If anyone can solve this or at least point me in the right direction I'll give you 1 million internets.
Two charged, parallel, flat conducting surfaces are spaced d=1.00cm apart and produce a potential difference (delta)V = 625V between them. An electron is projected from one surface directly toward the second. What is the initial speed of the electron if it stops just at the second surface?
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Anonymous2008-09-22 22:24
W = qV
W = .5mv^2
qV = .5mv^2
2qV/m = v^2
v = sqrt(2qV/m)
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Anonymous2008-09-22 22:32
i must be thick because the only thing I know how to solve for in there is m. What is V and what is q?
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Anonymous2008-09-22 22:37
also so I don't just write this down without understanding it, where did you get W = .5mv^2?
I understand how you derived everything else
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Anonymous2008-09-23 1:07
V is the voltage, or 625 volts. q is the charge of the electron.
.5 m v^2 is the equation for kinetic energy. The kinetic energy was converted into potential energy (qV).
V is the electric potential, and is measured in Volts.
The reason >>2 was using energy is because the electron moves across an electric potential difference. By definition, a charged particle moving across a potential difference gains/loses an amount of energy equal to its charge * the difference in electric potential.
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Anonymous2008-09-26 9:12
q = -1.6x10^-19 C for an electron,
E = V/d = 62500NC^-1
F = Eq = 1x10^-14N
a=F/m = ...
0 = u^2 + 2ad
sub in and do the math
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Anonymous2008-09-26 9:26
>>1
Can't be done. You're black. >>2-7
All black...Especially Redtran...
>>8
I know it must be tough sitting in your physics class all day with your iPod on full blast, trying to stifle the blue balls you have from looking at that fine bitch in front, yeah, the one with the BANGIN' ass, but can you please keep it off here?
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Anonymous2008-09-26 22:16
>>9
Look, buddy, don't start taking shots at me because you're black. If you have a problem with your negrosity then take it up with your parents.