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JEE Main 2022 · 24 Jun · Shift 2, Physics question 45

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)

Choose one option
Answer and worked solution

Official answer: A

Energy U=Q22CU=\dfrac{Q^2}{2C}, so (Q+2)2Q2=1.44\dfrac{(Q+2)^2}{Q^2}=1.44.

Solve. Q+2=1.2QQ+2=1.2Q, so Q=10Q=10 C.

(A)

Topics:Energy stored in a capacitor
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