Question
Find the charge on the capacitor shown in figure.
✓
Answer
In steady stage condition no current flows through the capacitor. 
$\text{R}_\text{eff}=10+20=30\Omega$
$\text{i}=\frac{2}{30}=\frac{1}{15}\text{A}$
Voltage drop across $10\Omega$ resistor = i × R$=\frac{1}{15}\times10=\frac{10}{15}=\frac{2}{3}\text{V}$
Charge stored on the capacitor (Q) = CV$=6\times10^{-6}\times\frac{2}{3}=4\times10^{-6}\text{C}=4\mu\text{C}.$
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