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(A) The electric field $E$ produced by a large charged plate is given by $E = \frac{\sigma}{2\epsilon_0}$.Given $\sigma = 2 	imes 10^{-6} \, C/m^2$ a

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$1.77 \, mm$

Vedclass · Answer

(A) The electric field $E$ produced by a large charged plate is given by $E = \frac{\sigma}{2\epsilon_0}$.Given $\sigma = 2 	imes 10^{-6} \, C/m^2$ and $\epsilon_0 = 8.854 	imes 10^{-12} \, C^2/(N \cdot m^2)$.$E = \frac{2 	imes 10^{-6}}{2 	imes 8.854 	imes 10^{-12}} \approx 1.13 	imes 10^5 \, N/C$.The electron will not strike the plate if its initial kinetic energy $K$ is completely converted into potential energy before it reaches the plate. Thus,$K = e \cdot E \cdot r$,where $r$ is the distance.$r = \frac{K}{eE} = \frac{200 \, eV}{e \cdot E} = \frac{200}{E} = \frac{200}{1.13 	imes 10^5} \approx 1.77 	imes 10^{-3} \, m = 1.77 \, mm$.

$A$ negatively charged plate has a surface charge density of $2 \times 10^{-6} \, C/m^2$. Find the minimum initial distance of an electron moving toward the plate such that it does not strike the plate,given that its initial kinetic energy is $200 \, eV$.

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