The surface charge density of a conducting pl | vedclass

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Question

(A) The electric field $E$ near a conducting plate is given by $E = \frac{|\sigma|}{\epsilon_0}$.Since the electron moves towards the plate and stops

Vedclass · Accepted Answer

$4.42 	imes 10^{-4} \ m$

Vedclass · Answer

(A) The electric field $E$ near a conducting plate is given by $E = \frac{|\sigma|}{\epsilon_0}$.Since the electron moves towards the plate and stops at the surface,the work done by the electric field equals the change in kinetic energy.$W = \Delta K = K_f - K_i = 0 - K = -K$.The potential difference between the initial position at distance $d$ and the plate is $\Delta V = -E \cdot d = -\frac{|\sigma|}{\epsilon_0} d$.The work done on the electron is $W = q \Delta V = (-e) \left( -\frac{|\sigma|}{\epsilon_0} d ight) = \frac{e |\sigma| d}{\epsilon_0}$.Equating the work done to the change in kinetic energy: $K = \frac{e |\sigma| d}{\epsilon_0}$.Given $K = 100 \ 	ext{eV} = 100 	imes 1.6 	imes 10^{-19} \ 	ext{J}$,$|\sigma| = 2 	imes 10^{-6} \ 	ext{C/m}^2$,and $\epsilon_0 = 8.85 	imes 10^{-12} \ 	ext{C}^2/(	ext{N} \cdot 	ext{m}^2)$.$100 	imes 1.6 	imes 10^{-19} = \frac{(1.6 	imes 10^{-19}) 	imes (2 	imes 10^{-6}) 	imes d}{8.85 	imes 10^{-12}}$.$100 = \frac{2 	imes 10^{-6} 	imes d}{8.85 	imes 10^{-12}} \Rightarrow 100 = \frac{2 	imes 10^6 	imes d}{8.85}$.$d = \frac{100 	imes 8.85}{2 	imes 10^6} = 4.425 	imes 10^{-4} \ 	ext{m}$.

The surface charge density of a conducting plate is $-2 \times 10^{-6} \ C/m^2$. An electron with energy $100 \ eV$ moves towards the plate and strikes it. What is the initial distance of the electron from the plate?

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