If the distance between the plates of a parallel plate capacitor is increased 10 times and the area is reduced to one-fourth, then its capacitance _________.

Correct option is C

$$\textbf{Explanation:} \\[4pt]\bullet \text{The capacitance } C \text{ of a parallel plate capacitor is given by the formula:} \\[6pt]C = \dfrac{\varepsilon_0 A}{d} \\[8pt]\text{Where:} \\[4pt]\varepsilon_0 = \text{the permittivity of free space,} \\[4pt]A = \text{the area of the plates, and} \\[4pt]d = \text{the distance between the plates.} \\[8pt]\text{According to the problem:} \\[4pt]\bullet \text{The distance between the plates is increased by 10 times, so } d_2 = 10d_1. \\[4pt]\bullet \text{The area is reduced to one-fourth, so } A_2 = \dfrac{A_1}{4}. \\[8pt]\text{Now, applying the formula to the new conditions:} \\[6pt]\dfrac{C_2}{C_1} = \dfrac{A_2}{A_1} \times \dfrac{d_1}{d_2} = \dfrac{A_1}{4A_1} \times \dfrac{d_1}{10d_1} \\[6pt]\dfrac{C_2}{C_1} = \dfrac{1}{4} \times \dfrac{1}{10} = \dfrac{1}{40} \\[8pt]\text{Thus, the new capacitance becomes } \dfrac{1}{40} \text{ times the initial capacitance.}$$​​