A parallel plate capacitor has capacitance of C farads. If the area of the plates is doubled and the distance between them is halved. then the capacitance of the capacitor is:

Correct option is C

The capacitance $C$ of a parallel plate capacitor is given by the formula:

$$C = \frac{\epsilon_0 \epsilon_r A}{d}$$

​​where $A$ is the area of the plates, and $d$ is the distance between them.
If the area of the plates is doubled ($A\to 2A$) and the distance between them is halved (d→d2d \to \frac{d}{2}d→d/2

​), the new capacitance C′C'C′ is:​

​$$C' = \frac{\epsilon_0 \epsilon_r (2A)}{\frac{d}{2}} = \frac{2 \epsilon_0 \epsilon_r A}{d / 2} = 4 \frac{\epsilon_0 \epsilon_r A}{d} = 4C$$

Therefore, the correct answer is: (c) 4C farads.​​