Question:
Two equal charges \(q\) are kept fixed at \(-a\) and \(+a\) along the \(x\)-axis. A particle of mass \(m\) and charge \(\dfrac{q}{2}\) is brought to the origin and given a small displacement along the \(x\)-axis, then
Two equal charges \(q\) are kept fixed at \(-a\) and \(+a\) along the \(x\)-axis. A particle of mass \(m\) and charge \(\dfrac{q}{2}\) is brought to the origin and given a small displacement along the \(x\)-axis, then
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For two like charges, the midpoint along the axis is unstable equilibrium, while the perpendicular bisector is stable equilibrium -- giving SHM for transverse displacements.
Updated On: May 2, 2026
- the particle executes oscillatory motion
- the particle remains stationary
- the particle executes, SHM along \(x\)-axis
- the particle executes SHM along \(y\)-axis
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
At the midpoint (origin), the forces from both charges cancel. For displacement along \(x\)-axis, it is unstable equilibrium; for displacement along \(y\)-axis, restoring force acts.
Step 2: Detailed Explanation:
If displaced along \(y\)-axis by small \(y\), the net force is directed back towards origin (restoring), producing SHM. Along \(x\)-axis, displacement leads to a repulsive force away from origin (unstable).
Step 3: Final Answer:
The particle executes SHM along the \(y\)-axis.
At the midpoint (origin), the forces from both charges cancel. For displacement along \(x\)-axis, it is unstable equilibrium; for displacement along \(y\)-axis, restoring force acts.
Step 2: Detailed Explanation:
If displaced along \(y\)-axis by small \(y\), the net force is directed back towards origin (restoring), producing SHM. Along \(x\)-axis, displacement leads to a repulsive force away from origin (unstable).
Step 3: Final Answer:
The particle executes SHM along the \(y\)-axis.
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