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If a person touches a single line of AC(Only on the live line, and no where else. Say he/she is hanging on the power line using one hand) as shown in the attached image, Do the ions in his/her body move to one side of his/her body in first half cycle of AC(Like electrons getting pulled from plate of capacitor when positive terminal of a battery is connected to capacitor even though a capacitor does not complete a physical loop in the circuit) and pushed further away in the next half cycle? If this to and fro oscillation of charges takes place then doesn't that create an AC current in the persons body and the person get electrocuted even if he touch a single line of AC?

Thanks in advance.

enter image description here

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    $\begingroup$ Some time ago there was a great article in the American Journal of Physics that answered a similar question in the context of birds sitting on power lines. $\endgroup$
    – Julius
    Commented Nov 7, 2018 at 16:50

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The pictures in the post are more consistent with the induced charge, while, under the described test conditions, the charge will actually flow in and out of the body.

Regardless, the AC current ($120$V/$60$Hz) flowing through the body will be insignificant to be dangerous (has to be on the order of tens of milliamps) or even to be felt (has to be on the order of $0.5$mA).

If we assume the capacitance of the body to be $100$pF, the charge required to charge it to a peak voltage of $170$V $(Q=CU)$, will be about $16$nC, yielding the average current over a half period of about $2$uA - too small to be perceived.

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  • $\begingroup$ @Nicolas A human body has capacitance on the order of tens of picofarads even if it is far from the ground. $\endgroup$
    – V.F.
    Commented Nov 7, 2018 at 16:58
  • $\begingroup$ You're absolutely right, I removed my comment. $\endgroup$
    – Nicolas
    Commented Nov 7, 2018 at 17:18
  • $\begingroup$ @Ajay You are welcome. $\endgroup$
    – V.F.
    Commented Nov 20, 2018 at 18:30
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If the body is not in contact with anything connected to the other wire (that is, ground), nor close enough to it to form one capacitor plate, then the electrons will not do anything, just as happens if you connect one side of a capacitor to one terminal of a battery.

In fact electrons react to electric field, not electric potential ; if the other wire and/or ground is very far, the electric field vanishes. That's why birds perched on overhead lines don't feel anything at all. On the other hand, just being close to the other wire may create a strong enough electric field to make electrons move as your picture shows for the case where the whole left edge of the picture is at ground potential)

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  • $\begingroup$ Hi @Nicolas, This may be a dumb question, When we connect the positive and negative of a cell to capacitor, electric field is formed inside capacitor only after some electrons are transferred from plate of the capacitor to positive terminal of the cell,aren't it? So how this transfer of electrons takes place first without an electric field being already present? $\endgroup$
    – Ajay
    Commented Nov 20, 2018 at 15:34
  • $\begingroup$ The electric field is immediately present across the battery ; the internal resistance of the battery (and that of your wires) limits the current that is pushed by this electric field and therefore the speed at which it appears across the capacitor plates. In other words, the current propagates "slowly" along the wire to the capacitor due to the resistive slowing of moving electrons in the wire/battery. $\endgroup$
    – Nicolas
    Commented Nov 21, 2018 at 10:11
  • $\begingroup$ Thanks@Nicolas. So, this is what I understood. Electric field is already present in the battery and this electric field pulled the electrons from plate of the capacitior. Did I understand correctly? $\endgroup$
    – Ajay
    Commented Nov 21, 2018 at 11:40
  • $\begingroup$ @Ajay Yes, that's correct. $\endgroup$
    – Nicolas
    Commented Nov 21, 2018 at 16:52

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