# Dipole Moment

Calculate the dipole moment of a spherical shell of radius 'a' whose surface charge density is σ=σ0(1+cos(θ)). The origin is at the center of the sphere.

What I know:
1) p=∫r'*ρ*dV'

I'm having trouble understanding how to transform this equation so that I can calculate the dipole moment knowing σ, instead of ρ.

-Thanks

i think you need to go to the 2 dimensional analogue.

you want to use a surface charge not a volume charge density.

try $\vec{p}= \int \sigma (r-r') \vec{dA'}$ remember that you'll need to use spherical polars and so $\vec{dA'}=r^2 \sin{\theta} d \theta d \phi \vec{\hat{r}}$

does that help

So integrate over θ and Ф and treat r and r' as constants?

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turin
Homework Helper
You can use a Dirac delta function. Are you familiar with those?

ρ(r,θ,φ) ~ δ(r-R) σ(θ,φ)

where R is the radius of the shell. I will leave the proportionality constant to you.

unfortunately I am not, my profesor skipped that portion of the book

delta functions act as follows in 3d:

$\delta(\vec{r}-\vec{r'}) f(\vec{r})=f(\vec{r'})$ for an arbitrary function $f$. if you use your original formula for the dipole moment and substitute the expression for rho given what will your new equation be? this one will involve $\sigma$ i.e. surface charge density and should therefore be applicable here.

so I get that $$\overline{p}$$=$$\int$$$$\overline{r'}$$*δ*(r-a)*dA'

$$\overline{r'}$$=a*$$\hat{r'}$$

so that...

$$\bar{p}$$=δ*a*r$$^{2}$$(r-a)*σ$$_{0}$$*$$\int$$[1+cos(θ)]siin(θ)]*dθ*dФ*$$\hat{r'}$$

does this look right...?

which is zero.... great

gabbagabbahey
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delta functions act as follows in 3d:

$\delta(\vec{r}-\vec{r'}) f(\vec{r})=f(\vec{r'})$ for an arbitrary function $f$

No,

$$\delta^3(\textbf{r}-\textbf{r}')f(\textbf{r})=\left\{\begin{array}{lr}0 &,\textbf{r}\neq\textbf{r}'\\\infty &,\textbf{r}=\textbf{r}'\end{array}$$

It is only when you integrate over a volume $\mathcal{V}$, enclosing $\textbf{r}=\textbf{r}'$, that you end up with a non-zero, finite result:

$$\int_{\mathcal{V}}\delta^3(\textbf{r}-\textbf{r}')f(\textbf{r}')dV'=f(\textbf{r})$$

In any case, it is a one-dimensional dirac delta function that is useful here...

gabbagabbahey
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so I get that $$\overline{p}$$=$$\int$$$$\overline{r'}$$*δ*(r-a)*dA'

$$\overline{r'}$$=a*$$\hat{r'}$$

so that...

$$\bar{p}$$=δ*a*r$$^{2}$$(r-a)*σ$$_{0}$$*$$\int$$[1+cos(θ)]siin(θ)]*dθ*dФ*$$\hat{r'}$$

does this look right...?

First start with something simpler....without using any dirac deltas (just the surface charge density), how would you calculate the total charge on the shell (don't actually carry out the integral, just set it up)

Once you done that, assume you have a volume charge density of the form $\rho(\textbf{r})=\delta(r-a)f(\textbf{r})$, where $r$ is the distance from the origin....what can you say about the integral

$$\int_{\text{all space}}\rho(\textbf{r'})dV'$$

?

ok... so

Q$$_{tot}$$=$$\int$$$$^{2л}_{0}$$$$\int$$$$^{л}_{0}$$$$\int$$$$^{a}_{0}$$ {ρ(r,θ,Ф)*r$$^{2}$$*sin(θ)*dr*dθ*dФ}

The integral becomes...

$$\int$$$$^{2л}_{0}$$$$\int$$$$^{л}_{0}$$$$\int$$$$^{a}_{0}$$ {δ(r-a)*σ(θ,Ф)*r$$^{2}$$*sin(θ)*dr*dθ*dФ}

So what happens to the $$\bar{r'}$$? $$\bar{r'}$$=a*$$\hat{r}$$?

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i give up on trying to get those integrals to have pi signs...

Anyways, is the reasoning above correct?

gabbagabbahey
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ok... so

Q$$_{tot}$$=$$\int$$$$^{2П}_{0}$$$$\int$$$$^{П}_{0}$$$$\int$$$$^{a}_{0}$$ {ρ(r,θ,Ф)*r$$^{2}$$*sin(θ)*dr*dθ*dФ}

The integral becomes...

$$\int$$$$^{2П}_{0}$$$$\int$$$$^{П}_{0}$$$$\int$$$$^{a}_{0}$$ {δ(r-a)*σ(θ,Ф)*r$$^{2}$$*sin(θ)*dr*dθ*dФ}

So what happens to the $$\bar{r'}$$? $$\bar{r'}$$=a*$$\hat{r}$$?

