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.