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Deflection of an electron

  1. Jan 23, 2008 #1
    1. The problem statement, all variables and given/known data

    An electron with an initial velocity of 4.7 x 10^7 m/s in the x-direction moves along a trajectory that takes it directly between the center of two 20 cm x 20 cm plates separated by 1 cm. The plates are connected to a high voltage power supply so that the potential difference between them is 500V. What is the magnitude of the y-component of the electron’s velocity when it leaves the plates? (Answer in units of ANSWER x 10^7 m/s.)

    2. Relevant equations
    E = V/d

    3. The attempt at a solution

    I think I found the electric field of the plates to be:
    E = 500V / .01m E = 50,000 V/m

    From there I have no idea where to go...Any help would be appreciated!

    Attached Files:

  2. jcsd
  3. Jan 23, 2008 #2
    The electric field is gonna apply a force to the electron

    Do you know the equation that relates electric field, charge, and force?
  4. Jan 23, 2008 #3
    just started EM stuff now, sorry cant help. but blochwave is right, there is an equation that relates to electric field, charge and force, the only one i can think of uses two particles and a separation between them, would not be useful for in this instance. best of luck. lol im curious actually...
  5. Jan 23, 2008 #4
    Well that's the equation FOR an electric field, which you can't quite use here exactly in that form, but he already used the equation for parallel plate capacitors and now KNOWS the electric field, and needs to find the force on a particle with known charge within that (conveniently constant) electric field
    Last edited: Jan 23, 2008
  6. Jan 23, 2008 #5


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    Homework Helper

    If F=EQ, and E=V/d, can you find F? What does Newton's 2nd law say now?

    (Hint: the velocity in the x-direction stays constant)
  7. Jan 24, 2008 #6
    So the force would be F = (50000 V)*(1.602 x10^-19 C)
    F = 8.01 x10^-15 N
  8. Jan 24, 2008 #7


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    Homework Helper

    Yes that is the force. Note that the units of E are either [itex]Vm^{-1} or NC^{-1}[/itex]
  9. Jan 24, 2008 #8
    Kinetic + potential = Emech
    (1/2m(Vi^2)) - q(E)*(Xi) = (1/2m(Vf^2)) - q(E)*(Xf)
    Then we changed the equation to this:
    qE(20cm) = 1/2(mass of electron)*((Vf^2) + (Vi^2))
    Then we solved for Vf:
    Vf = 3.61697 x10^7 m/s
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