# Does continuously moving dielectric changes capacitance?

Let's assume we have a standard two-plate capacitor and the space between its plates is filled with dielectric that has permittivity >> 1 . Will its capacitance change if we move such a diеlectric in the same plane, as capacitor's plates (i.e. parallel to and between its plates) at constant speed? Will it change if we move the dielectric with acceleration?

Again: dielectric in this case never leaves the capacitor's planes and continuously moves between them. For instance, it is a mechanically driven endless belt between the plates. Or dielectric fluid, pumped between the plates.

## 1 Answer

The answer depends on the velocity of the dielectric, for which the following cases can be distinguished:

1. The dielectric moves so slowly that the relaxation time $$\tau$$ of its atoms in the electric field is negligible compared to the time they need to enter it.

In this case, the capacitor and the dielectric behave as if the latter is not moving. Depending on the exact speed of the dielectric, the moving polarisation charges might lead to stronger or weaker currents which create magnetic fields. If you do not want to neglect these, you can use the dielectric function $$\varepsilon(\omega)$$ at frequency $$\omega = 0$$ of the dielectric to calculate the polarisation density as $$\vec P = \varepsilon_0 (\varepsilon(0) - 1) \vec E~,$$ where $$\varepsilon_0$$ ist the dielectric constant of the vacuum and $$\vec E$$ is the capacitor's electric field (without the dielectric). Taking a vector $$\vec n$$ perpendicular to a surface, this gives you the surface charge density $$\sigma = \vec n \vec P~.$$ Multiplying with the velocity of the dielectric yields the electric current density, which can be plugged into the law of Biot-Savard to calculate the magnetic field.

Once the equilibrium state has been reached, both the electric and the magnetic field will be static, so there is no need to account for any interaction between them, unless the dielectric accelerates.

2. The dielectric moves very fast and the time a single atom spends in the capacitor's field is negligible compared to $$\tau$$.

Since the dielectric is uncharged and has no time to become polarized before leaving the capacitor again, the latter will behave as if in a vacuum. This stays true if the dielectric accelerates without transition to case 3.

3. The time a single atom spends in the capacitor's field is comparable to $$\tau$$.

To cover this case quantitatively, I suppose you will have to use time dependent quantum mechanics to calculate the average (over time) density of polarized atoms at every point (in space) between the plates of the capacitor, while also taking into account the interaction with the charges on the plates using Maxwell's equations. This sounds messy and at least if the dielectric does not happen to be hydrogen the solution will only be an approximation because as far as I know for more complicated atoms there are no exact solutions known for the electrons' wave functions. An accelerating dielectric would only make this more complicated.

• Thank you very much for a great idea to transfer a problem into a frequency domain. From the answer it appears that at relatively low frequencies (dielectric speed) the capacitance fall will be proportional to dielectric speed, if I understood it right. Dec 25, 2021 at 16:00
• @bear2000 This is not what I wrote. According to my answer, in case 1 (verly low speed of the dielectric), the capacitance does not change at all with the dielectric's velocity (which is assumed to be constant during the time of observation and changes between measurements). In case 3, however, the dependence of the capacitance on the velocity is not necessarily linear or affine. You can of course try to Taylor expand it, but even then the coefficient of order 1 could be 0.
– sim0
Dec 25, 2021 at 17:31
• Thanks. So, there should be always a lag between applied field and time it takes dielectric to polarize. It shall apply to any frequency, even if the lag is femtoseconds – it does exist, isn’t it? If dielectric is not entirely uniform then any difference between polarization and relaxation time shall produce change in permittivity and hence capacitance. Besides: patents.google.com/patent/US5672831A/en Surely they don’t measure fluids flow at extreme cosmic speeds – so how does that work then? Dec 26, 2021 at 1:38
• IN the above patent... as long as it ever been implemented and worked, of course... Could it be a triboelectric effect for instance, if both the pipe and fluid are dielectrics, so the induced charge changes the capacitance? Hm... In any case, one more time thanks a lot for a given answer and explanation! Dec 26, 2021 at 1:53