Is current harmful when it doesn't flow out of you?

[Nogoodnames]

So say if you're in a vacuum, breathing via a space suit (so you're not grounded and are surrounded by perfect insulation) and you touch something with very high voltage relative to you, would you completely safe no matter how high the voltage is? Note I've made the setting a vacuum deliberately so issues like conducting via air to the ground (eg lightning) can be ignored, and so the charges you gain don't leave you.

In other words, is current only harmful when it can flow through and out of you? Are you safe when you receive an extremely high amount of charge if the charge can't flow out of you?

Similarly when static electricity damages the inside components of a computer, does there need to be a conducting path between the computer and the ground for the computer to be damaged? Or does simply discharging charges to computer components damage them even when the charges stay inside it and have no way of flowing out?

 

[PWiz]

Damage to the body is always caused because of the flow of electrons through it, i.e. when a current flows through you. Electrons only move through a body when a potential difference (voltage) is applied across it. You've already mentioned that there is a potential difference between you and an object in space, so electrons will naturally flow down the electric gradient until the net charge between you and the object is 0 if there is direct contact. So yes, if the p.d is large, it will damage your body.

Damage to electrical components occurs because there is a build up of charge in the component, and then it discharges when a low resistance pathway to the Earth opens up (due to contact with a bare wire for example). Unless the electrons have a pathway to move through, the built up charge won't usually damage the circuitry.

 

[Nogoodnames]

Damage to the body is always caused because of the flow of electrons through it, i.e. when a current flows through you. Electrons only move through a body when a potential difference (voltage) is applied across it. You've already mentioned that there is a potential difference between you and an object in space, so electrons will naturally flow down the electric gradient until the net charge between you and the object is 0 if there is direct contact. So yes, if the p.d is large, it will damage your body.

Damage to electrical components occurs because there is a build up of charge in the component, and then it discharges when a low resistance pathway to the Earth opens up (due to contact with a bare wire for example).

Thank you for the reply.

So in general, electrical components (even 'delicate' ones eg RAM inside the computer) can happily build up charge?

 

[PWiz]

Thank you for the reply.

So in general, electrical components (even 'delicate' ones eg RAM inside the computer) can happily build up charge?

Unfortunately, I'm not an expert on hardware. Hopefully an expert on that topic will stumble across to this thread :)

 

[Drakkith]

Thank you for the reply.

So in general, electrical components (even 'delicate' ones eg RAM inside the computer) can happily build up charge?

Sure, as long as the timeframe is long enough to keep the amperage (charges moved per second) low. The reason they usually get damaged is because a very large potential (thousands of volts) gets built up and you have a static discharge of a lot of current over a short amount of time, which burns and melts things.

 

[Nogoodnames]

Sure, as long as the timeframe is long enough to keep the amperage (charges moved per second) low. The reason they usually get damaged is because a very large potential (thousands of volts) gets built up and you have a static discharge of a lot of current over a short amount of time, which burns and melts things.

Sorry by 'discharge' do you mean charge going from a source to the electric component or from the electric component to some lower potential? In other words do electrical components get damaged if they get charged up to quickly?

 

[PWiz]

Sorry by 'discharge' do you mean charge going from a source to the electric component or from the electric component to some lower potential? In other words do electrical components get damaged if they get charged up to quickly?

Discharge means that a current flows. The direction is irrelevant.

 

[Nogoodnames]

Discharge means that a current flow. The direction is irrelevant.

A good point.

 

[CWatters]

Might be of interest if you haven't seen it already..

 

[CWatters]

Similarly when static electricity damages the inside components of a computer, does there need to be a conducting path between the computer and the ground for the computer to be damaged?

Even when two objects don't have an obvious visible electrical connection there may still be a capacitive connection.

Lets say you unplug your computer and lay it down on the floor to open the case. There will be a capacitance between the metal case and earth. Likewise your body has capacitance (http://en.wikipedia.org/wiki/Body_capacitance) to Earth and this can become charged up (http://en.wikipedia.org/wiki/Static_electricity). So when you reach into the PC to replace the memory card you may inadvertently discharge your body capacitance to Earth via the delicate memory card and the capacitance between the PC and earth.

 

[Nogoodnames]

Might be of interest if you haven't seen it already..

Interesting video, although I was asking whether or not a current that doesn't flow out of you would harm you, without any protection like a Faraday cage.

 

[PhanthomJay]

Interesting video, although I was asking whether or not a current that doesn't flow out of you would harm you, without any protection like a Faraday cage.

Did you ever see a bird resting atop a power wire energized at 20,000 volts to ground? They seem quite happy, no Faraday cage required. I have never seen a bird, though, on a power wire energized above 50,000 volts. The electric field voltage gradient is too much for them as they approach it, I guess.

 

[Drakkith]

Interesting video, although I was asking whether or not a current that doesn't flow out of you would harm you, without any protection like a Faraday cage.

