Question

A positively charged particle Q1 = +45 nC is held fixed at the origin. A second charge Q2 of mass m = 6.5 μg is floating a distance d = 25 cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth.

Answer 1

Answer:

Q₂ = (9.83 × 10⁻⁹) C = +9.83 nC

Explanation:

The force of attraction/repulsion between the two charges is equal to the force of gravity on the charge 2.

The force of gravity on the charge two = mg

= 6.5 μg × 9.8 m/s² = (6.37 × 10⁻⁵) N

Since the gravity force is directed downwards, the force on charge 2 due to charge 1 has to be directed upwards, hence it is a force of repulsion.

The magnitude of the force between the two charges is given according to the Coulomb's law.

F = kQ₁Q₂/r²

k = Coulomb's constant = (9.0 × 10⁹) Nm²/C²

Q₁ = 45 nC = (45 × 10⁻⁹) C

Q₂ = ?

r = 25 cm = 0.25 m

F = (6.37 × 10⁻⁵) N

(6.37 × 10⁻⁵) = [(9.0 × 10⁹) × (45 × 10⁻⁹) × Q₂] ÷ 0.25²

Q₂ = 0.00000000983 C = (9.83 × 10⁻⁹) C = 9.83 nC

Hope this Helps!!!