Question

A parallel-plate capacitor has plates of dimensions 2 cm × 3 cm separated by a 1 mm thickness of the paper. The capacitance of this device is (Given: dielectric constant of paper K = 3.7 and ϵ₀ = 8.85 × 10^-12 C² N^-1 m^-2)
1. 20 pF
2. 10 pF
3. 10 μF
4. 20 μF

Answer 1

Correct Answer - Option 1 : 20 pF

CONCEPT:

• Capacitance: The ability of an electric system to store the electric charge is known as capacitance.

C = Q/V

where Q is the charge on it, V is the voltage, and C is the capacitance of it.

For the parallel plate capacitor, the capacitance is given by:

\(C= \frac{KA\varepsilon _{0}}{d}\)

where A is the area of the plate, d is the distance between plates, K is the dielectric constant of material and ϵ₀ is constant.

CALCULATION:

Given that:

Dimensions of plate = area (A) = 2 cm × 3 cm = 6 cm² = 6 × 10^-4 cm²

Distance between plates (d) = 1 mm = 1 × 10^-3 m

Dielectric constant of paper (K) = 3.7

ϵ0 = 8.85 × 10-12 C2 N-1 m-2

For the parallel plate capacitor, the capacitance is given by:

\(C= \frac{KA\varepsilon _{0}}{d}\)

\(C= \frac{3.7× 6× 10^{-4} ×8.85 × 10^{-12}}{1 × 10^{-3}}\)

C = 2 × 10^-11 F = 20 pico farad

Capacitance (C) = 20 pF

Hence option 1 is correct.