Physics - Capacitor Question with Solution | TestHub

Consider the potential at D be 'V'. Potential drop across $C_1$ is $\left(V-V_1\right)$ and $C_2$ $\mathrm{i}$ $\left({\mathrm{V}}_2-\mathrm{V}\right)$ $\therefore {\mathrm{q}}_1={\mathrm{C}}_1\left(\mathrm{V}-{\mathrm{V}}_1\right),{\mathrm{q}}_2={\mathrm{C}}_2\left({\mathrm{V}}_2-\mathrm{V}\right)$ As ${\mathrm{q}}_1={\mathrm{q}}_2$ [capacitors are in series] $$\begin{array}{l}\therefore C_1\left(V-V_1\right)=C_2\left(V_2-V\right)\\ V=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}\end{array}$$