An electron is released from the bottom plate | vedclass

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(A) The force acting on the electron is $F = eE$,where $e$ is the charge of the electron and $E$ is the electric field.The acceleration of the electro

Vedclass · Accepted Answer

$0.85 	imes 10^7 \, m/s$

Vedclass · Answer

(A) The force acting on the electron is $F = eE$,where $e$ is the charge of the electron and $E$ is the electric field.The acceleration of the electron is $a = \frac{F}{m} = \frac{eE}{m}$,where $m$ is the mass of the electron.Using the kinematic equation $v^2 - u^2 = 2aS$,where $u = 0$ (initial velocity),$S = 2 \, cm = 2 	imes 10^{-2} \, m$ (distance between plates),and $a = \frac{eE}{m}$:$v^2 = 2 \left( \frac{e}{m} ight) E S$Given $\frac{e}{m} = 1.76 	imes 10^{11} \, C/kg$,$E = 10^4 \, N/C$,and $S = 2 	imes 10^{-2} \, m$:$v^2 = 2 	imes (1.76 	imes 10^{11}) 	imes 10^4 	imes (2 	imes 10^{-2})$$v^2 = 7.04 	imes 10^{13} = 70.4 	imes 10^{12}$$v = \sqrt{70.4} 	imes 10^6 \approx 8.39 	imes 10^6 \, m/s = 0.839 	imes 10^7 \, m/s$.Rounding to the nearest option,the velocity is approximately $0.85 	imes 10^7 \, m/s$.

An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^4 \, N/C)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to

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