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A parallel plate capacitor has $1\mu$F capacitance. One of its two plates is given $+2\mu$C charge and the other plate, $+4\mu$C charge. The potential difference developed across the capacitor is:
(1) 3 V
(2) 1 V
(3) 5 V
(4) 2 V
$V = \frac{Q}{C} = \frac{Q_1 - Q_2}{2C} = \frac{4 - 2}{2 \times 1} = 1$ V.
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