The separation between the plates of a parallel plate capacitor of capacitance 2 pF is 0.5 mm. The area of the plates of the capacitor is

(Given: $ε_0 = 8.85 × 10^{-12} C^2 N^{-1} m^{-2}$)

$1.13\, cm^2$

The correct answer is Option (4) → $1.13\, cm^2$

Capacitance of a parallel plate capacitor:

$ C = \frac{\epsilon_0 A}{d} $

Given:

$ C = 2 \times 10^{-12} \, F $

$ d = 0.5 \, mm = 5 \times 10^{-4} \, m $

$ \epsilon_0 = 8.85 \times 10^{-12} \, F/m $

Now,

$ A = \frac{C \, d}{\epsilon_0} $

$ A = \frac{(2 \times 10^{-12})(5 \times 10^{-4})}{8.85 \times 10^{-12}} $

$ A = \frac{1 \times 10^{-15}}{8.85 \times 10^{-12}} $

$ A = 1.13 \times 10^{-4} \, m^2 $

Area of the plates = $1.13 \times 10^{-4} \, m^2$