At steady state current does not flow in the branch of capacitor. ∴ we can replace all resistor connected in branch of capacitor with wire new circuit is
i=
6
(2+1)×103
⇒ 2 mA Potential drop across 2kΩ is same as potential drop across 1 μF & 2 μF. Potential drop across 2 kΩ = i × 2 × 103 = 2 × 10–3 × 2 × 103 = 4 volt. Charge on 1μF = Q = 1 × 4 × 10–6 = 4 μC Charge on 2μF = q = 2 × 4 × 10–6 = 8 μC