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Question

Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery.

A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected?

(a) The electric field between the plates of the capacitor

(b) The energy stored in thecapacitor

Justify your answer by writing the necessary expressions

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Solution

Consider a parallel plate capacitor connected across a d.c. battery as shown in the figure. The electric current will flow through the circuit. As the charges reach the plate, the insulating gap does not allow the charges to move further; hence, positive charges get deposited on one side of the plate and negative charges get deposited on the other side of the plate. As the voltage begins to develop, the electric charge begins to resist the deposition of further charge. Thus, the current flowing through the circuit gradually becomes less and then zero till the voltage of the capacitor is exactly equal but opposite to the voltage of the battery. This is how the capacitor gets charged when it is connected across a d.c. battery.

(a) The electric field between the plates is

E=Vd

The distance between plates is doubled, d' = 2d

E1=V1d1=(VK)×12d=12(EK)

Therefore, if the distance between the plates is double, the electric field will reduce to one half.

(b) As the capacitance of the capacitor,

C1=ϵ0KAd1=ϵ0KA2d=12C

Energy stored in the capacitor isU=Q22C......(1)

U=Q22C1=Q22(1/2)C=2(Q22C)2U (from1)

Therefore, when the distance between the plates is doubled, the capacitance reduces to half. Therefore, energy stored in the capacitor becomes double


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