
#1
Dec209, 08:11 PM

P: 202

1. The problem statement, all variables and given/known data
Show that the electric field of a "pure" dipole can be written in the coordinatefree 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 rp\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
Dec209, 08:58 PM

P: 383

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 



#3
Dec209, 09:16 PM

P: 202





#4
Dec209, 10:53 PM

P: 419

Griffiths E&M 3.33 write efield of dipole moment in coordinate free form
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. 



#5
Dec309, 06:31 AM

P: 383

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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