When an additional charge of 2C is given to a capacitor , energy stored in it is increased by 21 % . The original charge of the capacitor is :

Options

A. 30 C
B. 40 C
C. 10 C
D. 20 C

Answer

D

Explanation

Energy stored by a capacitor ` U = 1/2 (q_(2)^(2))/C`
So . Here ` :. " " (U_(1))/(U_(2)) = ((q_(1))/(q_(2)))^(2) = (q/(q+2))^(2)`
` :. " " (U_(1))/(U_(2)) = ((q_(1))/(q_(2)))^(2) = (q/(q+2))^(2)`
Given , `(U_(2)-U_(1))/(U_(1)) = 0.21 `
or `(U_(2))/(U_(1)) = 1.21 rArr (q/(q+2))^(2) = 1.21 `
On solving , we get
q = 20 C