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Please help with image method

  1. Apr 13, 2007 #1
    HI I dont understand the method of image.
    I have this problem:

    A uniform charged disk is suspended ( at a distance h) above an infinite conducting plane ( disk is parallel to the plane). The disk carries a charge Q and has a diameter d. Assume d>>h.
    1- Calculate the electric field vector near the center and below the disk.
    2- Calculate the capacitance of the system using the formula C=Q/V.
    3- Calculate the capacitance of the system using the energy formulas.


    I attach a picture.
    I get lost at some point

    the electric fiel for a disk is:
    E=Qy/(2*pi*epsilon)*(1/y-1/(sqrt(y^2+R^2)) y (unit vector)

    to answer the question 1-:
    electric field from charge Q:
    EQ=Q(h-y)/(2*pi*epsilon)*(1/(h-y)-1/(sqrt((h-y)^2+R^2)) (-y) (unit vector)

    electric field from charge -Q ( the image):
    E-Q=-Q(h+y)/(2*pi*epsilon)*(1/(h+y)-1/(sqrt((h+y)^2+R^2)) (-y) (unit vector)

    Here is my first problem: the direction of the two electric fields EQ and E-Q.
    I am not sure about the electric field of the expressions.

    (I suspect I get some mistakes because I tried a similar problem but with a more simple expression for the 2 fields and when I add them, I get zero)

    I would really have some help with this problem.

    please can someone help me?
    Thank you
     

    Attached Files:

  2. jcsd
  3. Apr 14, 2007 #2

    Mentz114

    User Avatar
    Gold Member

    Reverse the direction of the unit vector y for [tex]E_{-Q}[/tex] because the polarity is opposite to [tex]E_{Q}[/tex].
     
  4. Apr 14, 2007 #3
    thanks.
    I will try to continue from there.
     
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