# Why is there electric field OUTSIDE a battery?

Correct me if I'm wrong, but a battery's electric field is like an electric field of a capacitor consisting of two parallel plates. But we know that the electric field outside the two plates is zero.

So, in the context of electrostatics, why does an electric field form outside the battery and into the circuit?

• So you meant why is there an electric field in the wire, but outside the battery/capacitor? Again, there is no important physical difference between a capacitor and a battery. For you the primary difference between a capacitor and a battery is that people tell you different stories, (over)simplifications, and folklore about them. Mar 22, 2015 at 19:34
• Yes,that is what i meant.I think you are right.The only difference is that the battery also has some chemical forces acting in it. Mar 22, 2015 at 20:25
• There is a field outside the two plates of a capacitor, for the exact same reason there's a field outside a battery. The difference is that capacitors are typically wide and flat, while batteries are long and thin, so the field just happens to be smaller for capacitors. Mar 26, 2016 at 23:10

If the e-field outside the capacitor plates was zero, then the voltage on the plates would also be zero.

Another way to say this: The voltage across the capacitor plates is the same as the voltage across the connecting wires, but the space between the capacitor plates is very small. This means that the value of e-field located down between the two plates (the volts/meter) is much more intense than the e-field between the two, more separate lead-wires.

But, the e-field between the wires can never be zero except when the voltage between them is also zero. Or, we can reduce the spacing between the capacitor plates to decrease the external e-field, and if the capacitor gap goes to zero, that will also remove the external e-field (and remove the capacitor-voltage of course.)

The situation is like one very small capacitor (the two separate wires) connected in parallel to a much larger capacitor (that component with the actual microfarads.) The "dielectric" of the small capacitor is the empty space surrounding the two wires. If we connect a resistor across the far end of these wires, then their same e-field produces the driving force to create a current inside the resistor.

Note that with a capacitor connected to a resistor, the e-field inside the resistor doesn't extend lengthwise through the wires. The e-field inside the wires is nearly zero (and would be exactly zero for ideal superconductor wires.)

So, what's the origin of that strong e-field inside the resistor? It's from the connecting wires! The ends of the resistor have two opposite accumulations of charge on their surface, same polarity as the surface charges on the wires and on the capacitor plates.
The resistor itself is like a capacitor too. But it's a capacitor with a conductive dielectric. The e-field inside the resistive material is connected to the charges at the end of the resistor. The e-field flux does not extend inside the wires and back to the capacitor. Instead, the wires are the capacitor. In other words, the capacitor charges up the wires, and the wires charge-up the end terminals of the resistor, which gives an internal e-field to accelerate the mobile carriers inside the resistor. But (for ideal conductors,) the internal e-field is only inside the resistor, not inside the connecting wires.

It's also outside the resistor, of course, pointing in parallel to the resistor.

A great PDF paper on this is by the authors of the undergraduate physics series MATTER & INTERACTIONS. All your questions answered!

• The PDF link does not link to a PDF file. But there are several interesting PDF articles listed at matterandinteractions.org/articles-talks. Was one of these articles the one you originally linked to? Jan 13 at 17:07
• DOH they moved it again. It's matterandinteractions.org/wp-content/uploads/2016/07/… Jan 15 at 7:27
• Ah thanks. I wondered if it might be that one, I have been reading it already! Very clearly explained. I wonder how well-accepted is their model of surface charges? Jan 17 at 0:08
• @steveOw It's ancient stuff, going back to 1890 Oliver Heaviside. But was taught only in advanced EM courses, probably because students reject the correct idea that all electric circuits are based on "static electricity." It was a topic in 1980s science education, when student misconceptions were all the rage. Chabay/Sherwood decided to do something about it, teaching electrostatics-of-circuitry to undergrads. Earlier was MS Steinberg in '90s "The Physics Teacher," using supercapacitors to attack students' electrical misconceptions,scholar.google.com/scholar?q=ms+steinberg+capacitor Jan 18 at 1:49
• This is all very illuminating, thankyou. I'm currently learning about Inductance and I wonder if the present state of the art is able to model and visualize the spatial and temporal changes in surface charge distribution over a (DC) inducting coil during "charge up" and "charge release". Jan 18 at 11:55

In a parallel plate capacitor there's accumulation of electrons on one side and lack of them on the other. Since one plate is in front of another, the fields on each are equal in magnitude and opposite and therefore the field lines are straight (away from the boundaries) and cancel.

