As shown in the figure, a configuration of two equal point charges $(q_{0} = + 2 μC)$ 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 $\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2}, g = 10 {ms}^{- 2}$ )

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Detailed Solution

For equilibrium, $F_{C} = mgsinθ$

$\Rightarrow \frac{1}{4 π \in_{o}} \frac{q_{o} q_{o}}{l^{2}} = mgsinθ$
and $\frac{h}{l} = sinθ \Rightarrow l = \frac{h}{sinθ}$

$∴ \frac{q_{o}^{2}}{4 π \in_{0} h^{2}} \sin^{2} θ = mgsinθ$
$\Rightarrow h = \sqrt{\frac{q_{o}^{2} sinθ}{4 π \in_{0} mg}}$

Substituting, $q_{o} = 2 \times 10^{- 6} C; sinθ = \frac{1}{2}; \frac{1}{4 π \in_{o}} = 9 \times 10^{9}$
$mg = 20 \times 10^{- 3} \times 10$

We get, $h = 0.3 m = 300 \times 10^{- 3} m$

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