I think this is what you meant to write (just click on the image below to see the $\LaTeX$ code that generated it):

$$Q_{tot}=\int_{\text{all space}}\rho(\textbf{r'})dV'=\int_{0}^{\infty} \int_{0}^{\pi} \int_{0}^{2\pi}\rho(r')r'^2\sin\theta' dr' d\theta' d\phi'=\int_{0}^{\infty} \int_{0}^{\pi} \int_{0}^{2\pi}\delta(r'-a)\sigma(\theta',\phi')r'^2\sin\theta' dr' d\theta' d\phi'$$

(Remember, $\rho(\textbf{r})=\delta(r-a)\sigma(\theta,\phi)$ means that $\rho(\textbf{r}')=\delta(r'-a)\sigma(\theta',\phi')$ ; and when you integrate over all space, $r'$ goes from 0 to $\infty$)

Now, your radial integral encloses the point (technically it's a spherical shell, not a point) $r'=a$, so the delta function picks out that point and you have:

$$Q_{tot}=\int_{0}^{\infty} \int_{0}^{\pi} \int_{0}^{2\pi}\delta(r'-a)\sigma(\theta',\phi')r'^2\sin\theta' dr' d\theta' d\phi'=\int_{0}^{\pi} \int_{0}^{2\pi}a^2\sigma(\theta',\phi')\sin\theta' d\theta' d\phi'$$

Make sense?

If so, apply the same thing to your dipole moment integral and keep in mind that the spherical unit vector are position dependent and hence are not to be treated as constants under integration...

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Still missing how you get...

Q$$_{tot}$$=$$\int$$$$^{\infty}_{0}$$ (r'^2)*δ(r'-a)*dr=a^2

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gabbagabbahey
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It's basically the definition of the one-dimensional dirac delta function (If you are studying from Griffiths' text, it is just applying equation 1.92 ....with $r$ restricted to positive values for obvious reasons)

$$\int_0^\infty f(r')\delta(r'-a)dr'=f(a)$$

now I'm following, sorry for all the questions... this is the first time I've been introduced to Delta Dirac functions... I'm going to go at this equation some more and see what comes out, thanks for all the help

So finally, i can break the $$\bar{r'}$$ vector into r'*$$\hat{r'}$$, so that the final answer for $$\bar{p}$$, the dipole moment, points in the $$\hat{r'}$$ direction.

gabbagabbahey
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An important side note; the reason the Dirac Delta function appears at all is because you expect the volume charge density to be zero everywhere except at the shell $r=a$, and the shell has infinitesimal extent in the radial direction (it is infinitesimally thin), meaning that a small piece of charge $dq$ will have a localized volume charge density of

$$\rho(\textbf{r})\equiv\frac{dq}{dV}=\frac{dq}{drda}=\frac{\sigma}{dr}$$

which, in the limit that $dr$ (the extent of the charge distribution in the radial direction) goes to zero, becomes undefined.

However, you expect that the total charge $\int\rho(\textbf{r}')dV'$ is well defined, and the only function that has the properties; (1) zero everywhere except at a certain point ($r=a$ in this case) where it is undefined, (2) but having a well defined integral; is the dirac delta function.

Make sense?

gabbagabbahey
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Gold Member
So finally, i can break the $$\bar{r'}$$ vector into r'*$$\hat{r'}$$, so that the final answer for $$\bar{p}$$, the dipole moment, points in the $$\hat{r'}$$ direction.

Not quite. It is true that $\textbf{r}'=r'\mathbf{\hat{r}}'$, but the direction $\mathbf{\hat{r}}'$ is position dependent, and hence is not constant when integrating over $dV'$:

$$\mathbf{\hat{r}}'=\sin\theta'\cos\phi'\mathbf{\hat{x}}+\sin\theta'\sin\phi'\mathbf{\hat{y}}+\cos\theta'\mathbf{\hat{z}}$$

I realize this, but cannot think of anything else to do with that $$\bar{r'}$$ vector in the presence of the delta dirac function

gabbagabbahey
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I realize this, but cannot think of anything else to do with that $$\bar{r'}$$ vector in the presence of the delta dirac function

The Cartesian unit vectors are constant (position independent) and can be pulled outside of the integral, so just substitute the above expression for $\mathbf{\hat{r}}'$ in terms of them into your integral and then integrate...

\begin{aligned}\textbf{p} &=\int_{\text{all space}}\textbf{r}'\rho(\textbf{r'})dV'\\&=\int_{0}^{\infty}\int_{0}^{\pi} \int_{0}^{2\pi} \left(r'\sin\theta'\cos\phi'\mathbf{\hat{x}}+r'\sin\theta'\sin\phi'\mathbf{\hat{y}}+r'\cos\theta'\mathbf{\hat{z}}\right)\delta(r'-a)\sigma(\theta',\phi')r'^2\sin\theta' dr' d\theta' d\phi'\\&=\mathbf{\hat{x}}\int_{0}^{\infty}\int_{0}^{\pi} \int_{0}^{2\pi}\delta(r'-a)\sigma(\theta',\phi')r'^3\sin^2\theta'\cos\phi' dr' d\theta' d\phi'+\mathbf{\hat{y}}\int_{0}^{\infty}\int_{0}^{\pi} \int_{0}^{2\pi} \delta(r'-a)\sigma(\theta',\phi')r'^3\sin^2\theta'\sin\phi' dr' d\theta' d\phi'\\&\text{ }+\mathbf{\hat{z}}\int_{0}^{\infty}\int_{0}^{\pi} \int_{0}^{2\pi} \delta(r'-a)\sigma(\theta',\phi')r'^3\sin\theta'\cos\theta' dr' d\theta' d\phi'\end{aligned}

dear god

Should've seen that, but then again its far past bedtime. Thanks for the insight, anyway I can give you a kudos or something?

ooops! forgot the integral. my bad...