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The capacity of a parallel plate capacitor formed by the plates of same area A is 0.02µF with air as dielectric. Now one plate is replaced by a plate of area 2A and dielectric (k = 2) is introduced between the plates, the capacity is

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0.04 µF

0.01 µF

0.08 µF

2 µF

answer is A.

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Detailed Solution

Capacitance of a capacitor with dielectric of dielectric constant k is

$\begin{array}{l} C = \frac{k \in_{0} A}{d} = 0.02 μF \\ C_{2} = \frac{2 {kϵ}_{0} A}{d} = 0.04 μF \\ common area will be A only \end{array}$

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