# Induced charge on conducting sphere sliced by a plane

We are given a conducting solid sphere, and it is cut by a plane as shown.

A charge $$Q$$ is given to the smaller part of the conductor, and it is required to find the induced charge on the surface of the other hemisphere, and the force of interaction between them.

My initial approach was to apply Gauss law, as we know that electric field is always perpendicular to the conductor surface, and hence we can claim that field in both parts is radial and that the induced charges on the facing flat faces are equal and opposite. Then I assumed that since the gap between the faces was negligible, the potential throughout the sphere is constant, and thus the system is equal to a conducting sphere having a charge $$Q$$ distributed evenly over it.

Although this gives correct answers for finding the charge densities, I am not convinced that my reasoning is entirely correct.

Even if it is correct, how can the force of interaction between the two parts of the sphere possibly be found?

EDIT: As I understand, the exact distribution of charges is not easily calculable without solving Laplace equation; in that case, can the net charge on each surface of the sections be found easily?

• I showed what I tried, and also did not ask for a complete solution; yes it might be homework like, but this is not a very routine problem, and IMHO should be addressed accordingly. If relatively well written questions like this gets closed for being homework-like, I have no words. Commented Jun 12 at 13:46
• That is no problem: the OP is never asked for words, the question is simply closed! But if you understand that the total charge is distributed over the outer surface, then you'd probably also agree that the differential charge is distributed over the two flat "capacitor plates" that you create. And that should answer the question. Commented Jun 12 at 20:48
• @JosBergervoet Yes I can find out the force of interaction without any problems. What I am facing a problem with is finding the charge densities of the different parts of the sections. Although my 'reasoning' gives the correct answer, i.e q/4pir^2 for the larger curved surface, I myself am not thoroughly convinced of the credibility of it. Commented Jun 12 at 21:11
• In that case it definitely goes beyond a homework problem (for which such "reasoning" would likely be discouraged, these days...) Commented Jun 13 at 6:37