Question

A parallel plate capacitor has a plate area of 100 cm^2 and is separated by a distance of 20 mm. Find its capacitance.

(a) 6.425 × 10^-12 F

(b) 5.425 × 10^-12 F

(c) 4.425 × 10^-12 F

(d) 3.425 × 10^-12 F

I had been asked this question during an online exam.

I need to ask this question from Parallel Plate Capacitor topic in section Electrostatic Potential and Capacitance of Physics – Class 12

Answer 1

Correct choice is (c) 4.425 × 10^-12 F

To explain I would say: C = \(\frac {Q}{V} = [ \frac {\sigma A}{(\frac {\sigma d}{\varepsilon_0})} ] = \frac {A\varepsilon_0}{d}\).

C = \(\frac {100 \times 10^{-4} \times 8.85 \times 10^{-12})}{(20 \times 10^{-3})}\)

C = 4.425 × 10^-12 F.