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.
