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(C) The time taken by the electron to cross the plates of length $L = 10 \ cm = 0.1 \ m$ with a horizontal velocity $V_x = 10^6 \ m/s$ is:$t = \frac{L

Vedclass · Accepted Answer

$1.6 	imes 10^6 \ m/s$

Vedclass · Answer

(C) The time taken by the electron to cross the plates of length $L = 10 \ cm = 0.1 \ m$ with a horizontal velocity $V_x = 10^6 \ m/s$ is:$t = \frac{L}{V_x} = \frac{0.1}{10^6} = 10^{-7} \ s$The electric field $E = 9.1 \ V/cm = 9.1 	imes 10^2 \ V/m = 910 \ V/m$.The acceleration of the electron in the vertical direction is:$a_y = \frac{eE}{m} = \frac{1.6 	imes 10^{-19} 	imes 910}{9.1 	imes 10^{-31}} = \frac{1.6 	imes 10^{-19} 	imes 9.1 	imes 10^2}{9.1 	imes 10^{-31}} = 1.6 	imes 10^{14} \ m/s^2$The vertical component of velocity $V_y$ at the exit is given by $V_y = u_y + a_y t$,where $u_y = 0$:$V_y = 0 + (1.6 	imes 10^{14}) 	imes 10^{-7} = 1.6 	imes 10^7 \ m/s$. Wait,re-calculating: $E = 9.1 \ V/cm = 910 \ V/m$. $a_y = (1.6 	imes 10^{-19} 	imes 910) / (9.1 	imes 10^{-31}) = 1.6 	imes 10^{14} \ m/s^2$. $V_y = 1.6 	imes 10^{14} 	imes 10^{-7} = 1.6 	imes 10^7 \ m/s$. Given the options,there might be a typo in the question's provided options or the field value. If $E = 9.1 \ V/m$,then $V_y = 1.6 	imes 10^5 \ m/s$. If $E = 910 \ V/m$,$V_y = 1.6 	imes 10^7 \ m/s$. Given the options,$1.6 	imes 10^6 \ m/s$ is the closest intended answer if $E = 91 \ V/m$.

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