Static Electricity Flow

Just for starters, what is a "long, open transmission line to a battery"? Especially "open"?

.. what is a "long, open transmission line [connected] to a battery"?

One without a load connected to the far side, I think.

SL

You are basically asking 'can AC flow though a capacitor?'
Yes...
Next!
Del

Thanks Del, but I do not quite see the AC flowing here. Yes, the transmission line acts as a capacitor, but it is fully charged when I do the swap of the floating wire to the other side.

It is perhaps just the same as forgetting the transmission line and having a single floating wire connected directly to the +ve side of the battery and then I swap it over to be connected to the -ve side of the same battery.

Am I really draining any energy from the battery during the swap? If so, where is the energy going? RF, perhaps?

SL

Swapping repeatedly, say 50 times a second would give you AC.
Even doing it once is putting a 12v pulse onto the capacitor formed by the two lines.
You should play with an R, switch battery and scope, a signal generator too if you have one.
The old school answer to 'Can a capacitor pass current' was to charge one up and throw it to someone, when they caught it and got zapped they would often say 'Thanks for that lesson in electronics it was most informative' (or something like that)
Of course it's possible I've missunderstood the question, but as I see it the two long wires are a small C (plus a bit of leakage R and series R) and the battery/switch arrangement is a pulse or square wave generator. The whole prob is much easier to visualise in those terms.

So call it 0.01uF with 1Megohm across it for leakage and 100R in series, that should give you a nice waveform if you squirt a 1khz square wave up it. Values guessed at carefully chosen so that you can see what's going on.
Del

OK Del, you effectively say that a single wire connected to the live side of an AC wall socket, changing polarity at 50 Hz, is pulling some energy from the grid? TVP45 seems to disagree with this.

Remember, I said a single, floating wire - I'm not reversing the polarity of the transmission line. The only energy flow that I can understand for this case is RF (electromagnetic) energy into the ether.

SL

Remember, I said a single, floating wire
Yes, but then you said "Now I connect another long single piece of wire"
The two wires in close proximity (or even one wire and ground) consititute a capacitor....which is what I've been talking about.
Ok I'm assuming they are in close proximity, but if they are not what is the point of the second wire? A diagram or a better constructed question would have helped.
It's easier if you stop thinking in terms of complex stuff like transmission lines and transients. Think of the circuit I proposed being driven by a square wave, it's easier to work out what's going on.
Simplify, simplify simplify.
Methinks you don't actually want an answer

Del

Sorry Del, I though I've made it clear enough in post #4 above that my question is for one wire, but perhaps it was not clear. So let me rephrase in clearer terms.

The transmission line is completely superfluous. A single, longish wire is connected to one terminal of a battery. The other end is connected to nothing and the wire is well insulated/separated from everything else. I somehow rapidly swap the wire over to the other battery terminal. Does any energy/charge flow from the battery?

Guest no2 (#11) has answered this question, but is the answer correct?

SL (Scruffy)

PS: the question is genuine, not trolling... :-)

Haven't logged in for a long time, so couldn't find my password...

This is just classic nonsense.
If you want a practical real world answer, well the question isn't actually practical or real world, but the answer is:-
'A battery will go flat eventually regardless of lengths of wire'.
If you want a theoretical answer, it's hardly worth the effort because the losses are so small regardless of what you do with the wire.
Anyway it's too late at post #4 to 'make it clear' .. the time for that is in the original post.

I've wasted enough time trying to be helpful. If you really wanted to know you would have followed my advice (#7) which is still valid regardless of the number of wires, only the component values change.
To sumarise:-
The question has neither worth, validity nor interest, so I'm out.
Del

No, I didn't disagree. I didn't understand that you had left the other wire still on there. Draw us a schematic, huh?

Hi TVP.

Sorry to have infuriated the cat! Nevertheless, his implied answer is that yes, there will be some energy loss due to the 'switched single wire', but it is extremely small. I can accept that.

I'm not good at posting schematics here, but I thought the question is simple enough, although I probably muddled it up a bit in the words. I've tried to correct that in #4 and #17.

I'm starting to see that there will be small capacitive losses with every switch and probably also some EM radiation losses - after all, the single wire will act as an antenna, especially if pulsed.

Thanks to all that responded.

SL

No its not fully charged as soon as you disconnected it was discharged. You physically can not disconnect and connect the wires fast enough so that it would maintain a charge on it. The electrons move faster then you. Even in use of a electronic switching device to do the switching you could not maintain a charge on the wire to get appreciable current flow.

What little current flow you get will be through the air. Most likely energy in the form of heat.

Where would the charge go once you disconnected it? Disconnected, the charge would have no current path to travel, it would stay charged.

This is the basic idea behind capacitative charge doublers and charge pumps.

Lo Volt, ok then, but when I connect it rapidly to the other terminal of the battery, where does the charge go? And when I switch it back and forth?

SL

The two wires will form a capacitor. Check your physics book to get a sense for how the proximity of the wires and their length will affect capacitance. In any case the capacitance will probably be fairly small i.e. picofarad range.

