Voltage developed across a cooled crystal

[etotheipi]

Homework Statement

See below

Relevant Equations

N/A

Here I disagreed with my professor's solution, so I'd like to check with you guys. Each B⁴⁺ ion in a perovskite ABO₃ structure is displaced by $0.2 \mathring{A}$ parallel to the $c$-axis w.r.t. the centroid of the unit cell, after the cubic-tetragonal phase transition below the Curie temperature. We're asked to determine the voltage developed across a sample of thickness $0.1 \text{mm}$, given that $\kappa = 1000$, and assuming that the unit cell dimensions of the tetragonal phase are approximately the same as in the cubic phase, $4.0 \mathring{A} \times 4.0 \mathring{A} \times 4.0 \mathring{A}$.

The dipole moment per unit volume is then just $P_z = [(4e) \times 0.2 \mathring{A}]/[4.0 \mathring{A}]^3 = 0.2 \text{ Cm}^{-2}$, and then since $\vec{P} = \chi \varepsilon_0 \vec{E} = \varepsilon_0 (\kappa -1)\vec{E}$$$V = - E_z \Delta z = -\frac{P_z}{\varepsilon_0 (\kappa - 1)} \Delta z \approx -2300 \text{ V}$$i.e. the potential of the lower face w.r.t. the top face is approximately $2300 \text{ V}$. My professor, however, says that the correct equation should be$$V = -\frac{P_z}{\varepsilon_0 \kappa} \Delta z$$This doesn't make a difference to the answer, given that $\kappa \gg 1$, but conceptually I found that slightly odd, and haven't been able to figure out why the lower equation would be correct. I would like to see if anyone agrees with me, though!

 

[TSny]

$\vec{P} = \chi \varepsilon_0 \vec{E}$

This equation applies to a linear dielectric where the polarization is due to an applied electric field. But in your crystal, the displacement of the B⁺⁴ ions is not caused by an applied field. It is forced to occur by the change in crystal structure when the phase transition occurs. This polarization is "frozen in". You have an example of an "electret".

Maybe you can approach it as follows. The polarization due to the displacement of the B⁺⁴ ions causes a bound surface charge density $\pm \sigma_0$ to occur on the slab of material. This charge density produces an electric field E_o inside the material. All the atoms in the material respond to this field by becoming polarized. This alters the field inside the material to produce a resultant field E.

I think you can relate this to the familiar example of inserting a linear dielectric between the parallel plates of a capacitor where the plates have surface charge $\pm \sigma_0$. Before the dielectric is inserted, there is a field E_o between the plates. After the dielectric is inserted, the field between the plates is altered to become a resultant field E. What is the relation between E and E_o in terms of the dielectric constant $\kappa$?

 

[etotheipi]

Interesting, thanks! I wasn't aware that $\vec{P} = \chi \varepsilon_0 \vec{E}$ has a limited realm of applicability, but it makes some sense. I'll try my best to provide solutions to the questions you posed...

I think you can relate this to the familiar example of inserting a linear dielectric between the parallel plates of a capacitor where the plates have surface charge $\pm \sigma_0$. Before the dielectric is inserted, there is a field E_o between the plates. After the dielectric is inserted, the field between the plates is altered to become a resultant field E. What is the relation between E and E_o in terms of the dielectric constant $\kappa$?

Before the dielectric is inserted, the field between the plates is$$\vec{E}_0 = \frac{\sigma_0}{\varepsilon_0} \hat{z}$$Then you insert the dielectric, and the electric field $\vec{E}_0$ superposes with the electric field due to the dielectric. The displacement field hasn't changed [by Gauss' law, since the free charge density on the plates is constant], and satisfies$$\vec{D} = \kappa \varepsilon_0 \vec{E} = \sigma_0 \hat{z} \implies \vec{E} = \frac{\sigma_0}{\kappa \varepsilon_0} \hat{z}$$And that would imply that $V = -E_z \Delta z = - \frac{\sigma_0}{\kappa \varepsilon_0} \Delta z$, which is promising...

Maybe you can approach it as follows. The polarization due to the displacement of the B⁺⁴ ions causes a bound surface charge density $\pm \sigma_0$ to occur on the slab of material. This charge density produces an electric field E_o inside the material. All the atoms in the material respond to this field by becoming polarized. This alters the field inside the material to produce a resultant field E.

The difference now, is that the surface charge density is bound, and also on the surface of the material (and not on the conducting plates sandwiching the dielectric). I have the feeling that the field inside the material is going to come out to be, again, $\vec{E} = \frac{\sigma_0}{\kappa \varepsilon_0} \hat{z}$, but I'm struggling to justify it, not being able to make use of the displacement field here.

I wondered if you could give me a hint, as to how to calculate the field produced by the induced polarisation? (which I should then be able to add to $\vec{E}_0$ to get $\vec{E}$)

 

[TSny]

The difference now, is that the surface charge density is bound, and also on the surface of the material (and not on the conducting plates sandwiching the dielectric). I have the feeling that the field inside the material is going to come out to be, again, $\vec{E} = \frac{\sigma_0}{\kappa \varepsilon_0} \hat{z}$

Yes.

... but I'm struggling to justify it, not being able to make use of the displacement field here.

