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.)
Best Answer
A practical plate capacitor has a part of its field bulged out of the gap and that part becomes more significant as the distance of the plates grows. The bulged part doesn't stay constant if the distance of the plates oscillate. That generates oscillating magnetic field as J.C. Maxwell wrote in his famous compilation theory of the electricity. It can also cause significant radiation if the gap is long enough and the gap oscillation frequency is high enough. Otherwise a charged but unconnected capacitor (even no wires) acts like a tensioned spring which dissipates nothing when one pulls the plates further apart.
If there's wires the system is complex. The wires are a load and there's AC current between the plates and wires. That causes losses if there's resistance and the needed mechanical oscillation frequency for substantial radiation becomes lower although not necessarily practically realizable.