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Finding the angle of an electric field with respect to an axis

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data

    The electric potential at points in an xy plane is given by V = (2.7 V/m2)x2 -(4.4 V/m2)y2. What are (a) the magnitude of the electric field at the point (3.4 m, 1.6 m) and (b) the angle that the field there makes with the positive x direction.


    2. Relevant equations

    I have already taken the partial derivatives of V with respect to x and y to get the components to the field and then the magnitude. Ex=18.36 V/m and Ey=-14.08 V/m

    3. The attempt at a solution

    The magnitude of the field at the given point is 23.137 V/m. I absolutely cannot for the life of me figure out how to calculate that angle, and I know it's in front of my face. I've tried arctan, but it just doesn't work.
     
  2. jcsd
  3. Feb 20, 2009 #2

    Delphi51

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    Surely it has to be arctan(Ex/Ey) = -37.5 degrees.
     
  4. Feb 20, 2009 #3
    That's what I thought! Apparently it's not, hence my frustration.
     
  5. Feb 20, 2009 #4

    Nabeshin

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    Erm, are you sure about Ey?
     
  6. Feb 20, 2009 #5
    If I'm wrong it's the sign. I know this because I have verified the magnitude is correct.
     
  7. Feb 20, 2009 #6

    Nabeshin

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    Where did the 14.08 come from?
     
  8. Feb 20, 2009 #7
    I used the definition of the derivative: [tex]\frac{f(a,b+h)-f(a,b)}{h}[/tex]. h was .001.
     
  9. Feb 20, 2009 #8

    Nabeshin

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    Sorry my mistake. I agree with Delphi51 ^^
     
    Last edited: Feb 20, 2009
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