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: $ε_0 = 8.85 × 10^{-12} C^2 N^{-1} m^{-2}$)
Answer & explanation
Correct answer: option 4
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 = $1.13 \times 10^{-4} \, m^2$