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Old 01-13-2010, 04:54 PM
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Smile Parallel Plates Electric Field

Hello, I wonder if anyone could check my answer to this problem.

Thanks in advance

An electron is liberated from the lower of two parallel plates separated by a distance of 5mm. The upper plate has a p.d. of 1000v relative to the lower plate. How long does it it take for the electron to reach to upper plate.



I said F=ma=QE therefore QE=ma

and since E=V/d=1000/0.005=2x105

a=QE/m
a=3.513x1016m/s2!

Then using s=ut+1/2at2, t=5.335x10-10s.

Thanks again for any help

Mike
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  #2  
Old 01-14-2010, 05:33 AM
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clombard1973 will become famous soon enoughclombard1973 will become famous soon enough
Smile Look's fine to me

Quote:
Originally Posted by mikewhant View Post
Hello, I wonder if anyone could check my answer to this problem.

Thanks in advance

An electron is liberated from the lower of two parallel plates separated by a distance of 5mm. The upper plate has a p.d. of 1000v relative to the lower plate. How long does it it take for the electron to reach to upper plate.



I said F=ma=QE therefore QE=ma

and since E=V/d=1000/0.005=2x105

a=QE/m
a=3.513x1016m/s2!

Then using s=ut+1/2at2, t=5.335x10-10s.

Thanks again for any help

Mike
Hi there,
As long as your initial velocity, u, was zero, the mass you used was that of an electron, and lastly the charge , Q, was that of an electron, then your answer is correct. 1000V applied to such a tiny mass should cause it to undergo a great acceleration. Again this assumes that gravity has no effect on the electron, which is typically considered true. Then you should be able to set 0.005 = (1/2)at^2, plug in the value of, a, that you got, and then solve for the time, t.
Many Smiles,
Craig

Last edited by clombard1973; 01-14-2010 at 05:36 AM.
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  #3  
Old 01-14-2010, 06:47 AM
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clombard1973 will become famous soon enoughclombard1973 will become famous soon enough
Smile Yup you're correct

Hi there,
I just went through and solved the problem myself, I got both the same acceleration upon the electron that you did and the same time to traverse the distance from one plate to the other.

I feel pretty certain about my answer and since you have the same as I have, I'd feel pretty certain that you too also have the correct answer.

Nicely done!

Many Smiles,
Craig
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Old 01-14-2010, 08:06 AM
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Thanks Craig, as always most helpful!

Take care,

Mike
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