Answer to Question #179523 in Physics for Louis

Question

Three-point charges are arranged along the y – axis in a vacuum with distance 2×10^8 nm. The topmost charge bears a charge of –4 μC, the middle charge has a charge of +3 μC, and the bottom one carries a –7 μC charge. What is the magnitude and direction of the net electrostatic force that the middle charge experience?

Answer 1

The magnitude and direction depends on the forces caused by the other two charges. Find force from the upper charge:

$$F_{12}=k\frac{q_1q_2}{r^2}.$$

The force from the third charge:

$$F_{32}=k\frac{q_3q_2}{r^2}.$$

The resultant force on the middle charge:

$$F=F_{12}+F_{32}=\frac{kq_2}{r^2}(q_1+q_3),\\\space\\ F=\frac{9·10^9·3·10^{-6}}{0.2^2}((-4·10^{-6})+(-7·10^{-6}))=\\\space\\ =-7.41\text{ N}.$$

The force is directed downward (from the middle toward the lower charge) because the magnitude of the third charge is the greatest and charges 1 and 3 are separated by the same distance.