Consider two very large plates, one of which is lying on the $xz$ plane and sitting at a potential $+V$, and the other forming an angle $α$ with the aforementioned plate, attached to the other at the $z$ axis, only this one is sitting at a potential of $-V$.

My aim is to find the potential at every point in space for 3 cases, where the angle $\alpha$ is $\pi$, $\pi/2$, $\pi/4$. I started from the first case. The problem has cylindrical symmetry, so my idea was to use cylindrical coordinates. The general solution to the laplace equation in cylindrical coordinates with cylindrical symmetry is well known of course: enter image description here

However, I have major trouble with the conditions. Consider the case where $\alpha = \pi$. My first instinct would be ditching the terms with $\ln(s)$ and $1/s$ so as not to have trouble at the origin. That's all well and good, but then when I try to satisfy the condition that $V(s, 0) = V$, I'm forced to ditch all the Am's too, but then there is no way to make it so that $V(s, \pi) = -V$. Any ideas?


1 Answer 1


You general solution is inapplicable as the potential is not required to be $2\pi$ periodic in the polar angle $\theta$.

If you have a potential that is zero at at $\theta=0$ and at $r=R$, but takes proscribed values along as a function of $r$ along $\theta=\alpha$ then in the wedge between $0$ and $\alpha$ the potential is of the form $$ \varphi(r,\theta)= \int_0^\infty a(\nu)\sin(\nu \ln(r/R))\frac{\sinh (\nu\theta)}{\sinh (\nu \alpha)}\, d\nu. $$ You can relate $a(\nu)$ to the boundary data via the Mellin sine transformation.

If you do not specify that$\varphi$ is zero at some $R$ then you do not, in principle, have enough data to solve Laplace's equation as you do not have a data on the entire boundary. There is however the obvious solution $$ \varphi(r,\theta) = \frac{2V}{\alpha}(\theta - \alpha/2), $$ which satisfies Laplace equation and your boundary condition.

There is a brief discussion of this problem on page 235 my book with Paul Goldbart.

  • $\begingroup$ OOOH I see, that makes sense. However this was a question in a class where the Mellin transform hasn't been taught to us yet. Kind of a weird thing to ask then... $\endgroup$
    – Andreas C
    Jan 16, 2021 at 13:53

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