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Question about the magnitude and direction of an electric field.

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    The electric potential V in the space between two flat parallel plates 1 and 2 is given (in volts) by V = 1920x^2, where x (in meters) is the perpendicular distance from plate 1. What is the magnitude and direction of the electric field at x = 2.5 cm? (Take the direction perpendicular to and away from plate 1 to be positive.)


    2. Relevant equations

    E= V/r

    3. The attempt at a solution
    First, I converted the cm to m, and then I plugged everything in, by saying: (1920*0.025^2)/0.025= 48.0. The answer is incorrect, also the negative of 48 is incorrect. I am not sure where to go from here and there was no picture to assist. Any help would be appreciated. Thanks.
     
  2. jcsd
  3. Sep 19, 2009 #2

    rl.bhat

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    The relevant equation given by you is true for uniform field. In the given problem field is not uniform. Potential is a function of x^2. In that case the eelctric field is given by
    E = - dV/dx.
     
  4. Sep 20, 2009 #3
    When I put in my numbers using that equation, I get the same thing: E= -dV/dx = - 1920*(.025 ^2)/ .025 = 48V/m. I may be mistaken though... any thoughts?
     
  5. Sep 20, 2009 #4

    rl.bhat

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    It is not correct.
    E = -dV/dx = - 1920*d/dx(x^2).
    What is the derivative of x^2?
     
  6. Sep 20, 2009 #5
    Did you mean to say 1920*x/ dx? Then, I would get (1920*.025)/ (2*.025) = 960, does that sound correct?
     
  7. Sep 20, 2009 #6

    rl.bhat

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    No. d/dx(x^2) = 2x.
    E = -dV/dx = - 2*1920*x
     
  8. Sep 20, 2009 #7
    Oh, okay. Thanks a lot for all of your help. I am not that great at physics.
     
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