Question

37. A parallel-plate capacitor has plates of
dimensions 2 cm x 3 cm separated by a
mm thickness of paper. The capacitance
of this device is
(Given : dielectric constant of paper K = 3.7
and ε = 8.85 x 10-12 CPN-1 m2)
(A) 20 pF (B) 20 uF
(C) 10 uF (D) 10 pF​

Answer 1

Given:

A parallel-plate capacitor has plates of  dimensions 2 cm x 3 cm separated by a  mm thickness of paper.

To find:

The capacitance  of the device

Solution:

From the given information, we have the data as follows.

A parallel-plate capacitor has plates of  dimensions 2 cm × 3 cm separated by a  mm thickness of paper.

The area of the plates, A = 2 cm × 3 cm = 6 cm²

The distance between the plates, d = 1 mm = 0.1 cm

The capacitance  of this device is  given by the formula as follows.

C = k × εA/d

Substitute the values in the above equation.

C = 3.7 × (8.854 × 10^{-12} × 6) / 0.1

C = 1.965 nF.

The capacitance  of the device is 1.965 nF.