JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2024
- Medium
- Numerical
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 \(1 m s ^{-1}\). The maximum value of \(\sigma\), if the electron strikes \(S\) at \(t=1 s\) is \(\alpha\left[\frac{m \epsilon_0}{e}\right] \frac{ C }{ m ^2}\). The value of \(\alpha\) is __________ .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 8
Step-by-step Solution
Detailed explanation
\(u=1 m / s ; a=-\frac{\sigma e}{2 \epsilon_0 m}\)
\(t=1 s\)
\(S=-1 m\)
Using \(S=u t+\frac{1}{2} a t^2\)
\(-1=1 \times 1-\frac{1}{2} \times \frac{\sigma e}{2 \epsilon_0 m} \times(1)^2\)
\(\therefore \sigma=8 \frac{\epsilon_e m}{e}\)
\(\therefore \alpha=8\)
\(t=1 s\)
\(S=-1 m\)
Using \(S=u t+\frac{1}{2} a t^2\)
\(-1=1 \times 1-\frac{1}{2} \times \frac{\sigma e}{2 \epsilon_0 m} \times(1)^2\)
\(\therefore \sigma=8 \frac{\epsilon_e m}{e}\)
\(\therefore \alpha=8\)
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