The electric force is [tex]-3.6\cdot 10^8 N[/tex] (attractive)
Explanation:
The magnitude of the electric force between two charges is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=9\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
In this problem, we have the following:
[tex]q_1 = -0.0050 C[/tex] (charge 1)
[tex]q_2 = +0.0050 C[/tex] (charge 2)
r = 0.025 m (distance)
Substituting, we find the electric force between the two charges:
[tex]F=(9\cdot 10^9) \frac{(-0.0050)(0.0050)}{(0.025)^2}=-3.6\cdot 10^8 N[/tex]
And the negative sign means the force is attractive, since the two charges have opposite sign.
Learn more about electric force:
brainly.com/question/8960054
brainly.com/question/4273177
#LearnwithBrainly
