Question

A point charge A of +10\mu C and another point charge B of +20\mu C are kept 1m apart in free space. The electrostatic force on A due to B is $\overrightarrow{F}_1$, and the electrostatic force on B due to A is $\overrightarrow{F}_2$. Then

• A. $\overrightarrow{F}_2 = 2\overrightarrow{F}_2$
• B. $\overrightarrow{F}_1 = -\overrightarrow{F}_2$
• C. $2\overrightarrow{F}_1 = -\overrightarrow{F}_2$
• D. $\overrightarrow{F}_1 = \overrightarrow{F}_2$

Answer 1

The Correct Option is B

According to Coulomb's law, the electrostatic force between two point charges is:
$F = k_e \frac{|q_1 q_2|}{r^2}$ where: 
$- \ k_e$ is Coulomb’s constant, 
$- \ q_1$ and $q_2$ are the magnitudes of the charges, 
$- \ r$ is the distance between them. 

The forces on charges A and B due to each other are equal in magnitude but opposite in direction. This is a direct consequence of Newton’s Third Law, which states that every action has an equal and opposite reaction.