Griffiths E&M 3.33 write e-field of dipole moment in coordinate free form

  1. 1. The problem statement, all variables and given/known data
    Show that the electric field of a "pure" dipole can be written in the coordinate-free form

    [tex]
    E_{dip}(r)=\frac{1}{4\pi\epsilon_0}\frac{1}{r^3}[3(\vec p\cdot \hat r)\hat r-\vec p].[/tex]

    2. Relevant equations
    Starting from
    [tex]E_{dip}(r)=\frac{p}{4\pi\epsilon_0r^3}(2\cos \hat r+\sin\theta \hat \theta)[/tex]


    3. The attempt at a solution
    The equation immediately above assumes a spherical coordinate system such that p is oriented along z. We can therefore write
    [tex]\vec p=p\hat z[/tex]
    [tex]\hat z = \cos\theta \hat r - \sin\theta \hat \theta \implies \vec p=p\cos\theta\hat r-p\sin\theta\hat\theta[/tex]
    From equation 3.102 in the book we know that [itex]\hat r\cdot \vec p=p\cos\theta[/itex]

    Try as I might I don't know how to show, geometrically or via manipulation, that [tex]p\sin\theta\hat \theta=(\vec p \cdot \hat \theta)\hat \theta[/tex]. From there it's easy to get to the desired result.
     
  2. jcsd
  3. I find this problem is easier to see in reverse than forwards (that is, working from the solution to get the starting point)

    [tex]
    3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{p}=3p\cos\theta\hat{\mathbf{r}}-p\cos\theta\hat{\mathbf{r}}+p\sin\theta\hat{\boldsymbol{\theta}}=2p\cos\theta\hat{\mathbf{r}}+p\sin\theta\hat{\boldsymbol{\theta}}
    [/tex]

    which looks a lot like what you have in your relevant equations :wink:
     
  4. Well I guess that solves the problem. Seems a little anticlimactic. My professor recommended we draw a picture and show how things cancel out by dotting vectors together.
     
  5. I was able to do it by using pictures...backwards. Here's how to derive the given equation from the desired equation.

    Let theta be the polar angle and p point toward theta=0.

    First draw the p vector and the r hat vector intersecting at some angle theta. Project p onto the r axis.
    [tex]\overrightarrow{p}\cdot \widehat{r}=pcos(\Theta )[/tex]

    Then draw the p vector (parallel to the original p vector) at some point at vector r from the origin (where the dipole is actually located). Resolve the p vector onto the (r, theta) coordinates.
    [tex]
    \widehat{p}=cos(\Theta )\widehat{r}+sin(\Theta )\left ( -\widehat{\Theta } \right )
    [/tex]
    Plugging those into the problem statement equation and doing some algebra gives you the equation you were supposed to start with. I'm sure you could just work through it backwards.
     
    Last edited: Dec 2, 2009
  6. It is a little anticlimactic doing it backwards. I spent about 30 minutes trying to do the geometry forwards before I decided to do it backwards. When I saw the answer, my first thought was "That was it?"

    As Jolb said, you can do it graphically if you draw the appropriate vectors, but I personally think it's easier to do it mathematically.
     
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