Relevant diagram is available here.

The circular disk of radius $a$ lies in the $xy$ plane and carries surface charge density of

$\sigma (s, \phi) = s^{2}cos\phi $,

where $(s,\phi)$ are in cylindrical co-ordinates.

The problem is to find potential at a point which is slightly displaced from the $z$ axis at position $ r = z \hat z + \delta s \hat s = z\hat z + \delta x \hat x + \delta x \hat y$, and

$ r' = s cos\phi \hat x + s sin\phi \hat y $

Since potential is given by

$V(r) = \frac{1}{4 \pi \epsilon_{0}} \int \frac{1}{\bf {r} - \bf {r'}} dq$

Here, $dq = \sigma da = \sigma dl_{s} dl_{\phi} = \sigma sdsd\phi = s^3 cos\phi dsd\phi$


${r-r'} = z\hat z + \delta x\hat x + \delta x \hat y - (s cos\phi \hat x + s sin\phi \hat y) $


$|r-r'| = \sqrt{z^2 + s^2 - 2\delta x s(cos\phi + sin\phi)} $.

It can be assumed that $(\delta x)^2 = 0$ since $\delta x$ is infinitesimally small.

This means that the final integral for potential is given by

$V(r) = \frac{1}{4 \pi \epsilon_0} \int_S \frac{s^3 cos\phi}{\sqrt{z^2 + s^2 - 2\delta x s(cos\phi + sin\phi)}} dsd\phi$.

Any suggestions on how to proceed with evaluating this integral would be very much appreciated.


  • $\begingroup$ you question is not totally clear try to simplify it a little bit $\endgroup$
    – Deiknymi
    Aug 11, 2013 at 14:53

2 Answers 2


The problem is that you're ignoring the angular dependence of your probe point $\mathbf r$, and that is messing with your integral. If your probe point has cylindrical coordinates $(s,\phi,z)$ and your integration variables are $(s',\phi')$, then the distance between the two is $$\frac 1 {|\mathbf r-\mathbf r'|}=\frac{1}{\sqrt{s^2-2ss'\cos(\phi-\phi')+s'^2+(z-z')^2}}$$ by the cosine rule (draw it!). If you put this into your integral it will no longer vanish.

(A few pointers on the new integral: the new dependence on $\phi'$ is a bit more complicated. The standard practice is to change variables to $\varphi=\phi'-\phi$. This will leave a simpler denominator, and a factor of $\cos(\varphi+\phi)=\cos(\varphi)\cos(\phi)-\sin(\varphi) \sin(\phi)$ on the numerator. One of the two terms will vanish and the other will yield to a change of variables to $u=\cos(\varphi)$.)

  • $\begingroup$ I will edit the main post to reflect some corrections. However your answer (while correct), I think slightly overcomplicates the problem. $\endgroup$
    – achacttn
    Aug 11, 2013 at 15:39
  • $\begingroup$ Well, take it or leave it. My answer is correct and your revision is not. You cannot assume $\delta x$ is small unless $r\gg a$, in which case you might as well take a point dipole for your disk. Your expression for $|r-r'|$ is also incorrect, but since you do not provide appropriate definitions for the symbols you are working with it is impossible to point out the exact error. $\endgroup$ Aug 11, 2013 at 15:59
  • $\begingroup$ I've tried to include as much relevant information as possible, could you please let me know which definitions or symbols I have made unclear, and which revisions are incorrect? Also, I'm not sure what you mean when you say that I can't assume $\delta x$ is small? Isn't that the point of taking an integral over the infinitesimally small area? $\endgroup$
    – achacttn
    Aug 11, 2013 at 16:04
  • 1
    $\begingroup$ Choose one coordinate system and stick to it: do not mix Cartesian and cylindrical systems. Always use primed/unprimed versions of the same symbol for $\mathbf r'$ and $\mathbf r$. Do not use dots like $\cdot$ or $\mathbf .$ unless you mean dot products. Do not abuse the bold font. Do not print the same symbol in different fonts. And be careful when taking dot products for norms! $\endgroup$ Aug 11, 2013 at 16:16

Looks like you calculate $\boldsymbol{r}-\boldsymbol{r'}$ incorrectly: your $\boldsymbol{r'}$ does not depend on $\phi$, whereas you should integrate over the entire disk.


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