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In the problem of a dielectric slab being inserted between the plates of a parallel plate capacitor, it is usually argued that the origin of the inward force acting on the slab is the fringing electric fields near the ends of the capacitor. I fail to see how this explanation can be correct due to the following reasons:

  1. Fringing fields can only come in as an extra term to the force

First let's assume the capacitor is charged and then disconnected so that it carries a constant charge $Q$.

One can calculate the force on the dielectric slab by taking the gradient of the energy stored in the system. This energy is found by integrating the energy density over all space:

$$ W = \int_{\mathbb{R}^3} \frac12 \epsilon_0 \epsilon_r(\mathbf{r}) \mathbf{E}^2 \, d^3 \mathbf{r} = \left( \int_\mathrm{inside} + \int_\mathrm{outside} \right) \frac12 \epsilon_0 \epsilon_r(\mathbf{r}) \mathbf{E}^2 \, d^3 \mathbf{r},$$ where I've broken up all of space into two regions: inside the capacitor plates, and outside the capacitor plates. The second term accounts for the energy stored in the fringing fields.

Assuming the ideal capacitor model, the first term can be calculated so that the total energy becomes:

$$ W = \frac12 \frac{Q^2d}{\epsilon_0 w} \frac{\epsilon_r -1}{(\epsilon_r -1)x+l} + \int_\mathrm{outside} \frac12 \epsilon_0 \epsilon_r(\mathbf{r}) \mathbf{E}^2 \, d^3 \mathbf{r},$$ where the capactior plates measure $l \times w \times d$ and $x\in[0,l]$ is the length of the part of the dielectric inside the capacitor.

Taking the gradient of this, the first term gives the usual force (which depends on $x$) and the second one would be an extra term caused by the fringe fields.

  1. If the force were due to the fringe fields, it would not depend on $x$

We've already assumed that the slab is a homogeneous dielectric. From the point of view of the fringe electric field, translating the slab further inside the capacitor, doesn't change anything. Hence the force cannot depend on $x$ assuming it is purely due to fringe fields.

  1. The fringe fields do not exist in the model

How can fringe fields explain the force when they do not even exist in the model used to calculate the force?


NB. I've looked at similar questions on here, but none of them address the issues I've put out.

Dielectric slab inserted into a constant voltage capacitor

Capacitor in a dielectric fluid

Why does a dielectric slab move inside the capacitor?

Dielectric Slab: Force due to fringing field

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    $\begingroup$ The fringe fields do not exist in the model The fringing fields are not something that you can optionally have or optionally not have. They're required by Maxwell's equations. If the model omits them, then the model violates Maxwell's equations. $\endgroup$
    – user4552
    Dec 25, 2019 at 21:44
  • $\begingroup$ @BenCrowell The term which corresponds to the contribution to the energy $W$ from the fringe fields can be made arbitrarily small by choosing $l$ large enough. The first point I was making was that the force is generated by changing the energy stored in the not fringing field. $\endgroup$ Dec 25, 2019 at 21:54
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    $\begingroup$ When you find the usual field inside the capacitor you are neglecting all fringe effects by assumption. The neglected effects will include both "fringe" fields (i.e. outside-but-near-the-edges) and a modest reduction and redirection of the field inside-but-near-the-edges. And the same is true of the model with a partially inserted dielectric: the model assumes two distinct regions with no transition effects between them, but the real case will include such effects. In other words, as you insert the dielectric the "fringe" region will move inward with the edge of the slab. (cont.) $\endgroup$ Dec 25, 2019 at 22:39
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    $\begingroup$ Which means that saying "the simplified energy model gives the full results so there can't be a contribution from the fringe fields" is claiming that it is OK to ignore the the known incorrectness of the model with regard to the interior drop off and use the result to prove there is no effect due to the exterior non-zero fields; even though these two effects are of similar size. Not that I would want to attempt a full calculation with fringe fields. I haven't been that sharp on E&M since grad school. $\endgroup$ Dec 25, 2019 at 22:42
  • $\begingroup$ I don't see the problem here. There are two different ways of calculating the same thing, one by looking at force (which is solely due to fringe fields) and one by looking at derivative of energy (which, in the large capacitor limit, is solely due to non-fringe fields). Yes, the two different calculation methods have different intermediate steps. So what? Maxwell's equations are self-consistent, it doesn't matter how you got to the answer. $\endgroup$
    – knzhou
    Dec 26, 2019 at 0:13

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