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 _________ $\times 10^{-4} \mathrm{~V}$.
(we modified language of question to make it correct)

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 $\mathrm{K}_1=2, \mathrm{~K}_2=3$ respectively as shown in figure. If new force between the plates is $8 \mathrm{~N}$. The supply voltage is _________ $\times 10^{-4} \mathrm{~V}$.

(we modified language of question to make it correct)

Aicharya · Accepted Answer

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

A capacitor of $10 \mu \mathrm{F}$ capacitance whose plates are separated by $10 \mathrm{~mm}$ through air…

Solution