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

Just plug in the given equation for p into the desired equation and use the trig identity sin^2(x)+cos^2(x)=1 to cancel out the sin^2 term.
  • #1
naele
202
1

Homework Statement


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]

Homework Equations


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

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.
 
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  • #2
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:
 
  • #3
jdwood983 said:
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:

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.
 
  • #4
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:
  • #5
naele said:
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.

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.
 

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

What is Griffiths E&M 3.33 and what does it discuss?

Griffiths E&M 3.33 is a section in the textbook "Introduction to Electrodynamics" by David J. Griffiths. It discusses the mathematical representation of the electric field of a dipole moment in a coordinate-free form.

What is a dipole moment?

A dipole moment is a measure of the separation of positive and negative electrical charges in a system. It is a vector quantity that points from the negative charge to the positive charge.

Why is it important to represent the electric field of a dipole moment in a coordinate-free form?

Representing the electric field of a dipole moment in a coordinate-free form allows for a more general and universal approach to understanding and solving problems involving dipole moments. It eliminates the need for specific coordinate systems and makes the equations applicable in any situation.

What is the formula for the electric field of a dipole moment in coordinate-free form?

The formula for the electric field of a dipole moment in coordinate-free form is E = (1/4πε0)(3(p⃗⋅r̂)r̂-p⃗), where p⃗ is the dipole moment vector and r̂ is the unit vector pointing from the dipole moment to the point of observation.

How does the direction of the electric field change with distance from the dipole moment?

The direction of the electric field changes with distance from the dipole moment as the inverse cube of the distance. This means that as the distance increases, the electric field becomes more parallel to the dipole moment vector.

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