An electron is released from rest within a horizontal electric field E)) and pro

Question

An electron is released from rest within a horizontal electric field E)) and proceeds to accelerate to the right. If it accelerates at 5 x 10^9 m/s^2, what is the value of Eh? The length of the region with Eh is 10 mm. What is its final horizontal speed vh when it exits that region? The electron then enters a vertical electric field E, whose strength is 300 N/C upwards. If the electron spends 2.5 x 10^-7s in this new field, how far does it travel & to the right, what is its deflection Av from its original path and in what direction (up or down)?

Explanation / Answer

when the electron enters the electric field Eh the force

acting on the electron Fe = e*Eh

but from newtons law net force = F = m*a

e*Eh = m*a

Eh = m*a/e

Eh = (9.11*10^-31*5*10^9)/(1.6*10^-19) = 0.0285 N/C

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(b)

vx^2-vox^2 = 2*ax*x

vx^2 = 2*5*10^9*10*10^-3

vx = 10^4 m/s <----answer

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(c)

in the vertical field

along horizantal the speed vx remains same beause the
acceleration ax=0 in the vertical field

deltax = speed*time = vx*t = 10^4*2.5*10^-7

deltax = 0.0025 m = 2.5 mm

invertical direction

Fe = e*Ev

but Fe = m*ay

ay = e*E/m

ay = (1.6*10^-19*300)/(9.1*10^-31)

ay = -5.3*10^13 m/s^2

deltay/2 = voy*T - 0.5*ay*T^2

voy = 0

deltay/2 = 0.5*5.3*10^13*2.5*10^-7*2.5*10^-7 = 1.65625

deltay = 3.3125 m   <---answer

down <---answer

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