As I remembered, at the 2 poles of a battery, positive or negative electric charges are gathered. So there'll be electric field existing inside the battery. This filed is neutralized by the chemical power of the battery so the electric charges will stay at the poles.

Since there are electric charges at both poles, there must also be electric fields outside the battery. What happens when we connect a metal wire between the 2 poles of a battery? I vaguely remembered that the wire has the ability to constrain and reshape the electric field and keep it within the wire, maybe like an electric field tube. But is that true?


Yes Sam, there definitely is electric field reshaping in the wire. Strangely, it is not talked about in hardly any physics texts, but there are surface charge accumulations along the wire which maintain the electric field in the direction of the wire. (Note: it is a surface charge distribution since any extra charge on a conductor will reside on the surface.) It is the change in, or gradient of, the surface charge distribution on the wire that creates, and determines the direction of, the electric field within a wire or resistor.

For instance, the surface charge density on the wire near the negative terminal of the battery will be more negative than the surface charge density on the wire near the positive terminal. The surface charge density, as you go around the circuit, will change only slightly along a good conducting wire (Hence the gradient is small, and there is only a small electric field). Corners or bends in the wire will also cause surface charge accumulations that make the electrons flow around in the direction of the wire instead of flowing into a dead end. Resistors inserted into the circuit will have a more negative surface charge density on one side of the resistor as compared to the other side of the resistor. This larger gradient in surface charge distribution near the resistor causes the relatively larger electric field in the resistor (as compared to the wire). The direction of the gradients for all the aforementioned surface charge densities determine the direction of the electric fields.

This question is very fundamental, and is often misinterpreted or disregarded by people. We are all indoctrinated to just assume that a battery creates an electric field in the wire. However, when someone asks "how does the field get into the wire and how does the field know which way to go?" they are rarely given a straight answer.

A follow up question might be, "If nonzero surface charge accumulations are responsible for the size and direction of the electric field in a wire, why doesn't a normal circuit with a resistor exert an electric force on a nearby pith ball from all the built up charge in the circuit?" The answer is that it does exert a force, but the surface charge and force are so small for normal voltages and operating conditions that you don't notice it. If you hook up a 100,000V source to a resistor you would be able to measure the surface charge accumulation and the force it could exert.

Here's one more way to think about all this (excuse the length of this post, but there is so much confusion on this question it deserves appropriate detail). We all know there is an electric field in a wire connected to a battery. But the wire could be as long as desired, and so as far away from the battery terminals as desired. The charge on the battery terminals can't be directly and solely responsible for the size and direction of the electric field in the part of the wire miles away since the field would have died off and become too small there. (Yes, an infinite plane of charge, or other suitably exotic configurations, can create a field that does not decrease with distance, but we are not talking about anything like that.) If the charge near the terminals does not directly and solely determine the size and direction of the electric field in the part of the wire miles away, some other charge must be creating the field there (Yes, you can create an electric field with a changing magnetic field instead of a charge, but we can assume we have a steady current and non-varying magnetic field). The physical mechanism that creates the electric field in the part of the wire miles away is a small gradient of the nonzero surface charge distribution on the wire. And the direction of the gradient of that charge distribution is what determines the direction of the electric field there.

For a rare and absolutely beautiful description of how and why surface charge creates and shapes the electric field in a wire refer to the textbook: "Matter and Interactions: Volume 2 Electric and Magnetic Interactions" by Chabay and Sherwood, Chapter 18 "A Microscopic View of Electric Circuits" pg 631-640.

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    $\begingroup$ BTW, there is a paper by J.D. Jackson himself describing this process: ajp.aapt.org/resource/1/ajpias/v64/i7/p855_s1 $\endgroup$ – Joe Oct 23 '11 at 9:34
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    $\begingroup$ Good call on the Jackson paper. Good to hear it straight from the man himself. $\endgroup$ – David Santo Pietro Mar 12 '12 at 3:39
  • $\begingroup$ I was looking for a clear explanation of how exactly electric fields appear along a circuit connected to a battery. This explanation is exactly what I was looking for. $\endgroup$ – favq Jan 2 '13 at 14:18
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    $\begingroup$ @Joe The link of J.D. Jackson's article is broken. Could you help update it? Thanks. $\endgroup$ – smwikipedia Jun 12 '18 at 0:47
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    $\begingroup$ @smwikipedia updated link: doi.org/10.1119/1.18112 $\endgroup$ – Joe Jun 12 '18 at 6:59

When the 2 electrodes are at different potential an electric field will be established. The electric charges will gather at the two poles. Positive charges at the cathode and negative charges at the anode. If the two electrodes are not connected by an external conductor they will not be able to leave the surface of the electrodes and they simply accumulate over there producing an open circuit voltage. As soon as the two electrodes are connected by a conductor the charges will flow by the forces of the electric field in the appropriate direction. If the connecting wire has no resistance or almost zero resistance then it will be a short circuit and a huge current will flow only limited by the internal resistance of the battery. If the electrodes are connected by a conductor through a resistance then the current will be limited according to the Ohm's law.

