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consider the following situation. I parked my car underneath some high-voltage lines – let’s say 380 kV-lines. Suddenly, one of the lines breaks in the middle due to a storm and one end of the (still energized) line now touches the metal body of my car.

Now I know that it’s obviously best to stay in the car unless it’s absolutely necessary to get out. I’m also aware of touch and step potential. The thing that I’m interested in is this: If I’m still sitting in my car and NOT touching any part of the ground outside – Will I get shocked if I touch any metal part of my car?

Some sources that I checked claim that one cannot touch any metal parts of the car, while other videos that I’ve seen on youtube clearly show how the instructors touch metal parts of their vehicle before hopping out – for instance I remember an educational video where an instructor walks down the metal steps of his truck before jumping out.

My reasoning (I’m not a physicist/scientist by the way) goes like this: The metal frame of the car should be a very good conductor. So as soon as the car is contacted by the powerline, current will flow through the chassis, through the tires to ground. Now since the metal body of the car is a good conductor, all parts of it should roughly be at the same potential (ideally the potential difference between any two points on the car should be zero, much like in an ideal circuit where there are no voltage drops along the wire). So since all parts of the metal body are at the same potential, I should be good if I touch anything (again, while sitting completely inside the car and without touching the ground outside). I’m not sure if this reasoning is correct though. Could anyone explore this scenario from a physics point of view?

I guess another way to view this would be the following: If current flows through the car to ground and I’m touching, say, the ceiling and floor of the car simultaneously, there is now a network of parallel resistors through which current can flow. Since my resistance should be >> resistance of the car, no current (at least no lethal or harmful current) should pass through me. But again, I’m not sure if this reasoning is waterproof.

On a side note: If there are several passengers sitting in my car and one of them were to get out while touching both the car and the ground, would only this person get electrocuted or could this also electrocute everyone else sitting inside the vehicle?

Sorry, I guess I asked two questions now. But I’m kind of fascinated and intrigued by this whole topic.

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  • $\begingroup$ Extreme Jobs - High Voltage Power Line Inspection I don't know what to think of this video. I wouldn't have expected more than a momentary spark when making contact with the power line. $\endgroup$
    – mmesser314
    Commented Aug 5, 2023 at 0:22
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    $\begingroup$ @mmesser314, The power line is AC. When the helicopter is connected to the power line, it's frame must be at the same potential. As that voltage swings negative, electrons must flow in to the helicopter to give it a negative charge. When the voltage swings the other way, electrons must flow out. So long as the helicopter is connected to the line, there is a continuous AC current flowing in the connection. The same would happen to the lineman's body if he were not wearing a Faraday cage suit. (P.S., the reason he puts the clamp on the line before climbing across is so that his suit does not... $\endgroup$ Commented Aug 5, 2023 at 13:40
  • $\begingroup$ ...itself become the connection through which the current flows.) $\endgroup$ Commented Aug 5, 2023 at 13:42

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WARNING: This answer should in no way be considered electrical safety advice. Real cars (especially modern ones) are not ideal Faraday cages. If I were in a car in contact with a 380 KV line, I would try to move as little as possible until help arrived and the power line was deactivated, irrespective of the calculations below.

If there is no path to ground, then the current that flows when you touch the frame will the current required to move enough charge ($Q=CV$) to bring you to the same potential ($V$) as the frame. The current flow will be $$I=\frac{dQ}{dt}\sim C \frac{\Delta V}{\Delta t}$$ Your capacitance, $C$, depends on the detailed geometry of you, your car, and the ground, but an isolated human typically has a capacitance of about $100$ pf. The voltage change is $\Delta V \approx 380\,\mathrm{kV}$ in a time $\Delta t = 1/(2\times 60) \,\mathrm{s}$, assuming 60 Hz electrical power, so the voltage is switching 120 times per second. ($\Delta t$ is set by the mains frequency because the $RC$ time constant for the human body is typically only $\sim 10^{-7}$ seconds.) So we expect an AC current through you of $$I \sim (10^{-10} \mathrm{F}) (3.8 \times 10^{5} \mathrm{V})(120\,\mathrm{s^{-1}})\approx 5\,\mathrm{mA}$$ This shock is unlikely to be fatal, but as I note above, this is an idealized rough estimate. For example, what you are sitting on or touching, other people in the car, the construction of the car, the presence of water from the storm in the car and underneath it, … could increase the capacitance and hence the current. Also the shock could be enough to cause you to spasm which could make the situation worse. Even spherical cows can be deadly.

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  • $\begingroup$ Very informative, thanks! So to sum up, 1) the basic reasoning that all metal parts are at the same potential may be theoretically correct but in practice one shouldn't rely on this because cars aren't necessarily built that way and 2) we must also take into account the current that flows through the human body when he touches the car and their potentials are equalised $\endgroup$
    – ilovemaths
    Commented Aug 8, 2023 at 6:26

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