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Homework Help: Point charge above xy plane, find surface charge density!

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data
    A point charge, Q, is located @ (0,0,d) above infinite conducting plane that lies in xy plane and is maintained at ground potential. Find:
    a.) surface charge density as a function of x and y on conducting plane, and
    b.) total charge induced on conducting plane.

    2. Relevant equations
    See below.

    3. The attempt at a solution
    Well, I know that the E field is upward directed in the z-direction because the E-field is always perpindicular to a conducting surface. The answer is in the back of the book, and it is:
    a.) ρs = (-2Qd)/(4*pi*[ρ2 + d2](3/2)
    b.) Qsurface = -Q

    In part a, I do not see how the author got [ρ2 + d2](3/2) in the denominator of this formula. This equation shows up in the E-field equation for a uniform disc of charge, and how can we assume this?! If I were to follow this implementation, assuming I can follow the assumptions for an uniform disc of charge, r = daz, and r' = xax + yay. (r - r') = daz - xax - yay. The magnitude of this term to the third power (still following equations for uniform disc of charge) = sqrt(d2 + x2 + y2) = sqrt(d2 + ρ2).

    I get stuck here. If someone could let me know if I should continue to follow this method, and if so, how to continue. If I should try another way of thinking, please offer some tips to get me going in the right direction. Thank you for all help, it's most appreciated!
  2. jcsd
  3. Mar 30, 2010 #2


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  4. Mar 30, 2010 #3
    can you explain how they got the potential formula and then how they got to the surface charge density on the grounded plane (the next equation)? is there a formula for this or something that I am missing here. From my text, I see no relationship that incorporates the surface charge density and the potential function, involving the permittivity of free space constant.

    The closest thing I see is:
    surface charge density = (epsilon)*V/d = (epsilon*E) = D, because D = surface charge density at conducting planes.
  5. Mar 31, 2010 #4


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    If you are not familiar with the electric potential yet, see this:

    http://www.shef.ac.uk/physics/teaching/phy205/lecture_7.htm [Broken]

    Last edited by a moderator: May 4, 2017
  6. Mar 31, 2010 #5
    Thanks ehild, I figured it out. Going with your image charge suggestion, I determined that I could find the surface charge distribution by using the formula En = ps/permittivity of free space constant. I solved for the electric potential difference (V) by creating a "dipole" moment with the image charge. Then E = V/d, and then used E to solve for ps, as stated earlier.

    Thanks again.
  7. Mar 31, 2010 #6


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    Well done!

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