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Q.

Two identical sheets of a metallic foil, separated by a distance d and having capacitance C, are charged to a potential difference E. Keeping the charge constant, the separation is increased by $l$ . Then, the new capacitance and potential difference respectively will be :

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a

$\frac{ε_{0} A}{(d + l)}, (1 + \frac{l}{d}) E$

b

$\frac{ε_{0} A}{d}, (1 + \frac{l}{d}) E$

c

$\frac{ε_{0} A}{(d + l)}, E$

d

$\frac{ε_{0} A}{d}, E$

answer is C.

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Detailed Solution

$q = C V = C_{1} V_{1}; C = \frac{ε_{0} A}{d}$
$∴ q = \frac{ε_{0} A E}{d} = \frac{ε_{0} A E_{1}}{(d + l)}$
Or $\frac{ε_{0} A}{d} E = \frac{ε_{0} A}{(d + l)} E_{1}$
$∴ E_{1} = \frac{(d + l)}{d} E = (1 + \frac{l}{d}) E$ and $C_{1} = \frac{ε_{0} A}{d + l}$ .

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