Q.
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 JEEIIT JEE 2012Electrostatic Potential and Capacitance
Solution:
Between $3 \mu F $ and $2 \mu F $ (in parallel), total charge of $80 \mu C $
will distribute in direct ratio of capacity.
$\frac {q_3}{q_2}= \frac {3}{2} $
$q_3= \bigg (\frac {3}{3+2}\bigg )(80)=48 \mu C $
