A 2 microcoulomb charge is placed at a distance of 0.25 m away from a 3.6 microcoulomb charge. Describe the type of electrostatic force between the charges and calculate the magnitude of the electrostatic force between them.

The force between the charge is a repulsive force

The magnitude of the electrostatic force between them = 1.0368 N

Explanation:

Since both charges are positive charge, and they have the same sign, from the law of electrostatics, The type of force between the charges is Repulsive force.

Using

F = kqq'/r² ... Equation 1

Where F = Force between the charges, q = first charge, q' = second charge, r = distance between the charges, k = coulombs constant.

Given: q = 2 μC = 2*10⁻⁶ C, q' = 3.6 μC = 3.6*10⁻⁶ C, r = 0.25 m, k = 9*10⁹ Nm²/C²

Substitute into equation 1

F = 9*10⁹ (2*10⁻⁶) (3.6*10⁻⁶) / 0.25²

F = 64.8*10⁻³/0.0625

F = 1036.8*10⁻³

F = 1.0368 N

Hence the magnitude of the electrostatic force between them = 1.0368 N