How Can You Derive the Electric Field of a Dipole for r>>d?

[G01]

Homework Statement

Show that the electric field from an electric dipole for r>>d is:

\vec{E} = \frac{Qd}{4\pi\epsilon_0 r^3}(2\cos \theta \hat{r} + \sin \theta \hat{\theta})

Homework Equations

Electric Field of a Point Charge: \vec{E}=\frac{Q}{4\pi\epsilon_0r^2}

The Attempt at a Solution

First thing first, I can't use Electric potential to solve this, I need to use fields right from the start.

OK, here we go:

\vec{E}=\vec{E_+}+\vec{E_-}\vec{E}=\frac{Q}{4\pi\epsilon_0}(\frac{\vec{r_+}}{r_+^3}-\frac{\vec{r_-}}{r_-^3})

\vec{E}=\frac{Q}{4\pi\epsilon_0}(\frac{r_-^3\vec{r_+}-r_+^3\vec{r_-}}{(r_+r_-)^3})

Now assume r+ and r- are close to the same length, since r>>d:

\vec{E}=\frac{Q}{4\pi\epsilon_0r^3}(\vec{r_+}-\vec{r_-})

Ok, this is where I get stuck. I know the length of r+ - r- should be dcos(theta) in spherical coordinates, where theta is the angle from the +z axis., but I can't get the unit vector into terms of \hat{r} and \hat{\theta}.

Any hints would be appreciated. Thanks!

 

[Avodyne]

Your 2nd equation in part (3) is wrong, but I assume that's a typo, because your 3rd is correct. But then, you were too quick to set \vec{r_+}=\vec{r} and \vec{r_-}=\vec{r} in most places. In fact, \vec{r_+}-\vec{r_-} is just d\hat{z}, so your 4th equation is not correct. You have to expand everything consistently to linear order in d.

 

[G01]

Your 2nd equation in part (3) is wrong, but I assume that's a typo, because your 3rd is correct. But then, you were too quick to set \vec{r_+}=\vec{r} and \vec{r_-}=\vec{r} in most places. In fact, \vec{r_+}-\vec{r_-} is just d\hat{z}, so your 4th equation is not correct. You have to expand everything consistently to linear order in d.

Thanks for the reply. I fixed the typo. I hate to say though, I'm still a bit lost. Here's where I'm at now. I tried to take as much of your advice as I could, but I'm still confused. Please bear with me, I know I should be able to do this, being in a 400 level E&M course, but I've been looking at this problem for a long time, and I guess I'm frustrated. This is what I tried:

\vec{E}=\frac{Q}{4\pi\epsilon_0r^3}(\vec{r_+}-\vec{r_-}) = \frac{Qd}{4\pi\epsilon_0r^3} \hat{z}

Then, converting to spherical coordinates, I get:

\vec{E}=\frac{Qd}{4\pi\epsilon_0r^3} (\cos(\theta)\hat{r} - \sin(\theta)\hat{\theta})

I can't see how I can get this from what I have to what I'm trying to find. I think I'm still missing something important. Half of my problem may be the fact that I can do this EASILY working with potentials and then finding the gradient in spherical coordinates. I feel like this problem is asking me to dig a hole with a spoon when there is a shovel right there. Again, thanks for your time and help. I really appreciate it.

 

[Avodyne]

Your 2nd eq still has a typo; it should be
\vec{E}=\frac{Q}{4\pi\epsilon_0}\left(\frac{\vec{r}_+}{r_+^3}-\frac{\vec{r}_-}{r_-^3}\right).
But you knew that.

You can do it from here. Here's one way. Write out the x, y, and z components of \vec{r}_+/r^3_+ as functions of x, y, z, and d. Take d much less than x, y, or z, and expand each component in powers of d up through the first power. Then do the same for \vec{r}_-/r^3_- (which follows immediately from the previous result). Subtract. Convert to spherical coords. Or, take what you're supposed to get, and convert that to cartesian coords.

Or, here's another way. Time indenpendent electric fields can always be written as the gradient of a potential. What is that potential? (Hint: it's the sum of the potentials of two point charges!) Write it this potential in spherical coords, and then take the gradient in spherical coords. Then expand to linear order in d.

 

[G01]

Ok, so, I'm working on this some more. Here's where I'm at:
\vec{E}=\frac{Q}{4\pi\epsilon_0}\left(\frac{\vec{r}_+}{r_+^3}-\frac{\vec{r}_-}{r_-^3}\right)

Now, in Cartesian coordinates:

\frac{\vec{r_+}}{r_+^3}=\frac{1}{(x^2+y^2+(z-d/2)^2)^{3/2}}(x\hat{x}+y \hat{y} + (z-d/2)\hat{z}

and

\frac{\vec{r_-}}{r_-^3}=\frac{1}{(x^2+y^2+(z+d/2)^2)^{3/2}}(x\hat{x}+y \hat{y} + (z+d/2)\hat{z}

Now, expanding to first order in d, we get:

\frac{\vec{r_+}}{r_+^3}=\frac{1}{(x^2+y^2+z^2-zd)^{3/2}}(x\hat{x}+y \hat{y} + (z-d/2)\hat{z}\frac{\vec{r_-}}{r_-^3}=\frac{1}{(x^2+y^2+z^2+zd)^{3/2}}(x\hat{x}+y \hat{y} + (z+d/2)\hat{z}

Does this look correct so far? I have already solved this using potentials, here I have to use fields.

 

[learningphysics]

\frac{\vec{r_+}}{r_+^3}=\frac{1}{x^2+y^2+(z-d/2)^2}(x\hat{x}+y \hat{y} + (z-d/2)\hat{z}

You have r^2 in the denominator here instead of r^3.

 

[G01]

You have r^2 in the denominator here instead of r^3.

Thanks.

I'm really butchering this one. Thanks for pointing it out. I'll keep working at it.

Ok, I fixed the error in my last post. How do those lines look now? Anything else wrong with them?

 

[learningphysics]

Thanks.

I'm really butchering this one. Thanks for pointing it out. I'll keep working at it.

Ok, I fixed the error in my last post. How do those lines look now? Anything else wrong with them?

It's a tough problem. I was working on it when I saw it here, but I stopped because I was messing up.

For the expansion in d part, I think you should do the taylor series of:

\frac{1}{(x^2+y^2+(z-d/2)^2)^{3/2}}

with respect to d/2... around d/2 = 0. You just need two terms. And do the same for:

\frac{1}{(x^2+y^2+(z+d/2)^2)^{3/2}}

then you can get:
\frac{\vec{r_+}}{r_+^3}

and

\frac{\vec{r_-}}{r_-^3}

using those approximations...

 

[G01]

Ok thanks guys. I think I should be able to get it from here. If not, I'll be back.