QJEEM23Ph1eff: JEE Main 2023 Physics - Class 12 JEEM23Ph1eff Solution: As shown in the figure, a configuration of two equal point…

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QJEEM23Ph1eff: JEE Main 2023 Physics - Class 12 JEEM23Ph1eff Solution: As shown in the figure, a configuration of two equal point…

As shown in the figure, a configuration of two equal point charges $(q_{0} = + 2 μ C)$ is placed on an inclined plane. Mass of each point charge is $20 g$ . 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 $h = x \times 10^{- 3} m$ .

The value of $x$ is ____________.

(Take $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}, g = 10 m s^{- 2}$ )

JEE Main 2023 (Online) 11th April Morning Shift Physics - Electrostatics Question 59 English

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Accepted Answer

For equilibrium along the plane

\begin{aligned} mg \sin θ = \frac{1}{4 π ϵ_{0}} \times \frac{q_{0}^{2}}{{(h cosec 30^{\circ})}^{2}} \\ ∴ h^{2} = \frac{1}{4 π ϵ_{o}} \times \frac{q_{0}^{2}}{mg cosec 30^{\circ}} \\ = 9 \times 10^{9} \times \frac{{(2 \times 10^{- 6})}^{2}}{0.02 \times 10 \times 2} \\ ∴ h = 3 \times 10^{4} \times \frac{2 \times 10^{- 6}}{0.2} \\ = 0.3 m \\ = 300 mm \end{aligned}

QJEEM23Ph1eff: JEE Main 2023 Physics - Class 12 JEEM23Ph1eff Solution: As shown in the figure, a configuration of two equal point…

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