JEE MAIN - Physics (2024 - 6th April Evening Shift - No. 25)
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 $$K_1=2, K_2=3$$ respectively as shown in figure. If new force between the plates is $$8 \mathrm{~N}$$. The supply voltage is ________ V.
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Answer
80
Explanation
$$\begin{array}{ll} C_1=10 \mu \mathrm{F} & C_2=15 \mu \mathrm{F} \\ F=F_1+F_2 & \end{array}$$
$$\begin{aligned} & F=\frac{\left(10 \times 10^{-6} V\right)^2}{2 \cdot \frac{A}{2} \cdot \epsilon_0}+\frac{(15~~10)}{2 \cdot \frac{A}{2} \cdot \epsilon_0}=8 \\ & V^2=0.9 \times 10^{-4} \\ & V=0.95 \times 10^{-2} \mathrm{~V} \\ \end{aligned}$$
Data in consistent. Answer not matching.
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