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) 0.5
C) 2.0
D) 4.0
E) 8.0

Question

Quizplus · Accepted Answer

From the given information, we can write:

C1 = 1 pF    (capacitance with air as dielectric)
C2 = 2 pF    (capacitance with wax as dielectric)
k = ?       (dielectric constant of wax)

We know that the capacitance of a parallel-plate capacitor is given by:

C = εA/d

Where ε is the permittivity of the medium between the plates, A is the area of each plate, and d is the distance between the plates.

When the plate separation is doubled, the capacitance becomes half (because d is in the denominator). So, we can write:

C2 = (εwax * A)/2d     (with wax as dielectric)
C1 = (εair * A)/d      (with air as dielectric)

Dividing the two equations, we get:

C2/C1 = (εwax/εair)*2

2 = (εwax/εair)*2

εwax/εair = 2/2 = 1

Therefore, the dielectric constant of the wax is 1 times that of air, which is 4 (since air has a dielectric constant of approximately 1).

Hence, the answer is D (4.0).

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