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Electric Field Problem (why don't I multiply by sin and cos?)

  1. Feb 5, 2014 #1
    1. The problem statement, all variables and given/known data
    In the figure the three particles are fixed in place and have charges q1 = q2 = +e and q3 = 2e. Distance a = 6.00 µm. What is the magnitude of the net electric field at point P due to the particles?

    NAhIr74.gif

    2. Relevant equations
    E = k*q/r^2


    3. The attempt at a solution

    I looked in the back of the book and it shows that I shouldn't be multiplying by (cos45 + sin45) but I don't understand why since Electric Field is a vector value

    here is my work

    q1 and q2 cancel out so i only calculate q3

    F = 9*10^9 * (3.2/(6^2/2)) * 10^-13 * (cos45 + sin45)

    simplifies to
    F = 160 * sqrt 2 = 226.274N

    The actual answer is just 160N so that means I shouldn't multiply by (cos45 + sin45) even though that is the angle.

    Edit: picture should be viewable
     
    Last edited: Feb 5, 2014
  2. jcsd
  3. Feb 5, 2014 #2
    The image isn't displaying.
     
  4. Feb 5, 2014 #3
    can you see it now?
     
  5. Feb 5, 2014 #4
    It's important to know that r is the physical distance between the charge and the point where you want to calculate the field. You can't plug in vector components for r because it's a scaler. I think you assumed that r=6 um, but you actually don't have the value for r yet. You need to find the actual distance between q3 and P.
     
  6. Feb 5, 2014 #5
    i kind of skipped some of the steps but I did calculate r as sqrt(a^2/2) to get r^2 = 6um^2/2
     
  7. Feb 5, 2014 #6
    Double check that.
     
  8. Feb 6, 2014 #7
    a^2 = 2r^2

    (a^2)/2 = r^2

    sqrt((a^2)/2) = r
     
  9. Feb 6, 2014 #8
    Electric field is a vector, but the question indicates that it's only asking for the magnitude, so that's why you can disregard the direction.
     
  10. Feb 6, 2014 #9

    but isn't the magnitude effected by the direction or angle?

    If I'm applying a force at a 45 degree angle the amount it moves is different than if I apply it at a 90 degree angle.
     
  11. Feb 6, 2014 #10
    Depends. For torque it matters because the equation involves a vector product, which necessarily invokes direction. For electric field magnitude only depends on charges and distance, which are both scalars. Angle affects the amount of magnitude in the x and y directions, but not the total magnitude.
     
  12. Feb 6, 2014 #11

    haruspex

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    There is no logical basis for the term (cos45 + sin45). If the field is E (vector) it has X and Y components |E|cos45 and |E|sin45, but you cannot add those together. To combine perpendicular components of a vector to find its magnitude you use the root-sum-square formula, which, of course, produce the answer |E|.
     
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