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Q.
In the circuit given below, the charge in $\mu C$, on the capacitor having capacitance $5\, \mu F$ is
Electrostatic Potential and Capacitance
Solution:
Potential difference across the branch de is $6 V$.
Net capacitance of de branch is $2.1 \mu F$
So, $q = CV =2.1 \times 6 \mu C =12.6 \mu C$
Potential across $3 \mu F$ capacitance is
$V =\frac{12.6}{3}=4.2 \text { volt }$
Potential across 2 and 5 combination in parallel is $6-4.2=1.8 V$
$\text { So, } q^{\prime}=(1.8)(5)=9 \mu C$
