it once happened that when I touched the positive terminal of the Oscilloscope input probes letting its negative terminal open, I observed sinusoidal voltage signals with 0.5V peak to peak and of frequency nearly 50Hz. I don't understand the exact cause for these signals. When I touched the other, negative terminal output signal reached zero with a very little noise...(with the positive terminal connected to the body..) Is it because the power lines near the place I observed these are 50Hz and those induced a potential difference across me with respect to the supply ground which is also the ground of Oscilloscope... or something else.. When I touched a wire or a resistor with positive terminal of Oscilloscope.. there was no change in the Output.. Its because the equivalent model of our body includes capacitance and inductance with resistance.. If so then actually how we are being induced with such voltage signal..?
Actually the discontinuity in signal is because of the difference between the frequency of actual signal and sampling frequency of my camera.enter image description here

  • \$\begingroup\$ Change the trigger source to "Line" (at the right of your scope), you will trigger synchronous to the mains grid frequency. If you get a stable image, the 50Hz wave is related to the mains power supply, if the image rolls across the screen it is not related to mains power. Short answer is: Yes you are seeing mains power related signal (mostly capacitively coupled from the wiring all around you), but it is also a nice test since now you know what line triggering mode is. \$\endgroup\$
    – jippie
    Commented Mar 11, 2015 at 15:19

3 Answers 3


As far as I know, this effect has nothing to do with antennas and electromagnetic waves, unless you consider capacitive coupling as an antenna effect (which is unusual). Here is the complete explanation:

If you are sufficiently close to an electric wire connected to the main, there exists a small capacitance between the wire and your body. The closest you are, the more capacitance.

Let us assume, for example, you are at about 1m of the main wire. This may produce about 10pF of capacitance. So, the main is connected in series with a capacitance of 10pF, 9MOhm resistance of the tip of your probe, and 1MOhm resistance of the oscilloscope, to the earth (I've assumed you set the probe at 1:10).

Now, 10pF gives an impedance of 320 MOhm at 50Hz (approx.). Together with the resistances of the probe+oscilloscope, this gives a 10 MOhm : 330 MOhm = 1:33 voltage divider. So, the voltage indicated by the oscilloscope is 220V / 33 = 6V approx. (if your main is 220V as in the European standard).

This answers the question of the OP.

I add the following remark. If you touch the ground wire of the probe with one finger, and the tip of the probe with another finger, you'll still see a 50 Hz signal in the oscilloscope, but less stronger. The explanation is simple too:

The finger touching the ground wire offers a resistance of about 1 MOhm. So, now, the main is connected to the ground via a capacitance of 10pF = 320 MOhm impedance and a resistance of 1MOhm in series, hence we have a 1:321 voltage divider, which is about 10 times lesser than the previous voltage divider. So, the voltage seen by the oscilloscope will be of 0.5V approximately.

On the other hand, if you touch the ground wire with your tongue, then the resistance between your body and the earth becomes negligible and you observe no more signal if you touch the tip of the probe with your finger.

Of course, these computations were performed under the assumption that the capacitance between the main and your body is 10pF, but this may vary greatly according to the proximity of your body to the main or to a large conductive plate. This gives some insight though.

  • 2
    \$\begingroup\$ "So, the answers above are wrong." This will only be true if your answer gets down-voted enough so that it stays at the bottom. Answers float up and down by votes or user selected sorting preferences. \$\endgroup\$
    – Transistor
    Commented Jul 3, 2019 at 10:38
  • \$\begingroup\$ OK, it's not politically correct to say "other answers are wrong", so, I've suppressed that. \$\endgroup\$
    – MikeTeX
    Commented Jul 3, 2019 at 12:33
  • \$\begingroup\$ Nevertheless, I hope you agree that the right and the wrong have nothing to do with votes. \$\endgroup\$
    – MikeTeX
    Commented Jul 3, 2019 at 12:37
  • \$\begingroup\$ @MikeTeX But above and below is wring way to refer to other answers, since it can cause confusion. It would be better to, for example, use the permalinks instead. \$\endgroup\$
    – AndrejaKo
    Commented Jul 3, 2019 at 12:45
  • \$\begingroup\$ Thx for the advice. Anyway, I'll follow the advice of Transistor and won't give my opinion about other answers in my own answer. \$\endgroup\$
    – MikeTeX
    Commented Jul 3, 2019 at 12:48

Your body is a great big antenna.

Your body is surrounded by a 50Hz (or 60Hz in some countries) hum from the mains power that is all around you.

Put the two together and you get the effect that you are seeing.

  • \$\begingroup\$ How is my body (effective capacitance, Inductance,Resistance) different from a conducting wire which have less capacitance etc. in comparision with us .. because conducting wire or resistor of 150K ohms didn't produce such signal...? (That is I mean this effective transfer of signal has to do with Impedance.. how is it related) A resistor acts like resistor at 50Hz but we may have very different impedance...at such 50Hz frequency.. \$\endgroup\$ Commented Mar 11, 2015 at 15:32
  • 1
    \$\begingroup\$ Try it with a coil of wire the same size as you. \$\endgroup\$
    – Majenko
    Commented Mar 11, 2015 at 15:33
  • 1
    \$\begingroup\$ Your body is one plate of a capacitor. The other plate is the environment around you - the cables in the walls, the power lines passing by, etc. Your body has a very large surface area (some larger than others...). You can't compare it to a piece of wire a couple of inches long. \$\endgroup\$
    – Majenko
    Commented Mar 11, 2015 at 15:39
  • 4
    \$\begingroup\$ @Majenko "Your body has a very large surface area (some larger than others...)" - so maybe we should measure BMI in farads? \$\endgroup\$ Commented Mar 11, 2015 at 15:46
  • 1
    \$\begingroup\$ @MikeTeX That's your prerogative. Now I'm off to top myself :P \$\endgroup\$
    – Majenko
    Commented Jul 3, 2019 at 12:58

The 50Hz is a dead giveaway. Your body, which is somewhat conductive (due to being 80% water) is picking up electromagnetic waves emitted by your mains wiring, and a small current is induced in your body. When you touch the oscilloscope probe it picks up this current which will match the frequency (albeit not as clean a signal) of your mains (60Hz in the Americas, 50Hz in Europe and Asia).

  • \$\begingroup\$ DerStrom8. At 50Hz, there is no antenna effect with the human body (well, to be pedant there should be some effect, but it is so tiny that this cannot explain the large observed voltage). As explained in my answer, this effect has nothing to do with electromagnetic wave. \$\endgroup\$
    – MikeTeX
    Commented Jul 3, 2019 at 10:21

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