Consider the circuit given below and calculate the charge $\text{(in μC ) }$ on the capacitors.

$I=\frac{3}{6\times {10}^3}=0.5\times {10}^{-3}\text{A}$ 

$V_{AD}=IR=0.5\times {10}^{-3}\times (2+1)\times {10}^3=1.5\text{V}$
The equivalent capacitance of the given capacitors will be. 
  Charge on the capacitors.