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Calculate the induced charge on a conducting surface

  1. Nov 25, 2016 #1
    1. The problem statement, all variables and given/known data
    A very thin plane of length ##2a## is placed in the air at height ##a## above the conducting surface. The plane is charged on its surface and the expression for it's surface charge density (##ρ_s##) is given by ##ρ_s=ρ_{s0}*\frac{x}{a}## and ##ρ_{s0}## is some constant.
    a) Determine the surface charge density function inducted on the conducting surface.
    The image that goes with this is uploaded.
    2. Relevant equations
    $$ρ_s=ρ_{s0}*\frac{x}{a}$$
    3. The attempt at a solution
    I need help in starting to solve the problem. I have some theories as to how i would solve it and i would like a feedback on if it's ok to do so.
    I was gonna use a theory where we remove the conducting surface and add a mirror image of the object with an opposite charge. So at ##2a## distance from the first plane there is a second plane with a charge density $$ρ_{s2}=-ρ_{s0}*\frac{x}{a}$$. Now all i have to do is calculate the electric field right? I am confused as how im gonna do this.
    I know that the electric field of a plane is $$E_p=\frac{ρ_s}{2ε_0}$$ but how will i find the inducted charge and express it in a surface charge density function?
     

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  3. Nov 25, 2016 #2

    BvU

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    That's right. As if the conducting plane at z=0 were absent. From symmetry (that's why you use the mirror image method in the first place) you know you only need one component. You need a relevant equation for ##\vec E##, something like ##|E| = {kq\over r^2 }## and an integration.

    Once you have E, then you use ##E_p=\frac{ρ_s}{2ε_0}## to get the surface charge on the conducting plane at z=0.

    If the given charge distribution on the thin plane is causing you a problem, think of a negative charge on one end and a positive on the other: with the mirror you get a quadrupole field. So for this exercise you expect something in that ball park as well.
     
  4. Nov 25, 2016 #3
    Thanks, as for the relevant equation for ##E## in order to get what i said was the ##E## for the plane a used the Gausses's law :-). I understood this problem and m about to post another thats been bothering me :D. Thanks for the answer, your the kind that makes this site awesome (gives a big thumbs up)
     
  5. Nov 25, 2016 #4

    BvU

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    Gauss is valid for a pillbox close to the plane in case of non-uniform surface charge.
     
  6. Nov 25, 2016 #5
    Hmm..thinking about it now i can use the electric field of a line to replace this plane. What i mean is:
    $$E=\frac{ρ_l}{2πε_0r}$$ is the elekctric field of a line and ##ρ_l## is the line charge function. Here it's just ##ρ_l=ρ_sdx##. I guess this should work?
     
  7. Nov 25, 2016 #6

    BvU

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    It didn't appear in the problem statement, but I suppose the charged plane extends infinitely in the y+ and the y- direction, so: yes.
     
  8. Nov 30, 2016 #7
    Sorry for the delay, i was preoccupied but i did make some progress with this and want to verify if this is correct.
    So i used a mirror theorem in which i placed at distance ##2a## a plate that is exactly the same as the first one but with a negative charge on it. The next is my work:
    ##dE_1=\frac{ρ_l}{2\piε_0r}##
    ##dE_{total}=2dE_1cosθ##
    I have choosen some point ##x_0## in where i calculate the field
    ##cosθ=\frac{a}{r}=\frac{a}{\sqrt{(x_0-x)^2+a^2}}##
    when i apply the given surface charge density:
    ##E_{total}=\frac{ρ_{s0}}{πε_0}\int_{-a}^{a}\frac{xdx}{(x_0-x)^2+a^2}##
    This should be the electric field, right? But when i integrate i will get a ##x_0## term and that appears nowhere in the result that my book gives, which i can post if you want..
     
  9. Nov 30, 2016 #8

    BvU

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    ##x_0## is the x-coordinate of the point (line) on the conducting plate at z=0 where you want to express the surface charge density. I would expect that charge to depend on such an ##x_0## :confused: !?
     
  10. Nov 30, 2016 #9
    View attachment 109693
    Here is the solution, can you make something out of it?
     
    Last edited: Nov 30, 2016
  11. Nov 30, 2016 #10

    BvU

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    No, except: His/her ##x## is your ##x_0##.
     
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