A capacitor of capacitance $$10 \mu\text{F}$$ 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 figure. If new force between the plates is $$8 \text{ N}$$. The supply voltage is ___________ $$\text{ V}$$.

$$C_{\text{net}} = C_1 + C_2 = \frac{K_1 C_0}{2} + \frac{K_2 C_0}{2} = \frac{(K_1 + K_2)C_0}{2}$$

$$C_{\text{net}} = \frac{(2 + 3) \times 10 \times 10^{-6}}{2} = 25 \times 10^{-6}\ \text{F}$$

$$F = \frac{1}{2}\frac{C_{\text{net}} V^2}{d} \implies 8 = \frac{1}{2} \times \frac{25 \times 10^{-6} \times V^2}{10^{-2}}$$

$$8 = 1.25 \times 10^{-3} V^2$$

$$V^2 = \frac{8}{1.25 \times 10^{-3}} = 6400 \implies V = 80\ \text{V}$$