An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density +σ . The electron at t=0 is at a distance of 1 m from S and has a speed of 1 m / s . The maximum value of σ if the electron strikes S at t=1 s is α [frac m ε 0 e ] frac C m 2 the value of α is

Physics · Electrostatics

JEE Main 2024 — 29 January, Shift 1 — Question 57

An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density $+\sigma$ . The electron at $t=0$ is at a distance of 1 m from S and has a speed of $1 \mathrm{~m} / \mathrm{s}$ . The maximum value of $\sigma$ if the electron strikes $S$ at $t=1 \mathrm{~s}$ is $\alpha\left[\frac{\mathrm{m} \epsilon_{0}}{\mathrm{e}}\right] \frac{\mathrm{C}}{\mathrm{m}^{2}}$ the value of $\alpha$ is

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{u}=1 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=-\frac{\sigma \mathrm{e}}{2 \varepsilon_{0} \mathrm{~m}}$

$\mathrm{t}=1 \mathrm{~s}$

$S=-1 m$ Using $\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

$-1=1 \times 1-\frac{1}{2} \times \frac{\sigma \mathrm{e}}{2 \varepsilon_{0} \mathrm{~m}} \times(1)^{2}$

$\therefore \sigma=8 \frac{\varepsilon_{0} \mathrm{~m}}{\mathrm{e}}$

$\therefore \alpha=8$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Field & motion of charge