@PM,
You are posting results faster then my brain can follow 
Here is my reply to your post #422.
You said That is not true, you do measure some volts across C. The voltage across a capacitor is given by the time integral of the current divided by the C. You do measure the current but it is bug****d by the ringing. You do give the mean value of 1.348mA over a time period of 811nS (forget the rms value of the ringing, that is noise we do not want). The i*t integral value is 1.093E-9, and dividing this by your C value yields a voltage charge of 1.031mV that is compared to your measured value of 4.381mV. So your measured value that you say is really zero is not zero, it is quite significant. Dividing that i*t by the 680pF value gives a voltage charge of 1.608V which is close to your measured 1.478V. I think if you had a higher value for your Z component, maybe even the same value as C, you would get rid of that ringing noise and observe voltages that better agree with the measured i*t integral.
As regards your IMO that is entirely false reasoning. We do see current flow between them and that yields the different voltages. If you rotate the ring core 90 degrees so that both C and Z are outside the ring you still get the voltages across both due to non zero current flow, both capacitors get charged to different "voltage potentials". And the sum of those two potentials in series is the single turn voltage from the flux change in that closed loop.
Smudge
Smudge,
My previous test and the one that follows were taken in your #1 position with 'C' in the core and the 680pf is outside the core. The probe connections and ID's in the following are the same as before for that position. I apologize that some of my previous measurements were not as complete and detailed as they could have been.
So, I will respond to your post above by demonstrating the following.
P1 below shows the initial current seen in the probe while in the position between 'C' and the 680pf cap at the application of voltage to the 20T primary. From CH2(pnk) we see a peak voltage of 1.772 is reached with and avg current of 19.76ma over 94ns. If we assume we are charging the 680pf cap during this time, that would require a mean current of di=1.772*680e-12/94e-9=12.8ma. We see that we have measured an average current of 19.76ma which when considering the stray capacitance from leads, etc, this is reasonably close.
OTH, if we assume we are charging the 1.06uf cap during this time, that would require a mean current of di=1.772*1.06e-6/94e-9=19.98A.
P2 and
P3 show the mean voltage measurements across 'C' on CH2(blu) prior to and after stabilization of voltage application to the primary of 900.9uV and 1.17mV respectively. This is a differential of 269uV. These measurements were not taken properly in the initial posting.
Now we come to the elephant in the room! A 42uH inductor L1 is added in parallel to the 680pf cap. All other connection remain the same except the current probe on CH4(grn) now shows the current in L1.
P4 shows the voltage drop across the 680pf cap with CH3 of 130.3mV and
P5 shows the voltage drop across 'C' with CH2 of 138.6mV.
P6 shows us that L1 reaches a peak current of 91.79ma over this same time period of 2.914us. The energy in L1 is Ul1=.09179^2*42e-6/2=177uJ .
Now, let's use the starting and ending voltages taken on CH3 in
P4 of 1.446v and 1.315v respectively and apply first to the 680pf cap. U=(1.446^2-1.315^2)*680e-12/2=123pJ . Applying these voltages to 'C' yields UC=(1.446^2-1.315^2)*1.06e-6/2=192uJ .
It is obvious that the 680 pf cap is not supplying the energy to L1 but it appears that 'C' must be supplying the energy to L1. But how can this be when the voltage across 'C' appears to start the cycle at near zero volts as seen on CH2 in
P6? The answer is that 'C'
does contain the average voltage of 1.446v at the start of the cycle but the scope probe of CH2 has this voltage potential cancelled by the opposite polarity induced on the
ground lead probe via the E_Field.
The bottom line is, I stand by my claim that dielectric induction via the E-Field is real and the potential differential seen across said dielectric is produced by the aether.
Regards,
Pm