Answer:
[tex]1.02\cdot 10^{-8} N[/tex], repulsive
Explanation:
The magnitude of the electric force between two charged particles is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges of the two particles
r is the separation between the two charges
The force is:
- repulsive if the two charges have same sign
- Attractive if the two charges have opposite signs
In this problem, we have two electrons, so:
[tex]q_1=q_2=1.6\cdot 10^{-19}C[/tex] is the magnitude of the two electrons
[tex]r=1.5\cdot 10^{-10} m[/tex] is their separation
Substituting into the formula, we find the electric force between them:
[tex]F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N[/tex]
And the force is repulsive, since the two electrons have same sign charge.
