Capacitance and field of 2 eccentric spheres

[Janko1]

Homework Statement

Hi there!

So I have found trouble with the following problem. I have a system of two eccentric spheres, where one is inside another.

The Attempt at a Solution

I have found this answer:
https://physics.stackexchange.com/questions/16939/capacitance-of-non-concentric-spheres
and solved it also for the c, that is not close to the center of origin.

What my problem is, is that there I only obtain potential (field) between the electrodes, but I want it everywhere. How could I obtain it inside the smaller sphere and outside of bigger one? Then, there is a boundary condition of one of the spheres at 0, what if I put it on, let's say V1... what changes? How could I solve the problem for the capacitance after I made this change?

 

[kuruman]

Start with charge +Q on the outer sphere and -Q on the inner sphere. What is the electric field outside the outer sphere? What about the inner sphere? If the potential on its surface is V₀, what would it be inside it? More to the point, if you move the inner sphere around inside the outer sphere, will the surface charge distribution on the outer surface of the outer sphere change? How?

 

[Janko1]

Okay, so inside the inner sphere, I would say that there is no potential, according to the Gauss law. Also if the charge on the outer sphere is +Q and on the inner sphere is -Q, the electric field should be zero, again according to the Gauss law.
The charge distribution would change on the outer sphere. The closer the inner sphere would be to the outer sphere, more charge would there be.

So, to sum it up...field inside the inner sphere would always be zero. But I am still really not sure about the outer sphere, would Gauss law be sufficient tool?

 

[kuruman]

Okay, so inside the inner sphere, I would say that there is no potential, according to the Gauss law.

Not exactly. Inside the inner sphere the potential has the same value as on the surface. Remember that the conductor is an equipotential.

Also if the charge on the shell is +Q and on the inner sphere is -Q, the electric field should be zero, again according to the Gauss law.

Correct.

The charge distribution would change on the outer sphere. The closer the inner sphere would be to the outer sphere, more charge would there be.

Where do you think the charges would go? Answer: -Q on surface of the outer sphere and + Q on the inner surface of the outer shell. The result is that the electric field inside the conducting part of the outer shell will be zero according to Gauss Law and will remain zero no matter where you move the inner sphere inside the cavity. Moving the inner sphere changes the distribution of charges facing each other on the inner surface of the shell and the surface of the sphere, but does not move charges from the inside surface of the shell to the outside surface. That's because an electric field is needed to do that and there is none between the inner and outer surface of the shell.

So, to sum it up...field inside the inner sphere would always be zero.

Correct and it is zero also in the region between the inner and outer surface of the shell where there is conducting material.

But I am still really not sure about the outer sphere, would Gauss law be sufficient tool?

It should be when coupled with the fact the conductors are equipotentials which means that two points are on an equipotential, there can be no electric field between them.
Finally, remember that the capacitance depends only on the geometry, not on how much charge you put on the conductors.