A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surface of the solid sphere and the outer surface of the hollow shell is V. If the shell is now given a charge of -3Q, the new potential difference between the same two surfaces is:
a. V
b. 2V
c. 4V
d. -2V
ans is given (a)
1but, see
(lets assume smaller has radius r1 and larger r2)
pot on inner sphere is:
KQ/r1 - KQ/r2
(as an opp charge of -Q will be induced on outer shell)
for larger shell
pot=
KQ/r2 - KQ/r2=0
so pot diff intially is KQ/r1 - KQ/r2
in next case
for inner sphere
pot= KQ/r1 - 3KQ/r2
for outer shell
pot=KQ/r2 - 3KQ/r2
pot diff is still
KQ/r1 - KQ/r2
as pot due to outer shell cancels out
.
Shouldn't is be the correct logic??
1http://targetiit.com/iit-jee-forum/posts/electrostatics-help-14290.html
