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Homework Help: Calculating Electric Field due to a Dipole

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Two dipoles are oriented as shown in the diagram below. Each dipole consists of two charges +q and -q, held apart by a rod of length s, and the center of each dipole is a distance d from location A. If q = 4 nC, s = 1 mm, and d = 6 cm, what is the electric field at location A?

    As I am unaware of how to attach a file, here is a diagram

    the *s are there for formatting, or the else lines do not align
    2. Relevant equations
    (1/4*pi*epsilon)*(2qs/d^3) is the electric field of a dipole, on the dipole axis
    (1/4*pi*epsilon)*(qs/d^3) is the electric field of a dipole, perpendicular to the dipole axis

    3. The attempt at a solution
    Plugging the values given, I used the latter equation for the x component, and by doing (9e9)(4e-9*1e-3)/(6e-2)^3 I obtained 166 N/C.

    I used the first equation for the y component and obtained 333 N/C

    I know there is no z component, so I input <166, 333, 0> N/C, yet this is wrong. Can somebody explain to me what I am doing? I attempted submitting -166 for the x component in case I had somehow misinterpreted the direction of the field, but that was also wrong. Does anybody understand what I am doing wrong?
    Last edited: Aug 25, 2009
  2. jcsd
  3. Aug 25, 2009 #2


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    Homework Helper

    In the dipole above A, the negative charge is closer to A . So the attractive force is more than the repulsive force. So the electric field is towards the dipole.
    In the equatorial position the direction of the field is parallel to the dipole from positive charge to negative charge.
    Hence the electric field due to two dipoles are in the same direction. So the net field is the sum of them.
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