Question

A parallel plate capacitor of capacitance C has a dielectric slab between its plates. It is charged to a potential difference V by connecting it across a battery. The battery is then disconnected. If the dielectric slab is now withdrawn from the capacitor, how will the following be affected?

1. Сараcitance of the capacitor,
2. Energy stored in the capacitor, and
3. The potential difference between the plates of the capacitor.

Justify your answer in each case.

Answer 1

When the battery is disconnected, the charge (Q) on the plates remains constant because there is no path for it to flow.

Here is how the properties change when the dielectric (dielectric constant K > 1) is removed:

1. The capacitance (C) decreases. The capacitance of a parallel plate capacitor is C = `(K epsilon_0 A)/d`. Removing the dielectric means K drops to 1, so the capacitance decreases by a factor of `1/K`.
2. The energy stored (U) increases. Using the formula U = `Q^2/(2 C)`, we see that since Q is constant and C decreases, the energy stored increases. This extra energy comes from the work done by an external agent to pull the slab out against the electrostatic attraction of the plates.
3. The potential difference (V) increases. Since V = `Q/C` and Q is constant, the decrease in capacitance causes the potential difference to increase by a factor of K.