Question

Two point charges each carrying a charge of + 4.5 E - 6 C are located 4.5 meters away from each other. How strong is the electrostatic force between the two points, and is this force a repulsive force or an attractive force (k = 9.0 E9 Nm2/C2)

Answer 1

0.009 N, repulsive

Step-by-step explanation:

The electrostatic force between two electric charges is given by:

`F=k(q_1 q_2)/(r^2)`

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, we have

`q_1 =q_2 = +4.5\cdot 10^(-6)C` are the two charges

r = 4.5 m is their separation

Substituting into the equation, we find

`F=(9\cdot 10^9 Nm^2 C^(-2))((+4.5\cdot 10^(-6) C)(4.5\cdot 10^(-6) C))/((4.5 m)^2)=0.009 N`

Moreover, the force is repulsive. In fact, the following rules apply:

- When two charges have same sign, they repel each other

- When two charges have opposite signs, they attract each other