Question

Suppose that two point charges, each with a charge of +3.00 Coulomb are separated by a distance of 3.00 meters. Determine the magnitude of the electrical force of repulsion between them

Answer 1

Answer:

$9\cdot 10^9 N$

Explanation:

The magnitude of the electrical force between the two point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

$q_1 =q_2 = +3.0 C$ is the magnitude of each charge

r = 3.00 m is the separation between the two charges

Substituting the numbers into the formula, we find

$F=(9\cdot 10^9 Nm^2C^{-2})\frac{(+3 C)^2}{(3.00 m)^2}=9\cdot 10^9 N$