Griffiths, Electrodynamics Prob. 3.28a

[mzh]

Homework Statement

A spherical shell of radius R carries surface charge density \sigma=k \cos \theta. Whats the dipole moment of this distribution?

Homework Equations

The dipole moment is calculated as \bf{p} = \int \bf{r}' \sigma (\bf{r}') d\bf{a}' (Griffiths, Eq. 3.98).

The Attempt at a Solution

The Ansatz is supposed to be \int (R \cos \theta) (k\cos \theta) R^2 \sin\theta d\phi d\theta. I get every factor except for (R \cos \theta). Where does that come from?

Thanks for any help.

 

[SammyS]

Homework Statement

A spherical shell of radius R carries surface charge density \sigma=k \cos \theta. Whats the dipole moment of this distribution?

Homework Equations

The dipole moment is calculated as \bf{p} = \int \bf{r}' \sigma (\bf{r}') d\bf{a}' (Griffiths, Eq. 3.98).

The Attempt at a Solution

The Ansatz is supposed to be \int (R \cos \theta) (k\cos \theta) R^2 \sin\theta d\phi d\theta. I get every factor except for (R \cos \theta). Where does that come from?

Thanks for any help.

Which quantities are vectors?

 

[mzh]

Which quantities are vectors?

the bold ones, so \bf{r}' and \bf{p}.

 

[SammyS]

It looks to me like

\displaystyle \int (R \cos \theta) (k\cos \theta) R^2 \sin\theta d\phi d\theta​

is z - component of the dipole moment, p .

 

[vela]

That looks like the dipole term in the multipole expansion.
$$V_\text{dip}(P) = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \int (r'\cos\theta)\rho\,d\tau.$$ For the spherical shell of charge, this becomes
$$V_\text{dip}(P) = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \int (R\cos\theta)\sigma(\theta,\phi)\,R^2\sin\theta\, d\phi\,d\theta.$$ The $\cos\theta$ is from the Legendre polynomial $P_1(\cos\theta)$ in the dipole term of the multipole expansion.

The other way to look at it is that when you integrate over $\phi$, the other two components of $\mathbf{p}$ will vanish, so that's the only component you need to actually calculate.

 

[gabbagabbahey]

Homework Statement

A spherical shell of radius R carries surface charge density \sigma=k \cos \theta. Whats the dipole moment of this distribution?

Homework Equations

The dipole moment is calculated as \bf{p} = \int \bf{r}' \sigma (\bf{r}') d\bf{a}' (Griffiths, Eq. 3.98).

The Attempt at a Solution

The Ansatz is supposed to be \int (R \cos \theta) (k\cos \theta) R^2 \sin\theta d\phi d\theta. I get every factor except for (R \cos \theta). Where does that come from?

Thanks for any help.

The area element in this integral is not a vector, so it should not be bolded. The direction of \mathbf{r}'=R\mathbf{\hat{r}} varies over the spherical surface, so you need to express it in terms of Cartesian unit vectors (look inside the back cover of your textbook). When you integrate over \phi, the x- and y-components will vanish since \int_0^{2 \pi} \sin \phi d\phi = \int_0^{2 \pi} \cos \phi d\phi = 0 and you will be left with the second integral you wrote

 

[mzh]

That looks like the dipole term in the multipole expansion.
$$V_\text{dip}(P) = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \int (r'\cos\theta)\rho\,d\tau.$$ For the spherical shell of charge, this becomes
$$V_\text{dip}(P) = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \int (R\cos\theta)\sigma(\theta,\phi)\,R^2\sin\theta\, d\phi\,d\theta.$$ The $\cos\theta$ is from the Legendre polynomial $P_1(\cos\theta)$ in the dipole term of the multipole expansion.

The other way to look at it is that when you integrate over $\phi$, the other two components of $\mathbf{p}$ will vanish, so that's the only component you need to actually calculate.

I see, you're referring to the equation below Eq. 3.97 (which is unnumbered).

I tried to express the problem by a drawing. Does this make sense?

Here, $$P_z$$ is the z-component of the point $$P,$$ for which the dipole moment is evaluated, and it has arbitrarily a value of $$z'.$$ My problem is, that I can't see how $$(R \cos \theta)$$ translates to the distance marked as $$??.$$

 

[gabbagabbahey]

I see, you're referring to the equation below Eq. 3.97 (which is unnumbered).

I tried to express the problem by a drawing. Does this make sense?

Here, $$P_z$$ is the z-component of the point $$P,$$ for which the dipole moment is evaluated, and it has arbitrarily a value of $$z'.$$ My problem is, that I can't see how $$(R \cos \theta)$$ translates to the distance marked as $$??.$$

I think you need to go back to your textbook and look up how Griffiths defines \mathbf{r}'. The dipole moment is independent of where you measure it from (the field point), and this should be obvious to you when looking at equation 3.98, provided you understand what the primes on the variables means, and what you are really integrating over.

 

[mzh]

I think you need to go back to your textbook and look up how Griffiths defines \mathbf{r}'. The dipole moment is independent of where you measure it from (the field point), and this should be obvious to you when looking at equation 3.98, provided you understand what the primes on the variables means, and what you are really integrating over.

I know the primed variables indicate the source point, the unprimed variables the field point. This infact is the solution to my problem!

Now I realize that in Eq. 3.98 a primed r is used, so what (R \cos \theta) means is the z-coordinate of the source point on the spherical shell.

Thanks a lot guys, this forum rocks.

 

[gabbagabbahey]

Now I realize that in Eq. 3.98 a primed r is used, so what (R \cos \theta) means is the z-coordinate of the source point on the spherical shell.

Right, but just be careful here. The "primes" are unnecessary (the primed coordinates all get integrated over, so they are essentially just dummy variables in the same way that \int_a^b f(x)dx = \int_a^b f(u)du ) and many authors will leave them out. The fact that you integrate only over the source (since the charge density is zero elsewhere!) is what really tells you that the \mathbf{r}' in equation 3.98 is the location of each bit of source charge (it is the same as the argument of the charge distribution \sigma(\mathbf{r}')).

Equation 3.98 could just as readily be written as \mathbf{p} =\int \mathbf{r} \sigma(\mathbf{r}) da and still be valid.

 

[mzh]

Equation 3.98 could just as readily be written as \mathbf{p} =\int \mathbf{r} \sigma(\mathbf{r}) da and still be valid.

thats a good way of pointing it out, thanks again. also, i know now that my confusion arouse from the solution being written as \vec{p} = p\hat{\vec{z}}, p=\int z \rho d\tau, where I understood z to be the field-point. But that is so just a dummy variable, which is integrated out.