question_answer

A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of -4 Q, the new potential difference between the same two surfaces is:      [JEE Main 8-4-2019 Morning]

A) V

B) 2V

C) -2V

D) 4V

Correct Answer: A

Solution :

[a] As given in the first condition :

Both conducting spheres are shown.

\[{{V}_{in}}-{{V}_{out}}=\left( \frac{kQ}{{{r}_{1}}} \right)-\left( \frac{kQ}{{{r}_{2}}} \right)\]

\[=kQ\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)=V\]

In the second condition :

Shell is now given charge -4Q.

\[{{V}_{in}}-{{V}_{out}}=\left( \frac{kQ}{{{r}_{1}}}-\frac{4kQ}{{{r}_{2}}} \right)-\left( \frac{kQ}{{{r}_{2}}}-\frac{4kQ}{{{r}_{2}}} \right)\]

\[=\frac{kQ}{{{r}_{1}}}-\frac{kQ}{{{r}_{2}}}\]

\[=kQ\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)=V\]

Hence, we also obtain that potential difference does not depend on charge of outer sphere.

\[\therefore \] P.d. remains same