Question

A charge of $4Q$ is transferred to a solid metal sphere of radius $r$ . Where will this excess charge reside?
A) $3Q$ at the center, and $Q$ on the outer surface
B) $Q$ at the center, and $3Q$ on the outer surface
C) $ - 4Q$ at the center
D) $ - 4Q$ at the outer surface
E) $2Q$ at the center, $Q$ in a ring of radius $\dfrac{1}{2}r$ , and $Q$ on the outer surface

Answer 1

Hint:- The surface of a conductor is an equipotential surface. The electrons inside the conductor will move and are used to cancel the charges to avoid the repulsion. Hence the charge is neutralized inside the conductor.

Complete Step by step solution:
Given a charge of $4Q$ is transferred to a solid metal sphere of radius $r$ .
The total charge inside a conductor will be always zero. To attain equilibrium the charges used to come on the surface. This is also to minimize the repulsion between the charges also. Therefore the total charge will remain on the surface.
Therefore the charge inside the metal sphere is zero. The charge will reside on the outer surface. When charge of $4Q$ is transferred to a solid metal sphere, the charge $ - 4Q$ will be resided at the surface.

The answer is option D.

Note: We know that the charge inside the conductor is zero. Hence while we apply the Gauss’s law to find the electric field inside the conductor will be zero. This is because the charge is zero. But the electric field can be calculated at the surface of the sphere because the excess charge resides there.