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My question: At the very initial moment when the bird sits on the live wire, the bird and the live wire have a potential difference, which is normally of the magnitudes of $1000'$s. But why the current due to the charge build up build up doesn't kill the bird?

Okay, I have formulated my own theory for this. So please validate this too:

From google, the capacitance of human body is about $180pF-200pF$ and so birds should have a lesser value, since they are small. From this the charge build up in a human body would be(voltage of the live wire is $11kV$ in India)

$Q=180\times 10^{-12}\times 11000=0.00000198$

So such small charges cannot harm you.

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  • $\begingroup$ More on birds on wires. $\endgroup$
    – Qmechanic
    Feb 5, 2017 at 12:40
  • $\begingroup$ @Qmechanic♦ I couldn't find anything about this. But regarding the birds issue maybe i will change the question to superman with bare foot on live wire :-p hehe $\endgroup$
    – Allen
    Feb 5, 2017 at 12:45
  • $\begingroup$ @Allen you are on the right track. Yes, there's initially a potential difference, but because of the low capacitance it very quickly subsides. Consider that a typical static electricity shock of a person walking across a carpet (in the winter time--when humidity is low) and touching grounded metal is on the order of several thousand volts, but its so brief it doesn't cause any damage to the person. $\endgroup$
    – Digiproc
    Feb 5, 2017 at 14:39

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Answer by @Digiproc:

Yes, there's initially a potential difference, but because of the low capacitance it very quickly subsides. Consider that a typical static electricity shock of a person walking across a carpet (in the winter time--when humidity is low) and touching grounded metal is on the order of several thousand volts, but its so brief it doesn't cause any damage to the person.

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