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Direction of dipole's electric field

  1. Oct 24, 2011 #1
    Edit: This is probably suited for the lower-level homework section. But I wasn't sure, as this is part of my second year electromagnetism course.

    1. The problem statement, all variables and given/known data

    Two point charges, -q and +q are aligned along the x-axis, such that they are equidistant from the origin and the +q charge is upon the +ve x-axis. Obtain a formula for the electric field at any point from the dipole.

    2. Relevant equations

    This is what's given to me in my lecture notes;

    E(r)= [3(p [itex]\bullet[/itex] [itex]\widehat{r}[/itex])[itex]\widehat{r}[/itex] - p]/4[itex]\pi[/itex][itex]\epsilon[/itex][itex]r^{3}[/itex]


    [itex]\widehat{r}[/itex] is the unit vector pointing in the direction of the point in question, from the origin.

    p is the dipole moment, a vector pointing from the negative to the positive charge. It's magnitude is equal to qd. d is the vector pointing from the negative to the positive charge.



    3. The attempt at a solution

    I can derive this formula fine. What confuses me is the direction/signs of the whole thing.

    Let's say I wanted to look at a point P, situated on the positive x-axis a distance r from the origin.

    Then we'd have;

    [itex]\widehat{r}[/itex] = (1,0,0)

    If we sub in this [itex]\widehat{r}[/itex] and the dipole moment qd = (qd,0,0) then we get this overall;

    E(r) = (qd/2[itex]\pi[/itex][itex]\epsilon[/itex][itex]r^{3}, 0, 0)[/itex]

    So we get an electric field only in the x direction. Makes sense. I don't understand the sign though. This overall electric-field in the x-direction gives a positive result. Essentially saying the electric field is pointing in the positive x-direction, TOWARDS the positive charge. This defies all my logic, the electric field should be pointing in the negative x-direction, surely?
     
    Last edited: Oct 24, 2011
  2. jcsd
  3. Oct 24, 2011 #2

    vela

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    The expression is only valid for r>d, i.e. outside the dipole. So when you're on the positive x-axis, you're to the right of both charges. Since the positive charge is closer, its electric field will be stronger than the field due to the negative charge, so the net field will point to the right.

    P.S. You posted this problem in the correct section.
     
  4. Oct 24, 2011 #3
    Ah, you're quite right. Deriving the equation uses a Taylor expansion and simplifies everything on the assumption r >> a. Cheers, wouldn't have noticed that.
     
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