Question

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?

Answer 1

Step-by-step explanation:

E2020

Answer 2

F = 10.78 N

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

Step-by-step explanation:

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

`F = (kq_1q_2)/(r^2)`

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 = ((8.99\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^(-6)\ C)(2\ x\ 10^(-6)\ C))/((0.1\ m)^2)`

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.