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Electrical ground

  1. Aug 22, 2008 #1
    I do not understand the concept of electrical ground, or why it works. Let me explain.

    In elementary school physics, one learns that for a current to take place, you need a battery with positive and negative poles. This creates a voltage that makes charges run from plus to minus, because pluses would like to be with minuses. So therefore, if you connect wires and a lamp, you can make the lamp shine.

    This is how I learnt that an electric current arise. So I cannot understand, how there could be a current from, for instance the plus pole of the battery to the earth - there are no minuses in the earth that would attract the pluses, the earth is neutral (as opposed to the minus pole on the battery that is negative).
     
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  3. Aug 22, 2008 #2

    chroot

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    Voltage is a purely relative concept. A 9V battery, for example, is called a 9V battery because it produces a potential difference of 9V between its terminals. You are free to refer to the terminal voltages individually as, say, 3V and 12V, or 19V and 28V.

    The term "ground" is used to denote a reference potential, and all other voltages in the circuit will be measured with respect to it.

    In many systems, the potential of the earth itself is a reasonable reference potential, since we can all agree upon it (it's under all our feet). However, in many electrical systems -- those powered by batteries, for example -- the "ground" potential has nothing to do with the actual earth at all.

    - Warren
     
  4. Aug 22, 2008 #3

    turbo

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    Ground is useful for both AC and DC devices. When I was a kid there was a telegraph line from town to a remote logging camp and IIR, there was just a single copper wire -the return path was ground. In a more complicated DC circuit, it is often useful to reference parts of the circuit to ground. In an AC (household) circuit, the neutral leg is referenced to ground, and the two 120V hot legs vary with respect to that ground reference. The step-down transformer that feeds your house takes transmission-line voltage and steps it down to 240V. The secondary coil is center-tapped, and this is where the neutral line to your entrance panel originates. At the electrical entrance to your house there will be a local ground driven into the ground or clamped to incoming water pipes etc, and a heavy ground wire is run from this ground to the entrance panel and is clamped to the same buss as the incoming neutral wire.

    Edit: Warren jumped in ahead of me. He"s quick.
     
    Last edited: Aug 22, 2008
  5. Aug 22, 2008 #4

    tiny-tim

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    Hi Mårten! :smile:

    "pluses would like to be with minuses" … no … you're imagining that there are positive charges at one end of the battery, and that they are attracted to negative charges at the other end.

    All the charges are negative, aren't they? They're all electrons!

    The electrons drift "downstream" along the potential difference.

    Electrons don't need a "positive" plate to be attracted to … they only need to feel a greater pull in one direction than the other (which is what a voltage gradient is).

    So between a plus pole of a battery and the earth, there is a potential difference (though only roughly half that from the plus pole to the minus pole). :smile:
     
  6. Aug 22, 2008 #5
    Hi tiny-tim and you others! :smile:

    I thought it was an old convention to go from plus to minus, and that it didn't matter, even though it really is electrons that runs...

    Okey. So I've now done some experiments. I have three 1.5 V batteries in series (=4.5 V), a wire connected to the minus terminal on the last battery, and the other end of the wire connected to the metal on the side of a small 3 W bicycle lamp. I connect the bottom of the lamp to the plus terminal of the first battery, and the lamp is shining nicely and brightly. (I've also tried the lamp with just one battery (1.5 V), and it shines okey, not much, but it shines clearly). Now I try the following:

    1) I connect the bottom of the lamp to a big metal shelf here. Since there should be a potential of 2.25 V on the minus terminal of the last battery in the series, and the shelf should have 0 V potential, there ought to be a current - but no light. Why?

    2) (Not recommended) I connect the bottom of the lamp to the grounding pin in my AC outlet in the wall (it's a new house, so I know the grounding is properly installed here). But, still no light. Why?
     
  7. Aug 22, 2008 #6

    chroot

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    Because there is no continuous path for current to flow from one terminal of your battery pack to the other. Also, there's no reason why the shelf should be considered to have "0V potential." As I've said, voltage is a purely relative quantity. You cannot say "an object has a potential of 0V," only that "an object has a potential of 0V with respect to some other object."

    Again, because there's no continuous path for the current to flow.

