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Find the electric field at x and y coordinates?

  1. Jan 25, 2017 #1
    1. The problem statement, all variables and given/known data
    A 60 μC point charge is at the origin.
    Find the electric field at the point x1 = 50 cm , y1 = 0.

    2. Relevant equations
    Coulombs Law: F = (kQ1Q2)/r2
    Electric Field: E = kQ/r2
    k = 8.99*109 C

    3. The attempt at a solution
    So taking Coulombs Law and deriving the Electric Field equation I then took my charge particle and set that equal to Q (First converting it from micro to full Coulombs), then taking the distance x and setting that equal to the r2 I put that in the denominator to finally solve for the Ex.

    Note that k is 8.99*109

    Finally I do the same thing with the y-component, but this is my issue. When I take the numbers and divide by 0 I get the infamous infinity.

    So What Am I doing wrong here?

    Thanks in advance!!!
     
  2. jcsd
  3. Jan 25, 2017 #2

    gneill

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    Staff: Mentor

    The r in the denominator is the total distance from the charge, and includes the contributions of both x and y offsets at the same time. This version of the Coulomb's law equation is a scalar version, yielding the magnitude of the electric field. It's up to you to assign the direction (i.e. break it into x and y components) as required.
     
  4. Jan 25, 2017 #3
    Wait so the answer I get when taking .5+.0 and squaring it is the same answer for both the x and y components? This is where I am starting to get confused.
     
  5. Jan 25, 2017 #4

    gneill

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    Yes. But think Pythagoras: ##r^2 = x^2 + y^2##.
     
  6. Jan 25, 2017 #5
    Ok so r2 = (0.5)2+02 = 0.25
    Then taking this I simply get the equation (8.99*109*6*10-5)/0.25 = 2.1576*106.
    So I guess I am confused again on how I differentiate between x and y components. Is it the same answer for both?
     
  7. Jan 25, 2017 #6

    gneill

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    What you have calculated is the magnitude of the electric field (be sure to include the units!). You need to break that into x and y components depending upon the direction of the vector it's associated with. This case is pretty simple given the direction of the electric field vector at the location you were given.
     
  8. Jan 25, 2017 #7
    Ok, so I get 2.1567*106 N/C and since initially x = 50 cm or 0.5 m and y = 0 cm or 0 m then my x-component is 2.1567*106 N/C in the positive x-direction, and 2.1567*106 N/C in the positive y-direction? I might be confusing myself.
     
  9. Jan 25, 2017 #8

    gneill

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    You are definitely confusing yourself :smile: If you were to sketch the E vector on a diagram that shows the charge and the location in question on XY axes, what direction would it point?
     
  10. Jan 25, 2017 #9
    It would point to the right. Correct? The charge is positive and the E vector is positive so they would be repulsive.
     
  11. Jan 25, 2017 #10

    gneill

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    Well, the E vector doesn't have a sign per se, but yes, it would point to the right in the +x direction. What does that tell you about the magnitude of the y-component for such a vector?
     
  12. Jan 25, 2017 #11
    The y vector magnitude is 0 since it does not point in either direction on the y-axis.
     
  13. Jan 25, 2017 #12

    gneill

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    Correct. So the x-component must account for the entire magnitude of the vector.
     
  14. Jan 25, 2017 #13
    Thank you so much for sticking with me! That was my entire issue was the y-component and once I drew a sketch then it just pieced together. Thanks again!!!
     
  15. Jan 25, 2017 #14

    gneill

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    Staff: Mentor

    Cheers! :smile:
     
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