Energy Stored In Capacitor Assignment Help

Electrostatics - Energy Stored In Capacitor

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We know that a capacitor is a system of two conductors carrying charges Q and – Q. to calculate the energy stored in this charge configuration, suppose the conductors 1 and 2 are initially uncharged. Let positive charge be transferred from conductor 2 to conductor 1 in every small installments of dq each till conductor 1 gets charge + Q. by charge conservation, conductor 2 would acquire charge – Q.

At every stage of charging, conductor 1 is at a higher potential than conductor 2. Therefore, work is done externally in transferring each installment of charge.

At any intermediate stage, suppose charge on suppose charge on conductor 1 is +q and charge on conductor 2 is – q.

∴ Potential difference between conductors 1 and 2 is q/C, where C is the capacity of the capacitor. 

Suppose the condenser is charged gradually, and at any stage, the charge on the conductor is q.

∴ Potential of condenser = q/C

Small amount of work done in giving an additional charge dq to the condenser is

dW = q/C × dq

total work done in giving a charge Q to the condenser

W = 1/C Q2/2

As electrostatic force is conservative, this work is stored in the form of potential energy (U) of the condenser.

U = W = ½ Q2/C

Put Q = CV

∴ U = ½ CV2/C = ½ CV2                                               (1)

Put CV = Q ∴ U = ½ QV

Hence U = ½ Q2/C = ½ CV2 = ½ QV                              (2)


When Q is in coulomb, V is in volt, C is in farad, Energy U is in joule.

The potential energy stored in the capacitor is independent of the manner in which charge configuration of the capacitor is built up.

The potential energy of a capacitor can be viewed as stored in the electric field between the plates.

The potential energy of capacitor lies in the dielectric medium between the plates.

The potential energy of capacitor is obtained at the cost of chemical energy stored in the battery – used for charging the capacitor.

Example: calculate the energy stored in a capacitor of 5  F when it is charged to a potential of 250 volt.

Solution: here, C = 5 u F = 5 × 10-6 F, 

V = 250 volt, E =?

As E = ½ CV2

E = ½ × (5 × 10-6) (250)

= 0.156 joule.

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