    - Warren
     
  8. Aug 23, 2008 #7

    atyy

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    Let's think of a slightly simpler situation, dealing with a capacitor, instead of a battery.

    A capacitor is just two metal plates separated by air.

    If you put 10 positive charges on one plate, and 10 negative charges on the other, they will stay there, because the charges cannot flow through the air. If you now connect the plates with a metal wire, then the 10 positive charges will flow to the other plate and neutralize the 10 negative charges, exactly as you conceived.

    Now what happens if you put 10 positive charges on plate 1, and zero negative charges on plate 2? If you don't connect the plates, nothing will happen. If you connect the plates, will the positive charge flow from plate 1 to plate 2? Yes - you don't need negative charge on plate 2 for the charge on plate 1 to flow. Positive charges like to get as far from each other as possible. So if you give them a way to distribute themselves 5 on each plate, they will try to do that. Thus 5 charges will flow from plate 1 to plate 2 once you connect the plates with metal wire. The first charge to go will get to plate 2 quickly. The second charge to get there will take a little longer, because the first charge that just got to plate 2 will push him back a bit, but he will still get there, because the 8 remaining charges on plate 1 are pushing him away much harder than the 1 charge on plate 2. You can make up the rest of the story yourself.

    The main point about the earth is not that its potential is zero. Potential itself is meaningless. If you have a 9 V battery, you can call the negative terminal 10987 V, in which case the positive terminal will be 10987+9 V, and that will be completely correct; you can also call the negative terminal -926.765 V, in which case the positive terminal will be -926.765+9 V, and that will also be completely correct. Earth is usually given a potential of 0 V, but that's convention, it contains no physics, it is like deciding whether to call your negative terminal 10987 V or -926.765 V.

    The important part about earth is that it is like a very, very, very, very, very big plate with zero charge. So if you connect plate 1 with 10 charges to earth, the first charge to go will go quickly, just like if you connect it to plate 2. Unlike when you connect it to plate 2, the second charge to go will not take a longer time to reach the earth, because the first charge has lots of room on the earth, so much room that it will not push back on the second charge. In fact, there is so much room on the earth, that all 10 charges will be gone from plate 1 to the earth - quite unlike having 5 charges remaining on plate 1 when you connect it to plate 2. Furthermore, there is so much room on the earth, that although there are 10 more charges on the earth, we can still pretend that the earth has zero charge. That is the important idea about earth - it always has zero charge, no matter how much you charge add to it. It's obviously an approximation, but it's an excellent one as long as you don't have access to a Death Star.
     
  9. Aug 23, 2008 #8

    atyy

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    Since you can never change the number of charges in the earth, you can never change its potential.

    This means that whatever you connect to earth will have the same potential as earth very quickly. Let's adopt convention and call earth's potential 0 V. With that convention, the plate with 10 charges has a positive potential. But once you connect it to earth, the charge on the plate quickly becomes zero, just like earth, so the plate's potential will be 0 V, just like earth.
     
  10. Aug 23, 2008 #9

    tiny-tim

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    … capacitors confuse me …

    Hi atyy! Hi chroot! :smile:

    capacitors confuse me. :confused:

    Aren't you ignoring the electrostatic force in the capacitor?

    And is the potential of the shelf relevant?

    An excess of Q electrons on one plate will repel Q electrons from the other plate.

    For example, if you have a compete circuit with a battery and two capacitors in series, the emf of the battery will (loosely speaking) force Q electrons from the outer plate of one capacitor all the way round the circuit onto the outer plate of the other capacitor.

    But the emf isn't connected to the inner plates, so it can't influence them and yet we know that the inner plates will have charges ±Q also.

    This is because the Q on each outer plate repels Q electrons from one inner plate to the other.

    Now charge up a capacitor, disconnect it, and connect the plus plate plate to the earth, through a lamp.

    Do Q of the inexhaustible pool of electrons in the earth rush over to fill up the Q holes?

    No, because the other plate still has Q electrons, and so it will still repel any electrons trying to fill those holes.

    No electrons will flow, and the lamp will not light.

    And the potential of the earth is irrelevant.

