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Electrostatics

JEE Main level, numerical answer, medium

75 of 311

Question 75 of 311

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 $1m$ from $S$ and has a speed of $1m{s}^{-1}$ . The maximum value of $\sigma$ , if the electron strikes $S$ at $t=1s$ is $\alpha \left[\dfrac{m{ϵ}_{0}}{e}\right]\dfrac{C}{{m}^{2}}$ . The value of $\alpha$ is _____.

Show answer and solution

Answer: 8

The value of the electric field due to the uniformly charged plane sheet is given by

E=\dfrac{\sigma }{2{\epsilon }_{0}}...\left(1\right)

Thus, the force on the electron is given by

\begin{matrix}F & = & -eE \\ & = & -\dfrac{\sigma e}{2{\epsilon }_{0}}...\left(2\right)\end{matrix}

So, the acceleration of the electron is given by

a=-\dfrac{\sigma e}{2{\epsilon }_{0}m}

Given,

u=1m{s}^{-1}

t=1s

and

S=-1m

Using the equation of motion, it can be written that

-1=1\times 1-\dfrac{1}{2}\times \dfrac{\sigma e}{2{\epsilon }_{0}m}\times (1{)}^{2} \Rightarrow \sigma =8\dfrac{{\epsilon }_{0}m}{e}C{m}^{-2}

∴\alpha =8

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