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? (b) The energy stored in the capacitor Justify your answer by writing the necessary expressions

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#### Solution

As the capacitance of the capacitor,

`C'=(in_0KA)/(d')=(in_0KA)/(2d)=1/2C " ...1"`

Energy stored in the capacitor is

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

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

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

Concept: Capacitors and Capacitance

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