    I'm honestly not sure how a battery works :redface:, but I assume the potential pressure inside it is something like that "inside" a capacitor, and so connecting only one terminal to the earth will not allow electrons to flow from the earth because the other terminal will still repel them.

    But where does potential difference come into this? :cry:

    Afterthought: connect two capacitors of different capacitances in a circuit. Then connect a battery across opposite sides of the circuit, until the capacitors are fully charged. Now disconnect the battery.

    The unequal charges on the two capacitors will now even out, with charge moving from one capacitor to the other.

    This is despite the fact that both + plates are at the same potential, and so are both - plates, at all times.

    So charge moves, and current flows (for a short time) without any potential difference. Have I got that right? :confused:
     
  11. Aug 23, 2008 #10

    atyy

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    Re: … capacitors confuse me …

    The capacitor is just a pair of metal plates. If there is no charge on the metal plates, there is no electrostatic force. Electrostatic forces are forces between charges - it's equivalent to the idea that positive charges repel positive charges, whereas negative charges and positive charges attract each other. In general, the force between charges is not purely electrostatic, but the electrostatic approximation is good as long as the charges aren't giving off high frequency electromagnetic radiation.
     
  12. Aug 23, 2008 #11

    atyy

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    Re: … capacitors confuse me …

    OK, I guess this is the situation where you have 10 positive charges one plate of a capacitor, and 10 negative charges on the other plate. If you connect the positive plate to a bulb, and the bulb to the earth, the inexhaustible pool of electrons will rush up and neutralize the positive plate. Or if you prefer, you can think that the positive charges want to get as far away from each other as possible, so they will all go into the huge room that earth provides. Of course, the positive charges like to be near the negative charges on the other plate, but then they have to be all squeezed together on one small plate. They don't like 10 negative charges on the other plate that much to resist the infinite room that earth supplies for them to get away from other positive charges. So the bulb will light (in principle, to get this effect experimentally you may need more than 10 charges)- but only very briefly, because the current will stop once the positive charges have gone into the earth.
     
  13. Aug 23, 2008 #12

    tiny-tim

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    Re: … capacitors confuse me …

    Hi atyy! :smile:
    Yes, but I was talking about a charged capacitor, with one plate connected to earth …

    surely the charge on that plate will be determined only by the electrostatic force from the (fixed) charge on the other plate …

    and so it won't change, whatever the potential of the earth? :smile:
     
  14. Aug 23, 2008 #13
    Hi all!

    Thank's for some very thorough replies! I'm getting closer... :smile:

    Let me first say this, which I forgot in the first post here: In my elementary school physics understanding, I also learnt that you cannot have a current, if you don't have a closed circuit. So that's also something that has confused me, since when you have a ground in the circuit, it (the circuit) doesn't seem to have to be closed in order to make a current run. Now:

    I think I understand the relativity here pretty well. But - as atty said - 10 positive charges on one plate, and zero on the other plate, will make some of the ten charges to flow to the other plate. So why doesn't my excess of charges on the terminal of the battery flow to the other plate (i.e. the shelf, or the ground pin in my outlet)? Is it some property of the battery here that needs to be considered to explain this? Tiny-tim, you seemed to say that this experiment should produce a current from the battery terminal to the ground, but you may have changed your mind?
     
  15. Aug 23, 2008 #14

    atyy

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    Re: … capacitors confuse me …

    I think the unequal charges on the capacitors will not even out, so there will be no current flow. The capacitance kind of measures how big the metal plates of the capacitor are. So if you have two capacitors with different capacitances, they have plates of different sizes. So let's say one plate has capacitance 10 and the other plate has capacitance 3. When you charge the capacitors up with a battery, you will put 100 charges on the bigger capacitor, and 30 charges on the smaller capacitor. When you disconnect the battery, the 100 charges will not rush to even out the 30 charges on the smaller capacitor, because the plates of the big capacitor have 10/3 times more room than the plates of the small capacitor, so 100 charges on the big plate to 30 charges on the small plate already corresponds to all 130 charges being as far away from each other as possible.

    You will find that capacitance also depends on distance, so it could be that the plates of the big and small capacitor have the same size, but the plates of the big capacitor are further apart than the plates of the small capacitor.

