A) electric field near A in the cavity = electric field near B in the cavity
B) charge density at A = charge density at B
C) potential at A = potential at B
D) None of the above
Correct Answer: C
Solution :
The charge distribution on the inner surface of cylinder (cavity) is non-uniform, so E near A is not equal to E near B in the cavity. The conductor is an equipotential surface, so potentials at A and B are same. Charge density depends upon radius of curvature.
You need to login to perform this action.
You will be redirected in
3 sec
