A)
B)
C)
D)
Correct Answer: B
Solution :
[b] The potential of inner shell should be zero. \[V=\frac{Kq}{a}-\frac{Kq}{2a}+\frac{K(Q+q)}{3a}=0\] \[\frac{q}{2}+\left( \frac{Q+q}{3} \right)=0;\,\,\,q=\frac{-2Q}{5}\] \[V(a<r<2a)=\frac{2KQ}{5}\left[ \frac{1}{a}-\frac{1}{r} \right]\] Hence potential will increase up to \[r=2a\]. \[V(2a<r<3a)=\frac{KQ}{5a}\] For \[2a<r<3a,\] the potential will be constant. \[V(r>3a)=\frac{K\left[ \frac{3Q}{5} \right]}{r}\]You need to login to perform this action.
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