Answer:
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:
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
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,
[tex]F = \frac{(8.99\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(2\ x\ 10^{-6}\ C)}{(0.1\ m)^2}[/tex]
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.
