Question

From what distance should an electron of energy $E$ be fired on the charged metal plate of charge density '$\sigma$' so that it's velocity becomes zero just before it reaches the plate.

Answer 1

Given the electron $\left(e\right)$ has initial energy $E$, given charge density of plate $=\sigma$

Also by the time ${}^{\prime }{e}^{\prime }$ reaches the plate its velocity is ${}^{\prime }{o}^{\prime }$. Final $KE=0$.

$\Rightarrow$ Entire energy ${}^{\prime }{E}^{\prime }$ is used to do work against potential, through the distance it covered.

Let the distance be ${}^{\prime }{x}^{\prime }$.

$\Rightarrow =\frac{\sigma }{2{\epsilon }_0}{\int }_0^{x}xdx(i.e.Electronenergy=workdone)$

$\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}{\int }_0^{x}xdx(Fieldthroughplateofchargedensit{y}^{\prime }{\sigma }^{\prime }=\frac{\sigma }{2{\epsilon }_0})$

$\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}.e.\frac{x^2}{2}(\because {\int }_0^x{x}^n=\frac{x^{n+1}}{n+1}{\int }_0^x)$

$\Rightarrow =\frac{x^2}{2}=\frac{2E{\epsilon }_0}{\sigma e}$

$\Rightarrow x^2=\frac{4E{\epsilon }_0}{\sigma e}$

$\Rightarrow x=\sqrt{\frac{4E{\epsilon }_0}{\sigma e}}$

$\Rightarrow x=2\sqrt{\frac{E{\epsilon }_0}{\sigma e}}$

So $x=2\sqrt{\frac{E{\epsilon }_0}{\sigma e}}$ is required distance through which it should be fired.