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Help with electromagnetics problem

  1. Feb 12, 2005 #1
    find the electric field at (0,0,h) from the plane of charge
    -a < x < a, -b < y < b. electric density is Ps.
    the answer is E = Ps/(pi * Eo) * acrtan[ab/h(a^2 + b^2 + h^2)^1/2] in the z direction.
    ive got it down to this form

    Ps double integral with limits < x < a, -b < y < b [-xdxdy/(h^2 + x^2 + y^2)^3/2 in the x direction
    +
    double integral with limits < x < a, -b < y < b [-ydxdy/(h^2 + x^2 + y^2)^3/2 in the y direction
    +
    Ps double integral with limits < x < a, -b < y < b [h*dxdy/(h^2 + x^2 + y^2)^3/2 in the z direction

    i dont know how to evaluate these integrals. ive tried, but i end up with a log function when i do trig substitution.
     
  2. jcsd
  3. Feb 12, 2005 #2

    dextercioby

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    Are u sure about the validity of those integrals...?

    How would you end up with logarithm when making an apparently useless circular trigonometric substitution...?

    Daniel.

    P.S.Post you work...
     
  4. Feb 12, 2005 #3
    you can use symmetry argue the E field pointing to x and y direction are zero... so you don't need to evaluate the first 2 integral

    did you try the last integral?use trig substitution to do the last one...
    if you have any problem on the last integral... tell me what did you get and hopefully, i could point out your error
     
  5. Feb 12, 2005 #4

    dextercioby

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    For the same reason of symmetry,you can set the result as 2-ice the onne for intervals [itex] (0,a) [/itex] and [itex] (0,b) [/itex]

    Daniel.

    P.S.I'd still vote for [itex] \sinh [/itex] as the substitution.
     
  6. Feb 12, 2005 #5
    i started with E = integral of Ps dot Ds/4*pi*Eo*r^2 in the Ar direction.
    than i went to Ps Ds (r -r')/4*pi*Eo*|r-r'|^3
    r - r' = (0,0,h) - (x,y,0) = (-x,-y,h).
    i plugged that, and than i found |r-r'|^3 to be (x^2+y^2+h^2)^3/2. than i muliplied (-x,-y,h) by dydx to get double integrals in each direction.

    now, on evaluating the integrals in the z direction. our prof solved some similar integrals and said we just could the solved form instead of evaluating the integral everytime. so, on the first one y as a constant. i end up with [1/y^2+h^2]*[x/(x^2+y^2+h^2)^1/2]. i cant follow the steps to solve this integral. im getting lost in all the algebra.
     
  7. Feb 12, 2005 #6

    dextercioby

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    Then don't do it...Nobody is asking you to...

    Daniel.
     
  8. Feb 12, 2005 #7
    huh?
    i dont understand.
     
  9. Feb 12, 2005 #8

    dextercioby

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    You said that u couldn't do it,because u couldn't handle the algebra behind it.And i advised you not to do it...Or do it when u feel like going through those calculations...

    Daniel.
     
  10. Feb 12, 2005 #9
    the algebra is not tricky, it is messy..... If you got stuck on some step, we can pull you out, but it seems like you are scare by the algebra.... I'm quite sure you made a minor mistake for this one:
    because you add 1/y^2 to h^2, their units are different, you can't add them together...
    Patient is the only way to solve this kind of problem...
     
  11. Feb 12, 2005 #10

    dextercioby

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    Maybe they're both in the denominator...It would make a bit of sense.

    Daniel.
     
  12. Feb 12, 2005 #11
    they are both supposed to be in the denominator. how am i supposed to plug in the limits when the denominator has (x^2 + ..... )^1/2
    when i do the integral for y, do i use the equation i found from evaluating x?
     
  13. Feb 12, 2005 #12
    [tex] \int_{-b}^b \int_{-a}^a \frac{h}{(h^2+x^2+y^2)^{3/2}} dx dy [/tex]
    [tex] =\int_{-b}^b \frac{hx}{(h^2+y^2)\sqrt{h^2+y^2+x^2}}|_{-a}^a dy [/tex]

    where as

    [tex]\frac{hx}{(h^2+y^2)\sqrt{h^2+y^2+x^2}}|_{-a}^a[/tex]
    [tex] = \frac{ha}{(h^2+y^2)\sqrt{h^2+y^2+a^2}}-\frac{h(-a)}{(h^2+y^2)\sqrt{h^2+y^2+(-a)^2}} [/tex]
    [tex] =\frac{2ha}{(h^2+y^2)\sqrt{h^2+y^2+a^2}} [/tex]
    and plug it back in the original integral....

    EDIT:
    simple substitution.... don't tell me you've never learned that b4
     
    Last edited: Feb 12, 2005
  14. Feb 12, 2005 #13
    [tex] =\frac{2ha}{(h^2+y^2)\sqrt{h^2+y^2+a^2}} [/tex]
    started with that. i knew how to do the substitution i just wrote something down wrong and couldnt figure out what i did. anyways.

    i started with c^2 = a^2 + h^2
    y = c^2tan(o)
    dy = c^2*sec^2(o).

    this leaves
    c^2*sec^2(o)/ (numerator)
    (c^2-a^2 + c^2*tan^2(o))*(c^2+c^2tan^2(o))^1/2 (denominator)
    factoring the c^2 and transforming 1+tan^2(o) into sec^2(o)
    i get
    c^2*sec^2(o)/
    (-a^2+c^2*sec^2(o)*c*sec(o).
    the last two terms are after i took the square root of the squared values.
    simplifying some i get this

    c*sec(o)/
    (-a^2 +c^2*sec^2(o))

    im stuck here.
     
  15. Feb 12, 2005 #14
    I don't think this integral is solvable,
    i'll give you another approach... you can find the potential V first, then take the derivative.... I've try this method and it really works.....
     
  16. Feb 13, 2005 #15

    Gokul43201

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    Alternatively, you could try the problem in cylindrical co-ordinates.

    [tex]E_z(h) = 4\rho _s \cdot [\int _0 ^{\pi /4} d \phi \int _0 ^{a~sec \phi} \frac {hr~dr}{(r^2 + h^2)^{3/2}}~ + \int _0 ^{\pi /4} d \phi \int _0 ^{b~sec \phi} \frac {hr~dr}{(r^2 + h^2)^{3/2}}][/tex]
     
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