Question

Two electrons are separated by a distance of 3.00 x 10^-6 meter. What are the magnitude and direction of the electrostatic forces each exerts on the other?

Answer 1

The electrostatic force between two charges q1 and q2 is given by
$F=k_e \frac{q_1 q_2}{r^2}$
where 
$k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant
$r=3.00 \cdot 10^{-6} m$ is the distance between the two charges.

In our problem, the two charges are two electrons, so their charges are equal and equal to 
$q_1=q_2=q=-1.6 \cdot 10^{-19}C$

By substituting these values, we find the intensity of the force between the two electrons:
$F=(8.99 \cdot 10^9 N m^2 C^{-2}) \frac{(-1.6 \cdot 10^{-19}C)(-1.6 \cdot 10^{-19}C)}{(3.00 \cdot 10^{-6} m)^2}=2.6 \cdot 10^{-17}N$

This is the magnitude of the force each electron exerts to the other one. The direction is given by the sign of the charges: since the two electrons have same charge, they repel each other, so the force exerted by electron 1 is toward electron 2 and viceversa.