Question

The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

• A. P
• B. Q
• C. R
• D. S

Hint:

In \( V \) vs \( x \) graphs: - Electric field = negative slope of the graph. - Maximum force occurs where the graph is steepest (largest slope magnitude).

Answer 1

The Correct Option is D

Concept: The electric field is related to electric potential by: \[ E = -\frac{dV}{dx} \] Force on a charge: \[ F = qE \] Thus, the magnitude of force depends on the slope of the \( V \) vs \( x \) graph.

Steeper slope \( \Rightarrow \) larger electric field
Flat region \( \Rightarrow \) zero force

Step 1: Analyze each point. At P: The graph is horizontal (constant potential). \[ \frac{dV}{dx} = 0 \Rightarrow E = 0 \Rightarrow F = 0 \] At Q: The graph has a moderate negative slope. This means a finite electric field and moderate force. At R: Again, the graph is flat (constant potential). \[ E = 0 \Rightarrow F = 0 \] At S: The graph rises very steeply (large positive slope). Since electric field magnitude depends on slope: \[ |E| = \left|\frac{dV}{dx}\right| \text{ is maximum here} \] Thus, the force magnitude is maximum at S.
Step 2: Conclusion. Maximum force occurs where the potential changes most rapidly with position. This happens at point S.

Answer 2

The force on a charged particle at any point equals its charge times the local electric field, and the electric field itself equals the negative slope of the potential-versus-position graph, \( E = -\dfrac{dV}{dx} \). So the point with the steepest slope on the V-x graph is where the force is largest. Let's check each labelled point.

1. P: At P, the potential curve runs flat (horizontal), meaning \( V \) is not changing with \( x \) at that point. A flat curve gives \( \dfrac{dV}{dx} = 0 \), so the field, and therefore the force, is zero here.
2. Q: At Q, the curve has only a gentle slope, the potential is changing with position, but slowly. This gives a small, non-zero electric field and a correspondingly small force, far from the maximum.
3. R: At R, the graph flattens out again into a plateau, exactly like at P, so the slope is zero there too, and the force on the particle is again zero.
4. S: At S, the potential rises very sharply over a short distance, i.e. the steepest section of the whole graph. A steep slope means a large \( \left|\dfrac{dV}{dx}\right| \), which means the strongest electric field, and hence the largest force on the charge.

Comparing all four, only S has a slope steep enough to produce a large field; P and R contribute no force at all, and Q contributes only a small one.

Therefore, the correct answer is that the particle experiences the maximum force at S.