What do you mean by 'a current that doesn't flow out of you'? If I were to somehow induce a large current that flows in loops in your body, you would certainly be harmed. This large current would generate lots of heat and could cause severe burns to any tissue it flows through.

 

[Nogoodnames]

What do you mean by 'a current that doesn't flow out of you'? If I were to somehow induce a large current that flows in loops in your body, you would certainly be harmed. This large current would generate lots of heat and could cause severe burns to any tissue it flows through.

An unsteady current (or electrostatic shock if you prefer) induced by touching a conductor of higher potential, that quickly dissipates inside you as the charges move to reach equilibrium.

 

[Nogoodnames]

Did you ever see a bird resting atop a power wire energized at 20,000 volts to ground? They seem quite happy, no Faraday cage required. I have never seen a bird, though, on a power wire energized above 50,000 volts. The electric field voltage gradient is too much for them as they approach it, I guess.

I assume it's because the bird has the same voltage as the power line which is why it lives. Do you know how the bird reaches the same voltage as the 20,000 V power line though? I am guessing there must be a brief exchange of charges between the power line and the bird to achieve this? If so shouldn't the bird still get a short shock? I know Van de Graaff generators don't shock you because it has high internal resistance (unless you let it build up charge) but power lines have low resistance right?

 

[mfb]

I am guessing there must be a brief exchange of charges between the power line and the bird to achieve this?

There is an exchange of charges with a frequency of 50 or 60 Hz (depending on where the bird lives) unless the bird hits one of the few DC transmission lines (but those normally operate at even higher voltages). The capacitance of birds is very small.

A space suit is probably a good Faraday cage, by the way.

 

[Nogoodnames]

There is an exchange of charges with a frequency of 50 or 60 Hz (depending on where the bird lives) unless the bird hits one of the few DC transmission lines (but those normally operate at even higher voltages). The capacitance of birds is very small.

A space suit is probably a good Faraday cage, by the way.

I don't know much about capacitance but if the birds cannot store much charge, how does it achieve the same high voltage as the power line? From what I understand voltage=potential difference is the amount of energy needed to bring a unit positive charge from one point to another (in this case from ground to the powerline), and I am struggling to imagine it takes the same amount of energy to bring a unit positive charge from the ground to something barely charged (the bird) compared to something highly charged (the powerline)?

 

[mrspeedybob]

The voltage to which something is charged and the amount of charge on an object are 2 different things.
Voltage is measured in volts, charge is measured in coulombs. In the water analogy, volts are like water pressure measured in PSI, and coulombs are like the volume of water measured in gallons. When charge (coulombs) flow into or out of an object they can damage the object in proportion to the amount of energy involved. That energy is proportional to both the voltage and the amount of charge. In the case of a bird on a wire, the voltage of the bird varies rapidly with the voltage on the wire, cycling from -20,000 volts to +20,000 volts 50 or 60 times per second. This voltage (electrical pressure) pushes charge (coulombs of electrons) into and out of the bird, but, since the bird cannot store very many excess electrons,only a small amount of charge flows in and out of the bird. Even though the voltage is high, the charge is so low that voltage x charge is small, so the amount of electrical energy dissipated inside the bird is small. It is in fact so small that the bird either does not notice it, or finds it pleasant.

 

[mfb]

I don't know much about capacitance but if the birds cannot store much charge, how does it achieve the same high voltage as the power line?

Q=CU - charge is capacitance times voltage. The capacitance of birds is very small.

 

[Nogoodnames]

The voltage to which something is charged and the amount of charge on an object are 2 different things.
Voltage is measured in volts, charge is measured in coulombs. In the water analogy, volts are like water pressure measured in PSI, and coulombs are like the volume of water measured in gallons. When charge (coulombs) flow into or out of an object they can damage the object in proportion to the amount of energy involved. That energy is proportional to both the voltage and the amount of charge. In the case of a bird on a wire, the voltage of the bird varies rapidly with the voltage on the wire, cycling from -20,000 volts to +20,000 volts 50 or 60 times per second. This voltage (electrical pressure) pushes charge (coulombs of electrons) into and out of the bird, but, since the bird cannot store very many excess electrons,only a small amount of charge flows in and out of the bird. Even though the voltage is high, the charge is so low that voltage x charge is small, so the amount of electrical energy dissipated inside the bird is small. It is in fact so small that the bird either does not notice it, or finds it pleasant.

Yes but doesn't voltage or potential difference depend on the amount of charge? Potential difference is the line integral of electric field and electric field depends on the charge density by Coulomb's law or superposition principle?

 

[Nogoodnames]

Q=CU - charge is capacitance times voltage. The capacitance of birds is very small.

I think I get it now, the bird feels an energy of E=QV=CV^2 and because capacitance is small, the energy is small?

 

[PWiz]

I think I get it now, the bird feels an energy of E=QV=CV^2 and because capacitance is small, the energy is small?

You forgot a half there: $E=½QV=½CV^2$. (the actual equation is $E=\int_0^Q V dq$)

 

[mfb]

I think I get it now, the bird feels an energy of E=QV=CV^2 and because capacitance is small, the energy is small?

