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 Homework Statement:

An electron is thrown halfway between two large plates, with an angle Theta and an initial speed v0. The resultant electric field caused by the plates is vertical and its magnitude is noted by E. A neutral and insulating vertical membrane located to the right of the electron has an opening halfway between the two plates.
V0 = 5 * 10^6 m/s
E = 550 N/C
L = 25 cm
 Relevant Equations:

a) Calculate the value of the parameter s ( sigma )
b) Say if the electron should be projected at an up angle (theta> 0) or down (theta <0) to pass through the opening in the membrane. Justify your answer.
c) Consequently to your answer in b), calculate the two possible angle values in order to that the electron can exit through the opening.
a) E = s / E0 so s is 4.87E9
b) The electron will be projected at up angle since its charge is negative ( not sure if there's another reason behind it)
c)
Initial speed:
V0 = 5 * 10^6 * cos(theta) + 5 * 10^6 * sin(theta)
The force suffered by the electron is:
Fy = q*Ey
Fy = 1.602*10^19 * 550 N/C
Fy = 8.811E17
The acceleration of the electron:
Fy = m*ay
8.811E17 = 9.11*10^31 * ay
Ay = 9.6718E13
The time it takes for the electron to exit
X= x0 + v0x * t
0.25 = 0 m + 5 * 10^6 * t
t = 5 * 10E8
The speed that the electron quits the plates
Vy = v0y + ay * t
Vy = 0 + 9.6718E13 * 5 * 10E8
Vy = 4835894.62
I’m not sure if I’m on the right path or not.
I think the way I need to find the angle is like:
Theta = artcan ( vy / vx )
Thank you !