106V = kQ/r (r=radius of inner sphere)
New pd = V + k(-3Q)/R (R=radius of shell)
So ans shud be none of these.
Hence the answers (options are wrong)
Maybe the Q-setter has assumed r=R (i donno how that can be) then it wud be -2V
A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting spherical shell. Let the potential between the surface of the solid sphere and that of the outer surfacel be V.If the shell is now given a charge of - 3 Q,the new P.d between the same 2 surfaces will be ___??
a.V
B.2V
c.4V
d.-2V
-
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6 Answers
106
11thats what asish..i gt the q. from a friend..i think they IN ALL PROBALITY meant outer surface instead of shell.
1yes ,it will remain V.
because by the presence of charge on outer shell ,potential everywhere inside and on the surface will change by same amount and hence potential difference between sphere and shell will remain unchanged