voltage to current converter using op-amp and MOS
Your paragraph 1 is correct, as it is a feedback loop. MOSFET gate voltage cannot be calculated because it is in the middle of a feedback loop, and it depends on the MOSFETs uS value.
For the same reason the Vout of the op-amp is unknown without testing and recording its behavior. The MOSFET chosen will directly affect this value. Voltage-to-current tracking is very accurate as long as the op-amp inputs are in its common mode range and the MOSFET gate voltage is within specified Min - Max range. R1 at the source-to-ground has a limit of range. Too low and too much current will flow, thus a heat or power supply issue. Too high and you might exceed the MOSFET cut-off current.
All of this works very well when compensation is good enough and R1 has a tolerance of 0.1% if possible, so Vin (AC + DC) can be associated with a current value as accurate as a good DVM.
Because there is a 10 nF (typical but arbitrary) capacitor from op-amp output to the (-) pin, a capacitor across R1 could make the op-amp unstable and oscillate. If Vin is an AC signal plus a DC offset (to keep the loop stable) then accurate current modulation is possible, limited mostly to NOT cutting OFF the MOSFET, as this kills the feedback loop. HF modulation is possible, limited to the pass-band of the op-amp, including its output to (-) capacitor, which create a reaction (to changes in load or Vin) delay of 5 us to 100 us.
The output to (-) capacitor and a 10 K resistor in the feedback path are the compensation network. In fact it will NOT be stable until you add these parts. In any case HF modulation is limited to a few KHZ. The following schematic shows a simple compensation RC network, normally all that is needed. R1 100 ohms is placed at the gate of the MOSFET so trace is very short.



simulate this circuit &ndash; Schematic created using CircuitLab