A capacitor of \(10 \mu \mathrm{F}\) capacitance whose plates are separated by \(10 \mathrm{~mm}\) through air and each plate has area \(4 \mathrm{~cm}^2\) is now filled equally with two dielectric media of \(\mathrm{K}_1=2, \mathrm{~K}_2=3\) respectively as shown in figure. If new force between the plates is \(8 \mathrm{~N}\). The supply voltage is _______ \(\times 10^{-4}\)V.

\(\mathrm{C}_1=\frac{2 \epsilon_0 \mathrm{~A}}{2 \times \mathrm{d}}=10 \mu \mathrm{F}\)

\(\mathrm{C}_2=\frac{3 \epsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=15 \mu \mathrm{F}\)

\(\mathrm{C}_{\mathrm{eq}}=25 \mu \mathrm{F}\)

Now the charge on

\(\mathrm{C}_1=10 \mathrm{~V} \mu \mathrm{c}\)

\(\mathrm{C}_2=1.5 \mathrm{~V} \mu \mathrm{C}.\)

Now force between the plates \(\left[\mathrm{F}=\frac{\mathrm{Q}^2}{2 \mathrm{~A} \epsilon_0}\right]\)

\(\frac{100 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \epsilon_0}+\frac{225 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \epsilon_0}=8\)

\(325 \mathrm{~V}^2=8 \times 4 \times 10^{-4} \times 8.85\)

\(\mathrm{~V}^2-\frac{32 \times 8.85 \times 10^{-4}}{325}\)

\(\therefore \mathrm{V}=\sqrt{\frac{283.2 \times 10^{-4}}{325}}\)

\(\mathrm{~V}=0.93 \times 10^{-2}\)