2V
V
−2V
4V
Correct option is A. V
As given in the first condition:
Both conducting spheres are shown.
Vin−Vout=(kQ/r1)−(kQ/r2)=kQ(1/r1−1/r2)=V
In the second condition:
Shell is now given charge −4Q.
Vin−Vout=(kQ/r1−4kQ/r2)−(kQ/r2−4kQ/r2)
kQ/r1−kQ/r2=kQ(1/r1−1/r2)=V
Hence, we also obtain that potential difference does not depend on charge of outer sphere.
∴P.d remains same.