JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Capacitance

Let the junction potential be $x$, with the outer plates held by the sources. Charge is conserved at the junction, so the signed charges on the three capacitors sum to zero. The $2\mu$F and $4\mu$F capacitors carry $2(x-6)$ and $4(x-6)$, while the $5\mu$F capacitor carries $5x$. Setting the sum to zero, $2(x-6)+4(x-6)+5x=0$, which gives $11x=36$ and $x=\frac{36}{11}$ V. The charge on the $5\mu$F capacitor is then $5x=\frac{180}{11}\approx 16.36\mu$C.