Physics - Capacitor Question with Solution | TestHub

At the steady state, there will be no current through the capacitor. Therefore, potential difference across capacitor $V_C=\frac{10}{R+r}R=\frac{10\times 100}{100+10}=\frac{100}{11} \mathrm{V}$

Charge on capacitor is $q=C{V}_C$

$=\left(1.1\times {10}^{-6}\right)\left(\frac{100}{11}\right)$

$\Rightarrow q=10 \mu \mathrm{C}=10\times {10}^{-6} \mathrm{C}$.