1. Limited time only! Sign up for a free 30min personal 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!

Electric Field Calculation (Square Wire)

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the electric field a distance ##z## above center of a square loop of wire,each of whose sides has lenght ##l## and uniform charge per lenght ##λ##

    2. Relevant equations
    ##dE=\frac {1} {4πε_0} dq \frac {1} {r^2}## (magnitude)

    3. The attempt at a solution
    I have a pic.
    Adsız 1.png

    So ##dq=λdx##
    ##dE=\frac {1} {4πε_0} dq \frac {1} {r^2}##
    ##r^2## is here
    ##x^2+(\frac {l^2} {4})+z^2##

    ##E_x=E_y=0##
    so the equation becomes,
    ##dE=\frac {1} {4πε_0} dq \frac {1} {x^2+(\frac {l^2} {4})+z^2}sinθ## which its
    ##dE=\frac {1} {4πε_0} dq \frac {1} {x^2+(\frac {l^2} {4})+z^2}\frac {z} {\sqrt {(\frac {l^2} {4})+x^2}}##

    From there I get

    ##E=\frac {zλ} {4πε_0}\int_{\frac {-l} {2}}^{\frac {l} {2}}\frac {8} {(4x^2+l^2+4z^2)(\sqrt {l^2+4z^2})} \, dx##

    I stucked at the integral part.I tried calculator but it didnt work out.And before that, are my equations true ? Of course for result I have to mulitply this by ##4## so the answer is
    ##E_{total}=4E##
    It will be in the vertical direction
     
    Last edited: Feb 26, 2017
  2. jcsd
  3. Feb 26, 2017 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    See if you can find the mistake in setting ##\sin\theta = \frac {z} {\sqrt {(\frac {l^2} {4})+x^2}}##
     
  4. Feb 26, 2017 #3
    Looks right for me ?
     
  5. Feb 26, 2017 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    What is the length of the hypotenuse of the right triangle which has the angle θ indicated in your sketch?
     
  6. Feb 26, 2017 #5
    I see yeah thanks
     
  7. Feb 26, 2017 #6
    I found but I used calculator to find ##\int_\frac {1} {(4x^2+l^2+4z^2)^{\frac {3} {2}}} \ dx ## is it too bad ?
     
  8. Feb 26, 2017 #7

    TSny

    User Avatar
    Homework Helper
    Gold Member

    The denominator should be raised to some power other than 1.
     
  9. Feb 26, 2017 #8
    I fixed sorry my typo
     
  10. Feb 26, 2017 #9
    Its hard to write such things in here
     
  11. Feb 26, 2017 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I'm not sure what you are asking here. If you get a final answer, I can check to see if I get the same answer.
     
  12. Feb 26, 2017 #11
    No I found the answer,Just For solving the integral.I used calculator.Then I put the numbers and I found the answer.

    Just I couldnt see the Solution of integral.And asked you Is is bad to not solve such integral.Its stupid question I know :/
     
  13. Feb 26, 2017 #12

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I think it's OK to use a calculator to do the integral.

    To do the integration without a calculator, note that $$\int \frac{dx}{\left(4x^2 + l^2+4z^2 \right)^{3/2}} = \frac{1}{8} \int \frac{dx}{\left(x^2 + a^2 \right)^{3/2}}$$ where ##a^2 = \frac{l^2}{4}+z^2##. Let ##u = \frac{x}{a}## to put the integral into the form $$\frac{1}{8a^2} \int \frac{du}{\left(u^2 + 1\right)^{3/2}}$$ Then try an appropriate trigonometric substitution.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted