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Electric field, need help fast!

  1. May 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Two tiny objects with equal charges of 63.0 µC are placed at two corners of a square with sides of 0.260 m, as shown. How far above and to the left of the corner of the square labeled A would you place a third small object with the same charge so that the electric field is zero at A?

    p16-27.gif

    2. Relevant equations
    E = KQ/r^2


    3. The attempt at a solution
    I first solved for E net on point A. My values were
    Enetx = 2.962 x 10^6
    Enety = 1.13 x 10^7
    Magnitude of Etotal = 1.17 x 10^7

    angle formed : 75.3 degrees

    Now I tried two alternate solutions, but in the end I got the wrong answer still for both tries.
    E = kq/r^2
    r = sqrt(kq/E)
    r = sqrt(k(63e-6)/1.17e6)
    r=0.22m

    dx=0.22cos75.3=0.0558
    dy=0.22sin75.3=0.213

    That one was wrong, so I tried finding the value of r using the components of the electric field
    rx= sqrt(k(63e-6)/2.962e6)
    ry=sqrt(k(63e-6)/1.13e7)

    rx =0.43m
    ry=0.22m

    still wrong, i need help because I'm not sure what I'm doing wrong.
     
  2. jcsd
  3. May 15, 2013 #2

    Simon Bridge

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    The first method is the one I'd have used.
    Check units and arithmetic.

    I first solved for E net on point A. My values were
    Enetx = 2.962 x 10^6
    Enety = 1.13 x 10^7
    Magnitude of Etotal = 1.17 x 10^7

    angle formed : 75.3 degrees

    ... that angle looks bad - the resulting field should point up and to the left, so the angle from the +x axis will be bigger than 90deg.

    I don't see how you got the x and y components either.
     
    Last edited: May 15, 2013
  4. May 15, 2013 #3
    I ussed the equation

    E1 = kq/r^2
    = kq/0.26^2
    = 8.378e6 going south

    E2 = kq/(0.26^2+0.26^2)
    = 4.189e6 going south and to the right
    E2cos45 = 2.96e6 (to the right)
    E2sin45 = 2.96e6 (south)
    therefore Eynet = 1.13e7
    Exnet = 2.96e6
    Magnitude sqrt ((2.962e6)^2+(1.13e7)^2)
    =1.17e7
    angle = tan inverse (Ey/Ex)
    = 75.3 degrees

    So the electric field will go in the opposite direction of this to cancel out of the third charge.
    I might be doing something wrong, but I don't know what it is. This is due in 42 minutes, if anyone can check my values quickly or if there's a logic error that I missed i'd appreciate it.
     
  5. May 15, 2013 #4

    Simon Bridge

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    How do you kow it's wrong - is it computer moderated?
    If so - careful with rounding - make sure you have entered the correct format and units.
    i.e. does it want an angle? If so - in degrees or radians?
    How many sig fig is it expecting?

    You realize that 75.361° is clockwise from the -x axis - if they want to measure angles anticlockwise from the +x then this will be marked wrong. Also notice that it rounds to 75.4 not 75.3.

    You gotta check the details.

    I'm getting the same angle and field strength as you, but a different distance, using k=8.99e9(SIU)
     
  6. May 15, 2013 #5

    Simon Bridge

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    [edit]nope - I just forgot to invert.
    I'm getting the same as you - check for fiddly stuff.
     
    Last edited: May 15, 2013
  7. May 15, 2013 #6
    edit: nevermind, thanks simon
     
    Last edited: May 15, 2013
  8. May 15, 2013 #7
    Ok this helped me out. I guess I just wasn't being detailed with my calculations, it was off by 0.02 and that was enough to make it wrong on the computer. Thanks so much!
     
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