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?

(a) The capacitance of the capacitor.
(b) The electric field between the plates of the capacitor.
(c) The energy stored in the capacitor.

Justify your answer in each case.

Hint:

The charge stays constant. \(C' = kC/2 = 0.9C\), \(E' = E/k\), and \(U' = Q^2/(2C') = U/0.9\).

Answer 1

Step 1: Understanding the Concept:
The battery is disconnected, so the charge \(Q = CV\) on the plates stays constant. We then change the gap and add a dielectric, and track \(C\), \(E\) and \(U\) using the fixed charge.

Step 2: (a) Capacitance:
Initially \(C = \dfrac{\epsilon_0 A}{d}\).
After doubling the gap and filling it with the slab: \(C' = \dfrac{k\epsilon_0 A}{2d} = \dfrac{k}{2}C = \dfrac{1.8}{2}C = 0.9C\).
So the capacitance decreases by 10 percent, to \(0.9C\). The doubled gap halves \(C\), and the slab raises it by 1.8 times, which gives a net factor of 0.9.

Step 3: (b) Electric Field:
Initially \(E = \dfrac{\sigma}{\epsilon_0} = \dfrac{Q}{\epsilon_0 A}\).
The charge \(Q\) and the area \(A\) are unchanged. With the dielectric, the field becomes \(E' = \dfrac{Q}{k\epsilon_0 A} = \dfrac{E}{k} = \dfrac{E}{1.8} \approx 0.56E\).
The field does not depend on the plate separation when the charge is constant. So the field decreases to about 56 percent of the original value. The dielectric slab reduces it by polarisation.

Step 4: (c) Energy Stored:
Initially \(U = \dfrac{Q^2}{2C}\). Finally \(U' = \dfrac{Q^2}{2C'} = \dfrac{Q^2}{2(0.9C)} = \dfrac{U}{0.9} \approx 1.11U\).
So the stored energy increases by about 11 percent. The work done in pulling the plates apart is more than the energy returned when the slab is inserted, and the difference is stored in the field.

Final Answer:
(a) The capacitance falls to \(0.9C\). (b) The field falls to \(E/1.8\). (c) The energy rises to about \(1.11U\).