The plates of a parallel plate capacitor of capacity ' $\mathrm{C}_1$ ' are moved closer together until they…
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
$\mathrm{C}_2=\frac{\mathrm{C}_1}{2}$
$\mathrm{C}_2=\mathrm{C}_1$
$\mathrm{C}_2=2 \mathrm{C}_1$
$C_2=4 C_1$
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.
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