Question

particles q1, q2, and q3 are in a straight line. Particles q1= -28.1 uC, q2= +25.5 uC, and q3= -47.9 uC. Particles q1 and q2 are separated by 0.300m. Particles q2 and q3 are separated by 0.300m. What is the net force on q3?

Answer 1

Calculate the force acting on q3 by q1:

$$F_{13}=k\frac{q_1q_3}{r_{13}^2}.$$

Calculate the force acting on q3 by q2:

$$F_{23}=k\frac{q_2q_3}{r_{23}^2}.$$

The net force will be

$$F=kq_3\bigg(\frac{q_1}{r_{13}^2}+\frac{q_2}{r_{23}^2}\bigg)=\\\space\\ =9\cdot10^9\cdot (-47.9\cdot10^{-6})\bigg(\frac{(-28.1\cdot10^{-6})}{(0.3+0.3)^2}+\frac{25.5 \cdot10^{-6}}{0.3^2}\bigg)=\\\space\\ =-88.4\text{ N}.$$

The third charge will be "attracted" toward charges 2 and 1.