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Electric field's due to continuous charge distributions

  1. Jan 7, 2009 #1
    I'm currently at uni, but have difficulty doing problems involving continuous charge distributions. Say theres a charge distribution dl, dA or dV on a length surface or volume respectively at a distance R away from a point i know i must integrate over total length area or volume (depending on if the charge is distributed in 1 2 or 3 dimensions) but i get stuck here.
    I do not know how to set up the intergrand. could anyone elp me with some examples link or explanation?? thanks very much :)
    spoony
     
  2. jcsd
  3. Jan 7, 2009 #2

    gabbagabbahey

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    What text(s) are you using in your E&M course? Different authors use different notation, so to avoid confusion it is probably best if you post a couple of example problems from whichever text you are using.
     
  4. Jan 7, 2009 #3
    Well trouble is the course lecturer doesnt use online notes and i was a bit ill during that time, but ok i shall grab one from a past paper. (text book briefly covers this topic)

    A charge q is located at a perpendicular distance B from an infinite earthed plate. For a point distance R from charge q, the induced charge per unit area is given by

    [tex]\sigma[/tex] = -[tex]\frac{qa}{{2} {\pi} {r^{3}}}[/tex]

    Use this to verify total charge on plate is -q

    This question is like a capacitor i spose then, but if it can be solve with taking consideration to the electric field at point P (distance B from plate) then itd help greatly

    thanks, spoony
     
  5. Jan 7, 2009 #4

    gabbagabbahey

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    This question doesn't seem to be asking you what the electric field is. Rather it is asking you to compute the total charge on the plate.

    That being said, the induced charge on the plate will create an electric field (so will the point charge!) and if you like, I will help you through trying to calculate that field. First, I recommend you try to find the total charge on the plate. Do you know how to do that?
     
  6. Jan 7, 2009 #5
    intergrate the expression for charge density sigma = charge/area with respect to area over all area. But i dont know how to do the mathematics of this :S.
     
  7. Jan 7, 2009 #6

    gabbagabbahey

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    Okay, I would recommend your first step to be drawing a picture of this and choosing an appropriate coordinate system. My choice would be to use Cartesian coordinates and let the plate be the z=0 plane (i.e. the x-y plane) and place the point charge on the z-axis at [itex](0,0,a)[/itex].

    Now, the plate is infinite, so you need to integrate the charge density from x=-infinity to +infinity and likewise for y. In order to do this integration, you need to express [itex]\sigma[/itex] in terms of 'x' and 'y'. That means you need to express the distance [itex]r[/itex] from the point charge at [itex](0,0,a)[/itex] to a general point on the plane [itex](x,y,0)[/itex] in terms of x and y. Do you know how to do that? What is the general formula for the distance between to points?
     
  8. Jan 7, 2009 #7
    Ok so the distance between a point on the plane x-y and z=a is sqrt((x-0)^2+(y-0)^2+(a)^2)) which is:
    [tex]\sqrt{x^{2} + y^{2} + a^{2}}[/tex]
    so this is equal to r in the expression for charge density.
    so i end up with:

    area integral of [tex]\frac{-qa}{2 \pi (x^{2} + y^{2} + a^{2})^{\frac{3}{2}}[/tex]
    (for some reason its giving me a intergral of polar co-ords :S)

    But if it still isnt displaying correctly it should be -qa/(2*pi*(x^2+y^2+a^2)^(3/2)) intergrated with respect to dx then dy with limits of infinity and minus infinity.
    Question is now how to intergrate? normally i would express limits of x in terms of y then intergrate the first then the second... :S
     
  9. Jan 7, 2009 #8

    gabbagabbahey

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    x and y are independent for every point on the plane, so you can treat y as a constant and integrate with respect to x first, and then integrate with respect to y or vice versa.
     
  10. Jan 7, 2009 #9
    now to intergrate 1/(x^2+y^2+a^2)^3/2 with respect to x. trouble is i cant see a way of doing it unless it was by inspection. tried substitution and parts doesnt work because its a composite function.
     
  11. Jan 7, 2009 #10

    gabbagabbahey

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    The easiest way is to use polar coordinates [itex]\rho=\sqrt{x^2+y^2}[/itex], [itex]\phi=\tan^{-1}(y/x)[/itex] and [itex]da=\rho d\rho d\phi[/itex] and integrate [itex]\rho[/itex] from 0 to infinity and [itex]\phi[/itex] from 0 to 2*Pi.
     
  12. Jan 7, 2009 #11
    but the expression contains an extra a^2 term, so how does that compare to the expression for p?
     
  13. Jan 7, 2009 #12

    gabbagabbahey

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    [itex]a^2[/itex] is just a constant, you should be able to integrate [tex]\frac{\rho d\rho}{(\rho^2+a^2)^{(3/2)}}[/tex] easily.
     
  14. Jan 7, 2009 #13
    i cant seem to intergrate that :$ having an attack of the stupid today i really am, ok substitute u = p^2 then du = 2p so times the whole integrand by 1/2 then
    = 1 / (u+a^2)^(3/2) du
    still cant think how to :S
     
  15. Jan 7, 2009 #14

    gabbagabbahey

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    Try substituting [tex]u=(\rho^2+a^2)^{(-1/2)}[/tex] instead :wink:
     
  16. Jan 7, 2009 #15
    DONE! thanks for all your help mate :) more questions to follow though :P
     
  17. Jan 7, 2009 #16

    gabbagabbahey

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    You're welcome:smile: Feel free to post more of these problems in the homework forum and I'll try to help with them.
     
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