Question

A point charge A of +10µC and another point charge B of +20µC are kept Im 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. \(\overrightarrow{F_1}=-2\overrightarrow{F_2}\)
• 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

We are given the following information:

• Charge on A: \( +10 \, \mu C \)
• Charge on B: \( +20 \, \mu C \)
• Distance between A and B: \( 1 \, \text{m} \)

Step 1: Formula for electrostatic force

The electrostatic force between two point charges is given by Coulomb's law:

\[ F = k \frac{|q_1 q_2|}{r^2} \]

Step 2: Use of Newton's Third Law

By Newton's Third Law, the forces are equal in magnitude but opposite in direction. Since the charges on A and B are different, the magnitude of the force on each charge will differ.

Step 3: Calculation of the ratio of forces

The ratio of the forces is directly proportional to the ratio of the charges:

\[ \frac{F_1}{F_2} = \frac{|q_A|}{|q_B|} = \frac{10}{20} = \frac{1}{2} \]

Thus, the relation between the forces is: \( F_1 = -F_2 \)

The correct answer is (B) \( F_1 = -F_2 \).