Verified Solution • 2025–26 Session moderate

QJEEM24Ph5b0b: JEE Main 2024 Physics - Class 12 JEEM24Ph5b0b Solution: An electron is moving under the influence of the electric field …

An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S\mathrm{S} having surface charge density +σ+\sigma . The electron at t=0t=0 is at a distance of 1 m1 \mathrm{~m} from SS and has a speed of 1 m/s1 \mathrm{~m} / \mathrm{s} . The maximum value of σ\sigma if the electron strikes SS at t=1 st=1 \mathrm{~s} is α[mϵ0e]Cm2\alpha\left[\frac{m \epsilon_0}{e}\right] \frac{C}{m^2} , the value of α\alpha is ___________.

Step-by-Step Verified Solution

amp;u=1 m/s;a=−σe2ε0 mamp;t=1 samp; S=−1 mamp; Using S=ut+12at2amp;−1=1×1−12×σe2ε0 m×(1)2amp;∴σ=8ε0 meamp;∴α=8\begin{aligned}& \mathrm{u}=1 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=-\frac{\sigma \mathrm{e}}{2 \varepsilon_0 \mathrm{~m}} \\& \mathrm{t}=1 \mathrm{~s} \\& \mathrm{~S}=-1 \mathrm{~m} \\& \text { 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\end{aligned}

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