Physics - Capacitor Question with Solution | TestHub

$V_2-V_1$ will be divided between $C_1$ and $C_2$. Let $V_D$ be potential at point D. Let $\mathrm{Q}$ be the charge on each capacitor. Then the potential drop can be written as: $\begin{array}{rl}& V_2-V_1=V_2-V_D+V_D-V_1\\ & \Rightarrow V_2-V_1=\frac{Q}{C_2}+\frac{Q}{C_1}\\ & \Rightarrow Q=\frac{C_1{C}_2\left(V_2-V_1\right)}{C_1+C_2}\end{array}$ Potential difference across $C_1$ is $V_D=V_1+\frac{C_2\left(V_2-V_1\right)}{C_1+C_2}=\frac{C_2\left(V_1+V_2\right)}{C_1+C_2}$