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Griffiths Problem 3.34

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data
    This question refers to Griffiths E and M book.

    If they had not told me that the charge moves, I would have guessed that it just sits there since it seems like there is nothing to exert a force on it. Am I supposed to calculate the induced charge on the conducting plate? How do I use the fact that the plate is grounded? That means that the potential on the plate is 0 just like at infinity, but doesn't that imply there is no charge on the plate either?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 10, 2008 #2


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    No, the charge on the plate is needed to keep it at V=0.
    The charge is supplied by the ground.
    Read the part in Griffith's about images for grounded plates.
  4. Mar 10, 2008 #3
    A related question: on page 123, Griffiths says "Evidently the total induced charge on the plane is -q, as (with the benefit of hindsight) you can perhaps convince yourself that it had to be."

    I cannot convince myself unfortunately. How do you know this a priori? Is it related to the fact that the plate is grounded? Would it be different if the plate were not grounded?
  5. Mar 10, 2008 #4
    And also when I try to apply eqn 3.12 to this problem I get a differential equation which I have no idea how to solve:

    [tex]\ddot{z} = C/z^2[/tex]

    where C is a constant. What am I doing wrong?
  6. Mar 10, 2008 #5


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    1. The image charge is -q. By Gauss's law, that has to equal the induced surface charge on the grounded plane.

    2. Use dv/dt=v(dv/dx).
  7. Mar 11, 2008 #6
    Calculate the force on the charge at a height 'x' above the plane and equate it with the equation of motion of the charge 'q'.
  8. Mar 11, 2008 #7
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