Problem Statement:
A thin conducting shell having charge $Q$, radius $R$, and three point-charges $Q$,$-2Q$, and $3Q$ are also kept at points $A$, $B$, and $C$ respectively as shown. Find the potential at any point on the surface of the shell. (potential at infinity is zero)

enter image description here

My Attempt and Understanding :

As the Electric Field inside a conducting body is of necessity zero, the charge $Q$ given to the conducting shell must get distributed at the inner surface of the shell. (Applying Gauss Law on a spherical surface inside the meat of the body)

This would mean that there is no effective charge on the outer surface of the sphere.

Also, as potential due to the charges in a cavity and inner surface have no net potential at any part outside the cavity, the potential due to charges kept at $A$, $B$ and the charge $Q$ given to the shell have no role in the contribution of potential on the shell.

This leaves only charge $3Q$ kept at point $C$. But the potential due to this charge alone would vary with the distance of different points on the body.

I tried a few variations of this problem and came up with the following:

  1. If there was no charge inside the cavity (removing charges at $A$ and $B$), then although the charge distribution on the surface of the shell would change because of the charge at point $C$, the potential due to the shell charge at the center of the shell, being a scalar sum would still be $\frac{kQ}{R}$. And since there is no electric field inside the shell this time, the potential at the geometrical center and the surface would be the same And the net potential would be $\frac{5kQ}{2R}$. But in the given case there is an electric field in the cavity.

  2. If the charge $3Q$ was absent from the situation then the answer becomes zero as there is no net charge outside the shell and that within it does not contribute towards potential at its surface.

I have no idea what else am I missing so any guidance would be very helpful.


2 Answers 2


Most of what you said is perfectly correct, including that the net charge on the exterior is zero. However it must still be an equipotential, and as you say, the potential due the exterior charge varies over the surface. Hence there must be a zero sum distribution of charge on the exterior surface.

Have you met the idea of image charge?

  • $\begingroup$ I don't think i know about it, would be very useful if you could provide some source for introduction to the concept $\endgroup$
    – Hardik
    Commented Aug 7, 2021 at 5:16
  • $\begingroup$ You should check out your electrostatics textbook for the method of images. Wikipedia appears to be reliable on this subject but a proper textook would give more detail. $\endgroup$
    – CWPP
    Commented Aug 7, 2021 at 16:43

So after a bit of head-scratching, I figured it out.

As the potential due to charges present in a conducting body does not influence the potential at any external point, neither would the size of the cavity.

Also, due to the shielding property of a conductor, any change performed on the cavity would not change the phenomenon outside the material of the conductor.

So if we shrink the cavity down to an infinitely small size, the potential at the surface would not change. But as CWPP's answer already pointed out, there would be a net-zero charge distribution on the outer surface, but the potential due to these charges at the center would be zero as it is a scalar sum, so now in this new case we just need to determine the potential at the center of the sphere, which comes out to be


Which in fact is the correct answer


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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