Physics 2006 JAMB Past Questions - Question 32: The capacitance of a parallel-plate capacitor when in air is 3µf and in the presence of a dielectric material 6 µf. The dielectric constant is

Correct Answer

C

Explanation

The capacitance of a parallel-plate capacitor is determined by the formula:

C = (ε * A) / d

Where:
C is the capacitance.
ε is the permittivity of the material between the plates (in this case, it's air or a dielectric material).
A is the area of one of the capacitor plates.
d is the separation distance between the plates.

When the capacitor is in air, you have C1 = 3 µF.

When the capacitor is in the presence of a dielectric material, you have C2 = 6 µF.

The dielectric constant (also known as the relative permittivity), denoted by εr, is a measure of how much the capacitance of a capacitor is increased when a dielectric material is inserted between the plates. It's defined as:

εr = C2 / C1

In this case:

εr = (6 µF) / (3 µF) = 2

So, the dielectric constant (relative permittivity) of the material is 2.