Physics - Capacitor Question with Solution | TestHub

Potential difference across the branch de is $6\mathrm{V}$. Net capacitance of de branch is $2.1\mu \mathrm{F}$ So, $q=CV$ $\Rightarrow \mathrm{q}=2.1\times 6\mu \mathrm{C}\Rightarrow \mathrm{q}=12.6\mu \mathrm{C}$ Potential across $3\mu \mathrm{F}$ capacitance is $\mathrm{V}=\frac{12.6}{3}=4.2\mathrm{volt}$ Potential across 2 and 5 combination in parallel is $6-4.2=1.8\mathrm{V}$ So, $q^{\prime }=(1.8)(5)=9\mu C$