I little bit confused while I'm trying to convert a cylindrical charge distribution to spherical. The question is:

A uniformly charged thin disk of radius $a$ and surface charge density $\sigma$ placed at the $xy$-plane with its center at the origin. The disk is enclosed in a grounded conducting sphere of radius $b (b>a)$. Using the appropriate Green’s function, find the potential for all points inside the sphere.

First I need to convert that cylindrical distribution to spherical


$f(r)$ for r dependence and $\Theta$ is heavyside func. I'll integrate it over volume in spherical coordinates and equalize to $\sigma$$\pi$$a^2$ to decide $\rho(x).$

Now, am I have to add $1/r$ because of the metric from the beginning? And then that question came to my mind, what if I put a sphere with radius $a$ with $\rho $ inside a cylinder wiyh radius $b$, how can I convert this?


1 Answer 1


The important thing you need to take into account is

$$ \int{\rm d}^3\mathbf{r}\;\rho(\mathbf{r}) = \int{\rm d}^2\mathbf{r}\;\sigma(\mathbf{r}) \tag{1} $$

I am going to call $x = \cos \theta$ so that the disk is located at $x = 0$, and then Eq. (1) becomes

\begin{eqnarray} \int{\rm d}^2\mathbf{r}\;\sigma(\mathbf{r}) &=& \int{\rm d}r{\rm d}\phi\; r\sigma(r,\phi) = \int {\rm d}r{\rm d}\phi\; r\sigma(r,\phi) \underbrace{\int {\rm d}x\delta(x)}_{=1} \\ &=& \int{\rm d}r{\rm d}\phi{\rm d}x\; r^2\frac{1}{r}\sigma(r,\phi)\delta(x) \\ &\stackrel{(1)}{=}& \int{\rm d}r{\rm d}\phi{\rm d}x\; r^2\rho(r, x, \phi) = \int{\rm d}^3\mathbf{r}\;\rho(\mathbf{r}) \end{eqnarray}

You can then conclude that

$$ \rho(r,\theta,\phi) = \frac{1}{r}\sigma(r,\phi)\delta(\cos \theta) $$


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