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Electrostatics Problem (Calculating Electric Field)

  1. Mar 9, 2006 #1
    Hi all,

    I have this problem in my physics homework, which is not trivial. This is not a graded homework fortunately but I would like to see the solution for it and learn. I have attached my attempt at solving this problem. Not sure if this is the right way to solve it, and so I would like your personal input in this one.

    ===============================================
    Problem:

    If a charge is fixed above a large, flat, grounded, conducting floor, the electric field above the floor would be the same as that produced by the original charge plus a "mirror image" charge, equal in magnitude and opposite in sign, as far below the floor as the original charge is above it. The effect of this "image charge" is produced by the surface charge distribution induced on the floor by the presence of the original.

    a. Find the electric field at (just above) the floor, as a function of horizontal distance p from the point on the floor directly under the original charge. Take the original charge to be a point +Q, a distance a above the floor. Ignore any effects of "walls" or "ceiling."

    b. Find the surface charge density sigma(p) induced on the floor.

    c. Calculate the total surface charge induced on the floor. HINT: Divide the floor into concentric rings, centered on the point directly under the original charge. Assume the floor extends "to infinity," i.e., very far.

    =============================================
    Solution:
    Net field = 2ESinθ
    Where E=electric field due to each charge
    E=k*q/(a^2 + p^2)
    Sinθ=a/(a^2 + p^2)^(1/2)

    Total Electric field is given by:
    E1 = k*q/(a^2 + p^2) *a/(a^2 + p^2)^(1/2)
    =2Kqa/(a^2 + p^2)^(3/2)

    As no field is there on the floor,
    Therefore, field due to floor = E1
    E1= σ / 2ε (σ = surface charge density)
    σ =2ε (E1)
    =2ε 2qaK/(a^2 + p^2)^(3/2)
    = qa/Π(a^2 + p^2)^(3/2)

    Now, to calculate total charge, divide the whole floor into concentric rings centered at point just below the charge.


    Then the total
    Q=∫σ ds
    =∫ qa2 Π pdp/ Π(a^2 + p^2)^(3/2)
    =2aq ∫ p dp/ (a^2 + p^2)^(3/2)
    On integrating it we get,
    Q=2q
    ====================================================

    Thank you for passing by :)
     
  2. jcsd
  3. Mar 9, 2006 #2

    siddharth

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    Homework Helper
    Gold Member

    The total surface charge induced on the floor has to be -Q. Can you see why?

    For the first part, can you elaborate on what your symbols mean? For example, what's "p"?
     
  4. Mar 10, 2006 #3
    Is it because the surface charge is induced by the upper charge and thus has the same amount of charge as the upper one, but has an opposite sign ?

    p is a parameter that defines the horizontal distance (in meters), from right under the upper charge.

    This diagram may help clarify some things:

    +Q
    .
    .
    . p
    . _________|___________
    . | |
    ====================================== <- Floor
    .
    .
    .
    .
    .
    -Q <== "Mirror image charge" (virtual)
     
  5. Mar 10, 2006 #4

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    Ok, I understand your symbols now.

    I think that is where you are wrong.
    The boundary condition on [itex] \vec E [/itex] is

    [tex]E_{above} - E_{below} = \frac{\sigma}{\epsilon_0} \hat{n} [/tex]

    Since the field inside the conductor is zero, this means that the field just outside is
    [tex] \vec{E} = \frac{\sigma}{\epsilon_0} \hat{n} [/tex]

    Also, you might want to check out the uniqueness theorem for Laplace's equation. It's very interesting and explains how you get the correct field for the charge and grounded plane even though you solve a completely different problem.
     
    Last edited: Mar 10, 2006
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