Two charges that are 1 meter apart repel each other with a force of 2 N. If the distance between the charges is increased to 2 meters, the force of repulsion will be:

a) 1 N b) 0.5 N c) 8 N d) 4 N

b) 0.5 N

Explanation:

From coulomb's law,

F = kq'q/r² ... Equation 1

Where F = force of repulsion between the charges, q' = first charge, q = second charge, r = distance between the charges, k = proportionality constant.

q'q = Fr²/k ... Equation 2

Given: F = 2 N, r = 1 m, k = 9.0*10⁹ Nm²/C²

Substituting into equation 2

q'q = 2 (1) ² / (9.0*10⁹)

q'q = 2/9.0*10⁹ C².

If the distance between the charges is increased to 2 meters,

r = 2 m, q'q = 2/9.0*10⁹ C².

Substitute into equation 1

F = 9.0*10⁹ (2/9.0*10⁹) / 2²

F = 2/4

F = 1/2 = 0.5 N.

The right option is b) 0.5 N