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Find the Expression of an Electric Field Produced by a Dipole

  1. Jul 7, 2013 #1
    1. The problem statement, all variables and given/known data
    An electric dipole is located along the y axis as shown in (picture attached). The magnitude of its electric dipole moment is defined as [itex]p=2aq[/itex].

    (a) At a point P, which is far from the dipole [itex](r>>a)[/itex], show that the electric potential is
    [itex]V=\frac{k_{e}pcosθ}{r^{2}}[/itex]

    (b) Find an expression of the electric field produced by the dipole in terms of the vectors [itex]\hat{r}[/itex] and [itex]\vec{p}[/itex] (the dipole moment).


    2. Relevant equations
    [itex]p=2aq[/itex] (Magnitude of the Electric Dipole Moment)
    [itex]\vec{E}=k_{e}\frac{q}{r^{2}}\hat{r}[/itex] (Electric Field at P created by a q charge)

    3. The attempt at a solution
    Part (a):
    [itex]V=\frac{k_{e}q}{r_{1}}-\frac{k_{e}q}{r_{2}}=\frac{k_{e}q}{r_{1}r_{2}}(r_{2}-r_{1})[/itex]
    Since [itex]r>>a[/itex] and [itex]r_{2}-r_{1}≈2acosθ[/itex], then [itex]v≈\frac{k_{e}q}{r_{1}r_{2}}2acosθ≈\frac{k_{e}pcosθ}{r^{2}}[/itex]

    Part (b):
    I need help with this part.

    So far, I got that both +q and -q is [itex]E=k_{e}\frac{|q|}{r^{2}}[/itex]
    Not sure how to complete the rest since point P does not lie on the x-axis.

    NOTE:
    An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance of 2a. The electric dipole moment [itex]\vec{p}[/itex] is directed from -q toward +q.
     
    Last edited: Jul 7, 2013
  2. jcsd
  3. Jul 7, 2013 #2

    BruceW

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    Homework Helper

    in part a) what happened to r1 in the denominator? p.s. I think you have written down the electric potential V incorrectly, it should not be:
    [tex]V=\frac{k_{e}pcosθ}{r_{2}}[/tex]
    And for part b), you have written down the correct equation for the magnitude of the electric field due to a point particle, but what you need is the direction as well!
     
  4. Jul 7, 2013 #3

    vanhees71

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    Science Advisor
    2016 Award

    The idea is to perform a Taylor expansion.

    The full expression for the potential reads
    [tex]V(\vec{r})=q k_e \left (\frac{1}{\sqrt{(\vec{r}-a \vec{e}_y)^2}}-\frac{1}{\sqrt{(\vec{r}+a \vec{e}_y)^2}} \right ).[/tex]
    The point is that [itex]a/|\vec{r}|\ ll 1[/itex] if you look at the field from a distance far from the charges. Then in leading non-vanishing order in the expansion in powers of [itex]a/|\vec{r}|[/itex] you find
    [tex]V(\vec{r})=q k_e \left [-2a \vec{e}_y \cdot \vec{\nabla} \frac{1}{r} \right ]=k_e \frac{\vec{p} \cdot \vec{r}}{r^3}.[/tex]

    Note that the expression given in your posting cannot be correct for dimensional reasons!

    You find the electric field by taking the gradient of the potential
    [tex]\vec{E}=-\vec{\nabla} V(\vec{r}).[/tex]
     
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