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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 . . . .. . .V.
$50$
$80$
$60$
$30$
Solution
$\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2$
$\mathrm{C}_1=\frac{2 \epsilon_0 \mathrm{~A}}{2 \times \mathrm{d}}=10 \mu \mathrm{F}$
$\mathrm{C}_2=\frac{3 \epsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=15 \mu \mathrm{F}$
$\mathrm{C}_{\mathrm{eq}}=25 \mu \mathrm{F}$
Now the charge on
$\mathrm{C}_1=10 \mathrm{~V} \mu \mathrm{c}$
$\mathrm{C}_2=1.5 \mathrm{~V} \mu \mathrm{C}.$
Now force between the plates $\left[\mathrm{F}=\frac{\mathrm{Q}^2}{2 \mathrm{~A} \epsilon_0}\right]$
$\frac{100 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \epsilon_0}+\frac{225 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \epsilon_0}=8$
$325 \mathrm{~V}^2=8 \times 4 \times 10^{-4} \times 8.85$
$\mathrm{~V}^2-\frac{32 \times 8.85 \times 10^{-4}}{325}$
$\therefore \mathrm{V}=\sqrt{\frac{283.2 \times 10^{-4}}{325}}$
$\mathrm{~V}=0.93 \times 10^{-2}$
