Question

An isolated, parallel‑plate capacitor carries a charge Q . If the separation between the plates is doubled, the electrical energy stored in the capacitor will be:a. Unchanged.b. Halved.c. Doubled.d. Quartered.e. Quadrupled.

Answer 1

option C

Step-by-step explanation:

Given,

Change on the capacitor = Q

Separation is doubled

Energy stored in the capacitor,E = ?

we know,

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

and
`C = (\epsilon_0A)/(d)`

now,

`E = (Q^2)/(2\epsilon A)\ d`.......(1)

where d is the separation between the two plates.

now, when the separation is doubled

`E' = (Q^2)/(2\epsilon A)\ (2d)`

`E' = 2((Q^2)/(2\epsilon_0 A)\ d)`

From equation (1)

E' = 2 E

Hence, the energy stored in the capacitor is doubled if the separation is increased.

The correct answer is option C.