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Coulomb's Law And Electric Field between 2 Parallel Plates

  • Thread starter Yosty22
  • Start date
  • #1
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Homework Statement



A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.59cm distant from the first, in a time interval of 1.60×10−6s .

A). Find the magnitude of the electric field.
B). Find the speed of the proton when it strikes the negatively charged plate.

Homework Equations



F=ma
E=F/q
y=y_0+v_0t+1/2at^2

The Attempt at a Solution



Usually, to solve for the E field between two parallel plates, you use sigma/epsilon_0. However, I am not given sigma nor am I given anything about the dimensions of the plate, so I took the old kinematics route. First, I used the equation y=y_0+v_0t+1/2at^2. I said it starts from y_0=0 and it says it starts from rest, so I am left with y=1/2at^2. I plugged in the distance between the plates (.0159m) and the time (1.6*10^-6s) and solved for the proton's acceleration, which I found to be 1.24 m/s^2 in the negative y direction (towards the negative plate).

Next, I used F=ma - using the acceleration I just found (1.24 m/s^2) and the mass of the proton (1.6*10^-27kg) and solved for F. I got 1.984*10^-27N

Lastly, for part A, I used the equation E=F/q. Substituting the numbers in, I got that E=1.24*10^-8 N/C and it said that it was incorrect.

For part B, i used the acceleration I found above and plugged it into v=v_0+at and got V=1.984*10^-6, which was also wrong.

Any ideas what I did wrong?
 

Answers and Replies

  • #2
3,740
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The value you obtained from acceleration seem to be off by many orders of magnitude.
 
  • #3
rude man
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The value you obtained from acceleration seem to be off by many orders of magnitude.
'Many' is right! :smile:
 

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