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(A) The electron enters the electric field perpendicular to the field lines. The horizontal velocity $u$ remains constant as there is no force in the

Vedclass · Accepted Answer

$1.76$

Vedclass · Answer

(A) The electron enters the electric field perpendicular to the field lines. The horizontal velocity $u$ remains constant as there is no force in the horizontal direction.The time taken to travel a horizontal distance $x$ is $t = \frac{x}{u}$.The vertical force on the electron is $F = qE$,so the vertical acceleration is $a = \frac{qE}{m}$.The vertical deflection $y$ is given by $y = \frac{1}{2}at^2 = \frac{1}{2} \left( \frac{qE}{m} ight) \left( \frac{x}{u} ight)^2$.Given: $E = 3200 \ V/m$,$x = 0.10 \ m$,$u = 4 	imes 10^7 \ m/s$,$q = 1.6 	imes 10^{-19} \ C$,$m = 9.1 	imes 10^{-31} \ kg$.Substituting the values:$y = \frac{1}{2} 	imes \frac{1.6 	imes 10^{-19} 	imes 3200}{9.1 	imes 10^{-31}} 	imes \frac{(0.1)^2}{(4 	imes 10^7)^2}$$y = \frac{1}{2} 	imes \frac{5.12 	imes 10^{-16}}{9.1 	imes 10^{-31}} 	imes \frac{0.01}{16 	imes 10^{14}}$$y = \frac{5.12 	imes 10^{-16} 	imes 10^{-2}}{2 	imes 9.1 	imes 10^{-31} 	imes 16 	imes 10^{14}}$$y \approx 0.001758 \ m \approx 1.76 \ mm$.

An electron enters an electric field of intensity $3200 \ V/m$ perpendicular to the field with a velocity of $4 \times 10^7 \ m/s$. If it travels a horizontal distance of $0.10 \ m$,then its deflection from its path is ........ $mm$.

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