# Sinusoidally squeezing a capacitor

Assuming no load, how does continuously changing the distance between the plates of a capacitor (in a sinusoidal fashion for simplicity) affect the output voltage? Does electromagnetic induction come into play in this scenario?

I understand there will be an immediate change in the output voltage since $$V = Q/C = \frac{Q}{\varepsilon}\frac{d}{A}$$ and clearly the permittivity constant will change sinusoidally itself.

But what about the fact that there is a time-changing electric field? We must have a perpendicular time-changing magnetic field as a result. Does this magnetic field affect the voltage across the plates?

If not, I believe that would mean the energy spent in changing the distance would be dissipated purely thermally.

(For clarity, this is not a question from an assignment or such.)

• In practical terms that depends on how fast you do it... eg, yes, there is, but if you do this with any sort of ordinary motor driven apparatus it's probably not significant. Given one can pick up mains AC fairly easily it's probably detectable, with careful mechanical design you could even reach VLF frequencies, but beyond that... Oct 21, 2020 at 14:31
• Not sure - but this is the operating principle of electret microphones, so possibly the modelling of those may have some hints Oct 21, 2020 at 14:40
• Back in the day, there were rotary variable capacitors (usually used for trimming or mechanical tuning). These varied A instead of d, to the same result. Remember farads are coulombs per volt, so if the charge on the cap isn't changing, the voltage will vary inversely with the capacitance. Interestingly, if you reduce C by changing the distance, the voltage will increase, but the field strength will stay the same (volts per meter). Oct 21, 2020 at 15:53