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Q. 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:-

JEE MainJEE Main 2019Electrostatic Potential and Capacitance

Solution:

Charges at inner plates are $1 \mu C$ and $- 1 \mu C$
$\therefore $ Potential difference across capacitor
$ =\frac{q}{c} = \frac{1 \mu C}{1 \mu F} = \frac{1\times10^{-6}C}{1 \times10^{-6} Farad} = 1V $

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