Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
    
Did you manage to get the answer to this? I'm writing a simulation involving charged disks and this would be really helpful. –  Josh Calahan May 6 '13 at 16:23
    
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. –  David Z May 6 '13 at 18:07
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.