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I was recently looking at a couple of Q&A's on this site for a description of capacitive pickup of 60 Hz by an oscilloscope probe:

I wanted to do a quantitative analysis of this, so I made the following model:

schematic

V is 110 V. V1 is the voltage measured by the scope. Z2 is the capacitive impedance of the coupling of the probe to the power lines. For example, if I'm touching the probe, then this is coming from the capacitor consisting of my body as one plate and the wiring inside the wall as the other. Z1 is the impedance of the probe. I take Z2 to be made out of a capacitance C=1 pF, while Z1 is a resistor R=10 MΩ in parallel with a 10 pF capacitance.

So if I analyze this as a voltage divider, I get V1/V=1/(1+Z2/Z1)≈Z1/Z2. At 60 Hz, the probe's impedance comes out to be almost purely its resistance R, so V1=iωRCV, or ~400 mV, which is the right order of magnitude compared to what I observe.

The reason I'm posting this as a question is that I got into a discussion in comments where we don't seem to be agreeing on the analysis, and it doesn't seem like the kind of thing that is going to be easy to clarify within a comment thread.

In this answer, wbeaty says:

You're seeing a capacitive divider effect. One capacitor is within the scope probe and essentially is connected from the tip to the scope ground. The other much smaller capacitance is across empty space: from the probe tip to any distant 220vac wires in the walls. The empty space around the probe is the dielectric of this capacitor.

In a comment, I said:

I could be wrong, but it seems to me that this is not really a capacitive divider. The scope probe acts like a 1 Mohm resistor in parallel with, say, a 20 pF capacitor. So at 60 Hz, the probe is acting like a resistor, not a capacitor.

Wbeaty replied in a comment:

The Thevenin Equiv of the capacitive divider is 20pF in series with distant 220VAC source, all in parallel with 10Megs as you say ...AS WELL AS THE DIVIDER EFFECT. Perform calcs for three components. Suppose there is 0.01pF between the distant 220VAC line and the probe tip. That gives a 1/2000 divider effect, in addition to a highpass-filter attenuation of 1:15.9 at 50Hz. (In other words, what's the Thevenin equiv. for a capacitive divider using 0.01pF and 20pF? Then place 10Meg on its output, to create a hipass.) Apply 220VAC, and the scope w/floating probe will measure 6.9mV

I don't understand this comment. Can anyone clarify whether one of us is the right and the other person is wrong? Are we using different models, or the same model? I can't tell from wbeaty's verbal description whether he has in mind the same schematic I do.

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  • \$\begingroup\$ Why are you making a question out of this when your discussion is with @Wbeaty ? \$\endgroup\$ – Andy aka May 26 '20 at 13:43
  • \$\begingroup\$ Is the scope connected to a grounded outlet? \$\endgroup\$ – Justme May 26 '20 at 13:49
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    \$\begingroup\$ @Andyaka: For example, it's not possible to draw schematics in comments. There's just a limit on how much discussion you can do in comments. \$\endgroup\$ – Ben Crowell May 26 '20 at 14:12
  • \$\begingroup\$ @Justme: Yes, but I don't think that matters for this analysis, does it? \$\endgroup\$ – Ben Crowell May 26 '20 at 14:12
  • \$\begingroup\$ @BenCrowell Good thing it is grounded. It would matter a lot if it is ungrounded, and it would because it changes the analysis. A grounded scope has it's chassis at 0V potential, so connecting probe tip to ground measures 0V. An ungrounded scope (which is not safe to you or devices you are measuring) has half of mains voltage AC capacitively coupled to chassis, so connecting probe tip to ground measures half of mains voltage AC. \$\endgroup\$ – Justme May 26 '20 at 18:39
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AT LOW FREQUENCIES, capacitive coupling encounters the DC_blocked nature of the "schematic" that included the resistance of the scope probe (or the resistance of the circuit), and the High Pass Filter response causes additional attenuation.

AT HIGH FREQUENCIES, capacitive coupling produces a broadband waveshape-preserving behavior.

ALSO

A serious confusion arises when nearby pieces of metal exist.

The coupling changes from dipole (wire to wire) coupling, to wire over a sheet.

Or to wire to wire over aheet.

And if you have large masses of metal in the aggressor, and large masses of metal in the victim, the coupling changes again (to something well predicted by parallel_plate math.)

summary: the frequency matters, and the size and shape of metal matters.

================================================

assume parallel_plate model ofcoupling

assume the probe is over the metal in your bench

size is 1mm by 40mm (1.6 inches)

assume the aggressor/transmitter is 1 meer awy, in the ceiling overhead

Let us compute a capacitance, and then a displace mnt current.

C = eo * er * area / distance

C =~~~ 9 pf/meter * (1mm * 40mm) / 1,000 mm

C = 360 pf/meter * 1/1,000 * 1mm * (1meter/1,000mm)

C = 360/1,000,000 pf = 0.36 femto farad [ 0.36 e-15 farad]

with whatever pretenses to accuracy we wish to proclaim.

Now I = C * dV/dT = C * ( 120 * 1.414 * 377 radians/second )

I = 0.3e-15 * (200 * 300) = 0.3e-15 * 60,000 = 18 pico Amps at 60 Hz

(we'll forbid any florescent light spikes on the power waveform)

Now you have to decide on the system frequency response. High Pass Filter?

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  • \$\begingroup\$ Note that the original UK situation was: grounded scope with 1X probe being waved in the air, with 220VAC distant wires at 50Hz. The scope 'mysteriously' displays a 50Hz sine wave when the probe-tip is floating. I seat-of-pants speculate a coupling capacitance to line voltage through 100cm of air as 0.01pF, and a scope Zinp of 10Meg and 20pF. The scope then sees ~7mV, which is a combination of 1:2000 capacitive divider, plus a high-pass filter effect which further attenuates the 50Hz. OP doesn't understand "capacitive divider." \$\endgroup\$ – wbeaty May 27 '20 at 0:23
  • \$\begingroup\$ @wbeaty: I do understand what a capacitive divider is. I simply think your analysis is wrong, for the reasons explained in the question. \$\endgroup\$ – Ben Crowell May 28 '20 at 13:30
  • \$\begingroup\$ @BenCrowell >reasons explained... Which reason specifically? Please explain. (e.g. do you believe that adding a 10M load to a capacitive divider circuit will eliminate the capacitive divider effect?) \$\endgroup\$ – wbeaty May 28 '20 at 14:18

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