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Two thin conducting concentric shells of radii $\mathrm{R}$ and $2 \mathrm{R}$ are shown in the figure. The outer shell carries a charge $+Q$ and the inner shell is neutral. Then the correct statement/s is
(a) When the switch is closed, the potential on the inner shell becomes zero
(b) With the switch closed, the charge on the inner sphere is $\frac{Q}{2}$
Options:
(a) When the switch is closed, the potential on the inner shell becomes zero
(b) With the switch closed, the charge on the inner sphere is $\frac{Q}{2}$
Solution:
2418 Upvotes
Verified Answer
The correct answer is:
(a) is correct, (b) is wrong
Let charge on inner shell after switch is closed be $q$ then, $\mathrm{V}_{\text {inner }}=\mathrm{V}_{\text {outer }}$
$$
\frac{K q}{R}+\frac{K(Q-q)}{2 R}=\frac{K q}{2 R}+\frac{K(Q-q)}{2 R}
$$
$$
\therefore \mathrm{q}=0
$$
$$
\frac{K q}{R}+\frac{K(Q-q)}{2 R}=\frac{K q}{2 R}+\frac{K(Q-q)}{2 R}
$$
$$
\therefore \mathrm{q}=0
$$
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