,\theta = \frac{\pi}{2}-\beta\right) = 0\\

&\phi\left(r,\theta = \frac{\pi}{2} + \beta\right) = 0\\

&\phi\left(a, \frac{\pi}{2}-\beta <\theta < \frac{\pi}{2} + \beta\right) = 0

\end{align*}

## The Attempt at a Solution

In order to satisfy the boundary condition at ##\theta = \frac{\pi}{2}-\beta## and ##\theta = \frac{\pi}{2}+\beta##,

\begin{align*}

G_{\alpha}\left(\theta\right) = \cos\left(\alpha_m \theta\right) = \cos\left(\dfrac{m\pi}{2\beta}\theta\right), \mbox{ for } m=1,3,5,...

\end{align*}

This function is zero at the locations of the grounded plates. Hence,

Carrying out the derivative,

\begin{align*}

\left(\dfrac{\partial^2 }{\partial r^2} + \dfrac{1}{r}\dfrac{\partial }{\partial r} + \dfrac{1}{r^2}\dfrac{\partial^2 }{\partial \theta^2}\right)\phi\left(r,\theta\right)

&=\cos\left(\alpha_m \theta\right)\dfrac{d^2 R}{dr^2} + \cos\left(\alpha_m \theta\right)\dfrac{1}{r}\dfrac{dR}{dr} + \dfrac{R}{r^2}\left(-\alpha_m^2 \cos\left(\alpha_m \theta\right)\right) \\

&= -\dfrac{\rho_l}{\epsilon_0 r}\delta\left(r-d\right)\delta\left(\theta\right)

\end{align*}

Along the line connecting ##a## and ##d##, ##\theta = 0##

\begin{align*}

\dfrac{d^2 R}{dr^2} + \dfrac{1}{r}\dfrac{dR}{dr} - \dfrac{\alpha_m^2}{r^2} = -\dfrac{\rho_l}{\epsilon_0 r}\delta\left(r-d\right)

\end{align*}

Hence, for ##\alpha_m=0##,

\begin{align*}

R\left(r\right) = \begin{cases}

A\ln\left(r\right) + B, & r<d\\

C\ln\left(r\right) + D, & r>d

\end{cases}

\end{align*}

It's clear that the first of these equations is:

\begin{align*}

R\left(r\right) = -\dfrac{\rho_l}{2\pi\epsilon_0}\ln\left(\dfrac{r}{a}\right)

\end{align*}

For ##\alpha_m^2 \neq 0##,

\begin{align*}

R\left(r\right) = A_m\left[\left(\dfrac{r}{a}\right)^{\alpha_m} - \left(\dfrac{a}{r}\right)^{\alpha_m}\right]

\end{align*}

This function satisfies Laplace's equation and is equal to 0 at ##r=a##.

My question is: how can I find ##A_m##? I have no other boundary conditions to satisfy...

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