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Question #42922

Difficulty: 8/10 | 🟢 Free

The Question

A very large, isolated, thin non-conducting sheet has an initial area A0A_0 at temperature T0T_0 and carries a uniformly distributed total positive charge QQ. A point particle of mass mm and positive charge qq is projected from a point near the sheet with an initial speed v0v_0 at an angle θ\theta (0<θ<π/20 < \theta < \pi/2) with the normal to the sheet. The maximum perpendicular distance the particle travels from its projection point before turning back is x0x_0. The sheet is then uniformly heated to a temperature T0+ΔTT_0 + \Delta T. The coefficient of linear thermal expansion of the sheet material is α\alpha. If the same particle is projected again from the same point with the same initial velocity, what is the new maximum perpendicular distance xx it reaches? (Assume αΔT≪1\alpha \Delta T \ll 1, neglect gravity and edge effects).

Options

  • A.

    x0(1+2αΔT)x_0 (1 + 2\alpha \Delta T)

  • B.

    x0(1−2αΔT)x_0 (1 - 2\alpha \Delta T)

  • C.

    x0(1+αΔT)x_0 (1 + \alpha \Delta T)

  • D.

    x0x_0

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Field due to Infinite Sheet|Variation of Density with Temperature|Conservative Force Field

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple thermal expansion with electrostatics, specifically targeting the misconception that surface charge density (σ\sigma) is an invariant property of a charged sheet.

Students often incorrectly assume the electric field E=σ/2ε0E = \sigma / 2\varepsilon_0 remains constant during heating (leading to option D).

Furthermore, by projecting the particle at an angle, the problem verifies the understanding of conservative force fields; the work done depends only on the normal displacement, isolating the normal kinetic energy component.

The correct approach requires recognizing that for an isolated sheet, total charge QQ is conserved, so thermal area expansion (A=A0(1+2αΔT)A = A_0(1+2\alpha\Delta T)) decreases σ\sigma, thereby weakening the electric field and increasing the stopping distance.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptField due to Infinite Sheet
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree