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Question #42656

Difficulty: 8/10 | 🟢 Free

The Question

A small bead of mass mm (in kg\text{kg}) and charge +q+q (in C\text{C}) is threaded on a smooth, rigid, insulating wire shaped as a parabola y=αx2y = \alpha x^2 (where α\alpha is a positive constant in m−1\text{m}^{-1}) in the xx-yy plane. A uniform electric field E⃗=E0i^\vec{E} = E_0 \hat{i} (in V/m\text{V/m}) exists throughout the region. The bead is released from rest at the origin (0,0)(0,0). Assuming the entire setup is in a gravity-free space, find the magnitude of the normal force (in N\text{N}) exerted by the wire on the bead when it reaches the point where the tangent to the wire makes an angle of 60∘60^\circ with the positive xx-axis.

Options

  • A.

    334qE0\frac{3\sqrt{3}}{4} q E_0

  • B.

    332qE0\frac{3\sqrt{3}}{2} q E_0

  • C.

    3qE0\sqrt{3} q E_0

  • D.

    34qE0\frac{\sqrt{3}}{4} q E_0

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Neutral Equilibrium Points|The Velocity-Position Identity|Relative Velocity in 2D|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply Newton's Second Law and the Work-Energy theorem in a constrained 2D path.

A common trap (The Velocity-Position Identity trap) is assuming that because the external force is purely in the x-direction, the x-acceleration is simply ax=qE0/ma_x = qE_0/m.

Students then incorrectly use vx2=2axxv_x^2 = 2a_x x and v2=vx2+vy2v^2 = v_x^2 + v_y^2 to find the velocity.

This is fundamentally flawed because the normal force from the wire has an x-component, meaning ax≠qE0/ma_x \neq qE_0/m.

The correct approach requires using the Work-Energy theorem to find the true speed.

A secondary trap is failing to project the electric force onto the instantaneous normal axis when calculating the required centripetal force, leading students to equate the normal force solely to mv2/ρm v^2 / \rho.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptNeutral Equilibrium Points
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree