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X and Y Components of the Electric Field at the Origin

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Assume that you have a configuration of conductors with varying electric potentials applied to each. You make the following measurements of the values of the electric potential at different points in a two dimensional grid:

    X = 0.0 cm Y = 0.0 cm V = 2.0 volts
    X = 2.0 cm Y = 0.0 cm V = 2.25 volts
    X = 0.0 cm Y = 2.0 cm V = 2.0 volts

    What are the values of the x and y components of the electric field at the origin (Ex and Ey at (x, y) = (0,0))? Express the results in SI units.


    2. Relevant equations



    3. The attempt at a solution

    I'm not sure where to begin. Any suggestions?
     
  2. jcsd
  3. Sep 30, 2012 #2

    Dick

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    You can't, really. They haven't given you ANY information about the location of the conductors and their potentials. But the difference between the potential at the second point and the first should let you make an estimate of Ex and between the third and the first of Ey.
     
  4. Sep 30, 2012 #3
    The question references another question for help with this question. I'm told that:

    (Ex, Ey) = -(∂V/∂x, ∂V/∂y) = -(ΔV/Δx, ΔV/Δy)

    So, I calculated:

    Ex = -2.25 V/ 0.02m = -112.5 V/m

    Ey - 2 V/ 0.02 m = -100 V/m

    I'm not sure if this makes any sense, and if it does then where does the negative come from?
     
    Last edited: Sep 30, 2012
  5. Sep 30, 2012 #4

    Dick

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    ΔV means the DIFFERENCE between the potential at one point and another. I don't think you subtracting the potential at the two points.
     
  6. Sep 30, 2012 #5
    Ahh, yes you are correct. I shouldn't have missed that:

    Ex = -(2.25V-2.V0/(0.02 m))
    Ex = -12.5 V/m

    Ey = -(2.0V - 2.0V/(0.02 m))
    Ey = 0 V/m
     
  7. Sep 30, 2012 #6

    Dick

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    Yes, I think that's the answer they are looking for. You do know that the partial derivatives and the delta expressions aren't really equal, yes? The deltas are just an approximation to the true derivatives?
     
  8. Sep 30, 2012 #7
    Yes, my mistake; it should have been an approximation sign, not an equals sign.

    Thank you for your help.
     
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