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Potenial inside Wedge due to Line Charge

  1. Dec 8, 2014 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant 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*}

    3. 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...
     
    Last edited: Dec 8, 2014
  2. jcsd
  3. Dec 13, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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