Question 78 of 180

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 _________ $\times 10^{-4} \mathrm{~V}$.

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Answer: 93

$\begin{aligned}
& \mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2 \\
& \mathrm{C}_1=\frac{2 \epsilon_0 \mathrm{~A}}{2 \times \mathrm{d}}=10 \mu \mathrm{F} \\
& \mathrm{C}_2=\frac{3 \epsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=15 \mu \mathrm{F} \\
& \mathrm{C}_{\mathrm{eq}}=25 \mu \mathrm{F}
\end{aligned}$
Now the charge on
$\begin{aligned}
& \mathrm{C}_1=10 \mathrm{~V} \mu \mathrm{c} \\
& \mathrm{C}_2=1.5 \mathrm{~V} \mu \mathrm{C} .
\end{aligned}$
Now force between the plates $\left[F=\frac{Q^2}{2 A \in_0}\right]$
$\begin{aligned}
& \frac{100 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \epsilon_0}+\frac{225 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \epsilon_0}=8 \\
& 325 \mathrm{~V}^2=8 \times 4 \times 10^{-4} \times 8.85 \\
& \mathrm{~V}^2=\frac{32 \times 8.85 \times 10^{-4}}{325} \\
& \therefore \mathrm{V}=\sqrt{\frac{283.2 \times 10^{-4}}{325}} \\
& \mathrm{~V}=0.93 \times 10^{-2}
\end{aligned}$