- #1

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_{0}(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

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- Thread starter jmtome2
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- #1

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What I know:

1)

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

-Thanks

- #2

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

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

does that help

- #3

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

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- #4

turin

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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.

- #5

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

- #6

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[itex]\delta(\vec{r}-\vec{r'}) f(\vec{r})=f(\vec{r'})[/itex] for an arbitrary function [itex]f[/itex]. 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 [itex]\sigma[/itex] i.e. surface charge density and should therefore be applicable here.

- #7

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[tex]\overline{r'}[/tex]=a*[tex]\hat{r'}[/tex]

so that...

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

does this look right...?

- #8

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which is zero.... great

- #9

gabbagabbahey

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delta functions act as follows in 3d:

[itex]\delta(\vec{r}-\vec{r'}) f(\vec{r})=f(\vec{r'})[/itex] for an arbitrary function [itex]f[/itex]

No,

[tex]\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}[/tex]

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

[tex]\int_{\mathcal{V}}\delta^3(\textbf{r}-\textbf{r}')f(\textbf{r}')dV'=f(\textbf{r})[/tex]

In any case, it is a

- #10

gabbagabbahey

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[tex]\overline{r'}[/tex]=a*[tex]\hat{r'}[/tex]

so that...

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

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

[tex]\int_{\text{all space}}\rho(\textbf{r'})dV'[/tex]

?

- #11

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ok... so

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

The integral becomes...

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

So what happens to the [tex]\bar{r'}[/tex]? [tex]\bar{r'}[/tex]=a*[tex]\hat{r}[/tex]?

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

The integral becomes...

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

So what happens to the [tex]\bar{r'}[/tex]? [tex]\bar{r'}[/tex]=a*[tex]\hat{r}[/tex]?

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- #12

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Anyways, is the reasoning above correct?

- #13

gabbagabbahey

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ok... so

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

The integral becomes...

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

So what happens to the [tex]\bar{r'}[/tex]? [tex]\bar{r'}[/tex]=a*[tex]\hat{r}[/tex]?

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

[tex]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'[/tex]

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

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

[tex]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'[/tex]

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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- #14

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Still missing how you get...

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

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

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- #15

gabbagabbahey

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[tex]\int_0^\infty f(r')\delta(r'-a)dr'=f(a)[/tex]

- #16

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- #17

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- #18

gabbagabbahey

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[tex]\rho(\textbf{r})\equiv\frac{dq}{dV}=\frac{dq}{drda}=\frac{\sigma}{dr}[/tex]

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

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

Make sense?

- #19

gabbagabbahey

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

[tex]\mathbf{\hat{r}}'=\sin\theta'\cos\phi'\mathbf{\hat{x}}+\sin\theta'\sin\phi'\mathbf{\hat{y}}+\cos\theta'\mathbf{\hat{z}}[/tex]

- #20

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- #21

gabbagabbahey

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The Cartesian unit vectors

[tex]\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}[/tex]

- #22

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dear god

- #23

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- #24

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ooops! forgot the integral. my bad...

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