## Laplace’s equation – cylindrical shell with opposing charges

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Chapter 3, Post 39.

As another example of applying the solution to Laplace’s equation in cylindrical coordinates, we consider the following problem. We are given a cylindrical non-conducting shell or radius ${R}$ carrying a charge density of ${\sigma_{0}}$ on its upper half (${-\pi<\phi<0}$) and ${-\sigma_{0}}$ on its lower half (${0<\phi<\pi)}$. Find the potential everywhere.

We can begin in the same manner as for the other problem involving a cylindrical shell that we solved earlier. The solution is the same up until the point where we introduce the surface charge.

Thus, outside the shell, we have

$\displaystyle V_{out}=B_{out}+\sum_{n=1}^{\infty}\left[\frac{D_{n}}{r^{n}}\cos n\phi-\frac{C_{n}}{r^{n}}\sin n\phi\right]$

Inside the shell, we have

$\displaystyle V_{in}=B_{in}+\sum_{n=1}^{\infty}\left[A_{n}r^{n}\sin n\phi+B_{n}r^{n}\cos n\phi\right]$

Since the potential is continuous over a surface charge, we must have ${V_{out}(R)=V_{in}(R)}$, so we get

$\displaystyle B_{out}+\sum_{n=1}^{\infty}\left[\frac{D_{n}}{R^{n}}\cos n\phi-\frac{C_{n}}{R^{n}}\sin n\phi\right]=B_{in}+\sum_{n=1}^{\infty}\left[A_{n}R^{n}\sin n\phi+B_{n}R^{n}\cos n\phi\right]$

Equating coefficients of the sine and cosine, we get

 $\displaystyle B_{out}$ $\displaystyle =$ $\displaystyle B_{in}$ $\displaystyle C_{n}$ $\displaystyle =$ $\displaystyle -A_{n}R^{2n}$ $\displaystyle D_{n}$ $\displaystyle =$ $\displaystyle B_{n}R^{2n}$

The outward derivative of the potential is discontinuous across a surface charge, and we have

$\displaystyle \left.\frac{\partial V}{\partial r}\right|_{out}-\left.\frac{\partial V}{\partial r}\right|_{in}=-\frac{\sigma}{\epsilon_{0}}$

Plugging in the formulas for ${V_{out}}$ and ${V_{in}}$, we get

$\displaystyle \sum_{n=1}^{\infty}\left[\frac{-nR^{2n}A_{n}}{R^{n+1}}-nR^{n-1}A_{n}\right]\sin n\phi+\sum_{n=1}^{\infty}\left[\frac{-nR^{2n}B_{n}}{R^{n+1}}-nR^{n-1}B_{n}\right]\cos n\phi=\begin{cases} \frac{\sigma_{0}}{\epsilon_{0}} & -\pi<\phi<0\\ -\frac{\sigma_{0}}{\epsilon_{0}} & 0<\phi<\pi\end{cases}$

Since the surface charge is an odd function of ${\phi}$, we can eliminate the cosine terms, since the cosine is an even function. Therefore, we have ${B_{n}=D_{n}=0}$. We are free to choose the potential at infinity to be any constant, so we might as well take it to be zero, in which case we have ${B_{in}=B_{out}=0}$. We are therefore left with, after simplifying the term in brackets:

$\displaystyle -2\sum_{n=1}^{\infty}nR^{n-1}A_{n}\sin n\phi=\begin{cases} \frac{\sigma_{0}}{\epsilon_{0}} & -\pi<\phi<0\\ -\frac{\sigma_{0}}{\epsilon_{0}} & 0<\phi<\pi\end{cases}$

To find the ${A_{n}}$, we can use the fact that the set of ${\sin n\phi}$ functions is orthogonal over the interval ${[-\pi,\pi]}$. That is

$\displaystyle \int_{-\pi}^{\pi}\sin m\phi\sin n\phi d\phi=\begin{cases} 0 & n\ne m\\ \pi & n=m\end{cases}$

Therefore we can multiply both sides by ${\sin m\phi}$ and integrate to get

 $\displaystyle -2\pi mR^{m-1}A_{m}$ $\displaystyle =$ $\displaystyle \frac{\sigma_{0}}{\epsilon_{0}}\left[\int_{-\pi}^{0}\sin m\phi d\phi-\int_{0}^{\pi}\sin m\phi d\phi\right]$

The integrals in brackets on the right come out to

$\displaystyle \left[\int_{-\pi}^{0}\sin m\phi d\phi-\int_{0}^{\pi}\sin m\phi d\phi\right]=\begin{cases} 0 & m\;\mathrm{even}\\ -\frac{4}{m} & m\;\mathrm{odd}\end{cases}$

Therefore (changing the index from ${m}$ to ${n}$ for convenience):

$\displaystyle A_{n}=\begin{cases} 0 & n\;\mathrm{even}\\ \frac{2\sigma_{0}}{\pi\epsilon_{0}n^{2}R^{n-1}} & n\;\mathrm{odd}\end{cases}$

We thus get

$\displaystyle C_{n}=\begin{cases} 0 & n\;\mathrm{even}\\ -\frac{2\sigma_{0}R^{n+1}}{\pi\epsilon_{0}n^{2}} & n\;\mathrm{odd}\end{cases}$

The final formula for the potential is

$\displaystyle V(r,\phi)=\begin{cases} \frac{2\sigma_{0}}{\pi\epsilon_{0}}\sum_{n\; odd}^{\infty}\frac{r^{n}}{n^{2}R^{n-1}}\sin n\phi & rR\end{cases}$

• iman  On Friday, 22 June 2012 at 07:14

greens theorem would work too.

• Anonymous  On Thursday, 31 October 2013 at 01:12

shouldn’t dV/dr = -nR^(2n)/(R^(n-1)) not R^(n+1)?

• growescience  On Thursday, 31 October 2013 at 08:33

The derivative is

$\displaystyle \frac{\partial V_{out}}{\partial r}=\sum_{n=1}^{\infty}\left[-n\frac{D_{n}}{r^{n+1}}\cos n\phi+n\frac{C_{n}}{r^{n+1}}\sin n\phi\right]$
so I think the formula is correct as it stands. Remember that

$\displaystyle \frac{d}{dr}\left(\frac{1}{r^{n}}\right)=-n\frac{1}{r^{n+1}}$