Question

Two parallel plate capacitors whose capacities are C and 2C respectively, are joined in parallel. These arr charged to a potential difference of V. If the battery is removed and a dielectric of dielectric constant K isc filled in between there plates of the capacitor C, then what will be there potential difference across each capacitor ?

Answer 1

Dear Student,

$$\begin{gathered} Energyofthecapacitorbeforeremovingbattery: \\ {U}_i=\frac{1}{2}{C}_{eff}{V}^2=\frac{1}{2}(C+2C)V^2=\frac{3}{2}C{V}^2 \\ Finalenergyofthecapacitorafterremovingbattery: \\ {U}_f=\frac{1}{2}C{V}_1^2+\frac{1}{2}\left(2C\right)V_2^2 \\ where,V_{1}isthevoltagedropacross\text{'}C\text{'}and{V}_{2}bethatacross\text{'}2C\text{'},afterremovingthebattery. \\ now,fromtheconservationofenergy: \\ {U}_i=U_f \\ \frac{3}{2}C{V}^2=\frac{1}{2}C{V}_1^2+\frac{1}{2}\left(2C\right)V_2^2\left(1\right) \\ Also,weknowthat, \\ V=V_1+V_2\left(2\right) \\ Solvingequations\left(1\right)\&\left(2\right),wegetthevaluesof{V}_{1}and{V}_2 \\ \frac{3}{2}C{V}^2=\frac{1}{2}C(V-V_2{)}^2+C{V}_2^2 \\ 3C{V}^2=C(V^2+V_2^2-2V{V}_2)+2C{V}_2^2 \\ 2C{V}^2=3C{V}_2^2-2CV{V}_2 \\ 3C{V}_2^2-2CV{V}_2-2C{V}^2=0 \\ {V}_2=\frac{-(-2CV)\pm \sqrt{(-2CV{)}^2-4\times 3C\times -2C{V}^2}}{2\times 3C} \\ {V}_2=\frac{(1+\sqrt{7})V}{3},\frac{(1-\sqrt{7})V}{3} \\ Now,V_1=V-V_2=V-\frac{(1\pm \sqrt{7})V}{3} \\ {V}_1=\frac{(2+\sqrt{7})V}{3},\frac{(2-\sqrt{7})V}{3} \end{gathered}$$

Regards,
Manoj Singh.