A solid conducting sphere having a charge Q is surrounded by an uncharged concentric 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 -3Q, the new potential difference between the same two surfaces is : (a) V(b) 2 V(c) 4 V(d) – 2V

(a) The potential inside the shell will be the same everywhere as on its surface. As we add - 3Q charge on the surface, the potential on the surface changes by the same amount as that inside. Therefore the potential difference remains the same.

V inside= v surfaceLet radius of sphere be a and shell be bCase 1V sphere= kq/ aV Shell= kq/bPotential difference is vsphere- vshellCase 2V sphere= kq/ a- k3q/ bV shell= kq/b- k3q/bPotential difference is vsphere- vshellPotential difference observed in both cases is the same.

Dear Student


Charge=Q
Intial Potential Difference =V
Due to equipotential surface, new potential difference between the two surface is same.
Hence potential=V

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya