- #71

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Even 17Charles Link said:Yes=posts 44 and 49.

Ah! and ##E_{in} = -\frac{\vec{P}\hat{z}}{3 \epsilon_0}## ?

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In summary: Not trusting yourself could be a problem because you will be living 100% of the rest of your life with yourself.

- #71

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Even 17Charles Link said:Yes=posts 44 and 49.

Ah! and ##E_{in} = -\frac{\vec{P}\hat{z}}{3 \epsilon_0}## ?

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- #72

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@EpselonZero Please read the last couple sentences of post 63 by @kuruman where the boundary conditions are summarized for the case where ## q_{free}=0 ##.

With the first condition ## D_{out \, perpendicular}=D_{in \, perpendicular} ##,

we get, (note ## D_{in \, perpendicular}=\vec{D}_{in} \cdot \hat{n} ##, etc., and ## \vec{D}_{in}=\epsilon_o \vec{E}_{in}+\vec{P}_{in} ##, etc.)

## \epsilon_o E_{out \, perpendicular}=\epsilon_o E_{in \, perpendicular}+\vec{P} \cdot \hat{n} ##,

because ## \vec{P}_{out}=0 ##.

Note ## \vec{P} \cdot \hat{n}=\sigma_p ##.

This is consistent with everything else that we have calculated.

Working the problem with ## D's ## rather than ## E's ## really doesn't give us anything extra.

With the first condition ## D_{out \, perpendicular}=D_{in \, perpendicular} ##,

we get, (note ## D_{in \, perpendicular}=\vec{D}_{in} \cdot \hat{n} ##, etc., and ## \vec{D}_{in}=\epsilon_o \vec{E}_{in}+\vec{P}_{in} ##, etc.)

## \epsilon_o E_{out \, perpendicular}=\epsilon_o E_{in \, perpendicular}+\vec{P} \cdot \hat{n} ##,

because ## \vec{P}_{out}=0 ##.

Note ## \vec{P} \cdot \hat{n}=\sigma_p ##.

This is consistent with everything else that we have calculated.

Working the problem with ## D's ## rather than ## E's ## really doesn't give us anything extra.

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- #73

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Since, ##\vec{E}^{out}_\perp - \vec{E}^{in}_{\perp} = \frac{\sigma_p}{\epsilon_0}##

And we know that ##\vec{E}_{in} = -\frac{P \cdot \hat{z}}{3 \epsilon_0}##

Then, ## -\frac{P}{3 \epsilon_0} + \frac{\sigma_p}{\epsilon_0} = \frac{2P}{3\epsilon_0}##

Earlier, I found ##\vec{E}_{out} = \frac{2\vec{P} \cdot \hat{r} R^3}{\epsilon_0 r^3 3}##

Since we evaluate ##E_{out}## really close to the sphere where r --> R

I have

##\vec{E}_{out} = \frac{2\vec{P} \cdot \hat{r}}{3 \epsilon_0}##

It sounds fine, but I have to deal with those vectors now.

- #74

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@EpselonZero Looks like you are making progress. Please see post 61 with the "links" there. I think you also saw the calculations I added to post 72, but please take another look. In any case, I do think you are making progress. :)

See also again https://www.physicsforums.com/threa...iformly-polarized-sphere.877891/#post-5513730 especially posts 2 and 7. You should be able to start to follow these discussions in detail. You are really in about the same place these other students were a couple of years ago.

See also again https://www.physicsforums.com/threa...iformly-polarized-sphere.877891/#post-5513730 especially posts 2 and 7. You should be able to start to follow these discussions in detail. You are really in about the same place these other students were a couple of years ago.

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- #75

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For this equation (and the others that follow) you must write the left-hand side as a scalar quantity, not as a vector. Pick a point on the sphere at angle ##\theta## from the z-axis and relate the "in" and "out" normal field components (which are scalars) to the surface charge density at that angle. Same with the tangential components except that the right-hand side is zero.EpselonZero said:Ah, In this case

Since, ##\vec{E}^{out}_\perp - \vec{E}^{in}_{\perp} = \frac{\sigma_p}{\epsilon_0}##

##\dots##

- #76

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One thing that is very important in this problem of the polarized sphere is the result that the electric field when the sphere is polarized to the right is ## E_p=-P/(3 \epsilon_o) ##, i.e. it is uniform inside the sphere and points to the left with a factor of ## 1/3 ##. (This results from the surface polarization charge density of ## \sigma_p=\vec{P} \cdot \hat{n}=P \cos{\theta} ##).

This result can even be used to solve for the electric field when there is an applied field on a dielectric sphere of ## E_o ##. . The Legendre method is important, but this simple result that comes out of the Legendre method is a very important one.

It may be worthwhile to also look at the electric field inside a flat slab that is polarized, (e.g. spontaneous polarization). This does not need Legendre methods to calculate, and the result with ## \sigma_p=\vec{P} \cdot \hat{n}=\pm P ## is ## E_p=-P/\epsilon_o ##. Note that the ## E_p ## is the same from the ## P ##, even when the ## P ## is the result of an applied electric field ## E_o ##. We can write ## E_i=E_o+E_p ##, and ## P=\chi \epsilon_o E_i ## for linear materials. The result with a little algebra is ## (1+\chi)E_i=E_o ##, so that ## E_i=E_o/\epsilon_r ## for a dielectric slab with an applied ## E_o ##. Note that ## \epsilon_r=1+\chi ##, where ## \epsilon=\epsilon_o \epsilon_r ##.

A similar calculation can be done for the dielectric sphere with an applied electric field using the ## 1/3 ## factor result.

This result can even be used to solve for the electric field when there is an applied field on a dielectric sphere of ## E_o ##. . The Legendre method is important, but this simple result that comes out of the Legendre method is a very important one.

It may be worthwhile to also look at the electric field inside a flat slab that is polarized, (e.g. spontaneous polarization). This does not need Legendre methods to calculate, and the result with ## \sigma_p=\vec{P} \cdot \hat{n}=\pm P ## is ## E_p=-P/\epsilon_o ##. Note that the ## E_p ## is the same from the ## P ##, even when the ## P ## is the result of an applied electric field ## E_o ##. We can write ## E_i=E_o+E_p ##, and ## P=\chi \epsilon_o E_i ## for linear materials. The result with a little algebra is ## (1+\chi)E_i=E_o ##, so that ## E_i=E_o/\epsilon_r ## for a dielectric slab with an applied ## E_o ##. Note that ## \epsilon_r=1+\chi ##, where ## \epsilon=\epsilon_o \epsilon_r ##.

A similar calculation can be done for the dielectric sphere with an applied electric field using the ## 1/3 ## factor result.

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- #77

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I'm mentioning these results here and trying to emphasize them, because with the Legendre method it is so easy to get lost in the mathematics. It is important to take notice of these results that come out of the Legendre method=i.e.for the sphere and the cylinder. Hopefully the OP @EpselonZero sees this post and the previous one, and takes note of these important cases.

- #78

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I would like to add something about the original statement of the problem, and the boundary conditions. Normally for this type of problem, I think the boundary conditions are to have ## V_{in}=V_{out} ## everywhere on the boundary, and that ## E_{out \, perpendicular}-E_{in \, perpendicular}=\sigma_{total \, boundary}/\epsilon_o ##. (The electric fields are computed from derivatives of the potential ## V ##). Note ## \sigma_{total}=\sigma_{free}+\sigma_p ##.

Using ## D's ## instead of ## E's ##, we get ## D_{out \, perpendicular}-D_{in \, perpendicular}=\sigma_{free \, boundary} ##.

The tangential components for ## E ## will be the same if ## V_{in}=V_{out} ## everywhere on the boundary, and this last condition is very necessary. Perhaps @kuruman can add to this.

Using ## D's ## instead of ## E's ##, we get ## D_{out \, perpendicular}-D_{in \, perpendicular}=\sigma_{free \, boundary} ##.

The tangential components for ## E ## will be the same if ## V_{in}=V_{out} ## everywhere on the boundary, and this last condition is very necessary. Perhaps @kuruman can add to this.

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- #79

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I agree 100%. The only addition I have is that the continuity of the potential across the boundary is preferred when one crafts solutions by first finding the potential (e.g. the Legendre polynomial method) whilst the equivalent continuity of the tangential component of E is preferred when one has to deal with time-varying EM fields at interfaces (e.g. EM radiation).Charles Link said:

Using ## D's ## instead of ## E's ##, we get ## D_{out \, perpendicular}-D_{in \, perpendicular}=\sigma_{free \, boundary} ##.

The tangential components for ## E ## will be the same if ## V_{in}=V_{out} ## everywhere on the boundary, and this last condition is very necessary. Perhaps @kuruman can add to this.

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