Question

A charge \( q \) is placed at the centre of the line joining two equal charges \( Q \). The system of the three charges will be in equilibrium if \( q \) is equal to:

• A. \( -\frac{Q}{2} \)
• B. \( -\frac{Q}{4} \)
• C. \( \frac{Q}{2} \)
• D. \( \frac{Q}{4} \)

Hint:

To achieve equilibrium in a system of charges, the net force on each charge must be zero. Use Coulomb's law to calculate the forces.

Answer 1

The Correct Option is A

For the system to be in equilibrium, the net force on charge \( q \) due to the two charges \( Q \) must be zero. Since the charges are equal and placed symmetrically, the force due to each charge must cancel each other out. By Coulomb’s Law, the force between two charges is: \[ F = k \frac{Q q}{r^2} \] For equilibrium, the forces must balance, and this happens when \( q = -\frac{Q}{2} \).