Answer:
0.009 N, repulsive
Explanation:
The electrostatic force between two electric charges is given by:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
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
[tex]q_1 =q_2 = +4.5\cdot 10^{-6}C[/tex] are the two charges
r = 4.5 m is their separation
Substituting into the equation, we find
[tex]F=(9\cdot 10^9 Nm^2 C^{-2})\frac{(+4.5\cdot 10^{-6} C)(4.5\cdot 10^{-6} C)}{(4.5 m)^2}=0.009 N[/tex]
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
