Electrostatics conductor problem

[ehrenfest]

[SOLVED] electrostatics problem

Homework Statement

Two spherical cavities, of radii a and b, are hollowed out from the interior of a (neutral) conducting sphere of radius R. At the center of each cavity a point charge is placed--call these charges q_a and q_b.
Why is it necessarily true that the force on q_a and q_b is 0? Why is it true that the force is still zero no matter what kind of charge distribution you have outside the conductor.

Homework Equations

The Attempt at a Solution

You can use Gauss's Law to find the electric field in each cavity. And you find that it is as the charge were isolated in space. But that doesn't tell you the electric field at the charge...only around it, right?

 

[jacobrhcp]

*is a little rusty on the subject*

conducting sphere means that, if there is an electric field in the conducting sphere, then the electrons in the conducting sphere will move as to create an electric field to cancel the electric field in place, right?

That would answer you question, because E=0 => F=0

but I must admit I'm not completely convinced of my answer

 

[ehrenfest]

But q_a and q_b are in cavities, not in the meat of the conductor.

 

[pam]

Draw a Gaussian sphere in the conductor, just outside one cavity. The E field is zero everywhere on the sphere. This can only happen if the surface charge on the inside of the cavity is spherically symmetric, i.e. constant. Then the field at the center of the spherical cavity will be zero, giving zero force on the point charge. This only happens at the center of the cavity. There would be a force on a charge at other points in the cavity.

 

[kdv]

Draw a Gaussian sphere in the conductor, just outside one cavity. The E field is zero everywhere on the sphere. This can only happen if the surface charge on the inside of the cavity is spherically symmetric, i.e. constant. Then the field at the center of the spherical cavity will be zero, giving zero force on the point charge. This only happens at the center of the cavity. There would be a force on a charge at other points in the cavity.

I was going to say exactly the same thing.