As shown in the figure, a configuration of two equal point charges (q₀=+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 × 10⁻³ m.
The value of x is ____________.
(Take 1/(4 π ε₀)=9 × 10⁹ N m² C⁻², g=10 m s⁻² )

Solution

Physics sir · 4 steps

Along the incline, the Coulomb push must exactly cancel the mg sin 30° pull down the slope.

1. Forces on top charge. The lower charge stays at the foot of the incline. The upper charge feels its weight, the normal reaction and the Coulomb repulsion. The repulsion acts along the line joining the charges, which is along the incline. The normal reaction is perpendicular to the incline, so only the forces along the slope need to balance.

   1/4πε₀q₀²/r² = mg sin 30°

2. Put in numbers. Left side numerator: 9 × 10⁹ × (2 × 10⁻⁶)² = 0.036 N m². Right side: 0.02 × 10 × 1/2 = 0.1 N. So r² = 0.036/0.1 = 0.36 m², giving r = 0.6 m.
3. Distance to height. The separation r is measured along the incline, which is the hypotenuse. The height is the side opposite the 30° angle, so h = r sin 30° = 0.6 × 1/2 = 0.3 m.
4. So. h = 0.3 m = 300 × 10⁻³ m, so x = 300.

The common slip: 600. Taking the separation along the incline as the height gives h = 0.6 m, that is 600. The height is r sin 30°.

The idea to keep. On a smooth incline, balance forces along the slope only, where the weight contributes mg sin θ. Then check whether the question asks for the distance along the slope or the vertical height.

Answer from the NTA final answer key. Solution by PrepWiser.