Physics - Electrostatics Question with Solution | TestHub

Charge on $6 \mathrm{\mu F}$ capacitor $=$ Charge on $12 \mathrm{\mu F}$ capacitor.                 $\Rightarrow 6\times {10}^{-6}\times 2 \mathrm{V}=12\times {10}^{-6} \mathrm{C}$  $\therefore$    Potential difference across $12 \mathrm{\mu F}$ capacitor is                 $\mathrm{V}=\frac{12\times {10}^{-12}}{12\times {10}^{-12}}=1 \mathrm{V}$ Battery voltage$=2 \mathrm{V}+1 \mathrm{V}=3 \mathrm{V}$