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

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:
$325 V^2 = 8 \times 4 \times 10^{-4} \times 8.85$
$V^2 = \frac{32 \times 8.85 \times 10^{-4}}{325}$
$\therefore V = \sqrt{\frac{283.2 \times 10^{-4}}{325}}$
$V = 0.93 \times 10^{-2}$