Two charged particles are placed 2.0 meters apart. the first charge is +2.0 e-6 c and the second charge is +4.0 e-6
c. what is the electrical force between the two charges?

The Coulomb force between two charges is defined as: Fc=(k*Q₁*Q₂)/r² where k=9*10^9 N m² C⁻², Q₁ and Q₂ are charges and r is the distance between those charges. 
In our case:

Q₁=2*10^-6 C
Q₂=4*10^-6 C
r=2 m

Now we simply plug in the numbers into the equation:

Fc={(9*10^9)*(2*10^-6)*(4*10^-6)}/2^2=0.018 N

The Coulombs force between the two positive charges is Fc=0.018 N and it is a repulsive force because both charges are positive.

The answer is b hope this