Question

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? (k = 9.0 e9 nm2/c2)

Answer 1

We can calculate the electrical force between the two charges with Coulomb's law: 

Fc=(k*Q₁*Q₂)/r², where k=9*10^9 N m² C⁻², Q₁ and Q₂ are the two charges and r is the distance between the two charges.

Now we simply input the numbers and get:

Fc= { (9*10^9)*(2*10^-6)*(4*10^-6) } / 2²

Fc=0.018 N

The electric force between the charges Q₁ and Q₂ is Fc=0.018 N

Answer 2

Answer:

+1.8 E-2 N and it is repulsive

Explanation:

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