1
$\begingroup$

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

and

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

Therefore,

$|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.

Thanks.

$\endgroup$
  • $\begingroup$ you question is not totally clear try to simplify it a little bit $\endgroup$ – Dimensionless Aug 11 '13 at 14:53
1
$\begingroup$

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 $u=\cos(\varphi)$.)

$\endgroup$
  • $\begingroup$ 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 '13 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. $\endgroup$ – Emilio Pisanty Aug 11 '13 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 '13 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! $\endgroup$ – Emilio Pisanty Aug 11 '13 at 16:16
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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