In the given circuit, a charge of $+80 \ \mu C$ is given to the upper plate of the $4 \ \mu F$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \ \mu F$ capacitor is :
IIT 2012, Diffcult
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$q_3=\frac{C_3}{C_2+C_3} \cdot Q $
$q_3=\frac{3}{3+2} \times 80=\frac{3}{5} \times 80 $
$=48 \ \mu C$
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