Question

A parallel plate capacitor of capacitance C is charged to V volt by a battery. After sometime the battery is disconnected and the distance between the plates is doubled. A slab of dielectric constant k = 1.8 is then introduced to completely fill the space between the plates. How will the following be affected?

1. The capacitance of the capacitor.
2. The electric field between the plates of the capacitor.
3. The energy stored in the capacitor.

Justify your answer in each case.

Answer 1

a. Capacitance is defined as:

C = `(epsilon_0 A)/d`

Doubling the distance (2d) halves the capacitance `(C/2)`.

Adding a dielectric (k = 1.8) multiplies it by k.

C' = `k * (epsilon_0 A)(2 d)`

= `1.8/2 C`

= 0.9 C

It increases to 0.9 C (or effectively decreases slightly from the original C).

b. Since the battery is disconnected, the charge (Q) stays constant.

The electric field for a charged capacitor is:

E = `sigma/epsilon_0`

Introducing a dielectric reduces the field by a factor of k due to polarisation.

It decreases to `E/1.8` (or `V/(1.8 d)`)

c. We use U = `Q^2/(2 C)` because Q is constant.

The new energy is:

U' = `Q^2/(2 C')`

Since C' = 0.9 C,

then, U' = `Q^2/(2(0.9 C))`

= `U/0.9`

The increase in energy comes from the work done in pulling the plates apart, minus the energy released when the dielectric is sucked into the plates.