Question

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant K is inserted to fill the whole space between the plates with the voltage source remaining connected to the capacitor.

1. the energy stored in the capacitor will become 1/K times
2. the electric field inside the capacitor will decrease K²-times
3. the energy stored in the capacitor will become K-times
4. the charge on the capacitor will become 1/K-times

Answer 1

Correct Answer - Option 3 : the energy stored in the capacitor will become K-times

Concept:

The capacitance of a parallel plate capacitor is given by:

\(C=\frac{Aϵ}{d}\)

A = Area of the plates

d = distance between the plates

ϵ = permittivity of the medium

Also, the charge stored is given by:

Q = CV

The energy stored by a capacitor connected to a voltage source V is given by:

\(E=\frac{1}{2}CV^2=\frac{1}{2}QV=\frac{1}{2}\frac{Q^2}{C}\)

Analysis:

With no dielectric inserted, the capacitance is:

\(C_1=\frac{Aϵ_0}{d}\)   ---(1)

Q₁ = C₁V

\(E_1=\frac{1}{2}C_1V^2\)

After the dielectric is inserted, the capacitance will become:

\(C_2=\frac{AKϵ_0}{d}\)   ---(2)

Comparing Equations (1) and (2), we get:

C₂ = KC₁

Since the voltage applied is the same, the energy stored by the capacitor after the insertion of the dielectric wil be:

\(E_2=\frac{1}{2}C_2V^2\)

∴ E₂ = K × E₁

Notes: Since Q = CV, the charge becomes KQ as C becomes KC and V remains unchanged.