Solution
The potential outside the body is given by $V = A/r$. This implies that the electric field outside behaves exactly like that of a point charge or a spherically symmetric charge distribution located at the center.
However, this does not guarantee that the actual charge distribution inside is spherically symmetric.
Consider the case of a spherical conducting shell.
- If we place an arbitrary, non-uniform, or asymmetric charge distribution inside the cavity of the shell, electrostatic induction occurs.
- The charges on the inner surface of the conductor will redistribute to cancel the field of the internal charge within the conductor’s bulk.
- The charges on the outer surface will redistribute themselves to maintain the equipotential nature of the surface. For an isolated spherical conductor, the charge on the outer surface distributes uniformly, regardless of the distribution inside.
Consequently, the field outside the shell is purely radial and symmetric ($V \propto 1/r$), completely “hiding” the asymmetry of the charge distribution inside. This is known as electrostatic shielding.
Therefore, we cannot conclude that the internal charge distribution is symmetric; it could be uniform or non-uniform.
Answer: (c)