Question

A point charge \(q_1 = 4q_0\) is placed at origin. Another point charge \(q_2 = –q_0\) is placed at \(x = 12\text{ cm}\). Charge of proton is q₀. The proton is placed on x axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is _____ cm.

Hint:

For force equilibrium problems:
• Use Coulomb’s law to equate the forces due to charges on opposite sides.
• Solve for the distance r and adjust for the final position based on geometry.

Answer 1

Correct Answer: 24

1. Force Balance Condition: - The electrostatic force on the proton is zero when: \[ F_{q_1} = F_{q_2}. \]
   • Using Coulomb’s law: \[ \frac{k \cdot |4q_0|}{r^2} = \frac{k \cdot |q_0|}{(12 - r)^2}. \]
2. Simplify: - Cancel \( k \) and \( q_0 \): \[ \frac{4}{r^2} = \frac{1}{(12 - r)^2}. \]
   • Take the square root: \[ \frac{2}{r} = \frac{1}{12 - r}. \]
   • Cross-multiply: \[ 2(12 - r) = r \implies 24 - 2r = r. \]
   • Solve for \( r \): \[ 3r = 24 \implies r = 8 \, \text{cm}. \]
3. Position of Proton: - The proton is \( 12 + r = 24 \, \text{cm} \) from the origin.

Final Answer: 24 cm