Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic question regarding applied voltage and circuits

  1. Oct 16, 2011 #1
    So, in a physics lab I TA, we are doing a lab in which part of it, the students try to do Faraday's Ice Pail experiment. For part of it, they need to charge a piece of metal at the end of a insulating rod. To do this, the guy who designed the lab did the following.

    He has a voltage supply, and sets it to 2000V. Now, this is very important. He attaches the positive output of it to a large metal sphere (just like a Van de Graaf sphere, but not powered or anything. Just at the top of an insulating rod). And he doesn't attach the ground of the power supply to anything. He claims that some positive charge should move to the metal sphere. We then have another (insulated) metal sphere close to the first one, which is supposed to exhibit charge separation. Then, the students should be able to rub that piece of metal to various parts of this second sphere, and get different charges (positive if it's on the side of the second sphere farther from the first sphere, negative if it's on the closer side) on this piece of metal.

    Experimentally, this just wasn't working. The piece of metal just didn't seem to be getting charged (putting it inside metal containers with an electrometer attached).

    Now, this immediately set off alarm bells in my head. We've had it drilled into our heads since day 1 that in direct current circuits, no current flows unless there is a complete circuit. Of course, there is a way current can flow. If there is a capacitance between the first metal sphere and the ground of the PSU, it would act as a small capacitor and allow some positive charge to flow to the sphere, and a small bit of negative charge to flow out from the electrode of the PSU.

    The head guy's logic was thus. We know that when we attach the positive electrode to the metal sphere, the sphere must become the same potential (2000V here) as the PSU. He thinks that this means a "tiny" bit of charge flows from the PSU to the sphere, to put it at the same potential. This is what he thinks should allow us to do the experiment. I think that it puts it at the same potential, but no charge flows (because there is no complete circuit, minus the capacitance effect).

    I have a counterexample for thinking this isn't the case: Let's say a "little bit" of charge flows to a piece of metal, if we, say, just touch the positive end of a AA battery to the metal. Now, this might seem reasonable if the piece of metal is an inch long piece of wire. But what if we take the limit, and say the piece of metal is 100 miles long? Does the wire just completely drain the battery to put it at the same potential? This seems unlikely to me.

    So, what is the answer? Should any charge flow, theoretically? Practically?

  2. jcsd
  3. Oct 16, 2011 #2
    I'm not sure, but I'll throw this out there, I know that some power supplies use near ground potential positive and a large negative potential, supplying most of the energy through electrons. (I think this helps to make them more safe) But I know they make them the opposite way as well, so perhaps your power supply doesn't provide the potential on the positive side. So maybe try the negative?
  4. Oct 16, 2011 #3
    Errr, not sure what you mean by "supplying most of the energy through electrons." Isn't that normally how electricity works? We talk about "positive charge carriers" but they rarely exist.

    It shouldn't really matter though. I'm more worried about the theory side here.

    Any ideas?
  5. Oct 16, 2011 #4
    Thats what i'm trying to say here... if you have a low positive potential, you won't do much to your conductor. If thats the case, your negative terminal will do the job for you. Just because it says 2kV supply, doesn't mean you've got 2kV on the positive side of your supply. You've probably got 2kV negative potential, and a low positive. the difference between the terminals is 2kV, but there are two types.

    This is a famous experiment, so you gotta realize its your setup and not the theory thats wrong here.
  6. Oct 17, 2011 #5
    Oh, I definitely agree that our setup is wrong. In fact, I think the obvious solution is to attach the ground to something, so the first sphere will actually be a capacitor, and have charge build up.

    I'm just not really sure what you mean... I'm pretty sure voltage is just relative, so there's effectively no difference between saying the ground is 0V and the positive is 2kV, vs. saying the "ground" is -2kV and the positive is 0V, right? I mean...you can call them whatever you want, the important part is just that the positive is at a higher potential than the negative, right?
  7. Oct 17, 2011 #6
    If the power supply has a "ground" output I would expect that to be grounded internally, so you shouldn't need to attach it to anything. Does it have a positive and negative output terminal? If so, I think it is common practice to connect one of these to ground. On some power supplies I have used, there is a panel on the back of the supply where you can connect either the negative or the positive to ground with a jumper wire or something. That's all I can think of that might explain your problem.
  8. Oct 17, 2011 #7
    Here is a link to some company selling dual DC power supplies. Check out the picture. This 9V DC PSU shows what I mean. There is a terminal for +9v, 0v, and -9v. choosing 0v and -9v, gives you a positive potential of 0V relative to the earth and -9v on the negative terminal.


    I suspect you are touching your +0V terminal to your sphere. Try using the -2kV terminal.
  9. Oct 17, 2011 #8
    It definitely has a ground output, a black wire coming out of it. I'm sure it is grounded internally, to the ground jack in the wall the whole PSU plugs into. But what I'm getting at is that the metal sphere doesn't "know" about the rest of the circuit because the 2kV isn't relative to anything, because it's not connected to the rest of the circuit. It would be like saying something is 10 feet high, which is a pretty meaningless statement unless I also tell you where 0 feet high is.
  10. Oct 17, 2011 #9
    Ok, let's assume we are. That still shouldn't change my problem. Let's say this was my problem and now I touch the -2kV terminal to the sphere, and leave the 0V one totally unattached to anything.

    Do you expect any charge to flow to the sphere, now? This is essentially the same problem as before, I think...
  11. Oct 17, 2011 #10
    What do you mean 2 kV isn't relative to anything? It's relative to ground, since the power supply is grounded. The metal sphere may have a small charge on it, but you can bet it is very nearly at ground as well, especially since it's been touched recently.

    It is not the case that you need to have a "complete circuit" for charge to flow. All you need is a potential difference. That potential difference will quickly go away once charge starts flowing, hence the need for a circuit if you want any steady state current. But the initial charge transfer must occur. When you learn about DC circuits initially, you completely ignore any transient behavior and only focus on steady state.

    Use a voltmeter and make sure the various outlets of your power supply are really putting out what you think...you can do this at a lower voltage.
  12. Oct 17, 2011 #11
    Depends on the design of the supply, if it is similar to a battery or capacitor, ideally no charge would flow unless you connected both ends.

    However, with this power supply, I am implying the 0V potential is the same potential as the earth and its surroundings (your sphere for ex). When your circuit is grounded (The word comes from literally using the earth as a sink) current can flow from the higher or lower potential to anything, including your sphere.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook