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:

Whenever a graph of electric potential versus position is given, remember: \[ E=-\frac{dV}{dx}. \] The magnitude of electric field is equal to the magnitude of the slope of the \(V\)-\(x\) graph. The steeper the graph, the larger the electric field and hence the larger the force on a charged particle.

Answer 1

The Correct Option is D

Concept: The electric field is related to the variation of electric potential with position. Along the \(x\)-direction, \[ E=-\frac{dV}{dx}. \] The magnitude of electric field is therefore \[ |E|=\left|\frac{dV}{dx}\right|. \] The force experienced by a charged particle is \[ F=qE. \] Hence, for a given charge, the magnitude of force is maximum at the point where the magnitude of the slope of the \(V\)-versus-\(x\) graph is maximum. Therefore, our task is to identify the point where the graph has the steepest slope.

Step 1: Examine point \(P\). At point \(P\), the graph is horizontal. A horizontal graph means \[ \frac{dV}{dx}=0. \] Therefore, \[ E=0. \] Hence, the force at \(P\) is zero.

Step 2: Examine point \(Q\). At point \(Q\), the graph is sloping downward. Thus, \[ \frac{dV}{dx}\neq 0. \] Therefore, an electric field exists at \(Q\). The force is non-zero, but we must compare it with other points.

Step 3: Examine point \(R\). At point \(R\), the graph is again horizontal. Hence, \[ \frac{dV}{dx}=0. \] Therefore, \[ E=0. \] The force is zero at \(R\).

Step 4: Examine point \(S\). At point \(S\), the graph rises very steeply. This means the magnitude of the slope \[ \left|\frac{dV}{dx}\right| \] is maximum at \(S\). Since \[ |E| = \left|\frac{dV}{dx}\right|, \] the electric field magnitude is maximum at \(S\). Consequently, \[ F=qE \] is also maximum at \(S\).

Step 5: Compare all points. \[ P:\quad E=0 \] \[ Q:\quad E\neq0 \] \[ R:\quad E=0 \] \[ S:\quad |E| \text{ is maximum} \] Therefore, \[ \boxed{S} \] is the point where the charged particle experiences the maximum force. Hence, the correct answer is \[ \boxed{\text{(D) S}}. \]