In a battery the field is chemically produced inside the structure of the object. Due to the form of most batteries, geometry makes it impossible to put the positive and the negative poles in front of each other, and so the field lines must bend and escape.

If you could make a battery in the form of, e.g, a torus, and then take a very thin slice out of it, then the field inside this slice would be very close to the field of a parallel plate capacitor (that is, zero outside the slice).

Now, the reason why there's a current on the circuit has nothing to do with electric field. Since there's accumulation of charge on one side on lack on the other, there's an electric potential difference between the poles. The form of the capacitor/battery doesn't matter here: if you connect something to the poles/plates, a current will flow.

• For your third paragraph,you are saying that the field of the capacitor of two plates comes from the consideration that the two plates are much larger than the distance between them but in you fourth paragraph you state that because the battery does not have much larger edges from the distance between them then the field is not the same as a capacitor of two plates which is E=s/2ε.did i understand correctly? Mar 18, 2015 at 0:04
• The fourth paragraph is only about the nature of current. In the third paragraph I'm saying that if you bend a cylindrical battery in the form of a doughnut until the two poles are very very near, then the field in between will be almost like a plane capacitor's one (if you're veeery close to the poles). Mar 18, 2015 at 0:12

There are also electric fields outside of a real capacitor as well, any capacitor with finite-sized plates. The energy in a capacitor is stored in the electric field, and since some of the electric field is outside the plates, some of the energy is also already outside the plates already.

Imagine a bunch of surfaces everywhere in space that are orthogonal to the electric field. Electromagnetic energy actually travels along these surfaces (in the direction that is also orthogonal to the magnetic field). Energy can flow like traffic, some leaving to the right as a possibly equal amount comes in from the right, so there are like electric field lines in that energy flow field lines starts or stop is where energy density is changing (either changing to electromagnetic energy or changing from electromagnetic energy). Where they terminate is where energy starts and stops, so a resistor for instance has electromagnetic energy convert into heat, so the electromagnetic energy flow field lines will converge on places inside a resistor.

So if you have a parallel plate capacitor you can imagine a bunch of surfaces layered in between the plates. If you had, say, a resistor in your circuit, then there is an electric field inside the resistor driving the current, so those surfaces orthogonal to the electric field cross through the resistor, and the energy from the battery (or capacitor) actually flows along these surfaces. So the space outside and between the wires is actually essential for transporting energy from a battery or capacitor to a resistor.

Some good pictures are available at this website.

As your capacitor discharges, the electric field intensity gets smaller and that energy has flowed into the resistor, but the energy that flows into the resistor in a small moment in time is energy from right nearby, and just before the resistor there is a high conductive material with very low electric fields, so not much electromagnetic energy, so the energy the resistor turns into heat actually comes in from the sides from the "empty" space in between the wires.

edit to respond to question about fields in wires

If you look at figure four of the link you'll see that it isn't just the top plate of the capacitor that is charged, it is the whole wire all the way up to the resistor, similarly the whole wire on the bottom all the way to the bottom plate is negatively charged. Because they are basically one giant conductor with the charge distributed on the outside. So when you connect a wire to a capacitor it basically becomes part of the capacitor, and so it is no longer truly "just" a parallel plate capacitor.

This happens to/with a battery too when you connect perfect conductors to the terminals. If none of the conductors are perfect then they are more like resistors with a really really low resistance, so let's talk about the resistor.

In a resistor you have one equipotential surface at one end of the resistor and then many equipotential surfaces crossing through the resistor and then one last equipotential surface at the end. The electric field is orthogonal to the surfaces, so goes from one end of the resistor to the other.

