A parallel-plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change:

(i) electric field between the plates

(ii) capacitance, and

(iii) energy stored in the capacitor

#### Solution

**(i)**

*Q* = *CV*

`Q = ((epsi_0A)/d) (Ed)`

`Q = epsi_0AE`

`therefore E =Q/(epsi_0A)`

Therefore, the electric field between the parallel plates depends only on the charge and the plate area. It does not depend on the distance between the plates.

Since the charge as well as the area of the plates does not change, the electric field between the plates also does not change.

(ii)

Let the initial capacitance be *C* and the final capacitance be *C'.*

Accordingly,

`C = (epsiA)/d`

`C' = (epsi_0A)/(2d)`

`C/C' = 2`

`C' \ C/2`

Hence, the capacitance of the capacitor gets halved when the distance between the plates is doubled.

(iii)

Energy of a capacitor, *U *`=1/2 (Q_2)/C`

Since *Q* remains the same but the capacitance decreases,

`U' = 1/2 (Q^2)/((C/2))`

`U/U' = 1/2`

U' = 2U

The energy stored in the capacitor gets doubled when the distance between the plates is doubled.