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(C) Let the radius of the inner shell be $R$ and the outer shell be $2R$. Before connecting the shells: The potential of the inner shell is $V_{	ext{

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$V$

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(C) Let the radius of the inner shell be $R$ and the outer shell be $2R$. Before connecting the shells: The potential of the inner shell is $V_{	ext{inner}} = \frac{kQ}{R} + \frac{k(0)}{2R} = \frac{kQ}{R}$. The potential of the outer shell is $V_{	ext{outer}} = \frac{kQ}{2R} + \frac{k(0)}{2R} = \frac{kQ}{2R}$. The potential difference is $V = V_{	ext{inner}} - V_{	ext{outer}} = \frac{kQ}{R} - \frac{kQ}{2R} = \frac{kQ}{2R}$. Thus,$V = \frac{kQ}{2R}$,which implies $\frac{kQ}{R} = 2V$. When the shells are joined by a metal wire,the charge $Q$ flows from the inner shell to the outer shell until both shells reach the same potential. Since the outer shell is now connected to the inner shell,the entire charge $Q$ resides on the outer surface of the outer shell. The potential of the inner shell (which is now equal to the potential of the outer shell) is $V' = \frac{kQ}{2R}$. Since $V = \frac{kQ}{2R}$,the potential of the inner shell is $V' = V$.

Consider the shown system of two concentric thin metal shells. The inner shell has charge $Q$,while the outer shell is neutral. The potential difference between the shells is $V$. If the shells are joined by a metal wire,then the potential of the inner shell is

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