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Homework Help: Electric Field at the Center of a Square of Point Charges

  1. Jan 13, 2004 #1
    *sigh*, Another long semester of physics...

    Anyways, here we go ! Here's the problem:

    Calculate the magnitude of the electric field at the center of a square with sides 27.3 cm long if the corners, taken in rotation, have charges of 1.10 microC, 2.20 microC, 3.30 microC, and 4.40 microC (all positive).

    Using Pythagorean's thereom I find the radius to all these Point Charges to be 0.193040 m

    So then I try plugging All this info in the Eq'n for finding Electric field magnitude and direction.. E = kQ/(r^2) where k = 9.0*10^9

    At first, I tried E = (k/r^2)(Q1 + Q2 + Q3 + Q4)
    and got 2660000. Which, apparently, is wrong.

    From looking at the study guide and the book, I'm guessin I'm leaving out some very important trigonometric calculations necessary to go further.. So I decided to calculate the electric field produced by each point charge at the center

    E1 = 265668
    E2 = 531337
    E3 = 797005
    E4 = 1062674

    Then I Broke Each into their Components by arbitrarily assigning them corners of a square

    E1x = 187855
    E1y = -187855

    E2x = -375712
    E2y = -375712

    E3x = -563568
    E3y = 563568

    E4x = -751424
    E4y = 751424

    Then I Added up the components, and got -1 for the sum of x, 751424 for the sum of y. I then applied the pythagorean theorem and rejoiced in excitement as I got the wrong answer.. again.. after all that. Can someone help me out here ? ?
  2. jcsd
  3. Jan 13, 2004 #2


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    The correct formula you need is E = k (sum(i)) Q(i)R(1)

    where R(1) is the unit vector to the point charge and R^2 is the vector to the point charge correctly determined by pythagorean theory. Think of the square of charge as being on a graph with one charge at the orign to determine the relative vectors
  4. Jan 13, 2004 #3
    um.. what ?..lol, sum(i) Q(i) .. I'm guessin you mean sum the point charges? But can you explain the difference between R(1) and R^2 a lil bit further.. how do you find R(1) exactly ?
  5. Jan 13, 2004 #4


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    yeah sure

    The [tex]R^2[/tex] term is just determined by using the coordinates of your cartesian system. If you were using this vector form of the equation then you would have to add the extra term on top which is the unit vector of [tex]R^2[/tex] Which is defined as the vector

    [tex]\frac {\vec{R}}{|\vec{R}|}[/tex]

    I suggest that if you do not want to use this method then the value obtained by pythagoras theorem is just as viable for the value of R and you then do not have to use the unit vector R(1) so the equation is just

    [tex]\frac{1}{4\pi\epsilon_0}\sum_i \frac{Q_i}{R^2}[/tex]

    Due to the symmetry of the square R will be constant for all point charges. Any more Queries just ask :) I hope I have made it clear this time.
    Last edited: Jan 14, 2004
  6. Jan 13, 2004 #5


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    sorry that should be sum of Q/R^2

    I was trying to be clever
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