Physics 2011 JAMB Past Questions - Question 33: The capacitance of a parallel plate capacitor is 20μF in air and 60μF in the presence of a dielectric.What is the dielectric constant ?

Correct Answer

A

Explanation

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

C = (ε * A) / d

Where:
C = Capacitance
ε = Permittivity of the material between the plates
A = Area of the plates
d = Separation between the plates

In the case of a parallel plate capacitor in air (without a dielectric), the permittivity is ε₀ (epsilon naught), which is approximately 8.854 x 10⁻¹² F/m (farads per meter).

In the presence of a dielectric, the capacitance becomes:

C_with_dielectric = (κ * ε₀ * A) / d

Where:
C_with_dielectric = Capacitance with the dielectric
κ (kappa) = Dielectric constant
ε₀ = Permittivity of free space (8.854 x 10⁻¹² F/m)
A = Area of the plates
d = Separation between the plates

We are given that the capacitance in air (without the dielectric) is 20 μF, which is equal to 20 x 10⁻⁶ F, and the capacitance with the dielectric is 60 μF, which is equal to 60 x 10⁻⁶ F.

We can set up the following equation:

20 x 10⁻⁶ F = (κ * 8.854 x 10⁻¹² F/m * A) / d

60 x 10⁻⁶ F = (κ * 8.854 x 10⁻¹² F/m * A) / d

Now, let's divide the second equation by the first:

(60 x 10⁻⁶ F) / (20 x 10⁻⁶ F) = (κ * 8.854 x 10⁻¹² F/m * A) / d / ((κ * 8.854 x 10⁻¹² F/m * A) / d)

3 = κ

So, the dielectric constant (κ) is 3.