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Homework Help: Laplace equation problem

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A uniform line charge [tex]\lambda[/tex]is placed on an infinite straight wire , a distance above the conducting plane . (Lets say the wire runs parallel to the x-axis and directly above it, and the conducting plane is in the xyregion)

    a) Find the potential in the region above the plane

    b) find the charge density [tex]\sigma[/tex] induced on the conducting plane
    2. Relevant equations

    https://www.physicsforums.com/editpost.php?do=editpost&p=2388205 [Broken]

    3. The attempt at a solution

    a) V=([tex]\lambda[/tex]*l)/(4*[tex] \pi*\episilon_0[/tex])*1/(sqrt(x^2+(y+d^2))+sqrt(x^2+(y-d)^2))

    b)Q=[tex] \sigma*dA[/tex]. Q=[tex]\lambda[/tex]*l and dA=dxdy.therefore [tex] \sigma[/tex]= [tex]\lambda[/tex] *l/xy , l being the length ?

    Or I could just take the derivative of V with respect to x or y and then multiply the derivative of V by epilison_0
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 12, 2009 #2


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    Do you really mean what you've written here? If not, use the "preview post" and "edit post" options to help you type what you really meant. (What good is solving a problem if you can't effectively communicate your solution to others?)

    In any case, you should also show how you got your answer or explain your reasoning for it so that we can see where you are going wrong.

    This makes no sense! You are asked to find the induced surface charge on the conductor, not calculate the surface charge on the wire (which is a line, not a surface).

    More or less. There is a boundary condition that relates the normal derivative of the potential at a surface to the charge density of the surface....use that. (Once you've found the correct potential!)
  4. Oct 12, 2009 #3

    Is my expression now more readeable?

    . Oh , I see what you mean.

    Last edited: Oct 12, 2009
  5. Oct 12, 2009 #4


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    Yes, it is!:smile: (Although, there is no need to use the * symbol to represent multiplication...everyone who's taken a basic algebra class will recognize that [itex]x*y=xy[/itex])

    Unfortunately, your expression is incorrect:frown:...How did you arrive at it?
  6. Oct 12, 2009 #5
    Well , it is a method of image problem and it is a planar problem. Based on the information given in the book(griffiths) about the potential of a conducting planal, I know the potential of a plane wave is: [tex]V(x)=1/(4\pi e_0)*(\lambda*d/(sqrt(x^2+y^2+(z-d)^2)-(\lambda)d/(sqrt(x^2+y^2+(z+d)^2))[/tex]; Now since it is a 2 dimensional problem , I thought I would drop the z; Maybe for a charge traveling on wire the shape of a line, the wire is the shape of the line. So my expresion should be written as E rather than V and I then integrate E and I get a natural logarithm expression in my term for V;

    Then , [tex] E=\lambda/(4*\pi*e_0*s_-)-\lambda/(4*\pi*e_0*s_+)[/tex]

    [tex] s_-=\sqrt(x^2+(y-d)^2), s_+=\sqrt(x^2+(y+d)^2)[/tex]
    Last edited: Oct 12, 2009
  7. Oct 13, 2009 #6
    Am I headed in the right direction with my new solution
  8. Oct 14, 2009 #7


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    This is an electrostatics problem, it has nothing to do with plane waves (or waves of any kind!). :confused:

    I think you are trying to blindly apply Griffith's eq. 3.9, by just replacing [itex]q[/itex] with [itex]\lambda[/itex] without really understanding the method of images at all.

    No, this is a 3 dimensional problem.

    This is an electroSTATICS problem; there are no traveling charges!

    Start with a simpler problem. Forget about the conducting plane, forget about the method of images, and just calculate the potential of a uniform line of charge, running parallel to the x-axis at a height z=d....what do you get for that expression?
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