JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 23)
An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet $$\mathrm{S}$$ 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 $$\alpha\left[\frac{m \epsilon_0}{e}\right] \frac{C}{m^2}$$, the value of $$\alpha$$ is ___________.
Answer
8
Explanation
$$\begin{aligned}
& \mathrm{u}=1 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=-\frac{\sigma \mathrm{e}}{2 \varepsilon_0 \mathrm{~m}} \\
& \mathrm{t}=1 \mathrm{~s} \\
& \mathrm{~S}=-1 \mathrm{~m} \\
& \text { Using } \mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\
& -1=1 \times 1-\frac{1}{2} \times \frac{\sigma \mathrm{e}}{2 \varepsilon_0 \mathrm{~m}} \times(1)^2 \\
& \therefore \sigma=8 \frac{\varepsilon_0 \mathrm{~m}}{\mathrm{e}} \\
& \therefore \alpha=8
\end{aligned}$$
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