If you want to see how it got there, imagine a circuit with a capacitor (or battery) on the left, a switch on the top and a resistor on the right. When the switch is open the entire wire connected to the positive terminal/plate is at high voltage, and the entire rest of the circuit is at low voltage. In between the plates (or inside the battery) there are many equipotential surfaces. Think of that upper wire before the switch as a castle wall, all up high, between the plates (or inside the battery) there is a climb from the ground up to the heights. But leaving the wire in any direction would result in a fall. In particular, there are many equipotential surfaces between the two ends of the switch. As you start to close the switch, the metal ends start to get very close and the region between acts similar to a resistor with a very high resistance and a tiny current starts to flow and those equipotential surfaces flow with the current towards the actual resistor where they accumulate (since you need more current to push an equipotential surface through a finite resistance resistor). The closer the switch comes to being fully connected, the more current flows, and the more equipotential surfaces got pushed to the resistor, eventually the switch acts like no resistance and the current increased enough to have pushed all the equipotential surfaces into the resistor where they stay (until the battery or the capacitor loses voltage, and when that starts to happen the surfaces get pulled back into the battery/capacitor).

So there were always equipotential surfaces outside the wires and when you connected the final wire in a circuit, those surfaces got into the wire and hence into the circuit. Those equipotential surfaces getting into the wires is where and when the electric field enters the wires.

• can you explain it in simpler electrostatic terms?Because i know that the field from a capacitor of two plates is zero outside the capacitor but only because the length of the plates is much larger than the distance between them.But what differentiates the battery from that?Is it the criterea that the length of the ends of the battery are not much larger than the distance between them?I am asking you this simply because i can not understand how your explanation gives me the answer in terms of electrostatics Mar 18, 2015 at 0:30
• @LandosAdam Maybe you should just compare a capacitor and a battery that have an equal amount of useable stored energy. The E field is only approximately zero, so if you have a capacitor with lots of energy, an equal amount, maybe you'd see just as strong an E field outside it. But that nonzero part outside is actually essential to transporting energy to your circuit because energy flows out of the capacitor/battery to the empty space outside the capacitor/battery and then along the empty space near the wires. The surfaces orthogonal to the E field are equipotential surfaces in electrostatics. Mar 18, 2015 at 1:03
• @LandosAdam The electric field near the edge of a capacitor is a bit smaller than in the middle, but the amount it is smaller (energy wise) is exactly the amount that is outside the capacitor. You could make a capacitor out of a bunch of dipoles, so the field is definitely supposed to (even in theory) be nonzero outside the plates, you can just sometimes get the right answers by assuming it's zero out there (when it isn't) and assuming it is larger in some places than it really is. When you concentrate charge you increase the energy density outside the charges, this is a universal phenomena. Mar 19, 2015 at 15:17
• i did not say that i do not accept your answers.I know about energy,it is a very basic concept and the first thing that i analyze in a system before delving into forces and then microscopic analysis.The reason why i keep asking stuff is that i already know most of the stuff that most people mention,but i want a more microscopic analysis.When i say that this is what i concluded by my studies means that i have found many sources(like famous physics books(mainly for graduate studies in this occasion)) that support the things that i concluded.thank you for your different explanation Mar 21, 2015 at 23:15
• As for how MUCH micro,thermal is not considered to be affecting the things that i am questioning in a big way,so not accounting for them in the analysis is ok.The most micro analysis(if there is such term) is QM.But i haven't got the education that reaches QM yet.I am self educated(outside my academic career as a civil engineer which has nothing to do with QM and electromagnetism),so the next step in my self education is QM ;) Mar 22, 2015 at 2:19

Your understanding of the electric field in a capacitor is correct. In an ideal scenario, the electric field outside the two plates of a capacitor is indeed zero. However, a battery is not exactly the same as a capacitor.

In a battery, the electric field is created by the movement of electrons, and the battery creates a potential difference by moving electrons from the negative terminal to the positive terminal. This is due to a chemical reaction within the battery.

When a battery is connected to a circuit, the electric field extends into the circuit. This is because the electric field is created by the redistribution of charge so that there is a potential difference across its terminals. The negative terminal has a surplus of electrons and the positive terminal has a deficit of electrons.

In the case of a capacitor, the electric field is confined between the plates because the charges accumulate on the inner surfaces of the plates. But in a battery, the charges accumulate on both sides of the dielectric, which allows the electric field to extend outside the battery and into the circuit.

So, while a battery's electric field might seem similar to that of a capacitor, the presence of a chemical reaction and the accumulation of charges on both sides of the dielectric in a battery allow the electric field to extend outside the battery and into the circuit.