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Question #43779

Difficulty: 8/10 | 🟢 Free

The Question

An alpha particle (mass mm, charge +2e+2e) is projected from a distance dd directly towards a fixed, infinite non-conducting sheet with a uniform positive surface charge density σ\sigma. What is the minimum initial speed v0v_0 required for the alpha particle to just reach the surface of the sheet? (Assume the space is vacuum with permittivity ε0\varepsilon_0 and neglect gravity).

Options

  • A.

    2eσdmε0\sqrt{\frac{2e\sigma d}{m\varepsilon_0}}

  • B.

    eσdmε0\sqrt{\frac{e\sigma d}{m\varepsilon_0}}

  • C.

    2eσdmε02\sqrt{\frac{e\sigma d}{m\varepsilon_0}}

  • D.

    eσd2mε0\sqrt{\frac{e\sigma d}{2m\varepsilon_0}}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Distance of Closest Approach|Field due to Infinite Sheet|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between the electric field of a point charge (which follows an inverse-square law) and that of an infinite sheet (which is constant).

A common lethal mistake is assuming the potential energy takes the form U∝1/rU \propto 1/r or attempting to integrate a variable force, often by incorrectly modeling the sheet's effective charge as a finite disc Q=σπd2Q = \sigma \pi d^2.

This leads to the incorrect application of V=Q4πε0dV = \frac{Q}{4\pi\varepsilon_0 d}, resulting in the trap answer eσdmε0\sqrt{\frac{e\sigma d}{m\varepsilon_0}}.

The correct approach requires recognizing the constant nature of the field E=σ/(2ε0)E = \sigma / (2\varepsilon_0) and applying the Work-Energy Theorem for a constant force.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAtoms - Class 12
ConceptDistance of Closest Approach
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree