Electronic – Discharging a capacitor quicker, using an exterior time varying magnetic field

Okay, conceptually this is pretty easy, as I think you know.

A DDS chip from AD with sin/cos outputs, appropriate low pass filter, buffer amplifier. Apply a voltage much less than the bias voltage (but high enough to get good SNR) to the sample and measure the current multiplied by the sin and by the cos, low-pass filter the two results and calculate the real and imaginary components of the impedance from the (measured) voltage and (measured) current levels.

You should be able to add the bias voltage in the buffer amplifier, but you might want to capacitively couple the current input to keep the dynamic range of the mixer reasonable.

At 10MHz most precision analog multipliers are running out of steam, so I'd look at Gilbert cell mixers. Unfortunately the low-frequency and DC performance is seldom well specified.

Of course you could simply digitize the data at hundreds of MHz and digitally demodulate it with a fast FPGA, but that would be even more of a project.

The impedance of 0.156uF at 10MHz is only about 0.1 ohm, so the buffer should be able to handle tens of mA at 10MHz and your signal chain has to be happy with ~1mV total signal.

If you have access to a "lock-in amplifier" (the rack mount instrument), look at that to replace a chunk of the work. Same if you have a function generator with quadrature outputs.

I did something similar to characterize magnetic samples (there were some very special requirements) but the frequency had to be as close to DC as possible, so it was simply measured at low swept frequencies and curve-fit extrapolated to even lower frequencies (where there would be no SNR left).

It's not clear to me whether your model is primarily a series R-C or parallel R-C, of course the general measurement of Z gives you a complex number which could be applied to either model.

Your project also reminds me of some interesting work I did on conductivity cell measurements for dialysis water treatment. There were some heuristics involved.

Why is the resistor voltage initially equal to supply voltage? Is it because there is no voltage going across the capacitor yet? Therefore, as there is no voltage drop across the capacitor, all the voltage from the battery is across the resistor?

Sum of voltages on the passive elements must add up to the supply voltage.

$$ V_{supply}(t) = V_{switch}(t) + V_{resistor}(t) + V_{capacitor}(t) $$

Because of the fact that \$V_{switch}(t) = 0 \$ and \$V_{capacitor}(0) = 0 \$, \$V_{resistor}(0)\$ must be equal to \$V_{supply}(0)\$.

2.What exactly does "the voltage developed as the capacitor charges" refer to?

When you apply a voltage difference between capacitor plates, one plate has more positive potential with respect to the other one. This initiates an electric field field between the plates, which is a vector field, whose direction is from the positive plate the negative one.

There is an insulating material (dielectric material) between these capacitor plates. This dielectric material has no free electrons, so no charge flows through it. But another phenomenon occurs. The negatively charged electrons of the dielectric material tend to the positive plate, while the nucleus of the atoms/molecules shift to the negative plate. This causes a difference in the locations of "center of charge" of electrons and molecules in the dielectric field. This difference create tiny displacement dipols (electric field vectors) inside the dielectric material. This field makes the free electrons in the positive plate go away, while it collects more free electrons to the negative plate. This is how charge is collected in the capacitor plates.

3.Am i correct in assuming that the resistor voltage drops because the capacitor's voltage is increasing? (kirchoff's law where volt rise = volt drop).

As the capacitor voltage increases, the voltage across the resistor will decrease accordingly because of the Kirchoff's Law, which I formulated above. So, yes, you were correct.

1.If the capacitor's voltage is dropping(due to it being discharged), shouldn't the resistor's voltage be increasing due to kirchoff's law? Also,this should therefore INCREASE the current instead of decreasing it, which would then cause the capacitor to discharge even faster?

You are missing the fact that, the source voltage is zero (i.e.; the voltage source is missing) in the discharge circuit. Substitude \$V_{supply}(t)=0\$ in the formula above. The capacitor voltage will be equal to the resistor voltage in reverse polarities during the discharge. Together, they will tend to zero.