Electromagnetic field theory -- Potential inside a conducting sphere

[per persson]

Homework Statement

The potential of the hemisphere is just added to the potential inside the sphere and im wondering when it is possible to do this.

Relevant Equations

More conceptual.

The sphere with charge Q is of metall and the question is what its potential is . a and b I've already solved but c is difficult conceptually. The answer is down below but I don't understand it.

Here is the solution of c:

So the contribution of the potential of the hemisphere is simply added to the potential that is given in b. But this must mean the potential caused by the hemisphere must be equal everywhere inside the metallic sphere. Is this the case? And what if we replace the hemisphere with a point charge, lets say with charge -Q with distance 3a from the origin. Then you can't just add the potential of charge to the potential inside the metallic sphere since the potential of the charge will be different inside the metallic sphere.

So generally when can I add potential to a point and say its the same for the whole area?

 

[nasu]

Where is the problem statement?

 

[kuruman]

You haven't told what potential this $V$ is.

 

[per persson]

You haven't told what potential this $V$ is.

Potential of the hemisphere or the potential given in b? I don't think it matters though since the solution is already given. My problem is why can we just add the potential of the hemisphere to the potential inside the sphere and thats the answer.

 

[per persson]

Where is the problem statement?

Why is it legit to add the potential of the hemisphere to the potential inside sphere given by b and thus get the solution for c. The implications of this is really powerful and too good to be true. There has to be some restrictions when this method that solves c can be used.

 

[haruspex]

But this must mean the potential caused by the hemisphere must be equal everywhere inside the metallic sphere.

No, that clearly would not be true. But the charge on the inner shell will rearrange so that the potential is the same everywhere on and inside it.
If we consider the centre of the inner shell, it is a known and constant distance from every point on the inner shell and likewise wrt the dielectric and the hemisphere. Consequently we can immediately write down what the potential there is, regardless of the rearrangement of the charges. And due to that rearrangement, all other points on and inside the sphere will be at that same potential.

 

[Steve4Physics]

Why is it legit to add the potential of the hemisphere to the potential inside sphere given by b and thus get the solution for c. The implications of this is really powerful and too good to be true. There has to be some restrictions when this method that solves c can be used.

EDIT. Aha. I see @haruspex has replied. But since I've already drafted the following, I'll post it.

There are a set of surfaces having a common centre of curvature, C.

For a given surface, all points on that surface are the same distance from C. That's the key.

Every surface has a known total charge. We don’t know the distribution of the charge on any of the surfaces but that doesn’t matter.

That’s because, for a given surface of radius $R$, every elementary charge $dq$ on the surface is the same distance ($R$) from C. So it has the same contribution to the total potential at C as every other elementary charge on that surface.

If the surface has charge $Q$, it’s contribution to the total potential at C is $V = \int_{q=0}^{q=Q} \frac {dq}{4 \pi {\epsilon}_0 R} = \frac {Q}{4 \pi {\epsilon}_0 R}$.

 

[per persson]

No, that clearly would not be true. But the charge on the inner shell will rearrange so that the potential is the same everywhere on and inside it.
If we consider the centre of the inner shell, it is a known and constant distance from every point on the inner shell and likewise wrt the dielectric and the hemisphere. Consequently we can immediately write down what the potential there is, regardless of the rearrangement of the charges. And due to that rearrangement, all other points on and inside the sphere will be at that same potential.

But what is so special with the centre. I know it is easier to calculate but what makes the centre the point where we can just add potentials to get the total. Suppose b) gives potential Vo inside the sphere and the hemisphere gives V1 in the centre. Then the total potential is V=Vo+V1. But a bit below the centre the potential of the hemisphere will be greater. Does that mean the charges of the sphere and shell rearranges such that its potential is less than Vo in that point and thus keeping the total potential at V.

Similarly above the centre the potential by the hemisphere should be less so the potential of the sphere and shell are such that it is greater than Vo. Is this the case?

And what if we put a point charge -Q with distance 3a from centre. Will the potential inside the sphere be V_o-Q/(4epi3a)?

 

[haruspex]

But what is so special with the centre.

That it is at a known distance from every charge, no matter how those charges are distributed around the two shells. This means we can find the potential there without worrying about how those charges are distributed.

 

[nasu]

Why is it legit to add the potential of the hemisphere to the potential inside sphere given by b and thus get the solution for c. The implications of this is really powerful and too good to be true. There has to be some restrictions when this method that solves c can be used.

This is your question, not the statement of the problem.