A capacitor of 50 μ F is connected in a circuit as shown in figure. The charge on the upper plate of…

At steady state current through branch of capacitor is zero. Potential difference across capacitor is $2 V.$ So, $q=50\times 2=100\mu C$

$i=\frac{6}{(2+2+2)\times {10}^3}=1\times {10}^{-3}\mathrm{Amp}$
$V_{A,B}=$ potential across $50\mu \mathrm{F}$
$V_{A,B}=i\times 2\times {10}^3=1\times {10}^{-3}\times 2\times {10}^3=2$ volt
Charge on $50\mu \mathrm{F}$
$q=CV=50\times {10}^{-6}\times 2\Rightarrow 100\times {10}^{-6}C=100\mu C$