Q 12-02-179JEE MainJEE Main 2019 (8 Apr, Shift 2)Medium
A parallel plate capacitor has $1\ \mu\text{F}$ capacitance. One of its two plates is given $+2\ \mu\text{C}$ charge and the other plate, $+4\ \mu\text{C}$ charge. The potential difference developed across the capacitor is
Answer: (A) $1\ \text{V}$
Charges on the facing (inner) surfaces are $\pm\dfrac{q_2 - q_1}{2} = \pm1\ \mu\text{C}$; only these create the field between the plates.
$$V = \frac{|q_2 - q_1|}{2C} = \frac{2\ \mu\text{C}}{2\times1\ \mu\text{F}} = 1\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics