1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the electric field for a point above the center of a square loop

  1. Jun 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the electric field a distance z above the center of a square loop carrying a uniform line charge Q.


    2. Relevant equations



    3. The attempt at a solution

    So we want the electric field generated by the charges that lie on the boundary of the "square loop" - it's just a square or a rhombus with equal lengths.

    I attempted to solve this by splitting up the integral into 4 line integrals (each side of the square loop being an integral). Since the charge is uniform, I pull that out of the integral.

    Furthermore, all the horizontal components of the electric fields cancels out so we need to integrate the vertical components, i.e. [tex]cos(\theta) = \frac{z}{\sqrt( (x-x_{0})^{2} + (y-y_{0})^{2} + z^{2})}[/tex]

    Am I correct to assume that one of my integrals will look like this:

    [tex]\frac{Qz}{4\pi\epsilon_{0}}\int_{0}^{a}\frac{dx}{((x-x_{0})^{2} +(y-y_{0})^{2} + z^{2})^{3/2}}[/tex]

    Since [tex]R^{2} = (x-x_{0})^{2} + (y-y_{0})^{2} + z^{2}[/tex], i.e. assume the points on the square's boundary are (x, y, 0) and the point above the center of the square loop as coordinates [tex](x_{0}, y_{0}, z)[\tex]

    z is a fixed number!

    My questions are:
    1) Would this be a correct expression for one of the line integrals?
    2) How exactly do I integrate such an expression? Am I not seeing the antiderivative?

    Sorry if I'm not being clear or concise, this is my first attempt at learning electromagnetism (I'm learning out of Griffiths and I am having a little trouble with all the notation and what not).
     
    Last edited: Jun 21, 2008
  2. jcsd
  3. Jun 21, 2008 #2
    I assume we are talking about problem 2.4 in Griffiths?

    We put the origon of our coordinate-system in the center of the square. Now we look at one of the sides - we can choose the right one, whose parameter is:

    r'(t) = (a/2 ; t ; 0)

    where t runs from -a/2 to a/2.

    Our vector r (the one going from the origin to our field point) is just (0 ; 0 ; z). So we put this in the expression for the electric field:

    [tex]
    E = \frac{\lambda }{{4\pi \varepsilon _0 }}\int_{ - \frac{a}{2}}^{\frac{a}{2}} {\frac{z}{{\left( {\left( {\frac{a}{2}} \right)^2 + t^2 + z^2 } \right)^{\frac{3}{2}} }}} \,dt
    [/tex]

    This is the contribution in the z-direction, since all the other components cancel out. The other 3 sides give the same contribution, so just multiply the result by 4, and then you're set.

    EDIT: Actually, I see this is what you have already done.. perhaps I should start reading threads before commenting them.
     
    Last edited: Jun 21, 2008
  4. Jun 22, 2008 #3
    [tex]E = \frac{\lambda }{{4\pi \varepsilon _0 }}\int_{ - \frac{a}{2}}^{\frac{a}{2}} {\frac{z}{{\left( {\left( {\frac{a}{2}} \right)^2 + t^2 + z^2 } \right)^{\frac{3}{2}} }}} \,dt[/tex]

    Looks right. But it only looks like that because of symmetri so that E only has a komponent in the z direction. Now just substitute the constants together and it looks completely like ex 2.1 in griffith :)
     
  5. Oct 11, 2008 #4
    Hi.

    Regarding this question, my friend said all you have to do is use Example 2.1 and substitute L for a/2 and z for [tex] \sqrt{z^{2} + (\frac {a} {2})^{2}} [/tex]

    This gives the right answer... I understand the substitution for L however the substitution for z I can't get. Anyone care to shed some light?
     
  6. Oct 21, 2009 #5
    I believe that they want you to look at the previous example with the wire. If we are to chop the square loop into infinitely thin wires, then there would be pairs of wires with opposing electric fields, except for the wire directly under the charge. Therefore you simply refer to the problem before and subsititue a/2 (square loop has side a, they integrated from 0 to L and doubled the answer) for L.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Finding the electric field for a point above the center of a square loop
Loading...