A metal foil of thickness (3/4)d is introduced between two plates of a capacitor of capacitance C. The distance between the plates of the capacitor is d. The new capacitance of the capacitor will be

4 C

The correct answer is Option (4) → 4 C

Given: Parallel plate capacitor with plate separation $d$ and capacitance $C$. A metal foil of thickness $\frac{3}{4}d$ is inserted between the plates.

When a conducting foil of thickness $t$ is inserted, the effective separation becomes:

$d_{\text{eff}} = d - t$

Here, $t = \frac{3}{4}d$

$d_{\text{eff}} = d - \frac{3}{4}d = \frac{d}{4}$

New capacitance:

$C' = \frac{\varepsilon_0 A}{d_{\text{eff}}} = \frac{\varepsilon_0 A}{d/4} = 4 \frac{\varepsilon_0 A}{d} = 4C$

New capacitance = 4C