JEE Main 2022 · 29 Jul · Shift 2 · Physics Q33

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PhysicsSingle correct+4 / -1Question 33 of 90

Two identical thin metal plates has charge $q_1$ and $q_2$ respectively such that $q_1>q_2$ . The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

\dfrac{(q_1+q_2)}C

\dfrac{(q_1-q_2)}C

\dfrac{(q_1-q_2)}{2C}

\dfrac{2(q_1-q_2)}C

Answer and worked solution

Official answer: C

Facing surfaces carry $\pm\frac{q_1-q_2}2$ , so $V=\dfrac{q_1-q_2}{2C}$ .

(C)

Topics:Capacitors and capacitance

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JEE Main 2022 · 29 Jul · Shift 2, Physics question 33