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A large flat metal surface has uniform charge density $+ σ$ . An electron of mass m and charge e leaves the surface at point A with speed v, and return to it at point B. The maximum value of AB is

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$\frac{v^{2} e}{ϵ_{0} σm}$

$\frac{v^{2} σe}{ϵ_{0} m}$

$\frac{vm \in_{0}}{σc}$

$\frac{v^{2} m \in_{0}}{eσ}$

answer is B.

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Detailed Solution

For metal surface $E = \frac{σ}{ϵ_{0}}$
$Max . range = \frac{v^{2}}{a}$
$\begin{aligned} a = \frac{qE}{m} = \frac{cσ}{ϵ_{0} m} \\ R = AB = \frac{V^{2} {mϵ}_{0}}{eσ} \end{aligned}$

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