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This is a question regarding the physics behind the observation.

I have guessed the answer to the question, but I may be wrong, so I want to wait for the responses before posting it.

Some major considerations:

  1. A battery's positive terminal does have a positive potential. ie, a test positive charge will repel it and a test negative charge will attract it. Vice versa for negative terminal. From the paper below (Section 1.2.1), it seems abundantly clear that the battery will have positive and negative potential on respective terminals.

  2. Given 'point 1', above, connecting the positive terminal of battery A to negative terminal of battery B will lead to current flow in the conductor.

  3. If the potential difference only comes into effect when the circuit is closed using the terminals of the same battery (due to cell chemistry), then does that mean that 'point 1' will not be valid after this experiment (connecting positive terminal of A to negative terminal of B)?

  4. If it is only a tiny quantity of current that flows, then will it ever stop? And what determines this quantity? (Equation would be nice)

Related paper: http://www.astrophysik.uni-kiel.de/~hhaertel/Circuit/electric_circuit.pdf

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I think this picture answers the question CuZn battery

There battery's terminals are not charged. It's just a chemical reaction that starts the charge.

There's a Zn and a CuSO4. When they meet each other via a cable, the Zn gives 2 electrons to ChSO4.

Cu gets rid of SO4, leaves 2 extra electrons to SO4 and takes the 2 electrons it got from Zn.

Then we have SO4 with 2 extra electrons and Zn with two less electrons. SO4 moves to Zn2 to they bound to each other.

At the end, we have ZnSO4 and Cu.

With two different batteries, it's not possible. The SO4 has nowhere to go and make new connections. Thus the chemical and electrical process doesn't start.

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You need to realize that the terms positive and negative are relative. The positive side of a battery is only "positive" in relation to the "negative" terminal of the same battery. When you hook a wire from the positive terminal of the first battery to the negative terminal of the second, a very small amount of current will flow until the potential difference reaches zero.

Let's take an example with 2 nine volt batteries. If I hook the negative terminal of battery 1 to ground (which we will arbitrarily define as zero volts), and hook the negative of battery 2 to the positive of battery 1, then the negative of battery 2 will come quickly to equilibrium at 9V relative to ground. The positive of battery 2 is now at 18V relative to ground because it is always 9V above its own negative terminal at equilibrium.

As for a short circuit, in order to get a short circuit, I have to provide a complete circular path for current to flow through. I can do this by adding another wire between any two terminals.

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  • $\begingroup$ Sure, I get that negative/ground is merely a reference. So you say that the potential difference reaches zero after the connection. But the paper seems to imply that there is actually positive charge on the positive terminal and negative charge on the negative terminal. So are you saying that that is wrong? will a test charge not be attracted by one and repelled by the other? So okay, after disconnecting the connection, will the potential at the electrodes be different (lower on A+ and higher on B-)? If so, won't V(A+) < -V(A-)? $\endgroup$ – mehfoos Aug 24 '13 at 3:14
  • $\begingroup$ As for the example: let's say the battery connected to 'ground is A and the other one is B. The force on test charge near 'ground' is -F. ie, Force on A+ and A- will be +F and -F resepctively, if A were isolated. Force on B+ and B- if B were in isolation should be +F and -F respectively. Now, when they are connected, force on A+/B- should be zero because the cancel each other. but how will force on A- and B+ become -2F and +2F, which is what it should be, since they're in series, assuming the paper is correct? $\endgroup$ – mehfoos Aug 24 '13 at 3:19
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Suppose that we do as you suggest and connect the positive terminal of one battery to the negative terminal of another. Initially, I agree with your assessment that there is a potential difference between the two ends of the wire and so current will flow. However, charge is flowing from one side of the wire to the other and there is no way for that charge to flow back to the original battery, so there must be a buildup of charge either at the edges of the wire or in the battery's terminals. If we were just using one battery this charge would move through the chemical cells and out through the other terminal, but since we are using 2 it has nowhere to go. I don't think the cell chemistry is particularly important. Even if the charge makes it from the entering terminal of the battery to the unconnected terminal, it still has no place further to go and there will be a buildup.

This buildup of charge will start to equalize the potential difference across the wire. At some point, enough charge will have accumulated that current will stop flowing.

I can't say exactly how much charge will have to move before current stops flowing, but I believe that it will be a small amount. In a normal circuit, there is a surface charge that accumulates near/on places where the wire bends such that the electric field always points in the direction of current flow (this is the microscopic Ohm's Law). This surface charge is usually quite small, on the order of a few electrons worth of charge. I imagine that to calculate the current flow for some specific case would be quite difficult. You would need to know the potential difference and the geometry/configuration of the wire, I imagine.

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  • $\begingroup$ So current will flow initially? So during this flow, the potential of the electrodes have to be affected, right? From the diagram in the section of paper I cited, it seems like there already is a charge build-up on electrodes. If current flows, it will only weaken the charge accumulation, shouldn't it? $\endgroup$ – mehfoos Aug 23 '13 at 22:03
  • $\begingroup$ @mehfoos Yes, the potential difference between the terminals must be affected. If it were not, then current would continue to flow. The charge buildup on the terminals does exist as an equilibrium between chemical and electrical forces, but the charge buildup at the surface of an unconnected battery must be relatively small. How do we know? Because unconnected batteries do not repel/attract each other at large distances. $\endgroup$ – Kevin Driscoll Aug 24 '13 at 0:32
  • $\begingroup$ The charge buildup must be so small that from any macroscopic distance the terminals look effectively neutral. And so it will take only a small amount of charge flowing from one battery to the other to equalize the potential difference between the terminals. $\endgroup$ – Kevin Driscoll Aug 24 '13 at 0:34
  • $\begingroup$ If the potential difference between the terminals is affected, then upon disconnecting the setup, what would happen? Will charges appear back on the electrodes as shown in the figure I referred to, or will it remain neutral? As for macroscopic observation making the terminal looking neutral: Do you mean to say that the positve electrode will no longer attract electrons, for instance? $\endgroup$ – mehfoos Aug 24 '13 at 3:06
  • $\begingroup$ After current has stopped flowing, the system is in equilibrium. That is, the charges have found a new arrangement such that the repulsive forces between the electrons (since they are the charge carries in normal metals) and the chemical forces that sustain the battery exactly cancel. When you remove the wire, you take away some of those electrical forces that act on the charges near the terminals (there were charges on the wire after all, held there by a chain of charges leading back to the other battery where there are chemical forces). $\endgroup$ – Kevin Driscoll Aug 24 '13 at 5:03
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Simply it is because the circuit is not complete. There is no way for the electrons to flow back and without a valid pathway they will not flow.

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  • $\begingroup$ this was a correct observation and I do not know why it got a negative vote and evidently discourage a new user. $\endgroup$ – anna v Feb 22 '14 at 11:59

protected by Qmechanic Jul 30 '15 at 12:46

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