JEE MAIN - Physics (2023 - 11th April Morning Shift - No. 20)
As shown in the figure, a configuration of two equal point charges $$\left(q_{0}=+2 \mu \mathrm{C}\right)$$ is placed on an inclined plane. Mass of each point charge is $$20 \mathrm{~g}$$. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height $$\mathrm{h}=x \times 10^{-3} \mathrm{~m}$$.
The value of $$x$$ is ____________.
(Take $$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}, g=10 \mathrm{~m} \mathrm{~s}^{-2}$$ )
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Answer
300
Explanation
For equilibrium along the plane
$$ \begin{aligned} & \mathrm{mg} \sin \theta=\frac{1}{4 \pi \epsilon_0} \times \frac{\mathrm{q}_0^2}{\left(\mathrm{~h} \operatorname{cosec} 30^{\circ}\right)^2} \\\\ & \therefore \mathrm{h}^2=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \times \frac{\mathrm{q}_0^2}{\mathrm{mg} \operatorname{cosec} 30^{\circ}} \\\\ & =9 \times 10^9 \times \frac{\left(2 \times 10^{-6}\right)^2}{0.02 \times 10 \times 2} \\\\ & \therefore \mathrm{h}=3 \times 10^4 \times \frac{2 \times 10^{-6}}{0.2} \\\\ & =0.3 \mathrm{~m} \\\\ & =300 \mathrm{~mm} \end{aligned} $$
$$ \begin{aligned} & \mathrm{mg} \sin \theta=\frac{1}{4 \pi \epsilon_0} \times \frac{\mathrm{q}_0^2}{\left(\mathrm{~h} \operatorname{cosec} 30^{\circ}\right)^2} \\\\ & \therefore \mathrm{h}^2=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \times \frac{\mathrm{q}_0^2}{\mathrm{mg} \operatorname{cosec} 30^{\circ}} \\\\ & =9 \times 10^9 \times \frac{\left(2 \times 10^{-6}\right)^2}{0.02 \times 10 \times 2} \\\\ & \therefore \mathrm{h}=3 \times 10^4 \times \frac{2 \times 10^{-6}}{0.2} \\\\ & =0.3 \mathrm{~m} \\\\ & =300 \mathrm{~mm} \end{aligned} $$
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