Can a Faraday cage protect you from electrocution in a nano second?

[hokhani]

In the Faraday cage, after injecting the charge, it takes about a nano second for charge to be distributed over the external surface. Why doesn't one become electrocuted in this very little time (nano second)?

 

[Vanadium 50]

During this time how much charge is going through you?

 

[hokhani]

During this time how much charge is going through you?

Thanks, another question in this context arises about the Faraday cage which is made of grid lattice of wires (see: https://makezine.com/2007/06/20/tesla-faraday-cage). In the case of a cage made of wire, what is the external surface of the cage? Why don't the charges go inside the cage in contact with the person?

 

[osilmag]

Current is going to flow along the path of least resistance. So, the charges will flow through the metal wires to ground instead of the air gap and your flesh due to higher reactance/resistance.

 

[rcgldr]

Video example of a Faraday cage "suit" used to work on high voltage (500kv or 750kv) power lines:

 

[willem2]

In the Faraday cage, after injecting the charge, it takes about a nano second for charge to be distributed over the external surface. Why doesn't one become electrocuted in this very little time (nano second)?

It will also take time, before the effects of any charge outside the cage are felt by someone inside. This speed is limited by the speed of light, and the electric field will reach the cage first,and start to produce induced charges that cancel the field before it can reach anyone inside.

 

[hokhani]

Current is going to flow along the path of least resistance. So, the charges will flow through the metal wires to ground instead of the air gap and your flesh due to higher reactance/resistance.

How to realize the external surface of a grid-like conductor (which is like a chicken cage)? A chicken cage has two parts; inside and outside. One part of the wire, which is used for the cage, is inside the cage and the other part is outside. If we expect the Faraday cage be safe, the charge should be accumulated on the outside part otherwise one is electrocuted. My question is that how the outside part of the wire is taken as the external surface (on which the extra charge gather)?

 

[willem2]

How to realize the external surface of a grid-like conductor (which is like a chicken cage)? A chicken cage has two parts; inside and outside. One part of the wire, which is used for the cage, is inside the cage and the other part is outside. If we expect the Faraday cage be safe, the charge should be accumulated on the outside part otherwise one is electrocuted. My question is that how the outside part of the wire is taken as the external surface (on which the extra charge gather)?

I don't think chicken wire has an inside and an outside part. If you have a closed conducting surface, it's easy to see (gauss' law) that there can be no electric field on the inside, and therefore there can be no surface charge on the inside surface, unless you bring charges to the inside.
Chicken wire only acts like a closed metal surface if you consider the field averaged over a distance larger than the gaps.
To get some dangerous electric discharge from the cage through your body, you need an sufficiently large electric field between the cage and your body. The electric charges induced in the cage will cancel uit such a field, unless you get too close to the wire.

 

[Tom.G]

Hi hokhani,
Let's try a slightly different approach. (For all the purists here, a slightly tortured word picture.)

Start by remembering that a voltage is defined as a potential relative to the potential of something else. For instance to use a voltmeter to measure the voltage of a battery you connect the meter leads to the battery terminals. That gives you the voltage, or potential difference, between the two terminals of the battery. If you use only one meter lead, it won't read anything because there is no reference level or charge for the meter to relate to.

And you will get the same Zero reading if you connect both meter leads together and to one battery terminal, in this case because you are trying to measure the same thing you are using as a reference

When you are inside a Faraday cage, the only reference you have is the cage itself. If you try to measure the voltage that you are at, you measure from yourself to the cage; and will read Zero volts. If there is an electric field outside the cage, the entire cage, being conductive, will end up being at some uniform voltage dependent on that external field (and also on whether or not the cage is Grounded, but that can be ignored as the end result is the same). Again if you measure the voltage between you and the cage the result is still Zero, even though the cage itself may be at some crazy high voltage. Once more, that's because the cage is the only reference you have.

Now if you disconnect the meter lead that is connected to the cage and poke it thru a hole in the cage, you will read the voltage Difference between the cage and the end of the meter lead.

If you stick your finger thru that hole in the cage, the end of you finger will be at the voltage of the field at that point, while the rest of you is at the voltage of the cage. If it is a small difference you won't even notice it, above about 50V you will feel it.

All of this boils down to the rule that states: "Everything within a conductive enclosure is at the potential of that enclosure."

Lots of folks here can express a 'Proof' of this mathematically. However, for that you have to both understand the math and understand why it applies; not easy when you are first introduced to it!

Hope this helps.

Cheers,
Tom