JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 3)
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 :
$$\frac{\left(q_{1}+q_{2}\right)}{C}$$
$$\frac{\left(q_{1}-q_{2}\right)}{C}$$
$$\frac{\left(q_{1}-q_{2}\right)}{2 C}$$
$$\frac{2\left(q_{1}-q_{2}\right)}{C}$$
Explanation
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Charge on the left surface of plate $$\mathrm{A} = {{\mathrm{Total\,charge}} \over 2}$$
$$ = {{{q_1} + {q_2}} \over 2}$$
Let right surface of plate A has charge $$= x$$
And total charge on plate $$A = {q_1}$$
$$\therefore$$ $$q_1$$ = Charge on left surface of plate A + Charge on right surface of plate A
$$ = {{{q_1} + {q_2}} \over 2} + x$$
$$ \Rightarrow x = {q_1} - {{{q_1} + {q_2}} \over 2}$$
$$ = {{2{q_1} - {q_1} - {q_2}} \over 2}$$
$$ = {{{q_1} - {q_2}} \over 2}$$
Let potential difference between two plates $$= V$$
For capacitor we know,
$$q = CV$$
$$\therefore$$ $${{{q_1} - {q_2}} \over 2} = CV$$
$$ \Rightarrow V = {{{q_1} - {q_2}} \over {2C}}$$
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