An electron with a velocity of 3.00 x 10^6 m/s [horizontally] passes through two horizontal parallel plates.?
An electron with a velocity of 3.00 x 10^6 m/s [horizontally] passes through two horizontal parallel plates. The magnitude of the electric field between the plates is 120 N/C. The plates are 4.0 cm across. Edge effects in the field are negligible.
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the e travels straight through, and as it reaches the end, it is attracted toward the positive place at an angle theta.
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a) calculate the vertical deflection of the electron.
b) calculate the vertical component of the final velocity.
c) Calculate the angle at which the electron emerges.
d) If the magnitude of the electric field is increased will this angle increase, decrease or remain the same. Explain.
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We'll need some extra information for this: the charge of the electron. This is a pretty common number with these sorts of problems and we'll call it e = 1.6 * 10^-19 C. We'll also need the mass of the electron, which we'll say is m = 9.11 * 10^-31 kg (see link for these numbers).
A) Calculate Vertical Deflection
To calculate the vertical deflection (distance travelled vertically), we'll need to first figure out how long the electron remains in the electric field. Given its velocity of 3.00*10^6 m/s and the distance of 4 cm (0.04 m), we can quickly determine "t," the time of travel (because the E-Field is perpendicular to the direction of travel, it doesn't slow the electron down):
t = 0.04 / 3.00*10^6 = 1.33 * 10^-8 sec
To find the vertical deflection, we'll need to figure out how to find a function of distance in terms of time. To do this, we'll go all the way back to Newton's second law, F = ma. We will then find the vertical velocity, and then the vertical distance travelled. We calculate force by multiplying the E-Field strength by the charge of the electron, and then we'll find the acceleration by dividing by the mass:
a = F/m = E*e/m = (120 N/C) * (1.6*10^-19 C) / (9.11 * 10^-31 kg)
a = 2.1 * 10^13 m/s/s
To find the velocity, we just multiply the acceleration by the time the electron was in the field (keep in mind, we started at 0 m/s in the vertical direction).
v (vertical)= a*t = (2.1 * 10^13 m/s/s) * (1.33*10^-8 s)
v (vertical) = 2.80 * 10^5 m/s
Fast? Yes! However, how far did we travel? We'll have to multiply by time again to figure out the distance travelled (deflection):
d = v*t = (2.80*10^5 m/s) * (1.33*10^-8 s) = 3.72 *10^-3 m
d = 3.73 mm
Doesn't look like it travelled that far at all, but we have our answer for the first part, so we'll move on.
B) Calculate the vertical component of the velocity
Well, if you look above, we already solved for this:
v (vertical) = 2.80 * 10^5 m/s
Hooray, easy answers.
C) Calculate the angle (theta) at which the electron emerges
This angle is quickly calculated with some basic trigonimetry. If you look at your velocity vector components, you'll see that you have a horizontal component of 3.00*10^6 m/s and a vertical component of 2.80*10^5 m/s.
To find the angle, we simply take the inverse tangent of the ration of the vertical component to the horizontal component:
Theta = arctan(2.80*10^5 m/s / 3.00*10^6 m/s) = 0.093 radians
Theta = 5.31 degrees
A shortcut you can take is by noticing that the angle is going to be small considering the vertical velocity vs. the horizontal velocity, so the angle in radians is going to be pretty much equal to the ratio of the vertical and horizontal components. In other words, in this case:
arctan(v(vertical)/v(horizontal)) = v(vertical)/v(horizontal) in radians
D) If the magnitude of the electric field (E) increases, will the angle increase, decrease, or remain the same?
Well, the answer should be pretty obvious here. The higher the electric field, the more attracted the electron will be to the positive plate (or the more energy/force is put into the electron), and the greater the angle will become!
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