As shown in the figure, a configuration of two equal point charges (q₀=+2 μC) is placed on an inclined plane [the incline makes 30° with the horizontal; one q₀ rests on the incline at a vertical height h above the other q₀, which sits at the foot of the incline]. The mass of each point charge is 20…

Question

As shown in the figure, a configuration of two equal point charges (q₀=+2 μC) is placed on an inclined plane [the incline makes 30° with the horizontal; one q₀ rests on the incline at a vertical height h above the other q₀, which sits at the foot of the incline]. The mass of each point charge is 20 g. Assume that there is no friction between the charge and the plane. For the system of two point charges to be in equilibrium (at rest), the height h=x×10⁻³ m. The value of x is . (Take1/(4πε₀)=9×10⁹ N m² C⁻², g=10 m s⁻²)

Accepted Answer

300. Both charges lie on the smooth incline, separated by a distance d measured along the slope, so the Coulomb repulsion acts along the incline. For the upper charge, the only other along-incline force is the component of its weight, mg sin 30°, pointing down the slope. Equilibrium: (kq₀²)/(d²)=mg sin 30°, with m=0.020 kg. kq₀²=9×10⁹×(2×10⁻⁶)²=3.6×10⁻², and mg sin 30°=0.020×10×0.5=0.1 N. d²=(3.6×10⁻²)/(0.1)=0.36⇒ d=0.6 m. The vertical height is h=d sin 30°=0.6×0.5=0.3 m =300×10⁻³ m, so x=300.

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