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Electric Field & it's components

  1. Jun 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Screen Shot 2014-06-06 at 4.27.50 PM.png

    Two point charges, Q1 = -4.0 μC and Q2 = +3.0 μC, are placed as shown. In Figure 17.14b, the x-component of the electric field, at the origin O, is closest to:
    A) -3000 N/C
    B) 3800 N/C
    C) 9800 N/C
    D) -3800 N/C
    E) -9800 N/C


    2. Relevant equations

    kq/r^2


    3. The attempt at a solution

    i understand how to get the magnitude of both forces of Q1 and Q2 by using kq/r^2
    i found the hypotenuse to use as the length which is 2.57m .
    what i'm lost on is the direction vectors that i need multiply the forces by.

    in relation to Q2 i have found the inverse tangent and the angle comes out to be about 33. but when i multiply it by the force i get and get the difference with the force of Q1 it's not right.

    any help is much appreciated, this one problem has been giving me trouble for quite some time!
     
  2. jcsd
  3. Jun 6, 2014 #2
    Don't describe your calculation. Show it. If you don't show us your calculation how do expect us to find what's wrong with it?
     
  4. Jun 6, 2014 #3
    sorry here it is.
    kq/r^2

    Q1:
    (9 x 10^9) x (-4 x 10^-6) / (2.3^2) = -6810 N/C

    Q2:
    (9 x 10^9) x (3 x 10^-6) / (2.75^2) = 3570.3 N/C
    [the 2.75 distance i'm using is found - hypotenuse]

    i'm unsure about the direction vector for Q1 since it is just on the x axis ?
    direction of vector for Q2:
    atan(1.5/2/3) = 33.11 degrees
    so 3570cos(33.11) = 2990 N/C

    -6810 - 2990= -9800 N/C

    the right answer is 3800 but i'm not sure why?
     
  5. Jun 6, 2014 #4
    sorry! i'm new to this!!
     
  6. Jun 6, 2014 #5
    why -6810 in the last line? why not + 6810? After all Q1 is negative isn't it?
     
  7. Jun 6, 2014 #6
    well with the calculations you get the -6810 . why would you change it; Q1 is to the right of the origin so wouldn't it just stay the same?
     
    Last edited: Jun 6, 2014
  8. Jun 6, 2014 #7

    haruspex

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    Q1 is negative, as given. The x displacement of the origin from Q1 is negative. The sign of the field comes from the product of the two, so is positive.
    Just think about which way a positive test charge at the origin would be affected by Q1. It would be attracted to the right, yes?
     
  9. Jun 6, 2014 #8
    oh yeah that makes much more sense, thank you!
     
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