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I understand that every charged conductor is an equipotential, with electric field zero inside the conductor, and all excess charge on the surface of the conductor.

I am a bit confused as to how the excess charges on the conductor redistribute themselves so as to maintain the same potential throughout.

Consider this basic example: Suppose I have a conducting shell (in 2 dimensions to keep the math simple), shaped exactly like the unit circle with radius one in the $x,y$ plane.

Suppose I place exactly 2 identical point positive charges $q_1, q_2$, with each having charge q, on the conductor. My intuition tells me, they will try to get as far from each other as possible, so they will go to opposite ends of the circle (shown as red dots in the figure).

enter image description here $$$$ Consider the point A, the potential at A is $$V_A = \frac{kq}{\sqrt{2}} + \frac{kq}{\sqrt{2}} = kq\sqrt{2}$$

Now consider the point B which is extremely close to charge $q_1$, and at an approximate distance of 2 from $q_2$. The potential at B should be extremely large because if $r$, the distance from $q_1$, goes to zero, the fraction $\frac{kq}{r}$ gets arbitrarily large.

So how can the potential be constant throughout? It seems like it changes throughout the circle?

This question can easily be generalized to the more realistic of a 3 dimensional spherical shell. But this basic case demonstrates the root of my confusion.

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  • $\begingroup$ There is induced charge at outer and inner surface of the circle/sphere. The induced charge at inner surface is -2q and is concentrated near q1 and q2 (so as to maintain constant potential over the surface. If q1, q2 were at centre, induced -2q would have been uniformy distributed) The induced charge at the outer sphere = 2q is uniformly distributed over the sphere. I am not showing any calculations as i just want to give you a general idea of what happens. $\endgroup$ – Mitul Agrawal Mar 8 '20 at 3:24
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    $\begingroup$ Thank you thats helpful $\endgroup$ – user35687 Mar 8 '20 at 3:28
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Surface charges on a conductor won’t make the interior equipotential unless the charge density is continuous. If it is discrete, as in your example, then the best you can get is a sort of multipole approximation to an equipotential surface.

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    $\begingroup$ That's super helpful thanks! $\endgroup$ – user35687 Mar 8 '20 at 3:30

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