If one of the plates of the capacitor is removed, force acting on the same particle will become : Electric field between the oppositely charged plates of a capacitor is twice of that due to one plate. Hence, when one plate is removed, the electric force reduces to half of its earlier value. Was this answer helpful?
When the plate area of a capacitor increases, what happens to the capacitance? Capacitance is proportional to the plate area. Thus any increase on the plate area shall increase the capacitance.
The potential in the plate capacitor decreases linearly from the positively charged to the negatively charged plate. To express the electric field using the known voltage , the spatial derivative of the potential (gradient equation) is used (in the one-dimensional case):
This means that the force between the plates of the capacitor, which depends on the potential difference across the plates, is increased which in turns means more external work need to be done in separating the plates.
How will the magnetic field influence a capacitor rotating or not. At low f nada. Only the conductors with inductance can induce a voltage from the B field. But the inductance of free space is small so the coupling will also be small but rotating the conductors does not affect the inductance so the rotation has no effect expected.
Until you get to the 1/4 wavelength frequency of the path between the capacitor. Then any wobble will or modulation of capacitance will cause noise if erratic or if the gave is small enough to cause arcing. I discovered this using 928MHz Tx on a power meter with a rotating disc on an magnetic air bearing .