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 answer
    C
    $\mathrm{C}_2=2 \mathrm{C}_1$
  • 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'.

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