I wondered if you could give me a hint, as to how to calculate the field produced by the induced polarisation? (which I should then be able to add to $\vec{E}_0$ to get $\vec{E}$)

The following is the way I thought about it. There are probably better ways.

I find it easier to go straight for the net electric field $E$. Once you get an expression for $E$, you can then find $E_{\mathbf {ind}}$ produced by the induced polarization if you want. But to get the voltage, you only need $E$.

You have the polarization, $P_0$, due to the “frozen in” displacement of the B⁴⁺ atoms and you have the induced polarization, $P_1$, due to polarization of the atoms caused by the net electric field $E$ inside the material. You have already calculated $P_0$. $P_1$ is the “typical type” of polarization in which $P_1$ is related to $E$ by the susceptibility constant $\chi$.

$P_0$ and $P_1$ produce surface charge densities $\pm \sigma_0$ and $\pm \sigma_1$ which can be expressed in terms of $P_0$ and $P_1$, respectively. Being careful with signs and using the simple geometry of a thin slab, you should be able to write an equation that relates the net field $E$ inside the material to $\sigma_0$ and $\sigma_1$. If you can express $\sigma_1$ in terms of $E$ and $\chi$, then you have an equation that you can solve for $E$ in terms of $\sigma_0$ and $\chi$.

 

[etotheipi]

Awesome! I followed those steps and arrived at what I think is the answer we're expecting... for starters, I'll choose $\hat{z}$ to point upward (from lower face to upper face), such that the "frozen in" polarisation vector is $\vec{P}_0 = P_0 \hat{z}$. This causes a surface charge density $+\sigma_0$ on the top face, and a surface charge density $-\sigma_0$ on the lower face.

Taking it in steps, this surface charge sets up an applied electric field $\vec{E}_0 = -E_0 \hat{z}$ pointing downward, resulting in the rest of the material becoming polarised with $\vec{P}_1 = -P_1 \hat{z}$, causing a further charge density $+\sigma_1$ on the lower face and $-\sigma_1$ on the upper face, and thus setting up an induced field $\vec{E}_{\text{ind}} = E_{\text{ind}} \hat{z}$ pointing upward, "partially opposing" the field $\vec{E}_0$. [N.B. This discussion of $\vec{E}_0$ and $\vec{E}_{\text{ind}}$ is not important to get the answer, but I think it helps me to see what's going on].

The "typical" polarisation is related to the total electric field $\vec{E} = \vec{E}_{\text{ind}} + \vec{E}_0$ by the equation $\vec{P}_1 = \chi \varepsilon_0 \vec{E}$. We take a Gaussian pillbox through the top face to write down$$E = \frac{\sigma_0 - \sigma_1}{\varepsilon_0}$$And since $\sigma_0 = P_0$ and $\sigma_1 = P_1 = \chi \varepsilon_0 E$, this means$$E = \frac{P_0 - \chi \varepsilon_0 E}{\varepsilon_0} = \frac{P_0}{\varepsilon_0} - \chi E \implies E(1+\chi) = \frac{P_0}{\varepsilon_0}$$or, in other words,$$E = \frac{P_0}{\kappa \varepsilon_0} \implies \vec{E} = -\frac{P_0}{\kappa \varepsilon_0} \hat{z}$$which leads to the desired result, except now I realize that I also had a sign error! The potential of the top plate is actually higher than the potential of the lower plate.

Also, just for completeness, the induced field is$$\vec{E}_{\text{ind}} = \frac{\sigma_1}{\varepsilon_0}\hat{z} = \frac{\chi P_0}{\kappa \varepsilon_0} \hat{z}$$Thanks for your help, I hope I got that right! By the way, when I went to learn a bit more about electrets, I came across some treatments like this

https://web.mit.edu/6.013_book/www/chapter6/6.3.html

which look like they assume the "frozen in" polarisation is the only possible source of polarisation (and as such, don't worry about any further induced fields/polarisations). Is that usually a common assumption to make?

 

[TSny]

Your work looks very good to me.

Also, just for completeness, the induced field is$$\vec{E}_{\text{ind}} = \frac{\sigma_1}{\varepsilon_0}\hat{z} = \frac{\chi P_0}{\kappa \varepsilon_0} \hat{z}$$

Looks right.

By the way, when I went to learn a bit more about electrets, I came across some treatments like this

https://web.mit.edu/6.013_book/www/chapter6/6.3.html

which look like they assume the "frozen in" polarisation is the only possible source of polarisation (and as such, don't worry about any further induced fields/polarisations). Is that usually a common assumption to make?

I referred to the polarization $\vec P_0$ as the "frozen in" polarization. But, the total polarization $\vec P_0+\vec P_1$ can also be thought of as "frozen in" since it will be there even when no external electric field is applied to the material. So, I think that when they refer to the "frozen in" polarization, they are referring to the total $\vec P$. Thus, it includes $\vec P_0$ and $\vec P_1$.

Thanks for the link. It appears to be a very nice online book on EM.

 

[etotheipi]

I referred to the polarization $\vec P_0$ as the "frozen in" polarization. But, the total polarization $\vec P_0+\vec P_1$ can also be thought of as "frozen in" since it will be there even when no external electric field is applied to the material. So, I think that when they refer to the "frozen in" polarization, they are referring to the total $\vec P$. Thus, it includes $\vec P_0$ and $\vec P_1$.

Ah, I see! Thanks so much! ☺