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Homework Help: Problem on electric field due to dipoles

  1. Sep 29, 2004 #1
    look at the figure in the attachment

    The black dots at the ends of the square in the figure represent charges -q or +q and they are denoted + or - by the symbol next to the them.
    Now that line is drawn from the centre of the square to point P and is distance x
    The side of the square is 2a

    Show that Electric Field at P for x >> a is 3 (2qa^2) / 2 pi epsilon0 x^4

    so far i came up with this

    for the dipole closer to P the expression would be

    E = (1/4 pi epsilon) 2qa/ [(x-a)^2 +(2a/2)^2]^3/2

    and after a lot of algebra i get

    E = (qa / 2 pi epsilon ) * 1 / [ x^2 + 2xa + 2a^2]^3/2

    for the first closer dipole

    for the more distant dipole i get E = - (1/4 pi epsilon) 2qa / [ (x+a)^2 + a^2 ] ^3/2

    more algebra i get E = - (qa / 2 pi epsilon ) * 1 / [x^2 + 2xa + 2a^2]^3/2

    but it is after this point that i am stuck completely

    how do i get to the expression that is needed??

    please please help!
     

    Attached Files:

    Last edited: Sep 29, 2004
  2. jcsd
  3. Sep 30, 2004 #2
    nee some help here i'm not sure on how to proceed in this

    no replies.... still trying by myself but to no avail
     
  4. Sep 30, 2004 #3
    for x >> a , you can use binomial expansion.
     
  5. Sep 30, 2004 #4
    how would hte binomial expansion give me 3 (2qa) blah blah

    i'm not quite sure how to use the binomial expansion but wouldn't that be a little long expression... would it compress to form somethign nice and compact which si needed?
     
  6. Sep 30, 2004 #5
    binomial expansion of (x+a)^n = x[1 (a/x)]^n =x[ 1 + n(a/x) + [n(n-1)/2!](a/x)^2 + .... ]
    it will compress because x >> a, so the higher order terms go to zero.
    a/x = small ; (a/x)^2 = small times small= very small = << 1.
     
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