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

`9\cdot 10^9 N`

Step-by-step explanation:

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

`F=k(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))((+3 C)^2)/((3.00 m)^2)=9\cdot 10^9 N`