# Spherical capacitor and vector potential

1. Feb 7, 2014

### ShayanJ

There is a spherical capacitor with inner and outer radii of a and b respectively,has a dielectric material of small conductivity $\sigma$ between its concentric conducting spheres.Calculate vector potential and magnetic field arising from this configuration.
This is how I did it:
$\nabla^2\phi=0 \Rightarrow \phi=-\frac{C}{r}+B$
$R=\int_a^b \frac{dr}{4\pi \sigma r^2}=\frac{1}{4\pi\sigma}(\frac{1}{a}-\frac{1}{b})$
$I=\frac{\Delta \phi_{ab}}{R}=4\pi\sigma C$
$\vec{J}=-\frac{\sigma C}{r^2}\hat{r}$
$\vec{A}=\frac{\mu_0}{4\pi}\int_a^b \int_0^\pi \int_0^{2\pi} (-\frac{\sigma C}{r'^2}\hat{r})\frac{ r'^2 \sin{\theta'}d\varphi' d\theta' dr' }{ \sqrt{ r^2+r'^2-2rr' \cos{\gamma} } }$
Then using Legendre functions and their addition theorem and some manipulations,I've reached to the following:
$\vec{A}=-\frac{\sigma\mu_0 C}{r} \sum_{n=0}^\infty \sum_{m=-n}^n \frac{1}{2n+1}\frac{1}{r^n} Y^m_n(\theta,\varphi) \int_a^b r'^n dr' \int_0^\pi \int_0^{2\pi} \hat{r} Y^{m*}_n(\theta',\varphi')\sin{\theta'} d\varphi' d\theta'$
Then solved the integral above to get:
$\vec{A}=\frac{\sigma \mu_0 C(b^2-a^2)}{12r^2} \sin{\theta} e^{i\varphi}[(\hat{x}-i\hat{y})+2\hat{z}]$
I'm just skeptical about the result.I tried to find a similar calculation somewhere but found nothing.
Is it correct?

Last edited: Feb 7, 2014