A capacitor of 10 μF capacitance whose plates are separated by 10 mm through air and each plate has area ​4 cm^2

Question

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 \(K_{1}=2, K_{2}=3\) respectively as shown in figure. If new force between the plates is \(8 \mathrm{~N}\). The supply voltage is _______V.

Answer 1

Correct answer is : 80

\(\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}\)

\(\mathrm{C}_{1}=\frac{2 \epsilon_{0} \mathrm{~A}}{2 \times \mathrm{d}}=10 \mu \mathrm{F}\)

\(\mathrm{C}_{2}=\frac{3 \epsilon_{0} \mathrm{~A}}{2 \mathrm{d}}=15 \mu \mathrm{F}\)

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

Now the charge on

\( \begin{aligned} & \mathrm{C}_{1}=10 \mathrm{V} \mu \mathrm{c} \end{aligned} \)

\(\begin{aligned} \mathrm{C}_{2}=1.5 \mathrm{~V} \mu \mathrm{C} \end{aligned} \)

Now force between the plates \(\left[\mathrm{F}=\frac{\mathrm{Q}^{2}}{2 \mathrm{A} \ \in_{0}}\right]\)

\(\frac{100 \mathrm{V}^{2} \times 10^{-12}}{2 \times 2 \times 10^{-4} \ \epsilon_{0}}+\frac{225 \mathrm{V}^{2} \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \ \epsilon_{0}}=8\)

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

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

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

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