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JEE Main 202429 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2024 · 29 January · Morning Shift · Physics · Electrostatics

An electron is moving under the influence of the electric field of...

An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet SS having surface charge density +σ+ \sigma. The electron at t=0t = 0 is at a distance of 1m1m from SS and has a speed of 1ms−11m{s}^{- 1}. The maximum value of σ\sigma, if the electron strikes SS at t=1st = 1s is α[mϵ0e]Cm2\alpha\left[\dfrac{m{ϵ}_{0}}{e}\right]\dfrac{C}{{m}^{2}}. The value of α\alpha is _____.

Correct answer

0

Step-by-step solution

The value of the electric field due to the uniformly charged plane sheet is given by

E=σ2ε0...(1)E = \dfrac{\sigma}{2{ε}_{0}} . . . \left(1\right)

Thus, the force on the electron is given by

{F=−eE=−\sigmae2ε0...(2)\begin{cases}F & = & - eE \\ & = & - \dfrac{\sigmae}{2{ε}_{0}} . . . \left(2\right)\end{cases}

So, the acceleration of the electron is given by

a=−\sigmae2ε0ma = - \dfrac{\sigmae}{2{ε}_{0}m}

Given, u=1ms−1u = 1m{s}^{- 1}, t=1st = 1s and S=−1mS = - 1m

Using the equation of motion, it can be written that

−1=1×1−12×\sigmae2ε0m×(1)2⇒σ=8ε0meCm−2- 1 = 1 \times 1 - \dfrac{1}{2} \times \dfrac{\sigmae}{2{ε}_{0}m} \times ( 1{)}^{2} ⇒ \sigma = 8\dfrac{{ε}_{0}m}{e}C{m}^{- 2}

∴α=8∴ \alpha = 8.

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