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

C

The correct answer is Option (1) → C

C, capacitance of a parallel plate conductor = $\frac{ε_0A}{d}$

where,

$ε_0$ = permittivity of space

A = Area of each plate

d = distance between the plates

Introducing a metal foil of negligible thickness exactly at the center divides the capacitor into two capacitors in series.

Now, we have two capacitors:

Capacitor 1: With plate separation $d_1=d/2$. Its capacitance $C_1=\frac{ε_0A}{d/2} =\frac{2ε_0A}{d}=2C$

Capacitor 2: With plate separation $d_2=d/2$. Its capacitance $C_2=\frac{ε_0A}{d/2} =\frac{2ε_0A}{d}=2C$

These two capacitors ($C_1$ and $C_2$) are connected in series. The reciprocal of the equivalent capacitance ($C_{new}$) for capacitors in series is given by:

$\frac{1}{C_{new}}=\frac{1}{C_1}+\frac{1}{C_2}$

Substitute the values of $C_1$ and $C_2$:

$\frac{1}{C_{new}}=\frac{1}{2C}+\frac{1}{2C}$

$\frac{1}{C_{new}}=\frac{2}{2C}$

$\frac{1}{C_{new}}=\frac{1}{C}$

Therefore, $C_{new}=C$