Question

A capacitor of $10 \, \mu\text{F}$ capacitance whose plates are separated by $10 \, \text{mm}$ through air and each plate has area $4 \, \text{cm}^2$ is now filled equally with two dielectric media of $K_1 = 2$, $K_2 = 3$ respectively as shown in the figure. If new force between the plates is $8 \, \text{N}$, the supply voltage is ____ $\text{V}$.

Answer 1

Correct Answer: 80

The total capacitance \( C_{\text{eq}} \) of the capacitor is the sum of the two capacitances \( C_1 \) and \( C_2 \):
\[C_{\text{eq}} = C_1 + C_2\]
Capacitance \( C_1 \) with dielectric constant \( K_1 = 2 \) is:
\[C_1 = \frac{\varepsilon_0 A}{d} \cdot K_1 = 2 \cdot \frac{4 \times 10^{-4}}{10 \times 10^{-3}} \, \mu F = 10 \, \mu F\]
Capacitance \( C_2 \) with dielectric constant \( K_2 = 3 \) is:
\[C_2 = \frac{\varepsilon_0 A}{d} \cdot K_2 = 3 \cdot \frac{4 \times 10^{-4}}{10 \times 10^{-3}} \, \mu F = 15 \, \mu F\]
Thus, the total capacitance is:
\[C_{\text{eq}} = 10 \, \mu F + 15 \, \mu F = 25 \, \mu F\]
Now, the charge on the capacitors is given by:
\[C_1 = 10 \, \text{V}, \quad C_2 = 1.5 \, \text{V}\]
The force between the plates is:
\[F = \frac{Q^2}{2A\varepsilon_0}\]
Substitute the values:
\[\frac{100 V^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \varepsilon_0} + \frac{225 V^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \varepsilon_0} = 8\]
Thus, the supply voltage is:
\[325 V^2 = 8 \times 4 \times 10^{-4} \times 8.85\]
\[V^2 = \frac{32 \times 8.85 \times 10^{-4}}{325}\]
\[\therefore V = \sqrt{\frac{283.2 \times 10^{-4}}{325}}\]
\[V = 0.93 \times 10^{-2}\]