That's the energy stored in the birds, it is not the energy loss of currents inside birds.

The small charge Q allows to find the small current I, energy dissipated is then proportional to R*I² with some internal resistance R - you square that small current value.

 

[Nogoodnames]

That's the energy stored in the birds, it is not the energy loss of currents inside birds.

The small charge Q allows to find the small current I, energy dissipated is then proportional to R*I² with some internal resistance R - you square that small current value.

So it's the energy loss of current that potentially does the damage to the bird?

In calculus terms, Q is proportional to exp(-t/CR), you differentiate this with respect to time to work out the current I. Is the total energy dissipated from t=0 to t=T the integral with respect to time of (R*I^2) from t=0 to t=T?

 

[mfb]

Current and the associated energy loss, yes.

In calculus terms, Q is proportional to exp(-t/CR)

It is a bit more difficult as the voltage of the transmission line follows a sine curve, and you could also keep track of currents inside the bird (not just the connection point to the power lines), but all those things just give some numerical prefactors that are not too far away from 1.

 

[Nogoodnames]

Current and the associated energy loss, yes.
It is a bit more difficult as the voltage of the transmission line follows a sine curve, and you could also keep track of currents inside the bird (not just the connection point to the power lines), but all those things just give some numerical prefactors that are not too far away from 1.

You know how Q=(V/R)exp(-t/CR) so I=-(V/R)(1/CR)exp(-t/CR), so the integrand of RI²=0.5*V²(1/CR)exp(-2t/CR) How do these numbers work out to show the energy dissipated is small?. In particular, if the capacitance is small (ie not so different from radius of bird*4*pi*8.85x10^-12, permitivity of free space), and resistance is only about 100k ohms (taken from dry resistance of human), then the exponential term becomes 0 very quickly which leaves about 60000 gigajoules of energy dissipated in the first few milliseconds (assuming voltage is 20kV, radius of bird is 0.3m).

I am guessing the (V/R)² term is why birds are not so happy to approach 50k voltage power lines, and electricians wear Faraday cage suits?

 

[mfb]

Q=(V/R)exp(-t/CR)

That applies to a fixed outer potential only.

Let's take a bird radius of ~2cm for a capacitance of 2pF. That is an overestimate because the bird has a long power line close to it.
100 kOhm and 2 pF gives a time constant of 200ns, much shorter than the time constant of the AC. The potential of the bird will closely follow the potential of the power line, and I = C dV/dt. The maximal voltage gradient is of the order of 20kV/(10ms) which gives a current of just 4µA independent of the resistance of the bird (as long as it is not too large).

 

[Nogoodnames]

That applies to a fixed outer potential only.

Let's take a bird radius of ~2cm for a capacitance of 2pF. That is an overestimate because the bird has a long power line close to it.
100 kOhm and 2 pF gives a time constant of 200ns, much shorter than the time constant of the AC. The potential of the bird will closely follow the potential of the power line, and I = C dV/dt. The maximal voltage gradient is of the order of 20kV/(10ms) which gives a current of just 4µA independent of the resistance of the bird (as long as it is not too large).

If the line was 20000V but DC, would I = C dV/dt still work and would the bird still not fry? If so what would the maximal voltage gradient be? Also can you explain how you got the 10ms there.

 

[mfb]

My calculation was for a bird sitting on the line already - with DC, nothing would happen there. 10ms is a typical timescale (corresponding to one half-wave at 50 Hz).
For an initial approach, the current is higher, but also extremely brief.

 

[Nogoodnames]

My calculation was for a bird sitting on the line already - with DC, nothing would happen there. 10ms is a typical timescale (corresponding to one half-wave at 50 Hz).
For an initial approach, the current is higher, but also extremely brief.

Sorry to keep bothering you but what equations would you use for the current induced from when the bird first comes into contact with the line until the potential between them are equalised? You seemed to imply treating the bird like a capacitor using Q=(V/R)exp(-t/CR) isn't correct?

 

[mfb]

For the initial approach, that formula is correct, but I don't see where your GJ value comes from - the current won't flow for milliseconds.

 

[Nogoodnames]

For the initial approach, that formula is correct, but I don't see where your GJ value comes from - the current won't flow for milliseconds.

Yeah I don't really know how long the current flows for so I just plugged in a few small numbers. Do you know approximately how long that current flows usually?

Nvm I figured out how to work out the time.

 

[Nogoodnames]

For the initial approach, that formula is correct, but I don't see where your GJ value comes from - the current won't flow for milliseconds.

If I am not grounded (eg stand on a plastic stool) and touch a Van de Graaff generator, is it reasonable to model myself in this case as a capacitor? (using equations like I=(V/R)exp(-t/CR))

If I touch the Van de Graaf generator with a metal rod (still standing on the plastic stool), can I still model myself as a capacitor by simply taking the new resistance as my own resistance+the metal rod's resistance?

Thanks.

 

[mfb]

That should give a good approximation. But don't test it in practice...