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 & r<R\\ \frac{2\sigma_{0}}{\pi\epsilon_{0}}\sum_{n\; odd}^{\infty}\frac{R^{n+1}}{n^{2}r^{n}}\sin n\phi & r>R\end{cases}


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  • 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}}

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