An insulated disk, uniform surface charge density $\sigma$, of radius $R$ is laid on the $x,y$ plane. Deduce the electric potential $V(z)$ along the z-axis. Next consider an off axis point $p'$, with distance $\rho$ from the center, Making an angle $\theta$ with the z-axis. Expand the potential at $p'$ in terms of Legendre polynomials $P_l(\cos\theta)$ for $\rho < R$ and $\rho > R$

For the point on the z-axis, this is pretty easy. The differential Voltage from a differential ring of charge with radius $r$ is:

$$dV = \frac{1}{4 \pi \epsilon_o} \frac{dq}{ \mathscr{R}}$$

$$dq = \sigma dA = \sigma 2 \pi r dr$$

$$\mathscr{R} = \sqrt{r^2 + z^2}$$

$$ \Delta V(z) = \frac{ \sigma}{2 \epsilon_o}\int_0^R \frac{ r dr}{\sqrt{r^2 + z^2}} = \frac{ \sigma}{2 \epsilon_o} \left( \sqrt{R^2 + z^2} - |z| \right)$$

Which is obtained by using a U substitution.

As for the second part, The only thing that changes is the distance from the differential of charge and the point of interest so I have:

$$dV = \frac{ \sigma}{2 \epsilon_o} \frac{r dr}{ \mathscr{R}}$$

But now using the law of cosines, I use the angle between $r$ and $\mathscr{R}$, Note: this is not the angle recommended in the problem. Thus $\mathscr{R} = (r^2 + p^2 - 2rp\cos \phi)^{1/2} = r(1 - 2 \frac{p}{r}cos \phi + \frac{p^2}{r^2})^{1/2},$ using spherical polar coordinates, where $p$ is the distance from origin to the point of interest, $p'$.

This is the generating function of the Legendre polynomials,

$$\therefore \frac{1}{\mathscr{R}} = \frac1r G\left( \frac{p}{r}, \cos \phi\right)$$

$$dV = \frac{ \sigma}{2 \epsilon_o} G\left( \frac{p}{r}, \cos \phi\right) dr = \frac{ \sigma}{2 \epsilon_o} \sum_{l = 0} ^{\infty} p_l(\cos \phi) \left( \frac{p}{r} \right)^l dr$$

Okay, so my question is this, assuming all of this is correct (which I believe is not) How would possibly integrate this? Is it as simple as $\int_0^R \left( \frac{p}{r} \right)^l dr$? This creates an infinity. Any help would save me so very much.

  • $\begingroup$ Did you manage to get the answer to this? I'm writing a simulation involving charged disks and this would be really helpful. $\endgroup$ Commented May 6, 2013 at 16:23
  • $\begingroup$ Hi @Josh, and welcome to Physics Stack Exchange! If you look below, you'll see an answer with a green checkmark which indicates that it answered the question to Cactus BAMF's satisfaction. If it doesn't work for you, you're welcome to ask a followup question. $\endgroup$
    – David Z
    Commented May 6, 2013 at 18:07

1 Answer 1


Your formula for the generating function is wrong in a crucial sense. The formula you are after reads $$ \frac{1}{|\mathbf{r}-\mathbf{r}'|}=\sum_{l=0}^\infty \frac{r_<^l}{r_>^{l+1}}P_l(\cos\theta). $$ Note that the numerator and denominator of each term are powers of the lesser and greater, resp., of $r$ and $r'$. For the multipolar expansion your question asks about, you need the point of evaluation to be further away from the centre than the disk radius $R$, which means that the powers will be in $r/p$ instead of $p/r$. That will solve the divergence issues on the integrals.

I find the key to quickly seeing which way the expansion will go is seeing it as a Taylor series (for fixed $\theta$) in the relevant small paramenter.


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