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1. Electric Charges and Fields
hard
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 $1 \mathrm{~m}$ 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 $\alpha\left[\frac{\mathrm{m} \in_0}{\mathrm{e}}\right] \frac{\mathrm{C}}{\mathrm{m}^2}$ the value of $\alpha$ is
A$8$
B$5$
C$10$
D$45$
(JEE MAIN-2024)
Solution
$ \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$
$ \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$
Standard 12
Physics