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Electric Field Intensity. Finite sheet of charge.

  1. Nov 17, 2015 #1
    problem statement, all variables and given/known data

    A finite sheet of charge, of density ρ=2x(x2+y2+4)^3/2, lies in the z=0 plane for 0≤x≤2m and 0≤y≤2m.Determine E at (0,0,2)m

    Ans:(18x10^9)(-16/3ax-4ay+8az)
    2. Relevant equations
    E=kQ/R2

    3. The attempt at a solution
    dE=ρdA / R^2 aR
    dA=dxdy


    E=k 2x(x2+y2+4)^3/2 dy dx (-xax-yay+2az ) /x2+y2+4)^3/2

    E=k∫∫ 2x dy dx (-xax-yay+2az)

    E=(18x10^9)(-4ax-4ay+8az)

    How x and y component of E are different in the answer?
     
  2. jcsd
  3. Nov 17, 2015 #2

    BvU

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    Do you see a difference between the x- and the y-dependence of ##\rho## ?

    Or: Can you see x = y is not a symmetry line ?
    ChargeDist.jpg

    (edit: forgot the +4 in the first upload)

    Once again: making a drawing is soooooo useful !
     
    Last edited: Nov 17, 2015
  4. Nov 17, 2015 #3
    Thanks for the visual. Is there method to find a point of symmetry on that surface?
     
  5. Nov 18, 2015 #4

    BvU

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    What if there isn't ?
     
  6. Nov 18, 2015 #5
    Do I have to use mathematical theorem? If so I have to give up on this problem. I know how to use Coulomb law and multiple integration with help of Mathematica. This question no 2.50 from Schaum's Electromagnetics.
     
  7. Nov 18, 2015 #6

    BvU

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    Ah, didn't really read the small print :smile: and just answered the question "How x and y component of E are different in the answer?". So now it's time to go through your steps and see why you and Schaum end up with different answers.
    Let me please use ##\hat\imath,\ \hat \jmath ,## and ##\hat k## instead of ##a_x, a_y, a_z## as Cartesian unit vectors.

    Stumble on the first step:
    Your relevant equation only describes ##|\vec E|\ ##, so it doesn't help you find the components.
    Google (or look up in Schaum) the right equation and look at it carefully: the only vector that determines the direction of ##|\vec E|\ ## in there is ##\vec r-\vec r'## and you can't take it out of the integral because it varies with ##d\vec r' = dx dy##.

    Attempt at solution starts with
    which does not match with
    (it looks to me that even the brackets don't match...)

    My advice: Don't give up. This exercise is very important for your understanding and further curriculum. Make a drawing that shows ##\ \vec E\ ## at ##(0,0,2)## due to a small patch of charge ##\rho dxdy## located at ##\vec r' = (x, y)## in the x,y plane. You will see that the maker of the exercise chose a very nice expression for ##\rho##.

    The consider what you have to integrate to find the x-component of ##\vec E##, ##\ \vec E_x = \vec E \cdot \hat\imath \ ##, and remember that you can bring constant factors into the integral. Then idem the y component and the z-component.

    Sounds perhaps a bit cryptic. But you should really work this out yourself - if I do it for you it does not help you. So try a first step and post when you get stuck.
     
    Last edited by a moderator: May 7, 2017
  8. Nov 18, 2015 #7

    BvU

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    Nice of you to like my post. I hope it's for the encouragement, because I am far too critical about your contents. Let me repair the expression with an extra bracket and look at it again... :rolleyes:
     
  9. Nov 18, 2015 #8

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    OK, so it's all pretty minor. You're doing great and the only glitch is for the x component. All my well-intended teaching blabla doesn't help if you do the ##Ey = k\iint 2xy\, dx dy ## completely correctly and all you are missing is the ##Ex = k\iint 2x^2 \,dx dy ##. Look at this last one again and tell me it was in fact easy ! (and hopefully also that you could have done it without crutches like Mathematica (*) just as well !)

    I did find a worked out example for the first few steps. Later on he brings ##\sigma = \rho## outside the integral because it's constant -- you can't do that, but thanks to the nice ##\rho## function you could do other things to get simple integrals...and you did. It was just the last step that went off the rails a little bit.

    (*) I have nothing against Mathematica, but here it's just like using a calculator for 2 x 3
     
    Last edited: Nov 18, 2015
  10. Nov 18, 2015 #9
    Thanks. My fault is on the unit vector. I used to deal with constant z component.
    So the unit vector should be (- xi - yj+2k)/|R|. Maybe using aa, ay and az is confusing.
     
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