A particle of mass m and charge (-q) enters t | vedclass

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(C) Given:Velocity $v_{x} = 2.0 	imes 10^{6} \; m/s$Separation $d = 0.5 \; cm = 0.005 \; m$Electric field $E = 9.1 	imes 10^{2} \; N/C$Charge $q = 1

Vedclass · Accepted Answer

$1.6$

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(C) Given:Velocity $v_{x} = 2.0 	imes 10^{6} \; m/s$Separation $d = 0.5 \; cm = 0.005 \; m$Electric field $E = 9.1 	imes 10^{2} \; N/C$Charge $q = 1.6 	imes 10^{-19} \; C$Mass $m_{e} = 9.1 	imes 10^{-31} \; kg$The vertical deflection $s$ of the electron is given by $s = \frac{1}{2} a t^{2}$,where $a = \frac{qE}{m}$ and $t = \frac{L}{v_{x}}$.Substituting these,we get $s = \frac{q E L^{2}}{2 m v_{x}^{2}}$.To find the distance $L$ where the electron strikes the upper plate,we set $s = d = 0.005 \; m$.$L = \sqrt{\frac{2 d m v_{x}^{2}}{q E}}$$L = \sqrt{\frac{2 	imes 0.005 	imes 9.1 	imes 10^{-31} 	imes (2.0 	imes 10^{6})^{2}}{1.6 	imes 10^{-19} 	imes 9.1 	imes 10^{2}}}$$L = \sqrt{\frac{0.01 	imes 9.1 	imes 10^{-31} 	imes 4 	imes 10^{12}}{1.6 	imes 10^{-19} 	imes 9.1 	imes 10^{2}}}$$L = \sqrt{\frac{0.04 	imes 10^{-19}}{1.6 	imes 10^{-17}}} = \sqrt{0.025 	imes 10^{-2}} = \sqrt{0.00025} = 0.016 \; m = 1.6 \; cm$.

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