1. Parallel plates with capacitance C, charge Q, and plate separation x are loca

Question

1. Parallel plates with capacitance C, charge Q, and plate separation x are located just inside a uniform magnetic field, B (the shaded area shown), which is directed vertically upward. Protons of various speeds are sent in a stream directly southward on the path shown between the plates. Some of those protons continue on to strike a detection film perpendicularly, as shown. The detector film is placed at an angle of 60° west of north. When such a proton enters the capacitor, it takes 0.500 ms to reach the film (and it travels within the magnetic field for that entire trip). What distance does it travel within the capacitor?

The data: C = 3.00 mF Q = 640 nC x = 1.00 cm B = 127 mT

Directions You may either use the space provided here or you may supplement as needed with your own paper. (If you use your own paper, there is no need to re-state the questions.) But you'll always need to provide (and complete) the cover sheet (first page) of this file 1. Parallel plates with capacitance C, charge Q, and plate separation x are located just inside a uniform magnetic field, B (the shaded area shown), proton stream enters here which is directed vertically upward. Protons of various speeds are sent charged plates. in a stream directly southward on the path shown between the plates. detection Some of those protons continue on to strike a detection film perpendicularly, as shown. The detector film is placed at an angle of 60° west of north. some pro north When such a proton enters the capacitor, it takes 0.500 ms to reach tons hit film the film (and it travels within the magnetic field for that entire trip). perpendicu- east larly here What distance does it travel within the capacitor? 3.00 mF Q-640 nC 1.00 cm B 127

Explanation / Answer

the equations using is F = e E and F = e V B

then e E = e V B

V = E / B

and we have Q = C V

V = Q / C

the electro static force is E = V / x

E = Q / x C then we can write V = Q / B x C

the time spent on proton is in between capacitor plates

t1 = d / V

and the radius of the circular path R = mp V / e B

the distance travelled by proton of the outside capacirtor is

21 0 = 7 X 3.14 /6

so

t2 = ( 7 X 3.14 /6 ) X R / V

now t1 + t2 = 0.5 X 10-3

so we can write d / V + ( 7 X 3.14 /6 ) X R / V = 0.5 X 10-3

already we have V = Q / B x C and R = mp V / e B

substituting above values

so d / (Q / B x C) + ( 7 X 3.14 /6 ) X (mp V / e B) / V = 0.5 X 10-3

d (117 X 10-3 X 3 X 10-3 X 0.01 / 640 X 10-9 ) + ( 7 X 3.14 X / 6 ) ( 1.67 X 10-27 / 1.6 X 10-19 X 127 X 10-3) = 0.5 X10-3

d = 0.085 X 10-3 m

d = 0.085 mm

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