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.