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Magnitude & Direction of electric fields?

  1. Jan 31, 2008 #1
    1. The problem statement, all variables and given/known data
    What are the magnitude and direction (in degrees counterclockwise from the horizontal) of the electric field at the center of the square of Fig. 23-31 if q = 0.9*10^-8 C and a = 5.1 cm?

    [​IMG]

    2. Relevant equations
    E=kq/r^2
    k=8.99*10^9 N*m^2/C^2

    3. The attempt at a solution
    Calculated r using the pythagorean theorem:
    5.1cm--->.051m
    sqrt(.051^2+.051^2)=.07212m ...this is the length of the entire diagonal so r would be the diagonal divided by 2 which equals .036062445841 meters.

    Starting with E1 (the upper left hand corner of the square) and moving clockwise to E4...
    E1=+62214.53 N/C
    E2=-124429.07 N/C
    E3=+124429.07 N/C
    E4=-62214.53 N/C

    Here is where I get confused...do I resolve these in horizontal and vertical components? And, how do I find the angle? Thanks for your help!
     
    Last edited: Jan 31, 2008
  2. jcsd
  3. Jan 31, 2008 #2
    I believe you would do well to define an xy-coordinaate system, with origin at the center of the square. The electric field is a vector, and thus will have x and y components. You want to find all of the x-components of the field at the center of the square, and sum them, and do the same for the y-components. Once you have these two components, you can use the Pythagorean theorem to find the magnitude of the electric field. Also, at this point, it will be easy to find the angle at which the field vector points (because you will know which directions the components point), and you can use the arc-tangent to find this. Do you follow?
     
  4. Jan 31, 2008 #3
    Thanks for your tips. Here's what I got:
    Xnet= -87984.636 N/C
    Ynet= 0 N/C
    Using the pythagorean theorem:
    Enet= 87984.636 N/C
    This much is correct but when I take the arctangent(0/-87984.636) I get 0. Any ideas?
     
    Last edited: Jan 31, 2008
  5. Jan 31, 2008 #4
    If the x net is -87984.636, and there is zero net in the y-direction, then you simply have a vector pointing in the negative x-direction with the x-net magnitude. There will be no y-component and thus the vector cannot rise above the x-axis, and so the angle it makes with the horizontal is zero, right?
     
  6. Feb 1, 2008 #5
    That makes sense but I must have made a mistake somewhere along the way. The magnitude is correct but the angle isn't. Here's my math:

    +q E1=62214.53 N/C
    -2q E2=-124429.0657 N/C
    -q E3=-62214.53 N/C
    +2q E4=124429.0657 N/C

    E1
    X=62214.53sin135=43992.318 N/C
    Y=62214.53cos135=-43992.318 N/C
    E2
    X=-124429.0657sin45=-87984.636 N/C
    Y=-124429.0657cos45=-87984.636 N/C
    E3
    X=-62214.53287sin225=43992.318 N/C
    Y=-62214.53287cos225=43992.318 N/C
    E4
    X=124429.0657sin315=-87987.636 N/C
    Y=124429.0657cos315=87987.636 N/C

    Xnet and Ynet are the values listed above. Maybe I overlooked something? Thanks for your quick responses! The angle is 90 degrees above the horizontal...I don't really understand why though.
     
    Last edited: Feb 1, 2008
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