Question

A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the electrical force, Fe, applied by q1 on q2?

Answer 1

Theres the pic of answer

Answer 2

We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west

Substitute their values, 
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶)  / (3 * 10⁻²)²

F = 100 N  [ East of positive charge ]

Hope this helps!