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The common explanation of why birds don't receive a shock when standing on high tension cables is that there needs to be a considerable potential difference between both of his feet for him to receive the shock.

I understand this for a single cable, but not for two. When a bird stands on two cables, how is there a potential difference? I think it makes no sense to say one cable has a higher voltage than the other since voltage is $\int \bf{E}\dot\bf{dl}$ so it's therefore more a property of some trajectory than a single position (I do know you can say a certain point has a voltage, but I don't think it's important here). If there was some type of voltage between both cables, how was the necessary electric field generated?

Did one cable have more charges than the other? Is it because one cable goes to your house and the other goes back to the power station, such that if you were to imagine a representative circuit then this is a consequence of Kirchhoff's rules?

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  • $\begingroup$ An electric potential is the exact same thing as a mechanical potential: it describes the ability of the system to perform work. It would make no sense for the electricity company to put up two electric cables on a pole without giving them a different potential, since without that there would be no transfer of power. The electric field between two cables is indeed rather strong (usually on the order of tens of kV/m). The integral you mention can be reduced to a vector $\vec{E}$ pointing from one cable to the other, which is parallel to the vector $\vec{dl}$, hence the integral does not vanish. $\endgroup$
    – CuriousOne
    Commented Dec 17, 2014 at 5:34
  • $\begingroup$ But how is this electric field generated if they're only two cables conducing current? $\endgroup$
    – DLV
    Commented Dec 17, 2014 at 5:38
  • $\begingroup$ Is it because of the Kirchhoff thing I mentioned? $\endgroup$
    – DLV
    Commented Dec 17, 2014 at 5:38
  • $\begingroup$ The electricity company uses electromagnetic induction to generate the potential difference, but that doesn't matter for the potential difference, or the poor bird that makes the contact... $\endgroup$
    – CuriousOne
    Commented Dec 17, 2014 at 5:45
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    $\begingroup$ Birds usually sit on one cable. If the birds are large enough to touch two live wires they set themselves on fire. My son did his army duty at an outpost where three times crows shorted two cables and set fire to the forest and the soldiers had to rush and put it out. Now in Greece they are slowly replacing the four cables into one braided one and insulated for this reason ( trees too can short two wires) $\endgroup$
    – anna v
    Commented Dec 17, 2014 at 13:04

4 Answers 4

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One should really be careful when mixing circuit theory and electromagnetic theory...

Fact of the matter is, circuit theory is a much simplified version of electromagnetism. In circuit theory, we speak of the "voltage" of a point, in reference to the ground, and act like it is a well-defined quantity. In fact, what is called the electrical potential is properly defined only in the absence of all magnetic fields. In circuit theory, we assume magnetic fields are zero, except within "designated" circuit elements like the inductor.

After that lengthy introduction, let us get to the actual questions...

You state that $V_{12} = -\int \limits_1^2 \vec{E} \cdot \vec{dl}$ which is correct (the negative sign is there because $\vec{E}$ points downhill. This is correct. Also, as I stated above, defining this only makes sense in the absence of magnetic fields, in which case it is guaranteed that the integral is path independent, meaning, you will get the same value independent of the path you choose. This means that the above integral over a closed path will always give zero. By Maxwell's equations that happens when there is no magnetic field (to be more precise, a static magnetic field) by the third equation.

So, what about voltage of a point? Now, if we have a well-defined potential differences between any two points as discussed above we can assign a voltage to any point. We do this by picking an (arbitrary) point, and assign it the value zero (and call it ground). This is conveniently chosen to be the potential of the earth itself, and therefore called "ground". You can change your reference point for potential measurements from problem to problem; since it is truly arbitrary.

There is no "necessary" electric field so to speak; the electric field is a consequence, and not a reason. The real reason is, at one end of the cable we have set up a contraption which pumps charges around. Usually this is a contraption involving wires and magnets rotating, which pushes charges one way first, then the other, generating what we call an "alternating current". What is normally moved in a power circuit (as opposed to electronic, where it can be holes ... but I digress) are electrons. Electrons are pushed towards one wire from the other which has excess positive charges, making one wire positively and the other negatively charged. Nature then takes care of the rest, forming the potential and electric fields as necessary. Then the positive and negative charges move along the wire as a wave...

If the whole thing was a direct current circuit, and at the powerhouse you had a battery, the charges would be pushed apart by the battery from one wire to the other by means of a chemical reaction and not a generator contraption as the one I mentioned above. There would be no "wave", but the principle remains the same: You move the charges, nature forms the potential and electric fields (which are in fact the description of the same thing, and not two separate entities).