$I = \frac{V}{R+r}$ where $I$ is the Current, $V$ is the voltage between the electrodes, $R$ is the external resistance and $r$ is the internal resistance of the battery.

(Note: Electric field is not neutralized by the chemical reaction rather it is maintained by the reaction. No reshaping of the field takes place. The charges move through the conductor since it is the path of least resistance. It is similar to the flow of water through a pipe from a tank. The water flows since there exists a pressure difference. The flow of water does not depend upon the orientation of the pipe. it only depends on the pressure difference at the two ends. Gravitational field does not reshape itself through the pipe. Similarly no electric field reshaping takes place. Only the voltage difference matters between the two terminals.)

Additional comments for those who think electric field can reshape in a conductor along the conductor:

enter image description here

These are simply delta and star connected networks. Now if electric field could reshape then the lines of forces will intersect at three points for the delta and one point for the star networks. But we all know that lines of forces can not intersect. Hence no reshaping of field lines take place.

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    $\begingroup$ sb1, the E field always points along the direction of the wire. If you bend a wire the E field has to change directions to follow the new direction of the wire. Why would this not be considered "reshaping" of the E field in the wire? $\endgroup$ – David Santo Pietro Mar 12 '11 at 16:24
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    $\begingroup$ @sb1, I am not sure what you mean by "it never does". You correctly say that the field only has components along the wire. However, that means that if you change the direction of the wire, you change the direction of the field. For example, bend the wire into an S shape, and you will get S shaped field lines. Bend the wire into a spiral and you will get spiral shaped field lines. Why do you not consider that reshaping the field? BTW, I have read J.D. Jackson's Classical Electrodynamics cover to cover, so I know all about Maxwell's laws and what they imply. We can go as deep as you prefer. $\endgroup$ – David Santo Pietro Mar 12 '11 at 17:54
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    $\begingroup$ The structure of the electric field is determined by the distribution of charges and variation of magnetic fields, not by orientation of a conductor. This is clearly evident from Maxwell's laws. Is it so difficult to understand? $\endgroup$ – user1355 Mar 13 '11 at 3:51
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    $\begingroup$ @sb1: You are very mistaken. You keep bringing up things that are obvious, but do not address the question asked. Everyone knows fields can't cross. If you bend a wire into a figure 8 or circle, one part of the wire obviously has to be displaced on top of the other, not actually running through the point of intersection. i.e The metal of the wire can't touch or you have a short. If you did actually bend part of the wire into a figure 8 and actually ran the metal through the point of intersection the current would just skip the loop and there would be no electric field in the pinched off loop. $\endgroup$ – David Santo Pietro Mar 13 '11 at 6:47
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    $\begingroup$ Electric fields are a conservative field, but that also has no relevance here. Yes, an electron moving from the negative terminal to the positive terminal will have the same amount of work done regardless of the path or shape of the wire, but this does not mean the direction of or size of the field does not change when you change the length or direction of the wire. If you increase the length of wire connecting the terminals, the size of the electric field in the wire diminishes (E=-dV/dx). But the force will act over a larger distance and so, voila, same amount of work. $\endgroup$ – David Santo Pietro Mar 13 '11 at 6:48

Before you connect the wire, the electric field (at least theoretically, this requires that you ignore everything else in the universe, including the contents of the battery) consists of a + charge and a - charge. With a typical car battery these points are separated by about 10 or 20cm.

In reality, the charge is distributed in a very complicated manner. For example, if your car battery is rated for 1000 amp-hours, you can compute the "charge" by noting that an amp-second is a coulomb so an amp-hour is 3600 x 1000 coulombs. But actually these charges are held inside the battery chemically and do not appear at the electrodes until you take some of the charge that is there away. So to compute the initial charge you need to know only the voltage and the configuration of the leads. Then it's a matter of electrostatics to compute the charge distribution.

After you connect the wire, current begins flowing according to the resistance of the wire. The wire will have a steadily decreasing voltage from one end to the other. Since the electric field is given by the gradient of the voltage, this means that the electric field along the wire will point down the wire.

By the way, this happened accidentally to a battery in the back of his SUV. Unfortunately it happened next to a plastic can of gasoline. The resulting fire destroyed his vehicle seconds after he drove to the side of the road an jumped out of it.

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    $\begingroup$ An amp-second is a coulomb, not a farad. A farad is a coulomb per volt. $\endgroup$ – Keenan Pepper Feb 17 '11 at 1:31
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    $\begingroup$ You have repeated what was said in the question, but you didn't really answer it. $\endgroup$ – Joe Oct 6 '11 at 15:02
  • $\begingroup$ So you think that the answer stated that the wire will have a steadily decreasing voltage? The question could be abbreviated as "what happens to the electric field", which I answered. $\endgroup$ – Carl Brannen Oct 6 '11 at 18:57

protected by Qmechanic Nov 22 '13 at 14:05

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