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Physical basis of voltage

  1. Jun 11, 2017 #1
    An example of how trying to teach someone exposes gaps in your knowledge:

    I was talking to someone about why we have line, neutral and earth conductors. We got on to the fact that the earth is an enormous sink and source of charge, so anything referenced to it can be considered 0 volts.

    Then he asked what would happen if he connected a voltmeter between +ve of a 12V battery and ground. I said 0 volts, since the battery is not referenced to ground.

    But, he said, the battery as a whole is electrically neutral, but with separators so that an imbalance of charge is created between the two terminals. With this is mind, the positive should be +6V above ground, and the negative -6V below ground.

    I pointed out his confusion between voltage and charge, the former being the difference in energy stored per unit charge. Voltage is not a measure of charge, etc. But I must admit that I couldn't pinpoint the exact failure of his argument. What is the exact physical basis of voltage that, say, means that measuring between the +ve of a battery and the -ve of a seperate battery will give a reading of 0 volts? And I did just say that the ground can be considered 0 volts, even though measuring between ground and the +ve will give no voltage reading.
     
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  3. Jun 12, 2017 #2

    cnh1995

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    What is "ground" in this discussion? Actual earth ground?
     
  4. Jun 12, 2017 #3

    davenn

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    If the battery isn't connected to earth ( ground) in so way .... one terminal or the other, then there is no reference to ground
    That is ... THERE MUST be a connection for there to be a reference point else the battery is just
    a floating power source and any referencing is done to either the positive or negative terminal
     
  5. Jun 12, 2017 #4
    Yes, like a copper rod in the soil.
     
  6. Jun 12, 2017 #5
    I almost went to an example of a van der Graaf generator to dispel his confusion about voltage and charge. But then I realised I didn't fully get that - as you charge the machine, electrons, therefore charges, build up on the accumulator. More coulombs. But the voltage would also rise, so as the machine runs more charge accumulates but also each unit of charge would have more energy. I couldn't quite get that.
     
  7. Jun 12, 2017 #6

    cnh1995

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    So if I understood your friend correctly, what he's saying is:
    You have a 12V battery and a metal rod in the ground (soil) and they are not connected to each other. If you measure the voltage between the +ve terminal of the battery and the metal rod, it should be +6V and that between -ve terminal and the rod should be -6V?

    If so, perhaps he needs to read this thread from the beginning.
    https://www.physicsforums.com/posts/5723411/
     
  8. Jun 12, 2017 #7
    Very good thread, and video.

    I think he's thinking the +12V terminal is charged like a VDG generator. Perhaps the best way to explain it is: the terminals are electrically neutral - there is no static charge difference between them. But each coulomb of charge on the +ve has 12J more energy than on the -ve. Since the ground is not referenced to the battery, trying to see the energy/charge difference between either terminal and ground is like trying to say which of two astronauts floating in deep space is 'higher'. Without a common reference, like a planet, any height measurement is irrelevant.

    Voltage is clearly relative, as in potential DIFFERENCE, but the definition of J/C implies an absolute value - a coulomb is an invariant unit of charge, so you might think it possible to measure the energy each coulomb has and come up with an absolute measurement.

    Maybe an ideal multimeter could measure this? Perhaps you could measure the J/C of ground and compare it to the J/C of the +ve battery terminal? A real meter needs to take a tiny current, so trying to do this through the enormous internal impedance of the path from +ve terminal, through the case and along the ground might drop any measureable potential difference to zero?
     
  9. Jun 12, 2017 #8

    scottdave

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    With a battery not connected to the Earth, there is really no knowing what the voltage difference between each terminal and ground. It is possible that one terminal may be 20 volts, and the other is 32 volts, for example. If you hook up a voltmeter between one of the terminals and ground, you will read 0 volts. You need to understand how a voltmeter works. A voltmeter actually uses a very large resistance, then measures the current going through that then calculates the voltage between the test leads. By connecting them with your test leads, a small current will flow for a small amount of time, until the voltages are the same. What completes the circuit is a small capacitance, provided by you, the air, etc. If you have access to an oscilloscope, try just touching the probe to something (like your finger), you should see the display jump around momentarily, then settle out.
     
  10. Jun 13, 2017 #9
    So in theory, an 'acme' voltmeter with infinite internal impedance could measure a voltage, not necessarily 12V, between a battery terminal and ground, assuming ground to be baseline, 0V.

