I am trying to solve what seems like a simple problem but something is bothering me:

Imagine we have a sphere (1) with a charge $Q_1$ and at a distance $d$ we have another sphere (2) which is a perfect conductor with charge $Q_2 = 0$.

I am trying to calculate the potential that will be induced on this second sphere (2) by the first one.

We assume that the charged sphere (1) can be simplified to a point charge at the centre of the sphere (due to the charge being uniform).

I looked up the method of images in the case of a point charge in the presence of a insulated conductive sphere.

I understand that you have to combine the equations of potential of a grounded conductive sphere (2) and the equation of the potential of a point charge with $Q_3 = Q_2 - Q_1$ at the centre of the conductive sphere (2), but where should I calculate this potential so it's equal to the induced potential of the conductive sphere (2), on the surface ? At the centre?


1 Answer 1


The charged sphere induces surface charge density on the conductor. This is necessary because field lines are always perpendicular to the surface of the conductor. This induced surface charge density modifies electric potential in the region. It will no longer be $V(d)=Q/4\pi\epsilon_0.d$.
If you know the potential at the conductor(which I think is necessary for solving the problem), the potential everywhere else can be obtained by method of images.

  • $\begingroup$ The potential at the conductor is what I am looking for. What i know is the charge $Q$ of the first sphere and the charge $Q_2 = 0$ of the second sphere. Sorry I did not include enough information inside my question. $\endgroup$
    – Matazar
    Apr 20, 2015 at 16:01
  • $\begingroup$ I've updated my question with the new info. $\endgroup$
    – Matazar
    Apr 20, 2015 at 16:06
  • $\begingroup$ The charged sphere induces charge on the second conducting sphere. The net charge on the conductor will still be zero, but there is some charge distribution. $\endgroup$
    – Goobs
    Apr 20, 2015 at 16:13

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