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  • $\begingroup$ Potential difference is the thing that matter. NOT THE POTENTIAL $\endgroup$
    – Paul
    Commented Dec 17, 2014 at 14:39
  • $\begingroup$ @Paul Obviously once you know the potentials at the two points on the wire in contact with the resistance (bird, human, etc.), you already have the potential difference. $\endgroup$
    – Gaurav
    Commented Dec 17, 2014 at 14:46
  • $\begingroup$ I was actually saying ,'why to talk about the reference point(grounded or not)? potential may be different for different reference point but potential difference is always same in any reference point.. and here potential difference matter $\endgroup$
    – Paul
    Commented Dec 17, 2014 at 15:08
  • $\begingroup$ @safkan I find it weird that one wire would be positively charged and the other would be negatively charged. I think they're both electrically neutral (even if they conduce a current). I still think the electric field is a "reason" for there being potential differences, not just a "consequence". Perhaps I have misunderstood you? $\endgroup$
    – DLV
    Commented Dec 17, 2014 at 18:32
  • $\begingroup$ Yes.ofcourse electric field is the reason of potential. infact concept of potential is based on the fact that electric field is a conservative field. curl of electric field is zero $\endgroup$
    – Paul
    Commented Dec 17, 2014 at 19:00
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Hyper physics website says this

Electric current flow is proportional to voltage difference according to Ohm's law, and both the bird's feet are at the same voltage. Since currentflow is necessary for electric shock, the bird is quite safe unless it simultaneously touches another wire with a different voltage. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir2.html but why voltage should be same in two points on the wire?

my understanding says $J=σ E$ since for metals $σ$ is very huge so $E$ must be nearly zero inside the wire and hence line integral of $E.dl$ is zero and hence potential difference is zero. (here $J$ is current density and $σ$ is conductivity) the equation is Ohm's law in terms of current density and electric field.

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The current in overhead cables is Alternating Current. There are usually four cables - one for each of the three phases, and one for the shared neutral.

Since the voltage in the three live wires is constantly changing, and each is 60 degrees out of phase relative to the other, there is always a potential difference between any two wires, except at two points during the cycle (for phases 2 and 3 in the image below, it is at $90^\circ$ and $270^\circ$).

That's why you get shocked if you touch two different live wires.

Of course, you also get shocked if you touch one live wire and the neutral wire or the ground, but that's fairly obvious.

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  • $\begingroup$ @David: Each wire has a potential. The difference in potential of the two wires is the potential difference. I don't follow your question here. $\endgroup$
    – pho
    Commented Dec 25, 2014 at 18:23
  • $\begingroup$ Sorry, that post was posted somehow. I didn't finish writing it properly. $\endgroup$
    – DLV
    Commented Dec 25, 2014 at 23:34
  • $\begingroup$ Maybe I have to read more up on alternating currents, but I don't understand how a wire can have a potential (I only understand potential differences between two points via $\int E\cdot dl$). And perhaps I'm even more confused how there can be a potential difference between two neighbouring wires. Is there an actual electric field between both wires? If so, how is it generated? $\endgroup$
    – DLV
    Commented Dec 25, 2014 at 23:37
  • $\begingroup$ Let's consider the wires are part of two separate systems. The potential of wire A is its potential difference with respect to an arbitrary reference (in this case, the "common netural" wire). The same goes for wire B. Since the potential of A and B are both alternating between $-V_{\rm max}$ and $+V_{\rm max}$, but not in phase, their potentials are never the same, save for two instants (when their lines cross over in the diagram in my answer). Since their potentials aren't the same, there will be a potential difference across anything connecting the two wires. Clear now? $\endgroup$
    – pho
    Commented Dec 26, 2014 at 5:22
  • $\begingroup$ Maybe I'm being irritating but I don't see how this is a "fundamental" explanation. I feel it doesn't discuss the actual nuts and bolts of how an electric field will be generated between two different wires for there to be a voltage between them and how this will magically be equal to the difference in "potential" specified by the graph. It's probably I that doesn't understand, so please excuse me. Thanks. $\endgroup$
    – DLV
    Commented Dec 26, 2014 at 5:37
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If the bird is standing on a single bare wire with no place for the voltage to pass through the bird to return to the source. The bird is okay.

If the bird is standing on two separated bare wires containing enough voltage between them to kill the bird, the bird is toast.

In the case David has posed, I suspect he is talking about the wires from a power pole to a house. As with my house, they are a pair wires twisted around a supporting steel cable. Even though the pair of wires has sufficient voltage difference to toast the bird, they are insulated to keep the wires from contacting each other and shorting out. Since they have insulation around the wires, this keeps the bird (or squirrel) from coming in contact with the voltage and they are safe.

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