In the given circuit diagram, E = 5V, r = 1Ω, R _ 2 = 4Ω, R _ 1 = R...

Current will flow only through the branch containing resistance ${R}_{2}$

$∴ i = \dfrac{E}{{R}_{2} + r} = \dfrac{5}{4 + 1} = 1A$

Potential difference across ${R}_{2}$

$= 1 \times 4 = 4V$

If $q$ be the charge on each plate of capacitor, then

$\dfrac{q}{c} + \dfrac{q}{c} = 4$

or $\dfrac{2q}{c} = 4$

or $\dfrac{2q}{3 \times {10}^{- 6}} = 4$

$q = 6\mu C$