#### Question

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

#### 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.

Is there an error in this question or solution?

Solution 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. Concept: Capacitors and Capacitance.