Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has a charge of 4.42 nc. (a) What is the magnitude (in N) of the electric force that one particle exerts on the other? (b) is the force attractive or repulsive? O attractive O repulsive

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F = (Kq1q2) / r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[ (9*10^9) * (7.10*10^-9) * (4.42*10^-9) ] / (1.62^2)

F = (282.4*10^-9) / 2.6244

F=107.6*10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive