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Bound charges - Question about the maths. Did I calculate this correctly?

  • Thread starter phyzmatix
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  • #1
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Homework Statement



We know that a sphere of radius R carries a polarization P(r)=kr where k is a constant and r the vector from the centre. Calculate [tex]\sigma_b[/tex] and [tex]\rho_b[/tex]

The Attempt at a Solution



If we let the direction of polarization coincide with the z axis then:

[tex]\sigma_b=\textbf{P}\hat{n}[/tex]
[tex]\sigma_b=kr\cos\theta[/tex]

and

[tex]\rho_b=-\nabla\textbf{P}[/tex]
[tex]\rho_b=-\nabla kr\hat{r}[/tex]
[tex]\rho_b=-kr (\hat{x}\frac{d}{dx}+\hat{y}\frac{d}{dy}+\hat{z}\frac{d}{dz}) (\sin\theta\cos\phi\hat{x}+\sin\theta\sin\phi\hat{y}+\cos\theta\hat{z})[/tex]
[tex]\rho_b=0[/tex]

My query is mostly about these last three steps...are they correct? I'm still struggling to wrap my head around substituting the right infinitesimals for integration problems and, as in this case, the right coordinates for vectors.

Thanks for the help!
phyz
 

Answers and Replies

  • #2
diazona
Homework Helper
2,175
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If we let the direction of polarization coincide with the z axis then:
If the polarization is radial, how can the direction of polarization coincide with the z axis?

[tex]\sigma_b=\textbf{P}\hat{n}[/tex]
[tex]\sigma_b=kr\cos\theta[/tex]
Technically a dot product:
[tex]\sigma_b=\textbf{P}\cdot\hat{n}[/tex]
but you probably knew that ;-)

[tex]\rho_b=-\nabla\textbf{P}[/tex]
[tex]\rho_b=-\nabla kr\hat{r}[/tex]
[tex]\rho_b=-kr (\hat{x}\frac{d}{dx}+\hat{y}\frac{d}{dy}+\hat{z}\frac{d}{dz}) (\sin\theta\cos\phi\hat{x}+\sin\theta\sin\phi\hat{y}+\cos\theta\hat{z})[/tex]
Stop right there: you're mixing coordinate systems. Your derivative is expressed in Cartesian coordinates, but the radial vector is expressed in spherical coordinates. Now, you could keep going with this, by using the chain rule; you would have to calculate the coordinate derivatives
[tex]\frac{\mathrm{d}r}{\mathrm{d}x}, \frac{\mathrm{d}r}{\mathrm{d}y}, \frac{\mathrm{d}r}{\mathrm{d}z}, \frac{\mathrm{d}\theta}{\mathrm{d}x}, \frac{\mathrm{d}\theta}{\mathrm{d}y}, \frac{\mathrm{d}\theta}{\mathrm{d}z}, \frac{\mathrm{d}\phi}{\mathrm{d}x}, \frac{\mathrm{d}\phi}{\mathrm{d}y}, \frac{\mathrm{d}\phi}{\mathrm{d}z}[/tex]
but that'd be kind of a pain. It would be much easier to express the radial vector in terms of x, y, and z.
 
  • #3
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If the polarization is radial, how can the direction of polarization coincide with the z axis?
The thing is, I don't KNOW if the polarization is radial...I'm basing my (obviously very dangerous) assumption on a similar example (there is only one, can you believe it) in my textbook. So what do you suggest? Is there a way around this?

Technically a dot product:
[tex]\sigma_b=\textbf{P}\cdot\hat{n}[/tex]
but you probably knew that ;-)
Sorry, LaTex oversight on my part :smile:


Stop right there: you're mixing coordinate systems. Your derivative is expressed in Cartesian coordinates, but the radial vector is expressed in spherical coordinates. Now, you could keep going with this, by using the chain rule; you would have to calculate the coordinate derivatives
[tex]\frac{\mathrm{d}r}{\mathrm{d}x}, \frac{\mathrm{d}r}{\mathrm{d}y}, \frac{\mathrm{d}r}{\mathrm{d}z}, \frac{\mathrm{d}\theta}{\mathrm{d}x}, \frac{\mathrm{d}\theta}{\mathrm{d}y}, \frac{\mathrm{d}\theta}{\mathrm{d}z}, \frac{\mathrm{d}\phi}{\mathrm{d}x}, \frac{\mathrm{d}\phi}{\mathrm{d}y}, \frac{\mathrm{d}\phi}{\mathrm{d}z}[/tex]
but that'd be kind of a pain. It would be much easier to express the radial vector in terms of x, y, and z.
Now you see, this is exactly my problem. I simply don't get this stuff :redface:

I thought

[tex](\sin\theta\cos\phi\hat{x}+\sin\theta\sin\phi\hat{y}+\cos\theta\hat{z})[/tex]

was expressing spherical coordinates as Cartesian :cry:

So how do I start fixing this?
 
  • #4
Born2bwire
Science Advisor
Gold Member
1,779
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You are mixing the equations for the scalar components and the vector components.

If you want the scalar x coordinate converted from spherical coordinates, then [tex]x=R\sin\theta\cos\phi[/tex].

If you want the x component of a vector converted from spherical coordinates, then [tex]A_x = A_R\sin\thea\cos\phi+A_\theta\cos\thea\cos\phi-A_\phi\sin\phi[/tex].

I always keep a set of photocopies of various conversion charts hung up in my cube for handy reference.

But like diazona suggested, use the divergence operator that is for spherical coordinates, it will make this so much easier.
 
  • #5
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Thanks for the input so far! :smile:

OK, I'm still not sure if I fully understand this, but could you tell me if my revised attempt is at least on the right track?

[tex]\rho_b=-\nabla\textbf{P}[/tex]
[tex]\rho_b=-\nabla kr\hat{r}[/tex]
[tex]\rho_b=-k (\hat{r}\frac{d}{dr}+\hat{\theta}\frac{1}{3}\frac{d}{d\theta}+\hat{\phi}\frac{1}{r\sin\theta}\frac{d}{d\phi}) (r\hat{r})[/tex]
[tex]\rho_b=-kr[/tex]

I'm still not so sure if my vector expression [tex](r\hat{r})[/tex] is correct or if there's supposed to be a couple of theta's and phi's in there :frown:
 
  • #6
jtbell
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The thing is, I don't KNOW if the polarization is radial...
The problem statement says, "P(r)=kr where k is a constant and r the vector from the centre". A vector from the center of the sphere is radial, by definition.

[tex]\rho_b=-\nabla\textbf{P}[/tex]
You need the divergence, not the gradient:

[tex]\rho_b=-\nabla \cdot \textbf{P}[/tex]

Divergence in spherical coordinates

You do have the correct form for P:

[tex]\textbf{P}=kr\hat{r}[/tex]
 
Last edited:
  • #7
Cyosis
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The equation for [itex]\rho_b[/itex] is not correct. It should be [itex]\rho_b=-\nabla \cdot \vec{P}[/itex]. Now look up the divergence in spherical coordinates and write down the polarization like this [itex]P(\vec{r})=kr \hat{r}[/itex].
 
  • #8
jtbell
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Returning to the bound surface charge, you have

[tex]
\sigma_b=\textbf{P}\hat{n}
[/tex]

As diazona noted, this should be a dot product:

[tex]
\sigma_b=\textbf{P} \cdot \hat{n}
[/tex]

You have [itex]\textbf{P}=kr\hat{r}[/itex]. Now, what is [itex]\hat{n}[/itex] on the surface of a sphere, in spherical coordinates?
 
  • #9
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The problem statement says, "P(r)=kr where k is a constant and r the vector from the centre". A vector from the center of the sphere is radial, by definition.
*doh* :redface:

Thanks...

You need the divergence, not the gradient
Oh, ok...So I've been writing it as the dot product (divergence) but treating it as the gradient. Well done phyz! *sigh*

You have [itex]\textbf{P}=kr\hat{r}[/itex]. Now, what is [itex]\hat{n}[/itex] on the surface of a sphere, in spherical coordinates?
[itex]\hat{n}[/itex] at the surface of a sphere would be perpendicular to the surface...mmmm...so wouldn't it simply be [itex]\hat{r}[/itex]?
 
  • #10
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Let's try again...

[tex]\sigma_b=\textbf{P} \cdot \hat{n}[/tex]
[tex]\sigma_b=kr\hat{r} \cdot \hat{r}[/tex]
[tex]\sigma_b=kr[/tex]


[tex]\rho_b=-\nabla\cdot\textbf{P}[/tex]
[tex]\rho_b=-\nabla\cdot kr\hat{r}[/tex]
[tex]\rho_b=-(\frac{1}{r^2}\frac{d}{dr}(r^2 kr))[/tex]
[tex]\rho_b=-(\frac{1}{r^2}3kr^2)[/tex]
[tex]\rho_b=-3k[/tex]

If this is right, then I think I can finally say I'm beginning to understand this (fingers crossed) :biggrin:
 
  • #11
Cyosis
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Correct!
 
  • #12
jtbell
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15,505
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One more little thing... you have

[tex]
\sigma_b=kr
[/tex]

Lower-case [itex]r[/itex] is a variable. To get [itex]\sigma_b[/itex] at the surface of the sphere, you need to substitute the value of [itex]r[/itex] at the surface of the sphere, which is...

Finally, as a "sanity check" for this type of problem, note that when you polarize an object, you don't add or subtract any charge from the object as a whole. So the total bound charge should be zero.
 
  • #13
299
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Got it!

Thanks for all the patience and help! Slowly but surely confusion is being beaten into submission with the big stick of comprehension :tongue2:

Have an excellent day! :smile
 
Last edited:

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