Two identical capacitors are connected in parallel combination. Their equivalent capacitance is $C_1$. If the plate area is halved and separation between plates is doubled for both the capacitors, their equivalent capacitance now is $C_2$. Find the ratio $\frac{C_1}{C_2}$

4 : 1

The correct answer is Option (2) → 4 : 1

Let the original capacitance of each identical capacitor be $C$:

$C = \frac{\varepsilon_0 A}{d}$

For parallel combination of two identical capacitors:

$C_1 = C + C = 2C = 2 \frac{\varepsilon_0 A}{d}$

New configuration: Plate area halved ($A \to A/2$), separation doubled ($d \to 2d$)

New capacitance of each capacitor:

$C' = \frac{\varepsilon_0 (A/2)}{2d} = \frac{\varepsilon_0 A}{4d} = \frac{C}{4}$

Equivalent capacitance in parallel:

$C_2 = C' + C' = 2 \cdot \frac{C}{4} = \frac{C}{2}$

Ratio:

$\frac{C_1}{C_2} = \frac{2C}{C/2} = 4$

Answer: $C_1 / C_2 = 4$