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If two parallel metal are placed very close to each other and a saw-tooth voltage is applied in the plates, how will the electric field behave in the vacuum in between the plates?

As the separation between the plates is very small, I think we can assume the field will be homogeneous.

As we know, $\vec E = - \frac{\partial V}{\partial x}$ (assuming the parallel plates are placed along the x axis), then as the voltage $V$ is a linear function of time i.e. $V(x,t)= a \cdot t \cdot f(x)$ where $a$ is some positive constant); then it should be true that $\vec E(x,t) = t \cdot g(x)$, as the defining equation for $\vec E$ doesn't contain a derivative wrt time. Now, as we are assuming the field $\vec E$ being homogeneous as the separation between plates is very small; $\vec E(x,t) = k \cdot t$ (i.e. $g(x) = k$).

Is my analysis right?

I have one point of confusion on whether the electric field will change instantaneously inside the vacuum between the plates as the voltage changes linearly in time. If it does, I think my analysis is right.

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  • $\begingroup$ I don't know why there's a close vote. This seems a valid question to me. $\endgroup$
    – Mockingbird
    Commented Sep 3, 2019 at 12:34
  • $\begingroup$ Note that questions of the form "Is this right?" tend to be poor fits for this site because the answer, yes or no, is too short to be a valid answer. Consider making the question more open ended so a proper answer can be written $\endgroup$
    – Kyle Kanos
    Commented Sep 6, 2019 at 11:25

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As long as the quasi-static approximation is valid (the plate dimensions are much smaller than the wavelength $\lambda = c/f$, where $c$ is the speed of light and $f$ is the bandwidth of the waveform), you can safely assume that $\vec{E}$ has the same time variation as $V$, the potential difference between the plates. So if $V$ has a sawtooth time variation, so should $\vec{E}$.

Realistically, a bigger concern should likely be whether the potential difference between the plates (which act as a capacitor) is equal to the voltage waveform generated by whatever circuit/signal generator you are using. At high frequencies, the impedance of the capacitor is no longer infinite as it is at DC, so you may have significant potential drops along the wires connecting your signal generator to the plates, due their resistance and inductance. The output impedance of the signal generator might be an issue at high frequencies for the same reason.

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  • $\begingroup$ Can you suggest me any resource to study quasi-static approximation? $\endgroup$
    – Mockingbird
    Commented Sep 3, 2019 at 12:54
  • $\begingroup$ I think many (if not most) texts on classical electrodynamics should have at least a brief discussion of the quasi-static approximation. Unfortunately I don't have any with me right now. Still, a little Googling will take you a long way: see for instance here, here and here. $\endgroup$
    – Puk
    Commented Sep 3, 2019 at 13:14
  • $\begingroup$ Basically, if the above condition does not hold, you will need to treat wires etc. as transmission lines, worry about impedance matching and things will radiate. In particular, there will be waves propagating along the plates with sufficiently short wavelengths, and (setting aside the issue of what the potential even is and how the E-field is related to it) you won't be able to assume a constant potential on each plate at a given time. $\endgroup$
    – Puk
    Commented Sep 3, 2019 at 13:15

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