Griffiths, Electrodynamics Prob. 3.28a

Equation 3.98 could just as readily be written as \mathbf{p} =\int \mathbf{r} \sigma(\mathbf{r}) da and still be correct. The "primes" are just a mnemonic device to keep the two different \mathbf{r} vectors straight.
  • #1
mzh
64
0

Homework Statement


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

Homework Equations


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

The Attempt at a Solution


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

Thanks for any help.
 
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  • #2
mzh said:

Homework Statement


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


Homework Equations


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


The Attempt at a Solution


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

Thanks for any help.
Which quantities are vectors?
 
  • #3
SammyS said:
Which quantities are vectors?

the bold ones, so [itex]\bf{r}'[/itex] and [itex]\bf{p}[/itex].
 
  • #4
It looks to me like
[itex]\displaystyle \int (R \cos \theta) (k\cos \theta) R^2 \sin\theta d\phi d\theta[/itex]​
is z - component of the dipole moment, p .
 
  • #5
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.
 
  • #6
mzh said:

Homework Statement


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


Homework Equations


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


The Attempt at a Solution


The Ansatz is supposed to be [itex]\int (R \cos \theta) (k\cos \theta) R^2 \sin\theta d\phi d\theta[/itex]. 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 [itex]\mathbf{r}'=R\mathbf{\hat{r}}[/itex] 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 [itex]\phi[/itex], the [itex]x[/itex]- and [itex]y[/itex]-components will vanish since [itex]\int_0^{2 \pi} \sin \phi d\phi = \int_0^{2 \pi} \cos \phi d\phi = 0[/itex] and you will be left with the second integral you wrote
 
  • #7
vela said:
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?
9007238.png


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 $$??.$$
 
  • #8
mzh said:
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?
9007238.png


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 [itex]\mathbf{r}'[/itex]. 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.
 
  • #9
gabbagabbahey said:
I think you need to go back to your textbook and look up how Griffiths defines [itex]\mathbf{r}'[/itex]. 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 [itex]r[/itex] is used, so what [itex](R \cos \theta)[/itex] means is the z-coordinate of the source point on the spherical shell.

Thanks a lot guys, this forum rocks.
 
  • #10
mzh said:
Now I realize that in Eq. 3.98 a primed [itex]r[/itex] is used, so what [itex](R \cos \theta)[/itex] 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 [itex]\int_a^b f(x)dx = \int_a^b f(u)du[/itex] ) 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 [itex]\mathbf{r}'[/itex] in equation 3.98 is the location of each bit of source charge (it is the same as the argument of the charge distribution [itex]\sigma(\mathbf{r}')[/itex]).

Equation 3.98 could just as readily be written as [itex]\mathbf{p} =\int \mathbf{r} \sigma(\mathbf{r}) da[/itex] and still be valid.
 
  • #11
gabbagabbahey said:
Equation 3.98 could just as readily be written as [itex]\mathbf{p} =\int \mathbf{r} \sigma(\mathbf{r}) da[/itex] 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 [itex]\vec{p} = p\hat{\vec{z}}, p=\int z \rho d\tau[/itex], where I understood [itex]z[/itex] to be the field-point. But that is so just a dummy variable, which is integrated out.
 
Last edited:

1. What is Griffiths, Electrodynamics Prob. 3.28a?

Griffiths, Electrodynamics Prob. 3.28a is a specific problem in the textbook "Introduction to Electrodynamics" by David J. Griffiths. It is a problem related to the study of electromagnetic fields and their properties.

2. What is the difficulty level of Griffiths, Electrodynamics Prob. 3.28a?

The difficulty level of Griffiths, Electrodynamics Prob. 3.28a can vary depending on the individual's understanding of the subject matter. However, it is considered to be a challenging problem that requires a solid understanding of electromagnetic theory and mathematical skills.

3. Can you provide a brief overview of the problem?

The problem involves a charge distribution on a long thin wire and finding the electric field at a point outside the wire. The solution involves using the concept of line integrals and the equation for the electric field due to a charged wire.

4. Are there any tips for solving Griffiths, Electrodynamics Prob. 3.28a?

One tip for solving this problem is to carefully read and understand the given information and to draw a diagram to visualize the situation. It is also important to use the appropriate equations and to pay attention to units and signs.

5. How can solving Griffiths, Electrodynamics Prob. 3.28a be helpful in understanding electromagnetism?

Solving Griffiths, Electrodynamics Prob. 3.28a involves applying the fundamental concepts of electromagnetism, such as Coulomb's law and the superposition principle. By solving this problem, one can gain a better understanding of these concepts and their application in real-world situations.

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