JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Capacitance

The trap here is that both plates carry charge of the same sign, so you cannot simply divide a total charge by the capacitance. When two plates carry charges $Q_1$ and $Q_2$, the charge that actually faces across the gap and sets up the field between the plates is $\frac{Q_1-Q_2}{2}$. With $+2\mu \text{C}$ on one plate and $+4\mu \text{C}$ on the other, the inner facing surfaces carry $\frac{4-2}{2}=1\mu \text{C}$ (one surface $+1\mu \text{C}$, the other $-1\mu \text{C}$), while the outer surfaces share the remainder as $+3\mu \text{C}$ each. Only that $1\mu \text{C}$ on the inner surfaces produces the potential difference, so $V=\frac{q}{C}=\frac{1\mu \text{C}}{1\mu \text{F}}=1\text{V}$.