Answer to Question #164609 in Physics for Abdulrehman

Question

Two positively charged particles shown in figure are fixed in place on an x axis. The charges are q1 =1.60x10-19 C and q2 =3.20x10-19 C, and the particle separation is R=0.02 m. What are the magnitude and direction of the electrostatic force F12 on particle 1 from particle 2?

Answer 1

Let's find the magnitude of the electrostatic force $F_{12}$ on particle 1 due to particle 2:

$$F_{12}=\dfrac{kq_1q_2}{r^2},$$ $$F_{12}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot1.6\cdot10^{-19}\ C\cdot3.2\cdot10^{-19}\ C}{(0.02\ m)^2}=1.15\cdot10^{-24}\ N.$$

Since the two charges are both positively charged they will repel each other, therefore, the direction of the electrostatic force is repulsive (in other words, the electrostatic force is directed away from particle 2).