I am working on designing an antenna to detect 60Hz electric field. I came across the microstrip antenna being used to detect the field. Before designing the antenna I wanted to maek sure this is the right way, So I took a piece of 2 layer PCB(both the patch and ground plane were same size) of 10cmX15cm size and soldered a wire on one side and connected to a 2 Mohm resistor and measure voltage across the resistor through DAQ. I read approx. 1.7V when placed near a plugged heater. When unplugged I didn't read any. So I was positive that I am reading something on 60Hz. My question are as below.

1) Is this the right method? Am I reading 60Hz or some noise, static field, etc.?

2) I connected only one side of the PCB to resistor and measured voltage across the resistor. Do I need to necessarily ground the other plane(which is called ground plane)?

3) If I need to calculate the electric field(I have been researching on this from antenna-theory.com). By calculation, for 60Hz I would need a bigger size patch(like in kilometer range). Is this correct? If so, how come I am reading 60Hz in a small size?

4) I also looked on the other way considering the 2 sides as a plates of capacitor and FR4 material as a dielectric medium and measured the e-field. Is this the right method too? Considering that I have connected only one side of the plate and the other side is left open.

  • \$\begingroup\$ 60 Hz noise may be coming through the power lines, not through the air. Does the magnitude change when you move the board around physically (not changing where it is plugged in)? \$\endgroup\$
    – Justin
    Dec 3, 2018 at 16:47
  • \$\begingroup\$ When I keep in the air, there is a slight amount of voltage like 0.4V when I bring the plates closer(5cm gap or so) to heater wire, there is a significant voltage detected like 1.6V. \$\endgroup\$
    – Karthik
    Dec 3, 2018 at 17:07
  • \$\begingroup\$ Low frequency antennas are often designed for the magnetic field component. Some have 1 meter long coils of 10'000 turns of wire on a drain pipe. Search for VLF receiver. \$\endgroup\$
    – KalleMP
    Dec 4, 2018 at 11:53

1 Answer 1


The wavelength of 60Hz is huge. In your convenient local-lab measurements, you will be in the "near field" where the Efield acts independent of the Hfield (magnetic field).

Simply dangling a wire from a 10Meg Ohm scope probe should get you volts of 60Hz. Bring the wire near the scope's powercord. Or near any light fixture. Set the scope to 20 or 50 milliSeconds per division; you should see huge spikes near an older-style non-electronic-ballast fluorescent light, as the arc is re-struck at 120 times per second; expect 200 volts in 10 microsecond risetime.

I was thinking, over the weekend, about volts/meter of field strength for 60 Hertz. Because of such nearness, volts/meter is impossible to measure to extract power.

What you have, and what I have in my Efield interference modeling, is simply capacitive coupling between 2 wires, or from a wire in your wall or power cord, to your sensing plate.

Grounding the "other plate" will shield the backside. In this case, you have two capacitors in series, forming a voltage divider. The Efield flux will still impinge on the side facing the power wiring.


Lets model a coupling structure, of 2 wires that are 50 meters apart. Let each wire have 10cm diameter. The coupling capacitance is

EoErlength * 2*pi / ln(X * sqrt(X*X -1))

where that X is (separation ^2 - ... small terms) / (2*radius1*radius2)

and X is (50 * 50) / ( 2 * 5cm * 5cm)

X is 2,500 / (2 * 0.05 * 0.05) = 2,500*20*20/2

X is 500,000

natural_log (500,000 + 500,000) ~~ 2.3 * 6 = 14

The capacitance(per meter of receiver wire) is

~~ 50pF/meter / 14 = 3.5 picoFarad.

Technically, the mechanical topology is NOT 2 infinite wires, but one infinite (Power line) wire and your very short wire, thus more like the coupling between a wire and a small blob.

But lets use the 3.5 picoFarad we have computed. Why not? We are letting your receiver metal target be 10cm diam (5cm radius) and length 1meter.

I = C * dVdT = 3.5pF * 377rad/sec * 100,000 volts

I ~~ 1e-9 * 100,000 ~~ 1e-4 = 100uA

Across 10 Meg Ohm scope probe, expect 1e+7 * 1e-4


1,000 volts.

So its dangerous to you and to the scope, to use large pieces of metal.

  • \$\begingroup\$ Thanks for the explanation. I have a efield meter which uses pcb with a copper plate as the sensor. I dont see 2 plates though. But It can detect from 50meters though from the powerlines. Is this the same way the meter is working too? \$\endgroup\$
    – Karthik
    Dec 3, 2018 at 20:12
  • \$\begingroup\$ Without a schematic, and drawing of the sensor-plate, I cannot even guess. \$\endgroup\$ Dec 4, 2018 at 14:11
  • \$\begingroup\$ This makes sense as the wire dia is 10cm which is equivalent to power lines. I dont have a schematic though. \$\endgroup\$
    – Karthik
    Dec 5, 2018 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.