Okay, one statement from Purcell's book goes like:

[...]Thus, we find the potential at $P_2$: $$\phi = \frac{1}{4\pi\epsilon_0} \int_{-\pi/2}^{\pi/2} 2\sigma\cos\theta d\theta = \frac{\sigma a}{\pi\epsilon_0}$$. Comparing this with $\sigma a/ 2\epsilon_0$, the potential at the center of the disk, we see that, as we should expect, the potential falls off the center to the edge of the disk. The electric field therefore, must have an outward component in the plane of the disk. That is why, we remarked earlier that the charge , if free to move, would redistribute itself towards the rim. To put it another way, our uniformly charged disk is not a surface of constant potential, which any conducting surface must be unless charge is moving.

Actually Mr. Purcell after deducing the potential at the rim of the uniformly-charged disk, which is an insulator, is telling that if it were a conductor-disk, then there would be a redistribution of the charge towards the rim of the disk as the potential is lesser there & electric field facing in that direction from the center towards the rim. Really? But how?? The field is created by the charges, right? Even he deduced the potential of the rim by integrating the contribution from all the charges in the disk. Then how, if the disk were a conductor, could the charge move towards the rim in response to, although-outward-electric field? It seems ridiculous as the electric field is created by the charges & of course the charges can't get bothered by their own field, isn't it??

But it is true that in a conducting-disk, the charges remain more dense at the rim. Then he seems to be right:( But I am not getting the logic how the charges redistribute in response of the field they created? Can anyone please explain & help me sort out the confusion? Thanks.


1 Answer 1


Let's look at the simplest case: two like charges a distance d away. If free to move, they move away from each other and away from the center.

Now imagine three like charges on the corners of an equilateral triangle, if free to move they also move away from the center.

Now imagine four like charges on the corners of a square, if free to move they also move away from the center.

Similarly you can imagine n like charges on the corners of a regular n-gon, if free to move they also move away from the center.

Why? Each charge feels the force from all the other charges.

So if you have a uniformly charged disk each ring feels a push outwards from itself (like out example above each charge feeling all the other charges in the ring) and from the rings closer in. The rings farther out are more complicated since some parts push it in and other parts push it out.

Charges aren't on a sports team; they don't pick and choose which other charges to feel forces from; they feel a force from every other charge in the universe that depends just on where the other charge was and how it was moving (velocity and acceleration).

So in your disk, each charge feels the potential due to every charge.

If it were a conductor, the charge inside would be free to move & so it would move towards outside. The force the charges feel isn't because of some potential difference with infinity; it is more like a local effect based on how the potential is changing right where the charge is. Eventually when the charge is all on the rim there is no electric field inside and in fact the field inside the disk gets smaller and smaller as more and more charge moves to the rim. Eventually the potential is constant throughout the disk and only changes outside the disk.

  • $\begingroup$ First of all, a BIG thanks for responding! +1 for that!! Now am starting to read...... $\endgroup$
    – user36790
    Aug 11, 2015 at 16:08
  • $\begingroup$ I've pondered greatly but still couldn't digest one point: yes, you are right in your argument that the charges flow away from them if not being restrained. Coming to the disk, the potential at the rim is lower than that of the center as Purcell deduced & as you said, the charges would flow outwards but really, as Purcell said, is it due to the potential difference between that of rim & center?? $\endgroup$
    – user36790
    Aug 13, 2015 at 1:58
  • $\begingroup$ How can it be? Take a charge, for example, somewhere between center & rim. So, when I deduced the potential at the center & the rim, I've integrated over all the charges which means our charge also contributed in the potential of the center & the rim, isn't it? I'am greatly convined with your argument & also accepting that if it were a conductor, the charge would flow towards the rim. But is it really due to the potential difference between that of rim & that of the location where the charge is located? Yes, the charge wouldn't contribute to the potential at the point it is located but... $\endgroup$
    – user36790
    Aug 13, 2015 at 2:11
  • 2
    $\begingroup$ @user36790 The idea is that in a macroscopically sized disk the potential is due to many charges so the effect of just one elementary charge is small. You also deal with continuous charge distributions which aren't perfectly realistic, you look at the average over a small sphere and map how that average changes as you move the center. If you make the sphere large enough that it doesn't jump too much when you move it so a change comes in or out that is good, but you want the radius of the sphere to be small enough that you can resolve real changes in how densely the charge is packed. $\endgroup$
    – Timaeus
    Aug 13, 2015 at 2:26
  • 1
    $\begingroup$ @user36790 It is the same as with mass density. You don't want it to jump up at every nucleus and then down again for the space between each nucleus. So since you make the radius of the moving average large enough to contain enough charge that one electron isn't a huge deal, then the electron itself isn't a huge deal. You can react to the potential due to every other charge and whether you include that one charge out of many doesn't change it much. Of course near that one charge it contributes a lot but on the average over the sphere it isn't changing it that much. $\endgroup$
    – Timaeus
    Aug 13, 2015 at 2:29

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