An illustration with two positive spheres 0.1m apart. The one on the left is labeled q Subscript 1 baseline = 6 microcoulombs and the sphere on the right is labeled q Subscript 2 baseline = 2 microcoulombs. Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.

Recall that k = 8.99 × 109 N•meters squared per Coulomb squared.

What is the force applied between q1 and q2?

In which direction does particle q2 want to go?

F = 10.78 N

Hence q₂ will move away from the charge q₁ towards right side.

Explanation:

The force between two charged particles can be found by using Colomb's Law:

where,

F = Force = ?

k = Colomb Constant = 8.99 x 10⁹ N.m²/C²

q₁ = charge on first particle = 6 μC = 6 x 10⁻⁶ C

q₂ = charge on second particle = 2 μC = 2 x 10⁻⁶ C

r = distance between particles = 0.1 m

Therefore,

F = 10.78 N

Since both particles have a positive charge. Therefore this force will be the force of repulsion.

Hence q₂ will move away from the charge q₁ towards right side.

the answer would be   dignified - apex