Answer to Question #161673 in Physics for PUFF

Question

Three charges lie in the plane in a vacuum. The distance between q₁ and q₂ is 5.0 cm while the distance from q₂ and q₃ is 7.5cm. If q₁ = +3.0 nC, q₂= -1.5 nC and q₃= +2.0 nC, find the magnitude and direction of the net electrostatic force on q_{2. Note that 1nC = 1x10}^-9_C

Answer 1

Since the second charge is between the other two, it is subjected to the action of two forces. The first charge attracts it, the third charge repel it, which means that the force on the second charge is toward charge 1. Calculate the magnitude:

$$F=F_1+F_2,\\\space\\ F=k\frac{q_1q_2}{r_{12}^2}+k\frac{q_3q_2}{r_{23}^2}=\\\space\\ F=kq_2\bigg(\frac{q_1}{r_{12}^2}+\frac{q_3}{r_{23}^2}\bigg)=\\\space\\ =-2.1\cdot10^{-5}\text{ N}.$$