Question

What is the electric force acting between two charges of 0.0072 C and -0.0060 C that are 0.0040 m apart?

Answer 1

Answer:
F = -2.4 * 10¹⁰ N

Explanation:
To get the electric force between two charges, we would use Coulomb's law which states that:
F = $\frac{k * q1 * q2}{d^{2} }$

where:
F is the force between the two charges
k is Coulomb's constant = 9 * 10⁹ Nm²/C²
q1 is the first charge = 0.0072 C
q2 is the second charge = -0.006 C
d is the distance between the two charges = 0.004 m

Substitute with the givens in the above formula to get the force between the two charges as follows:
F = $\frac{9*10^9*0.0072*-0.006}{(0.004)^2}$

F = -2.4 * 10¹⁰ N

Hope this helps :)