Answer to Question #102730 in Electricity and Magnetism for Chanon

Question

A charged particle A exerts a force of 2.62 N to the right on charged particle B when the particles are 1.37mm apart. Particle B move straight away from A to make the distance between them 17.7mm. What vector force does particle B then exert on A.

Answer 1

Charged particles interact with each other according to Coulomb's law. 'Coulomb's law states that:

The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. The force is along the straight line joining them. If the two charges have the same sign, the electrostatic force between them is repulsive; if they have different signs, the force between them is attractive.' [1]

(1) $F\sim \frac{Qq}{r^2}$

Since the only the distance between the particles has changed we can write

(2) $\vec{F_2} \cdot R_2^2=\vec{F_1} \cdot R_1^2$ , where $F_1=2.62 N$, $R_1=1.37mm$, $R_2=17.7mm$, and $\vec F_2$ is the force particle A exerts on the particle B after event happened.

The direction of the force $\vec F_2$ coincides with $\vec F_1$

$\vec F_2=\vec F_1\cdot \frac{R_1^2}{R_2^2}\\F_2=2.62\cdot \frac {1.37^2}{17.7^2}N=0.0157 N$

According to Newton's third law, particle B acts on particle A with equal but opposite force. That is

$\vec F_{BA}=-\vec F_2\\ F_{BA}= 0.0157 N$

Answer: Particle B exert on A force equals 0.0157 N with direction opposite the force particle A exerted on particle B before its move straight away from A.

[1] https://en.wikipedia.org/wiki/Coulomb%27s_law