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Magnitude of Electric Field

  1. Jan 25, 2007 #1
    1. The problem statement, all variables and given/known data
    An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3.00\;\mu{\rm s} after it is released.

    What is the magnitude of the electric field?

    2. Relevant equations
    F = 1/4*pi*e_o * |q| / r^2 for single point charge

    e_o = 8.85 * 10^-12

    Mass of electron = 9.109 * 10 ^-31

    3. The attempt at a solution

    1/(4*pi*8.85 * 10^-12) * ( |9.109 * 10 ^-31| / (4.50)^2 )

    The online system said the solution was incorrect.
  2. jcsd
  3. Jan 25, 2007 #2


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    Note the problem hints at an acceleration; you'll need to calculate the aveage acceleration of the particle, and then use Newton's second law to relate it to the force applied by the field. You need to account for another force here too; do you know what it is? Then you need an equation that relates the net force to both charge and electric field. As an aside, your first equation gives the force between two point charges, which is not the situation here, and in any case it is incorrect as written.
  4. Jan 25, 2007 #3
    Can anyone answer my tension/friction question, please.
  5. Jan 25, 2007 #4
    Can you list the equations thats necessary? I'm not sure which equations you're talking about. Thank you
  6. Jan 25, 2007 #5


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    The electric part of the Lorentz force, and Newton's second law.
  7. Jan 26, 2007 #6
    I'm no familiar with Lorentz force, the section we're in is introduction to Electricity and the only equation I've seen so far is Coulomb's Law.

    F = k*|q1|*|q2| / r^2

    If you could start me off with an equation that would be most helpful. I can't see where I have to go with the givens I have so far.
  8. Jan 26, 2007 #7


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    The electric Lorentz force is
    in SI units.
  9. Jan 26, 2007 #8
    So we have the particle traveling 0.45m in the first 3us and you said I had to find the average acceleration?

    Velocity = 4.50m / 0.000003 seconds = 1.5*10^6 m/s

    Now we have to take the avg. acceleration which should be the change in velocity divided by the change in time? Would that just be that solution divided by 3us again to obtain that value?
  10. Jan 26, 2007 #9


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    You need to take a look in your book. The formula relating distance to acceleration and time is
    Did you find it? Make sense?
    Suggest you ask your teacher for some help.
  11. Jan 26, 2007 #10
    Makes sense, I just got it! Thanks again. It's been a very long time since I've done kinematics.
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