Question

If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C) :

(A) 10 

(B) 20

(C) 30

(D) 40

Answer 1

Correct option is (A) 10

 We know that

\(U = \frac{q^2}{2C}\)

\(\frac{U_1}{U_2} = \left(\frac{q_2}{q_1}\right)^2\)

\(\frac{U_2}{U_1} - 1 = \left(\frac{q_2}{q_1}\right)^2 - 1\)

\(\frac{U_2 - U_1}{U_1} = \frac{q_2^2}{q_1^2} - 1\)

\(\frac{U_2 - U_1}{U_1} \times 100\) \(= \left(\frac{q_2^2}{q_1^2} - 1\right) \times 100\)

\(44 = \left(\frac{q_2^2}{q_1^2} - 1\right) \times 100\)

\(\frac{q_2^2}{q_1^2} = 0.44 + 1\)

\(q_2 = 1.2 q_1\)

then,

\(q_2 - q_1 = 2c \)

\(0.2 q_1 = 2\)

\(q_1 = \frac{2}{0.2}\)

\(q_1 = 10 c\)