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my name is Amin and I'm a High school student. I have been struggling with the concept of the electric field produced by a battery for months; so I'd really appreciate it if I could get some assistance with understanding the answers to the questions below (Just to clarify, my knowledge in this topic is very limited: However, I know and understand the Maxwell's equations to an acceptable degree):

Q(1): Would it be reasonable to say that a battery behaves like an electric dipole?

Q(2): Is my depiction of the electric field produced by a battery correct? (The battery hasn't been connected to a wire)

enter image description here

Q(3): If the answer to the previous question is YES, how does the above illustration tie in with the chemistry behind the operation of a battery?

To understand this, I studied the topic about electrochemical cells. Using an electrochemical cell consisting of two half-cells containing a zinc and a copper electrode as an example, my understanding of this topic could be summarised in the illustration shown below:

enter image description here

(Based on my understanding, since Zinc favours oxidation the position of equilibrium for the reversible reaction shown on the left will be shifted to the right. Similarly since Copper favours reduction the position of equilibrium for the reversible reaction shown on the right will be shifted to the left. This would imply that the electron charge density on the Zinc electrode will be greater than that of the copper electrode)

Q(3)": Doesn't this mean that the electric field produced by the battery should look like the illustration shown below, instead of looking like a dipole?

enter image description here

(sorry if the picture is unclear; I was only able to show a general direction for the field lines in the picture.)

Q(4): Now, what ever the answer to the previous questions may have been, I understand that when a battery is connected across the ends of a circuit electrons would be pushed away from its negative pole and attracted towards the positive pole (It makes sense that the electrons would leave the electrode with the lower electric potential and move towards the one with a higher electric potential).

My question, here is that "Does the electric potential difference across the ends of the battery stay constant as electrons leave one electrode and move to the other one? (In other words, does the electron charge density on both of the electrodes stay unchanged when a current is present?So are there the same number of electrons present on each electrode at all times?)

Q(5): If the answer to the previous question is YES (meaning that the electron charge density on the electrodes doesn't change when a current is present), then does that mean that the electrons that will be going through the circuit are the ones that were present in the circuit in the first place and aren't from the battery? (Would it be reasonable to say that the electrons that have accumulated on each electrode only provide a force which directs the electrons in the circuit and that they do not take part in producing a current themselves?)

Q(6): If the electric charge on the electrodes behave in the way described in the previous question, would the following argument be correct?:

enter image description here

"Given that the circuit is open (shown above) there will be no current going through it. However, when such set-up is arranged there will be a brief current for an instant of time until the ends of the two wires become charged. This charging would continue until the electric potential of each of the ends of the two wires equals the electric potential of the battery terminal it is connected to.(The electron charge density at the end of each wire becomes equal to the electron charge density of the terminal it is connected to)"

If the above argument is incorrect, what's the correct behaviour of the charge in a wire at the very instant it is connected to a battery (in an arrangement such as the one shown above)? and why?

Q(7): What causes the current in the open circuit shown above to stop after a short instant of time (If the electric charge density of , say the negative pole, has stayed constant what opposes the further pushing of the electrons out of the negative terminal)?

That's all. I'd really appreciate it if you could help me with any of the above questions.

Thank You

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closed as too broad by JMac, Jon Custer, sammy gerbil, Kyle Kanos, M. Enns Apr 4 '18 at 19:52

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ this site does not accept multiple questions in one question. It should be seven different questions. $\endgroup$ – anna v Apr 3 '18 at 14:10
  • $\begingroup$ For tinkering around you might consider a high sensitivity electrometer: amasci.com/emotor/chargdet.html (or a more sophisticated implementation of the same idea edn.com/design/analog/4417628/…). $\endgroup$ – dmckee Apr 3 '18 at 15:11
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This answer was borrowed almost entirely from this COMSOL blog post: https://www.comsol.com/blogs/does-the-current-flow-backwards-inside-a-battery/

A battery is an electrochemical object, and it cannot in general be treated as an electric dipole. This is why you're having trouble reconciling the electric-dipole picture with the electrochemical picture - the dipole picture isn't at all realistic.

The potential inside a battery (and therefore the electric field) looks different depending on whether or not current is flowing. If the battery is disconnected from any circuit, then the electric potential within it looks like this*:

enter image description here

As you can see, the electric potential is flat for nearly all of the battery; this means that there is no electric field anywhere except in a small region of the electrolyte that is close to each electrode. In this region, the double layer (marked DL on the diagram) at the surface of the electrode is the only place that has any electric field, and that's because, for the negative electrode, the double layer consists of a layer of negative ions on the surface of the electrode, and then a layer of positive ions attracted by the negative ions (reverse everything for the double layer at the positive electrode). This double layer is extremely thin, on the nanometer scale for most batteries.

When the battery is discharging (producing current), its potential looks like this:

enter image description here

There are several things that are different here. Most prominently, there's a reference electrode inserted in this diagram; ignore it, as it doesn't change anything about the potential shown. The first important thing that's different: there is now an electric field across the electrolyte which allows a current to flow inside the battery (note that this diagram uses the electrical-engineering convention of current as the flow of positive charge; as such, it describes the motion of electrons inside the battery from the positive terminal to the negative terminal, "backwards" from the usual direction of flow outside the battery, but consistent with electron flow in a loop around the circuit).

The fact that the potential has a slope across the electrode means that the electrode potentials will also be at different levels. The double-layer potential differences are also slightly weaker, due to the fact that charge moving across the double layer interferes with the charge separation between the two layers. The potential within each electrode is also slightly sloped, because there is a small resistance associated with current passing through the electrode. So now there is an electric field everywhere in the battery, but the magnitude depends on exactly where you are, and it definitely doesn't look like a dipole.

*Most batteries nowadays have porous electrodes, rather than the solid ones that I'm using here. The referenced blog post covers the case of porous electrodes, but it's conceptually more complicated, so I used the solid-electrode diagrams to avoid confusion.

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  • $\begingroup$ Very useful – for me, anyway. You don't say anything about the electric field $outside$ the cell. Perhaps not that far off from dipolar? $\endgroup$ – Philip Wood Apr 3 '18 at 15:03
  • $\begingroup$ If you're far enough away, it might look like a dipole, but then again, when you're far enough away, nearly everything looks like a dipole (in fact, that's what the electric dipole moment measures: how much a particular potential looks like a dipole). But if you're close to the battery, there will be significant differences, as the potential above doesn't look very much like a dipole potential (en.wikipedia.org/wiki/…). $\endgroup$ – probably_someone Apr 3 '18 at 17:44

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