Explanation

(c) As continuous flow of DC do not take place through a capacitor, Therefore resistance of the circuit
R = 1+0.5 = 1.5
Current with circuit
$E = \frac{E}{R^{'}}$
$= \frac{2}{1.5} = \frac{4}{3}A$
Potential difference across capacitor =Potential difference across $1\ \Omega$ resistor
$= \frac{4}{3} \times 1 = \frac{4}{3}V$
$\therefore\ $Charge on capacitor$\ q = CV$
$= 1 \times \frac{4}{3} = 1.33\ \mu F$