If the charge on a capacitorn is increased by 2C, then the energy stored in it increases by 20 %. The original charge on the capacitor is

Options

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

Answer

B

Explanation

The energy stored in the capacitor is given by
`E_(1)=(1)/(2) QV=(1)/(2)(Q^(2))/(C) and E_(2)=(121)/(100)((1)/(2)((Q+2)^(2))/(C))`
[C will not change]
`therefore (E_(2))/(E_(1))=(121)/(100)xx((Q+2)^(2))/(Q^(2))`
Taking the square roots, we get `(Q+2)/(Q)=(11)/(10)`
`therefore 11Q=10Q=20" "therefore Q=20C`