A parallel plate capacitor having plate area 'A' and separation 'd' is charged to a potential difference ' V '. The charging battery is disconnected and the plates are pu

Question

A parallel plate capacitor having plate area 'A' and separation 'd' is charged to a potential difference ' V '. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between the plates is ( ϵ 0 ​ = permittivity of free space)

TestHub · Accepted Answer

Initial capacitance of the parallel plate capacitor is $C_{1} = \frac{ϵ _{0} A}{d}$ .

Initial energy stored in the capacitor is $U_{1} = \frac{1}{2} C_{1} V^{2} = \frac{ϵ _{0} A V ^{2}}{2 d}$ .

Charge on the capacitor is $Q = C_{1} V = \frac{ϵ _{0} A V}{d}$ .

Since the battery is disconnected, the charge $Q$ on the plates remains constant.

When the plates are pulled apart to a distance $d_{2} = 4 d$ , the new capacitance becomes $C_{2} = \frac{ϵ _{0} A}{4 d} = \frac{C _{1}}{4}$ .

Final energy stored in the capacitor is $U_{2} = \frac{Q ^{2}}{2 C _{2}} = \frac{Q ^{2}}{2 ( C _{1} /4 )} = 4 (\frac{Q ^{2}}{2 C _{1}}) = 4 U_{1}$ .

The work required to increase the distance between the plates is equal to the change in the potential energy of the capacitor.

$W = U_{2} - U_{1} = 4 U_{1} - U_{1} = 3 U_{1}$

$W = 3 (\frac{ϵ _{0} A V ^{2}}{2 d}) = \frac{3 ϵ _{0} A V ^{2}}{2 d}$

Answer: $\frac{3 ϵ _{0} A V ^{2}}{2 d}$

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