A capacitor of 50 is connected in a circuit as shown in figure. The...

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

$i = \dfrac{6}{( 2 + 2 + 2 ) \times {10}^{3}} = 1 \times {10}^{- 3}Amp$
${V}_{A, B} =$ potential across $50\muF$
${V}_{A, B} = i \times 2 \times {10}^{3} = 1 \times {10}^{- 3} \times 2 \times {10}^{3} = 2$ volt
Charge on $50\muF$
$q = CV = 50 \times {10}^{- 6} \times 2 ⇒ 100 \times {10}^{- 6}C = 100\muC$