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Electric field above square loop

  1. Mar 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Problem 2.4 from Griffiths Intro to Electro

    Find the electric field a height z above the centre of a square loop with sides a and linear charge density λ.

    height is given to be z and sides given to be a, ∴ distance from origin to side is given by a/2

    2. Relevant equations
    [tex] E = \frac{1}{4 \pi \epsilon_0} \int {\frac{1}{r^2} \hat {r}dq} [/tex]

    [tex]dq = \lambda dl [/tex]

    3. The attempt at a solution
    Considering the side of the square perpendicular to the positive y axis

    First using the position of point P (0, 0, z) and position on the side (x, a/2, z) where z and a/2 are constants the direction from origin to P is given by [itex]r=z\hat{z}[/itex] and direction from origin to side is given by [itex]r' = (a/2)\hat{y} + x\hat{x} [/itex].

    ∴direction of electric field is given by [itex] r = r - r' =z\hat{z}-(a/2)\hat{y}-x\hat{x} [/itex]

    [tex] \hat{r}=\frac{r}{|r|}=\frac{z\hat{z}-(a/2)\hat{y}-x\hat{x}}{\sqrt{z^2+\frac{a^2}{4}+x^2}} [/tex]

    considering the "right" side of the square loop y and z values are constant while the x goes from -a/2 to a/2
    ∴ [itex]dq=\lambda dx[/itex]

    Plugging these equations into the integral we get
    [tex] E = \frac{1}{4 \pi \epsilon_0} \int {\frac{1}{z^2+\frac{a^2}{4}+x^2} \frac{z\hat{z}-(a/2)\hat{y}-x\hat{x}}{\sqrt{z^2+\frac{a^2}{4}+x^2}}\lambda dx} [/tex]

    which can be simplified to
    [tex] E=\frac{\lambda}{4\pi\epsilon_0} \int{\frac{z\hat{z}-(a/2)\hat{y}-x\hat{x}}{\sqrt[3]{z^2+\frac{a^2}{4}+x^2}}} [/tex]

    I know I can split the integral up into three different integrals w.r.t x y and z directions then group the y and z integrals because z and (a/2) are constants.while leaving the integral in the x direction due to its dependence on x which changes from -a/2 to a/2 but I am unsure whether that is the correct way to proceeed/what to do next.

    Thanks for your help!
     
  2. jcsd
  3. Mar 2, 2015 #2

    TSny

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    Welcome to PF!

    Your work looks good to me. You now have three integrals that can be done separately. However, using symmetry arguments, you should see that you will only need to evaluate one of the three integrals to derive the answer for the entire square.
     
  4. Mar 2, 2015 #3

    BvU

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    Hello Vaentus, and welcome to PF :smile:

    First a few comments:

    "P (0, 0, z) and position on the side (x, a/2, z)"
    you mean (x, a/2, 0) I suppose ? (which is correct, and you do it correctly in the integral).​

    "I know I can split the integral up into three different integrals w.r.t x y and z directions then group the y and z integrals because z and (a/2) are constants.while leaving the integral in the x direction due to its dependence on x which changes from -a/2 to a/2"
    Weren't you integrating over x only ? and x appears in all three terms.
    What do you mean with the y and z integrals ? I can imagine and integral over dy for the side (a/2, y,0) but I can't imagine a z integral !​


    Then two tactical questions:

    1) Which way do you think the electric field will point when you've done all the integrals ?
    Or: is there a way to explore some of the symmetry here ?
    (Physicists love symmetries: they can save loads of dull work...)

    2) A slightly easier problem is the electric field at a distance ##s## above the midpoint of a straight line segment of charge.
    Again: which way will the result point ?
    (I use s because of course in your case ##s^2 = z^2 + {a^2\over 4}##).
    There's four of these that point in 4 different directions. Adding them up (as vectors) is relatively easy.

    [edit] I got beaten by TSny; my fault: too verbose. Your integral setup is indeed just fine and quite usable.
     
    Last edited: Mar 2, 2015
  5. Mar 2, 2015 #4

    BvU

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    Hey, in my 1999 version there's a very useful "[Hint: use the result of exercise 2.1]" ! Did you forget that part of the problem statement ?
     
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