Solve Proton Homework: Min Speed to Miss Lower Plate

AI Thread Summary
To determine the minimum speed of a proton that just misses the lower plate in an electric field, the mass of the proton is 1.67×10^−27 kg and the electric field strength is 7.8×10^5 N/C. The initial calculations incorrectly used the full plate separation of 0.20 cm instead of the correct midpoint of 0.10 cm for vertical displacement. The acceleration of the proton was calculated to be 7.47×10^13 m/s², leading to a time of flight of approximately 7.32×10^-9 s. Adjusting for the correct vertical distance yields a revised minimum speed of 7.4×10^6 m/s, confirming the importance of accurately interpreting the problem's parameters.
Vanessa Avila
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Homework Statement


In the figure, a proton, mass 1.67×10^−27 kg, is projected horizontally midway between two parallel plates that are separated by 0.20cm, with an electrical field with magnitude 7.8×10^5N/C between the plates. If the plates are 5.40cm long, find the minimum speed of the proton that just misses the lower plate as it emerges from the field.
YjE1TBR.jpg

Homework Equations


q(E) = ma
x = vt
y = 1/2a(t^2)

The Attempt at a Solution


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This is what I did:
1.6*10^-19(7.8*10^5) = 1.67*10^-27 (a)
solving for a I get: 7.47*10^13 m/s^2

then i solve for t
2*10^-3 = (1/2)(7.47*10^13)(t^2)
t = sqrt(7.47*10^13/((2)(2*10^-3))
I get 7.32*10^-9 s

then i plug that in the x = vt eqn
0.054/7.23*10^-9 = v
v = 7.4*10^6 m/s

But that is wrong
 
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It says the proton comes in midway, which means y is 0.1cm, not 0.2cm
 
Xiaochao Zheng said:
It says the proton comes in midway, which means y is 0.1cm, not 0.2cm
Oh I see.Thank you!
 
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