An electron is fired at a speed v0 =5.10x106 m/s and at an angle 0 = 45° between

Question

An electron is fired at a speed v0 =5.10x106 m/s and at an angle 0 = 45° between two parallel conducting plates that are D = 5.0 mm apart, as in the figure. If the voltage difference between the plates is V = 105.5 V, determine (a) the magnitude of the electric field between the two plates. (b) the acceleration of the electron in the y-direction.

(c) how close, d, the electron will get to the bottom plate.(d) the time of flight. (e) where the electron will strike the top plate.

Explanation / Answer

part a)

electric field between the plates = V/d

electric field between the plates = 105.5/0.005

electric field between the plates = 21100 V/m

part b) let the acceleration of the electron is a

m * a = e * E

9.11 *10^-31 * a = 21100 * 1.602 *10^-19

a = 3.711 *10^15 m/s^2

the acceleration of electron is 3.711 *10^15 m/s^2

part c)

for the distance d

d = D/2 - height reached by electron

d = 0.005/2 - (5.1 * 10^6 * sin(45 degree))^2/(2 * 3.711 *10^15)

d = 7.48 *10^-4 m

the distance d is 7.48 *10^-4 m

part d)

let the time of flight is t

Using second equation of motion

0.005/2 = -5.1 * 10^6 * sin(45 degree) * t + 0.5 * 3.711 *10^15 * t^2

solving for t

t = 2.49 * 10^-9 s

the time of flight is 2.49 * 10^-9 s

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