Hi Peiru Li

because every conductor has a capacity, even when surrounded by free space, the metal plate 8 has a capacity, too, which depends on the size of the plate and the surroundings. Accordingly, there is a load, and the current is not exactly zero but with 50 Hz, 1200 V probably of the order of a few 100 nA.

As a consequence, the magnetic field generated by the current in the wire as well as accompanying the changing electric field will be very weak, too, probably much weaker than the geomagnetic field. So, yes, for most practical purposes you can neglect it.

The same is true for the radiation: With 50 Hz, the wavelength in vacuum is 6000 km, so as long as both you and your setup remain on the same side of earth, the space under observation is always located inside the near field where the reactive part of the field is much stronger than the very weak radiated part.

Generally, the density of the charge on the plate will have the highest values at the corners (if any) and the edges. Accordingly, near the plate the highest field strength will occur near the corners. As the distance to the plate increases, the field will increasingly resemble that of a point charge located at the center of the plate. In this distant region, the field strength will decrease proportionally to 1/r2, with r being the distance between the center of the plate and the point in space under observation. However, objects in the neighborhood might distort the field considerably.

In order to determine the exact spatial distribution of the field, I would recommend using a simulator.

*** If the plate or other parts of this setup can be touched, you should handle it with great care because touching two parts of the setup or touching one part while being connected to GND (earth) could be fatal. ***

BTW, the presence of the "feedback" 5 makes a big difference: Without it, at point 3a of the secondary winding there is virtually no load while 3b is loaded with the metal plate. Consequently, the amplitude of the potential at 3a will be much higher than the amplitude at 3b. The voltage between plate and GND will be much lower than 1200 V. With the connection 5, 3a is clamped to the potential of L or N of the power grid, so the voltage of the plate to GND will be either 1200 V (5 -> N = GND) or 980 V (5 -> L).

Peiru Li

1.: If "slower" is a typo and should read "lower", then the opposite is true:

At 3a, there is just the end of the wire, that is, a small object with low capacitance; at 3b, there is the plate, a larger object with larger capacitance. Now represent these capacitances by common capacitors between 3a and GND as well as 3b and GND; so, both capacitors are connected together via GND. Seen this way, we have a closed circuit, consisting of the secondary winding and C3a and C3b in series.

As a numerical example, assume C3a = 1 pF, C3b = 10 pF. Then Ctotal is about 0.909 pF. In order to avoid dealing with the charge, we can apply a kind of "voltage divider rule" for capacitors in series:

V3a = Vtotal * Ctotal / C3a = 1091 V.

V3b = Vtotal * Ctotal / C3b = 109 V.

In this example, the RMS voltage between the plate and GND is less than 110 V.

In case "slower" wasn't a typo: The potential at 3a as well as at 3b changes with 50 Hz, i.e. each full wave takes 20 ms, and the phase angles differ by 180°. Of course, the rate of potential change is different, near the zero crossing in the numerical example above, we have about 485 V/ms at 3a, and about 48.5 V/ms at 3b.

2.: For a quick read, you can have a look at paragraph 3.6 of this:

https://web.archive.org/web/20090226225105/http://www.phys.unsw.edu.au/COURSES/FIRST_YEAR/pdf%20files/5Capacitanceanddielectr.pdf

If you would prefer a real book, take, for example, a look at subchapter 4.2 of Jean van Bladel: "Electromagnetic Fields", 2nd ed.

3.: Sorry, I have to admit that I myself never simulated low frequency EM fields using commercially available software. Hopefully, someone else here can answer your point 3. If not, you could put a new question on RG about this specific theme.