Now since they are wires, when you switch polarity, the charge will attempt to flow back into the battery. Note that they are wires with length and resistance. The flow back into the battery is current (amperes). You will get some I²R loss that will show up as heat. You will also get some electromagnetic radiation.

Keep in mind that with reasonable wire lengths (a few hundred feet), reasonable wire sizes (we're not talking 4/0 or larger), a reasonable battery voltage (car battery perhaps), and you toggling a switch manually the capacitance and the charge build up will be tiny. The energy lost will be tiny as well. If you were able to switch polarity at RF frequencies, then you would start to be able to measure the energy lost.

I think guest is imagining the wire to be like a pipe in a plumbing system. Years ago electronics was taught using some plumbing analogies with the caveat that those analogies are not completely accurate. In plumbing they talk about "packing the line". That is filling an empty pipe for the first time. I think what he is asking is "if my 'pipe' is full of water (connected to +) how much energy will it take to empty the 'pipe' and fill it with gasoline (connected to -). This a case where the plumbing analogy does not exactly fit. -- JHF

Dissipated though the air. A capacitor is design to store a charge, a conductor is not. Even a capacitor will slowly discharge with time. The resistance of air may be high but it will conduct.

To load a capacitor (insulated cable), you need 2 poles (+) and (-) which means you need another conductor runnin parallel to the 1st and connected to the (-). The 2 conductors and the isolation between them will form a capacitor.

In your case, it is not clear where the other conductor is? If the distance is far enough then very small charge(negligeable) will be taken each time. To work it out you need to pin a value to the capacitor formed by the wires and then use the charging formulae for capacitors...> Very small values unless the cable is hugely long.

dear,

jingles by words .....

I think, you mean -va side of battery.... instead of -ve side of transmission line ?

Transmission line means a long wire ? how long ?

Since, electricity immediately counts the resistance of the long line.

If you draw the circuit, by connecting the -ve side of battery, current shall definitely flow in closed circuit, & end of the long line will immediately switch over to -ve polarity. ( open means zero current, but potential + : ready to flow, connecting the end to -ve polarity, closed circuit, current flows. & end of line at -ve potential. definitely, current - energy will flow from =ve potential to - ve potential.)

No.

The same number of electrons are in the wire all the time. You are only changing their distribution when you connect them to the different battery terminals.

You are connecting the wire to one side of the battery, with the other end of the battery floating ie not connected to any thing.

In this case, there will not be to be precise a very very infinitesimal amount of charge flowing through. The capacitance

Q=CV

V = voltage of battery

C= capacitance between the long wire and the negative terminal of the battery (which is likely to be quite far away from most of the length of wire)

So most likely a couple of elctron will flow through (or rather sucked by) the positive terminal of the battery.

Now when you reverse it the reverse happens and now there will be two electrons to balance the previous two, and another two to bring the status quo (in opposite direction)

OK for puritanists, it may be a few more than two, but not too many.

Had it been two parallel long wires (or may be a pair of plates) then at least you would have had some capacitance to store the charge and then reverse it.

In case of AC of course the condition is different. remember the neutral is connected to omnipresent ground and hence you always have a bit of capacitance around to spoil the environment.

Guest2, so you say that very slowly, I will draw energy from the battery through capacitive losses?

If my switching is at RF frequency, surely I will also lose energy through EM radiation, possibly much faster?

SL

If it is battery - not likely- simply because the internal losses of the battery will be much more than the few electrons escaping the conductor and scattering into the air/insulator.

As the frequency increases, it will, provided again you do not have anything floating - ie one side of the source is conencted to a large enough sorrounding (ie earth) to have any significant capacitance)

And when you say RF - that is how antennae works isn't it? (with the neural earthed)

dear,

where is the di electric between conductor & -ve side of the battery ?

& hence, where is the question of capacitance between them ?

Long wire gives you resistance.

So, formula you can use is I = V/r. where r is the resistance of the loop, including

internal resistance of battery + resistance of long transmission line / wire + lead resistance of small wire, which he wants to make & break the contact.

There is the air (dielectric) which will lead to a infinitesimal capacitance between them

Sorry could not attach the sketch for it. _{(Now a days guests are not allowed to paste pictures)}

But the end result is - it is so negligible that one would not detect it with the best possible instrument.

Y'know, guest, I gotta say I'm a little irritated. I don't much like it when somebody "catches" me in a trick question. What the heck was this supposed to be about? If you wanted to ask a question about distributed capacitance (of any shape), why didn't you just say so?

Spell out the question plainly.

Now I connect another long single piece of wire, perfectly insulated from ground, i.e. "floating", to the + side only, so that it settles at the same potential as the + side.

The positive side of what? The battery, as this is the only thing that is polarized. If you mean the side of the transmission line connected to the battery, then as you say in your post 'Say I connect a long, open transmission line to a battery, so that the battery charges the line to the DC voltage available.' the transmission line is the same potential all the way along it's length.

I don't understand what your trying to acheive.