# Electrostatic Potential in a Hollow Cylinder

An electrostatic field $E$ (i.e. an electric field produced only by a static charge) is a conservative field, i.e. there exists a scalar potential $\psi$ such that $E=-\nabla\psi$. This is clear from Maxwell’s equations. Since there is no change of the magnetic field $B$ in time, $\nabla\times E=0$. If there is no charge present in a region, $\nabla\cdot E=0$. Together with $E=-\nabla\psi$, we obtain the Laplace equation $\nabla^2\psi=0$. Thus the Laplace equation can be used to find the electrostatic potential $\psi(\rho,\varphi,z)$ in a hollow cylinder with radius $a$ and height $l$ ($0\leq z\leq l$).

Using the separation of variables, we find the mode
\begin{align*}
\psi_{km}(\rho,\varphi,z)&=P_{km}(\rho)\Phi_m(\varphi)Z_k(z)\\
&=J_m(k\rho)[a_m\sin m\varphi+b_m\cos m\varphi][c_1e^{kz}+c_2e^{-kz}].
\end{align*}
The boundary conditions are:
$$\psi(\rho,\varphi,l)=\psi(\rho,\varphi),$$
where $\psi(\rho,\varphi)$ is a potential distribution. Elsewhere on the surface $\psi=0$. Now we find electrostatic potential
$$\psi(\rho,\varphi,z)=\sum_{k,m}\psi_{km}$$
inside the cylinder. From the boundary condition $\psi(\rho,\varphi,0)=0$, we find $c_1+c_2=1$. So we choose $c_1=-c_2=\frac{1}{2}$ and thereby $c_1e^{kz}+c_2e^{-kz}\sinh kz$. Since $\psi=0$ on the lateral surface of the cylinder, $\psi(a,\varphi,z)=0$. This implies that $J_m(ka)=0$. If we write the $n$-th Bessel zero as $a_{mn}$, then $k_{mn}a=a_{mn}$ or $k_{mn}=\frac{a_{mn}}{a}$. Hence,
$$\psi(\rho,\varphi,z)=\sum_{m=0}^\infty\sum_{n=1}^\infty J_m\left(\alpha_{mn}\frac{\rho}{a}\right)[a_m\sin m\varphi+b_m\cos m\varphi]\sinh\left(\alpha_{mn}\frac{z}{a}\right).$$
Finally using the boundary condition
$$\psi(\rho,\varphi)=\sum_{m=0}^\infty\sum_{n=1}^\infty J_m\left(\alpha_{mn}\frac{\rho}{a}\right)[a_m\sin m\varphi+b_m\cos m\varphi]\sinh\left(\alpha_{mn}\frac{1}{a}\right)$$ and the orthogonality of $\sin m\varphi$ and $\cos m\varphi$, we can determine the coefficients $a_m$ and $b_m$ as
\begin{align*}\left\{\begin{aligned}a_{mn}\\b_{mn}\end{aligned}\right\}=\frac{2}{\pi a^2\sinh\left(\alpha_{mn}\frac{1}{a}\right)J_{m+1}^2(\alpha_{mn})}\int_0^{2\pi}\int_0^a\psi(\rho,\varphi)&J_m\left(\alpha_{mn}\frac{\rho}{a}\right)\\
&\left\{\begin{aligned}
\sin m\varphi\\
\cos m\varphi
\end{aligned}\right\}\rho d\rho d\varphi.\end{align*}

## One thought on “Electrostatic Potential in a Hollow Cylinder”

1. hancy

This is solution for the electrostatic potential in a hollow cylinder.
What is going to change if we want to find the el. potential outside the cylinder?