How to determine electric field polarity from electric field mill signal?

Question

I have built an electric field mill like this, more info can be found in the (short) wiki article:

It works by exposing the sensing electrode to the ambient electric field, which (depending on E field polarity) will induce a positive or negative charge. The sensing electrode is the covered by the shutter (which is grounded,) thus blocking the field and allowing the collected charge to drain to some sensing instrument. In my case, I'm using an oscilloscope.

Here is what the signal looks like before and after I statically charge a PVC pipe and put it above the mill:

Pretty cool stuff. However, I have yet to figure out how exactly I can determine the polarity of the field. I have tried making a large plate capacitor out of sheet metal and applying both negative and positive high-voltage to simulate +/- electric fields. However, the signals do not seem to differ by their phase, which I'm assuming has something to do with the polarity of the field.

I have seen some field mill designs incorporate a photogate to determine when the sensing plate is exposed/not exposed, but no mention of how the position of the shutter relates to the sinusoidal output signal.

Answer 1

When the shutter is starting to uncover the sensing electrode, the voltage on that electrode will increase if the E-field is pointing down toward the mill (as shown in your diagram) and decrease if the E-field is pointing up away from the mill. This diagram on Wikipedia illustrates this visually.

As the sensing electrode is getting covered up, the observed voltage will show the opposite (plus/minus any DC offset).

You can align your photogate to mechanically coincide with the sensing electrode becoming uncovered - you should expect triggering pulses to coincide with positive peaks (for E-field pointing down) or negative peaks (for E-field pointing up).