Q112 / 150 · MHT CET 2021 (23 Sep Shift 1) · Physics
The plates of a parallel plate capacitor of capacity ' $\mathrm{C}_1$ ' are moved closer together until they are half their original separation. The new capacitance ' $\mathrm{C}_2$ ' is
- A$\mathrm{C}_2=\frac{\mathrm{C}_1}{2}$
- B$\mathrm{C}_2=\mathrm{C}_1$
- Correct answerC$\mathrm{C}_2=2 \mathrm{C}_1$
- D$C_2=4 C_1$
Solution
The capacitance of a parallel plate capacitor is given by
$$
\mathrm{C}=\frac{\mathrm{kA} \varepsilon_0}{\mathrm{~d}}
$$
If the distance between the plates $\mathrm{d}$ is decreased to half, the capacitance will become double.
Hence, the correct answer is option: 'C'.