What Determines Surface Charge Density Inside a Cavity: A or B?

[i_island0]

The figure shows a cuboidal shaped conductor inside which there is an ellipsoidal cavity. The center of the cavity is at the center of the cuboid.

Link to figure: http://www.picscrazy.com/view/1qtB9

A charge qo is kept at the center of the cavity. I am showing two positions 'A' and 'B'. MY question is in which location, among 'A' and 'B', (on the inner surface of cavity) the surface charge density will be more. At 'A' or at 'B'.

I have two arguments to present:
(i) The system will try to minimize its energy - so, in that case the surface charge density at B will be less than surface charge density at A. Or to say in other words, since the radius of curvature at B is more, so the charge at B will be less. A is sharper than B so charge density at A will be more.
(ii) now charge qo is closer to B than A. So shouldn't i conclude that charge density at B will be more than charge density at A. Since charge qo is closer to B, so electric field lines will be more dense near B, so the charge density at B will be more.

I m not sure which of the arguments is correct. And why? can anyone help me in this. ??

 

[Meir Achuz]

B has the higher surface charge density.
Your argument (i) is wrong because the curvature wrt the conductor is actually negative for this geometry.
For a charge outside a conducting ellipsoid, the SCD would be greater at the end.

 

[Manchot]

(ii) is correct. More rigorously, the conductor will be at a constant potential, and so the line integrals of the electric field from q0 to B and from q0 to A must be the same. Therefore, the magnitude of the field at B must be higher than A.

 

[i_island0]

wow, sometimes the explanations are so simple, but yet so tough to think
thanx to both of you