    See how trying to explain something makes you think? What I thought was a misconception about voltage and charge turned out to be perhaps a valid, if purely theoretical, point.
     
  11. Jun 13, 2017 #10

    scottdave

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    While voltage and charge are not the same, they are related. A charge creates an electric field around it. The electric field is a voltage gradient, which changes across distance.
     
  12. Jun 13, 2017 #11

    davenn

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    Again .... ONLY if the other terminal of the battery is connected to ground
     
  13. Jun 13, 2017 #12

    sophiecentaur

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    This is true but the 'imbalance of charge' you refer to would not correspond to 14Amp Hours worth of charge. The charge that the battery can dole out is only when the terminals connected via an electrical path. There is chemical energy (electrically based energy, of course, involving electrons, ions and neutral atoms in there). The Potential Difference corresponds to the Energy for each Coulomb that passes across the terminals. (Volts is Joules per Coulomb)
    If, instead of a battery, you had a petrol generator, the PD would be maintained by the energy from the burning petrol and the idea of 'charge imbalance' is less meaningful; it would (as always, basically) be the Energy that counts.
    It's a bit of Chicken and Egg and you can look at it in several ways but if you physically displace charges by moving two plates apart, you can increase the PD but the displaced charge doesn't change. As with all Physics, it's always dodgy to start talking about "what's really happening". It's all about the Relationships between Variables.
     
  14. Jun 13, 2017 #13

    scottdave

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    You might find this interesting. Note this is not "free energy" (any more than wind, solar, etc is).
     
  15. Jun 13, 2017 #14

    jim hardy

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    A couple of significant little points there.

    1. "Ground" is a widely misunderstood term. Most folks mean by it the point of reference they've selected , the place they intend to connect their voltmeter's black wire. I much prefer to call it "Circuit Common" or "Zero Volt Reference" because it may not be "ground" at all - consider the batteries in a satellite.
    I assume you meant the real earth ground Terra Firma in your scenario.

    2. A "Real" voltmeter with not infinite impedance l, in your scenario,
    will allow charge to flow from the object you're measuring to the reference point you've selected(Earth Ground?) .
    That will let your object come to same absolute potential as your reference, 'bleeding' down the voltage between them to zero. If the meter is fast enough it might show the bleed-down.
    An ideal one would , as you suggest, report the actual potential difference without bleeding it down.
    There's something called an "electrostatic voltmeter" that's mighty close to ideal .. Check Wikipedia.

    If you live up North where the air gets dry you have "felt" that voltage between a battery negative terminal and ground.
    When on a dry day you exit your car with one hand on the door and get that shock when your foot first touches the ground what happens ?
    The car is charged with static electricity and the tires are a decent insulator so there IS considerable voltage between its battery negative(chassis) and earth.
    Think of your hand as a 'digital voltmeter' , it reports that voltage as measured in "ouches" .

    Here's an old thread that's sort of related. I hope you find it interesting. It's good to get these basic concepts straight - saves a lot of unlearning later on.
    https://www.physicsforums.com/threads/what-does-more-less-voltage-mean.852532/#post-5347090
     
  16. Jun 13, 2017 #15
    Yes. In the UK, we generally call a connection to the soil 'earth', as in 'earth rod', and 'earthed metal chassis'; A local circuit reference is usually 'ground'. This is not totally consistent, however.

    I like your idea of a 'digital' voltmeter. I'll use that next time I get a nip from an HT lead: "Just using my in-built digital voltmeter - one f*** per hundred volts."

    So although for convenience we consider earth to be 0V, the true absolute potential would only be revealed by bringing a unit charge from infinity to it. It's just a useful, local, common reference.

    This chap (a Youtube commenter) also referred to a gold leaf electroscope - charge it up, and the leaves part; touch the plate, and they fall back. He was thinking a car battery would behave similarly, albeit at lower voltage. I've not seen one, but I'm guessing the simple answer is, the electroscope must be ground referenced.
     
  17. Jun 14, 2017 #16

    sophiecentaur

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    No. The electroscope just registers the charge imbalance on it - as long as it is not near another large object. The mutual Capacity between the scope and the Earth / bench / floor is too small to affect the charge distribution on the metallic parts. If you connect it to Earth via a large capacitor (maintaining the same 'charge' on it), the excess charges will all rush to the capacitor plate and the leaves will drop. They will rise when the capacitor is disconnected from Earth.
     
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