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In a parallel-plate capacitor filled with a dielectric, if the plates are kept at a potential $V$, and the dielectric has some (non-linear) polarization $P$ that is dependent on the magnitude of $E$ (so $P$ is a vector of which the magnitude is some function of $E$, for simplicity sake we can say it is in the direction of $E$), then the electric field applied to the space within the plates is $E = V/d$.

Then, from this the E-field inside the dielectric can be calculated, which opposes this applied $E$ field.

However, what I don't understand is, why is this not done recursively? Why is the resulting $E$ field not the one that again determines the polarization, which then again determines the dipole moments and thus how the E-field inside the dielectric behaves? Like finding an equilibrium? But in all exercises on this it is just done naively, once. I don't understand this physically.

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  • $\begingroup$ Well, one way to look at it is that you can perform the infinite sum and set the total difference to be the polarisation. $\endgroup$ Commented May 9, 2023 at 5:21

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This is a good question. It is not "recursive" because in the equation
$$\mathbf D= \epsilon_0 \mathbf E + \mathbf P \tag{1}\label{1}$$ and in the constitutive equation $$\mathbf P = \mathbf P(\mathbf E)\tag{2}\label{2}$$ the field $\mathbf E$ is defined to be the end-result of the implicit recursion you are alluding to. In other words, it implies an electrostatic & thermodynamic equilibrium that has been established after all the dipoles will have aligned properly so that the macroscopic fields $\mathbf {\{D, E, P\}}$ so that $\eqref {1}$ and $\eqref{2}$ are satisfied along with all the macroscopic boundary conditions. Note that the microscopic field of which there is only one, say $\mathfrak E$, with $\mathbf E$ being the macroscopic average of $\mathfrak E$, satisfying the Maxwell equations $\nabla \times \mathfrak E = 0$ and $\nabla \cdot \mathfrak E = \rho/\epsilon_0$ everywhere, does not have to satisfy the macroscopic boundary conditions on the dielectric, such as the continuity of $\mathfrak E_t$ or $\mathfrak E_n$.

When calculating the energy/work relationship in electrostatics it is "derived" that spatial energy density is $\mathbf E \cdot d \mathbf D$. In fact, this is not the internal energy but rather the (Helmholtz) free energy at some temperature where the temperature dependence is expressed in the temperature dependent constitutive relationship $\mathbf P = \mathbf P(\mathbf E, T)$ because it is related to the infinitesimal isothermal work needed to create and maintain the electrostatic configuration. Actually it is better to consider the full volume integral $\delta w =\int_{\infty} dV \mathbf E \cdot \delta \mathbf D$ as being representative of that work as the amount it takes to change the displacement vector field by $\delta \mathbf D$. Interestingly and significantly, using some standard vector analytical identities this integral taken over all space can be transformed to another one $$\delta w = \delta \frac{1}{2}\int_{\infty} dV \mathbf E_0^2 - \int_{\mathcal V}dV\mathbf P \cdot \delta \mathbf E_0 \tag{3}\label{3}$$ where $\mathcal V$ is the space occupied by the dielectric and $\mathbf E_0$ is the vacuum field without the dielectric, and in this form $\delta w$ is the electrostatic work it takes to maintain the sources (free charges) of $\mathbf E_0$ to be the same amount at the same places as it was before the dielectric was inserted.

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  • $\begingroup$ Thanks, this answer elucidates the matter, and surprises me. I have used a wrong interpretation where the E in P(E) is the vacuum field. $\endgroup$
    – buddhabrot
    Commented May 9, 2023 at 12:12
  • $\begingroup$ If you have a vacuum energy like E = V/d, because two plates are put at potential diff V at distance d, and I know the formula for polarization P(E), how then do I have to calculate the E field? $\endgroup$
    – buddhabrot
    Commented May 9, 2023 at 12:45
  • $\begingroup$ Many years ago I made the same mistake as the one you are alluding to, specifically, in the magnetic case ( H v. E and B v. D). While nowadays this issue is better discussed in most books it is rare to find a fully detailed analysis of it. Probably the first ones are in Stratton or Guggenheim or Heine but few read those, see similar discussion here and here $\endgroup$
    – hyportnex
    Commented May 9, 2023 at 12:45
  • $\begingroup$ If it is a "simple" material then there is a local macroscopic relationship between the field $E$ and the induced polarization $P$. The relationship can be linear (e.g., glass) and then there is a susceptibility coefficient $\chi = \chi (E)$, or relative permittivity $\epsilon_r = 1+ \chi$, that gives you P (or D). So the question is how to get $E$ from $E_0$? That is quite complicated problem unless the geometry is simple. The parallel plate capacitor with negligible fringing field has a simple geometry where you can make the assumption that $E\approx E_0 = V/d$ and then $P = \chi V/d$. $\endgroup$
    – hyportnex
    Commented May 9, 2023 at 13:17
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    $\begingroup$ So it is a nonlinear dielectric $P=\chi(E)E$ with $\chi = \chi_0 + \chi_2 E^2+....$. As long as the geometry is simple, such as a parallel plate condenser with negligible fringing fields, you can still write $E \approx E_0 = V/d$. $\endgroup$
    – hyportnex
    Commented May 9, 2023 at 13:51
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When puzzling over problems like this, you should always consider how you would do the experiment. Here, the dielectric constant is a parameter you measure by making a capacitor and determining the polarization by measuring the capacitance. So, the experiment finds the equilibrium you seek.

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  • $\begingroup$ I'll update the question because the polarization is a formula and I'm looking for a theoretical solution, not an experimental one. $\endgroup$
    – buddhabrot
    Commented May 8, 2023 at 17:50
  • $\begingroup$ @buddhabrot Physics is not mathematics. Every physical question is fundamentally about the results of experiments and observations. Polarization is a theoretical model crafted to match experimental results. $\endgroup$
    – John Doty
    Commented May 8, 2023 at 17:55

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