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Infinite lines of charge on 3 corners of square

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Infinite lines of charge are placed on three corners of a square of side 0.2 m, as shown in the figure (attached). If the linear charge density [itex]\lambda_{2}[/itex] = +1 uC/m, calculate the values of [itex]\lambda_{1}[/itex] and [itex]\lambda_{3}[/itex] so that the electric field at the fourth corner is 0.


    2. Relevant equations
    [itex]E = \int k\frac{dq}{r^{2}}[/itex]
    [itex]E = E_{1} + E_{2} + E_{3}[/itex]


    3. The attempt at a solution
    First question - the figure is confusing. If it's a LINE of charge, what does it mean that it's located at a corner i.e. POINT? There are four lines in the square.. which 3 are the 3 line charges the question is talking about... i.e. what exactly is the picture depicting?
    Thanks!
     

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    Last edited: Sep 28, 2011
  2. jcsd
  3. Sep 28, 2011 #2

    SammyS

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    Suppose that the square lies in the xy-plane. Then the line charges are parallel to the z-axis and pass through the xy-plane.
     
  4. Sep 28, 2011 #3
    So the dots depicted in the corners of the square are just the ends of a cylinder, the rest is going into the plane so as to not be seen? So the lines connecting the dots are just there for reference, nothing actually connecting them?
     
  5. Sep 28, 2011 #4

    SammyS

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    The lines connecting the dots merely indicate the edges of the square, nothing is there. The lines of charge extend from -∞ to +∞ in the z direction.
     
  6. Oct 8, 2011 #5
    Okay, so related to the post above but a different question... this one is the same diagram, except the corners just have charges, not inward cylindrical lines of charge.

    There is a square, with charges at all corners except for the top-right one. The top-left corner is Q1, the bottom-left corner is Q2 = +1 μC, and the bottom-right corner is Q3. The sides of the square are 0.2 m. The top-right corner is just called point P, and the question asks to calculate Q1 and Q3 so that the electric field at point P is 0.

    So, my attempts:
    At point P, the E field due to Q2 points away in the 45-degree direction (since it's a square).
    For the net field at point P to be 0, there would have to be an equal E field in the direction opposite of the one due to Q2. So, looking at point P, if I have the E field due to Q1 pointing left and the E field due to Q3 pointing down, the resultant vector from those two E field vectors would point in the direction opposite of the one due to Q2 and so could cancel it out. So that means Q1 and Q3 are negative charges. For my question, the net field due to Q1 and Q3 would have to equal the E field due to Q2. So E13 = E2.

    Is that okay to say?

    Also, the signs of E have me a bit confused. If I choose my coordinate system as point P being the origin, the E field due to Q3 would then point down, so should I make the value of that E field negative? I'm not sure of how to properly use the signs.
     
    Last edited: Oct 8, 2011
  7. Oct 8, 2011 #6
    Net Vertical E Field:
    EPy = -E3 + E2y

    E2y = E2 cos(45)
    E2 = k Q2/d2 = k Q2/sqrt(2*0.22) = 32,143 N/C N/C

    EPy = -E3 + 22,728 N/C

    Net Horizontal E Field:
    EPx = -E1 + E2x

    Since the angle is 45 degrees, E2x = E2y

    EPx = -E1 + 22728 N/C

    Net Field Due to Q1 and Q3
    -E2 = E13
    -32143 N/C = sqrt(E12 + E22)

    [itex]-32143 = \sqrt{(\frac{kQ1}{0.2^2})^2 + (\frac{kQ3}{0.2^2})^2}[/itex] k/0.22 = 2.25E11
    [itex]-32143 = \sqrt{((2.25E11)Q1)^2 + ((2.25E11)Q3)^2}[/itex]

    Then what? In similar problems online, the two charges end up being equal. But if I assume that here, then Q ends up equaling 0.03... where did I go wrong?
     
    Last edited: Oct 8, 2011
  8. Oct 8, 2011 #7

    SammyS

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    The field due to Q1 has to cancel the x component of the field (at P) due to Q2.

    Like-wise, the field due to Q3 has to cancel the y component of the field due to Q2.
     
  9. Oct 8, 2011 #8
    Wow, I way overcomplicated it.
    Thanks for the clarification.
    So then all I have to do is find the x and y components of E2, which I can do because I'm given the value of charge Q2 and can figure out the diagonal distance because I'm given the sides of the square. Since the angle is 45 degrees, both x and y components end up being equal though.

    So then my charge Q1 and Q3 are equal, both negative charges of 1 * 10-7.
    Correct?
     
    Last edited: Oct 8, 2011
  10. Oct 8, 2011 #9

    SammyS

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    Just doing the calculation in my head, I think Q1 & Q3 should be 1/4 of the magnitude of Q2, and be negative, of course.
     
  11. Oct 8, 2011 #10
    Oh, I just had a math error. My final result is -3.6E-7 for Q1 and Q3, which is close to 1/4th of Q2 like you said. So all must be well. Thanks so much for helping :)
     
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