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Two identical sheets of a metallic foil are separated by $d$ and capacitance of the system is $C_{0}$ and charged to a potential difference $V_{0}$ . Keeping the charge constant, the separation is increased by $l$ . Then the new capacitance and potential difference are $C$ and $V$ respectively. Then

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$C = \frac{C_{0}}{(1 + \frac{l}{d})}$

$C = C_{0} (1 + \frac{l}{d})$

$V = V_{0} (1 + \frac{l}{d})$

$V = \frac{V_{0}}{(1 + \frac{l}{d})}$

answer is A, C.

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Detailed Solution

$q = C V = C_{0} V_{0}$

where $C_{0} = \frac{ε_{0} A}{d} q = \frac{ε_{0} A V_{0}}{d} = \frac{ε_{0} A V}{(d + l)}$

$\Rightarrow \frac{ε_{0} A}{d} V_{0} = \frac{ε_{0} A}{(d + l)} V \Rightarrow V = \frac{(d + l)}{d} V_{0} = (1 + \frac{l}{d}) V_{0}$

and $C = \frac{ε_{0} A}{d + l} = \frac{ε_{0} A}{d (1 + \frac{l}{d})} = \frac{C_{0}}{(1 + \frac{l}{d})}$

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