An insulated disk, uniform surface charge density $\sigma$, of radius R is laid on the xy 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.

$\mathscr{R} = (r^2 + p^2 - 2rp\cos \phi)^{1/2} = r(1 - 2 \frac{p}{r}cos \phi + \frac{p^2}{r^2})^{1/2}$

Using Spherical Polar coordinates, where $p =$ distance from origin to point of interrest p'

This is the Generating function of the Legendre polynomials

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

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

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.

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  • $\begingroup$ Why the $\Delta V$? What is $V$ relative to? (i.e., what is your ground potential?) I know this isn't a typical worry in doing these problems, but your notation is all over the place, so it's hard to see if you really have mastery over the subject matter. Your solution to the 1st part looks OK, just figure out what quantity the function represents. Also, what makes an angle with the $z$-axis? $\endgroup$ – Ron Gordon Jan 9 '13 at 0:30
  • $\begingroup$ Relative to V($ \infty$) = zero $\endgroup$ – Cactus BAMF Jan 9 '13 at 0:48
  • $\begingroup$ As is customary in Electrostatics $\endgroup$ – Cactus BAMF Jan 9 '13 at 0:49
  • $\begingroup$ I agree (I am a physicist, too). In the Math section, I would use a little more care in defining terms. Not everyone who can possibly help you is a physicist who understands that you mean $V$ when you write $\Delta V$. $\endgroup$ – Ron Gordon Jan 9 '13 at 0:52

There are three variables involved, and it's important not to mix them up, or you'll go astray. There's the distance from the origin to the field point, call this $r$. There's the distance from the point on the surface of the disc being integrated to the field point, call this $\mathscr{R}$. And there's the distance from the origin on the disc to the point being integrated, call this $r'$. Assuming $\sigma$ is not a function of $r'$ the last equation will then look like:

$\frac{ \sigma}{4\pi \epsilon_o}\frac{1}{r} \sum_{l = 0} ^{\infty} p_l(\cos \phi) \left( \frac{r'}{r} \right)^l dt$,

for a general surface or volume element $dt$. You are integrating with respect to $r'$, so the $r$ comes outside the integral and you get (in polar coordinates):

$\frac{ \sigma}{4\pi \epsilon_o} \sum_{l = 0} ^{\infty} \frac{1}{r^{1+l}}\int p_l(\cos \phi) \left( r' \right)^l r'dr'd\phi.$

It's then just a matter of "pulling out" as many terms as you like, like:

$\frac{ \sigma}{4\pi \epsilon_or}\int r'dr' + \frac{ \sigma}{4\pi \epsilon_o r^2}\int r'^2\cos(\phi)dr'd\phi...$

to get an approximation for the potential to any accuracy you desire.

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  • $\begingroup$ That was beautiful. thank you $\endgroup$ – Cactus BAMF Jan 9 '13 at 2:13

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