A metal foil of negligible thickness is introduced between two plates of a capacitor at the centre. The new Capacitance of the capacitor would be:

C

The correct answer is Option (1) → C

When we insert a foil of negligible thickness between the parallel plates of the capacitor, the area will not change.

$C_1=C_2=\frac{2ε_0A}{d}$

$\frac{1}{C_{eff}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{d}{ε_0A}$

$⇒C_{eff}=\frac{ε_0A}{d}$

Hence, the capacitance remains the same.