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(A) The electric field $E$ produced by a large charged plate is given by $E = \frac{\sigma}{2\varepsilon_0}$.The work done by the electric field on th

Vedclass · Accepted Answer

$1.77$

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(A) The electric field $E$ produced by a large charged plate is given by $E = \frac{\sigma}{2\varepsilon_0}$.The work done by the electric field on the electron as it moves towards the plate is $W = F \cdot r = eE \cdot r$,where $e$ is the charge of the electron.For the electron not to strike the plate,its initial kinetic energy must be completely converted into potential energy before it reaches the plate. Thus,$KE = eEr$.Substituting $E = \frac{\sigma}{2\varepsilon_0}$,we get $KE = e \left( \frac{\sigma}{2\varepsilon_0} ight) r$.Rearranging for $r$: $r = \frac{2 \cdot KE \cdot \varepsilon_0}{e \cdot \sigma}$.Given $KE = 200 \ eV = 200 	imes 1.6 	imes 10^{-19} \ J$,$\sigma = 2 	imes 10^{-6} \ C/m^2$,and $\varepsilon_0 = 8.854 	imes 10^{-12} \ C^2/(N \cdot m^2)$.$r = \frac{2 	imes 200 	imes 1.6 	imes 10^{-19} 	imes 8.854 	imes 10^{-12}}{1.6 	imes 10^{-19} 	imes 2 	imes 10^{-6}} = \frac{400 	imes 8.854 	imes 10^{-12}}{2 	imes 10^{-6}} = 200 	imes 8.854 	imes 10^{-6} \ m = 1770.8 	imes 10^{-6} \ m \approx 1.77 	imes 10^{-3} \ m = 1.77 \ mm$.

$A$ negatively charged plate has a surface charge density of $\sigma = 2 \times 10^{-6} \ C/m^2$. An electron with kinetic energy $KE = 200 \ eV$ is projected towards the plate. If the electron does not strike the plate,find the minimum initial distance $r$ in $mm$ from the plate.

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