    How is this related to potential difference? Potential difference measures the tendency for charge to move in a certain direction. So if you have one + charge and one - charge, there will be a potential difference between them, since they will want to move together. But potential difference also depends on distance, since if the + and - charges are very far from each other, their mutual attraction will be smaller, and the potential difference between them will also be smaller. This is why it is possible to change capacitance by adjusting either the area of the metal plates, or the distance between the plates.
     
  16. Aug 23, 2008 #15

    atyy

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    Re: … capacitors confuse me …

    Oh, I see what you mean. If you have negative charge on plate 1, and no charge on plate 2, then you connect plate 2 to earth, the negative charge will attract positive charge from the earth to plate 2.

    OK, in this case, I would predict that if you have negative charge on plate 1, and no charge on plate 2, then you connect plate 2 through a bulb to the earth, the bulb will light when positive charge moves from the earth to plate 2.

    Which is the opposite of what I thought earlier ....

    Well, at least I'm predicting both times that the bulb will light;)

    I think we have to be careful by what we mean by 10 positive charges, or 10 negative charges ...
     
  17. Aug 23, 2008 #16

    tiny-tim

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    Hi Mårten! :smile:
    ooh … I'd forgotten that!

    Yes … I think I have changed my mind … subject to what everyone else says :rolleyes:

    I can see that if you connect an overhead supply ("high-tension") cable to the earth, a current will certainly flow, because the cable is kept at a large potential difference to earth. But is that only because the "battery" at the power station is still connected? :confused:

    If you disconnect that battery, will the potential difference drop to zero (unlike the charge on the plate of a capacitor)?

    So does current flow through the earth from a domestic power circuit because the power source is still in a circuit (a different one), and able to maintain a potential difference, but not from one terminal of a battery through your shelf because the power source (the battery itself) is not in any circuit? :confused:
    Hi atyy! :smile:

    Yes, you're right … no charge will flow … I wasn't thinking straight. Sorry! :redface:
    But can you start like that? Doesn't the electrostatic force mean that the charges on the two plates of the same capacitor are always equal (and opposite), whatever happens? :confused:
     
  18. Aug 23, 2008 #17

    atyy

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    OK, I think you are right that if you charge up a capacitor then connect it to a bulb then to ground, nothing will happen. But I'm somewhat confused about this. Am off to get some sleep, see you guys later!
     
  19. Aug 23, 2008 #18
    Hm, I'll try to organize a little bit here of what we have concluded...

    If we have a charged metal plate and connect it to a metal shelf, a wire or the earth, the charge on the plate will be evenly distributed on the new system that contains e.g. the metal plate and the metal shelf. This new system is still charged, but because the system has a bigger volume, the charge density is less. If the system happens to be the metal plate and the earth, the charge will be also evenly distributed, but because the earth is so enormously big, the charge distribution will be so low, so the amount of charges on the metal plate part of this new system, will almost be zero. Now, all this is ultimately due to Coulomb's law. - Have I understood the concept of grounding correct so far?

    If we talk about capacitors and batteries, it seems to be a little bit more complicated. The plus plate of a capacitor will maybe not lead current to the earth (if connected to the earth), since the negative charges on the other plate of the capacitor holds the plus charges in its grip. This due to electrostatic induction between the plates. Is that correct? Something similar applies to batteries.

    Now, when is the electrostatic induction between the metal plates of the capacitor, or between the plus and minus parts inside the battery (or whatever happens in the battery) less than the tendency for the charges to be evenly distributed in its conductor where they resides? If you connect one of the metal plates to the earth, the electrostatic induction seems to win. If you connect one of the metal plates to the other metal plate, the electrostatic induction seems to lose. - So why then? :confused:
     
  20. Aug 24, 2008 #19

    atyy

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    OK, I would like to change some of my answers.:smile: Here are my latest guesses. I am making my guesses under the approximation that the earth has an infinite supply of free positive and negative charges, and that no matter how much you add or subtract from it, the earth remains uncharged. So anything you connect to it will end up having the same potential as the earth. Thus there cannot be charge on anything that you connect to earth.

    1) Charge a capacitor, disconnect the battery. There are 10 + charges on plate 1, and 10 - charges on plate 2. Connect plate 2 via a bulb to earth, and the 10 - charges should flow to earth, lighting the bulb momentarily, and finally leaving plate 2 uncharged.

    2) Charge a capacitor, disconnect the battery. There are 10 + charges on plate 1, and 10 - charges on plate 2. Now we take away plate 2, and replace it with plate 3 which is uncharged. Connect plate 3 via a bulb to earth, and nothing should happen.

    So what about the rule tiny-tim cites about the charges on opposite plates of a capacitor always being the same? Perhaps this is a rule from circuit theory, and only works under conditions in which the opposite plates are connected within some closed circuit?

    I know there are times when circuit theory fails. For example, in circuit theory, you can change the capacitance of one capacitor without changing the capacitance of the other capacitor. Since changing the capacitance of one capacitor is equivalent to moving one metal plate about, that means that moving one conductor about doesn't affect the capacitance of the other capacitor. But there are times when this is not true. When you tune a radio, you are adjusting a capacitor in the radio. As we all know, if you tune the radio perfectly, then move away from it, the radio can become slightly untuned. This is because you are a conductor, and your position relative to the capacitor in the radio affects its capacitance, and hence the tuning of the radio. So that's an example where the rule from circuit theory doesn't work.

    BUT I am not sure whether my latest guesses are just wrong, or whether we are actually bumping against a limitation of circuit theory here.:confused:
     
  21. Aug 24, 2008 #20

    tiny-tim

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    cramming work?


    This electrostatic force has really been bugging me … :mad:

    I think you're right, atty, and exact equality occurs only in a "circuit", and then only from conservation of charge.​

    I think this is how it works:

    Imagine an isolated metal sphere N, already holding a negative charge, of 100 electrons.

    Those 100 electrons will be repelling each other, and will try to get as far away from each other as possible, so they'll be fairly uniformly distributed around the sphere.

    Attach a metal wire to N … only one or two of the electrons will go into the wire, because the wire is small, and any more would be repelled back onto the sphere.

    Connect the wire to the earth, E … now all 100 electrons will be able to go to E, because E is so large (or will 1 electron remain on N, simply because it has no reason to move? :confused:).

    In other words, the electrons move from N to E not because of any potential difference, but because they repel each other, and E is a great place to hide! :smile:

    Now start again, with the same negatively charged sphere N and a positively charged sphere P with 200 "holes".

    Bring them close together … the closer they are, the more the electrons on N will be attracted to the holes on P, so the electrons and holes will no longer be uniformly distributed.

    Connect the wire to N as before, and then connect E to the wire.

    This time, the 100 electrons will still be repelling each other, but they will also be attracted to the 200 holes on P. So not all will want to get away. :smile:

    How many will stay on N? Will any more arrive from E? :rolleyes:

    I dunno …

    Perhaps we can work it out like this … suppose P and N (in their final position) were originally charged, in the same circuit, so as to have 200 holes and electrons respectively.

    So the 200 electrons on N could have gone a long way away (to the battery, which we'll assume is as good as infinitely far away, so far as electrostatic force is concerned). But they didn't, 'cos their natural repulsion for each other was exactly balanced by their attraction for the holes on P plus the "pressure" from the battery.

    Now disconnect the battery, and connect N to E … the "pressure" has gone, but the attraction from P is still there! :smile:

    So some of the electrons on N will go to E, but some will remain crammed on N, attracted by P.

    So the questions seem to be,
    i] what is the work is done by moving a charge Q though a voltage V, compared with the work done by cramming a charge Q (made up of many mutually repelling electrons) into a small volume? and
    ii] what is the electrostatic attraction compared with the battery "pressure"?

    At this point, I have no idea how to calculate either.

    Though it seem obvious that the smaller the area A of the plate, the more the cramming work … and capacitor plates are very small compared with metal spheres … and that a "positive feedback" means that the larger Q is, the more the electrostatic attraction is "helping".

    I've never even seen cramming work mentioned … but it seems an essential part of the operation of a capacitor.

    Dimensionally, one would expect it to be something like Q²/ε√A (which is much larger than the standard work done, Q²d/εA). :confused:

    Or is it insignificantly tiny? :redface:
     
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