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A parallel plate capacitor C with plates of area 1 square unit and separation d is filled with a liquid of dielectric constant K=2. The level of liquid is d/3, initially. If the liquid level decreases at a constant speed v. Then, the capacity of capacitor after t second is

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$\frac{(15 d + 9 vt) ε_{0}}{2 d^{2} - 3 dvt - 9^{2} t^{2}}$

$\frac{6 ε_{0}}{5 d - 3 vt}$

$\frac{6 ε_{0}}{5 d + 3 vt}$

$\frac{(15 d - 9 vt) ε_{0}}{2 d^{2} + 3 dvt - 9 v^{2} t^{2}}$

answer is A.

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Detailed Solution

(a) After time t, thickness of liquid will remain $(\frac{d}{3} - vt)$

So, capacity after t second is $C = \frac{ε_{0} (1)}{(d - \frac{d}{3} + vt) + (\frac{d / 3 - vt}{2})}; C = \frac{6 ε_{0}}{5 d + 3 vt} F$

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