An air-filled parallel-plate capacitor has a capacitance of 1 pF.The plate separation is then doubled and a wax dielectric is inserted, completely filling the space between the plates.As a result, the capacitance becomes 2 pF.The dielectric constant of the wax is:
A)0.25
B)0.50
C)2.0
D)4.0
E)8.0

Question

Quizplus · Accepted Answer

The capacitance of an air-filled parallel plate capacitor is given by:

C = ε0A/d

where ε0 is the permittivity of free space, A is the area of the plates and d is the distance between them.

When the plate separation is doubled, the new distance is 2d. Let the thickness of the wax layer be x. Then, the distance between the plates is d + x. The capacitance of the wax-filled capacitor is:

C' = εrε0A/(d+x)

where εr is the relative permittivity (dielectric constant) of the wax.

We are given that the capacitance increases from 1 pF to 2 pF when the wax is inserted. Therefore:

2 pF = εrε0A/(d+x)

Substituting the given values:

2 x 10^-12 F = εr (8.85 x 10^-12 F/m) (A) / (2d)

Solving for εr, we get:

εr = 4.0

An Air-Filled Parallel-Plate Capacitor Has a Capacitance of 1 PF