Conductors with off-centered cavity

[raging]

Suppose we had an uncharged solid spherical conductor with radius R, and some spherical cavity of radius a that is located b units above R, where a+b<R. If we place a charge inside the cavity at the center, how would you quantitatively express the E-field lines outside of the conductor?

 

[raging]

This is NOT a homework. Concentric cavities are too easy. I'm wondering if we can somehow apply uniqueness here to show that it actually doesn't matter where the cavity is...

 

[cragar]

The E field is easy to find using Gauss's law . And it doesn't matter where the charge is inside the sphere or the shape of the cavity.

 

[thegreenlaser]

The E field is easy to find using Gauss's law . And it doesn't matter where the charge is inside the sphere or the shape of the cavity.

Exactly. You can kind of reason it out in a way, by knowing that charges like to stay on the very outside of a conductor. Think about it, say you've got a positive charge inside that off-centre cavity. That charge is going to pull negative charges towards the inner surface of that cavity to balance it out. Then, to balance out those negative charges in the conductor, positive charges are going to show up on the outside of the sphere (regardless of where the cavity is), and they'll create an electric field outside the sphere exactly the same as if you had just a charged sphere.

You can use Gauss' Law to prove every part of that by setting up surfaces just outside the cavity and just outside the sphere and using the relation:

\Phi_E = \int\vec{E}\cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}

Where you know \vec{E} = 0 inside the conductor.

 

[raging]

True. But to solve that surface integral simply one must apply a symmetry argument. How do you arrive about that symmetry? The induced charge will not be evenly distributed on the outside of the sphere, but is there a way to apply uniqueness to arguing that?

 

[Meir Achuz]

The outer surface of the conductor will be an equipotential surface of radius R. An outside potential phi=Q/r satisfies this boundary condition. It is thus the unique potential outside the sphere.
It doesn't matter what goes on in cavities within the conduciting sphere.