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How would I calculate the magnitude of the electric field?

  1. Oct 20, 2005 #1
    An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3 us (microsecond??) after it is released.

    Part A What is the magnitude of the electric field?

    How would I calculate the magnitude of the electric field? I am not sure which forumla to use, but I am guessing E = F/q

    E= F/q also equals 1/(4pi(8.854x10^-12)) * Q/r^2

    If that is the right equation to use, then I know we have r already (4.50m) but how do we solve for Q?


    Part B What is the direction of the electric field? Upward or Downward?


    Since it says the electron moves upward after released, will that mean the direction of the electric field is upward also?


    Thank you for your help.
     
  2. jcsd
  3. Oct 20, 2005 #2

    Tide

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    You're being asked to find the strength of the (uniform) applied electric field - not the electric field produced by the electron.
     
  4. Oct 20, 2005 #3

    vanesch

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    hotmail590, may I remind you that this kind of questions is to be posted in the homework/textbook help section of this forum ?
     
  5. Oct 20, 2005 #4
    @hotmail590: You see, the electron is moving vertically upwards, it is overcoming gravity and this can only be due to another force (which is stronger than the gravitational force). Now, the key question is: what is the source of this force? You've got the answer in your problem statement: There's an electric field (which exerts a force on any charged object within its reach).
    You already wrote down the right equation for the connection between the force on a charged object in an e-field and the e-field strength.
    How to proceed: Clearly you must find the magnitude of the force. This comes down to a kinematics/dynamics problem: You are given information about the acceleration of the electron (it starts at rest, I suppose, and will have travelled 4.5m in 3 microseconds). Using this you can calculate the total acceleration of the electron. Note though that this acceleration is the result of two forces (from the gravitational and the electric field), so (using the mass of the electron to calculate the gravitational force) you can calculate the force due to the e-field.
    Now, that's it (as you surely know the charge of an electron).
    I suggest you do the math and explain all steps involved and post it all here, so that we can check.
    Regards...Cliowa
    Btw, one more thing: For this to work it is absolutely mandatory that the e-field be homogenous. Do you know why?
     
  6. Oct 20, 2005 #5

    Tide

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    cliowa,

    Please hold off until he solves the original problem, if you get my drift! ;)
     
  7. Oct 21, 2005 #6
    Alright, I just felt this was the basic assumption which makes the calculation that straight-forward (You're implying it if your proceeding as I suggested).
     
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