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Deflection of electron

  • Thread starter eltel2910
  • Start date
  • #1
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An electron with a kinetic energy 5000.0 eV (1 eV = 1.602×10-19 J) is fired horizontally over a charged plate with surface charge density +2.0 μC m-2. Taking the positive direction to be upwards (away from plate), what is the vertical deflection of the electron after it has traveled a horizontal distance 2.0 cm?

Here is my work:

Find V of electron - KE = .5mV ---> 41934557 = Velocity

time = s/V ----> .02m/41934557 --------> 4.769e^-10 = time

electric field E = sigma/2*Eo -----> 2.0e^-6/(2*(8.85*10^-12) ---->112994 = E

Force = q*E ------>(1.602*10^-19)(112994) ------->1.810e^-14 = F

acceleration = F/m -------> 1.810e^-14/(9.11*10^-31) = 1.98e^16. 9.11*10^-31=mass of one proton.

vertical deflection = d=.5at^2 --------> .00226=.5(1.98e^16)(4.769e^-10)^2

I come up with .00226 meters, but "the man" still says I'm wrong

Can you see anything wrong with my set up??
 

Answers and Replies

  • #2
2,063
2
eltel2910 said:
Find V of electron - KE = .5mV ---> 41934557 = Velocity

time = s/V ----> .02m/41934557 --------> 4.769e^-10 = time

electric field E = sigma/2*Eo -----> 2.0e^-6/(2*(8.85*10^-12) ---->112994 = E

Force = q*E ------>(1.602*10^-19)(112994) ------->1.810e^-14 = F

acceleration = F/m -------> 1.810e^-14/(9.11*10^-31) = 1.98e^16. 9.11*10^-31=mass of one proton.

vertical deflection = d=.5at^2 --------> .00226=.5(1.98e^16)(4.769e^-10)^2
Can't see anything wrong except the one that has been highlighted.
 

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