Physics - Capacitor Question with Solution | TestHub

We know that in steady state the capacitor behaves like as an open circuit i.e., capacitor will not pass the current.
So, the potential difference across the capacitor $=45\mathrm{V}$ Hence, the final charge on the capacitor is $$q=CV$$ Here, $C=20\mu \mathrm{F},V=45\mathrm{V}$ $\therefore$ $$q=20\times 10^{-6}\times 45$$ or $$q=900\times 10^{-6}$$ Or $$q=9\times 10^{-4}\mathrm{C}$$