Q. 305.0( 3 Votes )

A particle A having a charge of 2.0 × 10–6 C and a mass of 100g is placed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium?

Answer :

Charge of the particle A and B: q1=q2= q = 2.0 × 10–6 C.
Mass of the particles A and B : m = 100 g =0.1 kg.
Inclination of the inclined plain : θ = 30°

Formula used:
The particle A and B will exert electrostatic repulsive forces on each other.
This can be given by Coulomb’s Law:
We use Coulomb’s law:
Where Fe is the electrostatic force on b due to A, k is a constant .
k =
= 9× 109 Nm2C-2 and r is the distance between two charges.
Particle B would face a friction force opposite to the rolling force due to inclination. It is given as:
F= mgsinθ

For particle B to remain in equilibrium with the inclination and particle A. The electrostatic force of repulsion and the friction force must be equal.
Fe = F

Hence, particle b must be place at a distance of 0.2701m from particle A to remain in equilibrium.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.