# Why does not a battery kept in empty space discharge on its own? If space between the terminals is empty, what prevents the electrons to flow?

Given here is a picture of a twisted battery kept in free space (vacuum). Will the battery discharge quickly? I mean, it should discharge faster than it would if we were to connect the terminals with copper wire.

The reason I give is simply that attraction from some opposite charge induces motion in an electron. The resistance to its flow is caused by the medium. If the medium was empty, there would be no resistance, hence empty space should conduct electricity better than copper wires!

So will the battery as shown discharge quickly if kept in vacuum?

I don't think that the battery will discharge as fast as it would if short-circuited with copper wire. But this does not means that empty space resists current more than copper wires. I explain it this way:

Electrons flow in a battery because of completion of a circuit-reaction within the battery. So, for an electron ejected from the negative terminal, there has to be an electron entering the positive terminal to complete the feedback. An imaginary electron ejected would take time to reach the other terminal, hence there is lack of feedback and so no quick discharge.

This means that if we somehow introduce a cloud of electrons between the terminals, there would be an instantaneous feedback, and space would start conducting better than copper wires! (For an electron ejected from the negative terminal, there would be an electron available to enter the positive.)

Another aspect that can be tested is to see if a parallel plate capacitor with oppositely charged plates discharges more quickly when it has vacuum between the plates or when the plates are connected by copper wire.

Will such capacitors with vacuum between the plates discharge quickly? If not, then why not?

Edit: My earlier explanation to the question is wrong. The correct explanation has something to do with photoelectric effect, as indicated in the comments and the answers here and also on this question as pointed by lan.

## 5 Answers

How quickly discharge will occur in the situation you sketch depends entirely on the surface properties of the negative electrode. For current to flow, electrons need to be released from the negative electrode; once they are free, they will accelerate unimpeded to the positive electrode. They will arrive there with 1.5 eV of energy, causing a small amount of heating on impact.

The medium (or lack of it) does not determine the discharge current - it is the surface properties of the cathode. More specifically, it is the local field strength right at the surface. Field strength is initially determine by the cathode-anode spacing, but there can be a (very) significant local enhancement due to surface roughness (asperities).

The "cold cathode" emission usually requires quite a high field strength; by looking at the potential difference actually needed you can learn a lot about the field enhancement, and this is actually used as a way to study surface roughness. Discharge current is described by the Fowler-Nordheim equations. Without going into too much detail, they basically describe the probability that an electron can tunnel through the barrier provided by the surface work function. This probability depends strongly on temperature (which is why hot cathodes work in cathode ray tubes etc), and the work function. At room temperature, at the kind of electric fields that can be created with an AAA battery, the leakage in the electrolyte of the battery is likely to be much more significant than that leakage through the vacuum.

A nice step by step walk through the calculations can be found in these lecture notes.

In vacuum there are no charge carriers like ions or electrons. With nothing to carry charge, i.e. current, such a battery would discharge much, much slower than when the battery poles are connected by something that can carry charge like a conductor or an imperfect insulator.

• And the electrons in the (-) terminal of the battery cannot simply jump into space because they are bound to the metal. Commented Aug 15, 2015 at 22:20
• Yes, the photoelectric effect clearly demonstrates that there is a work function to overcome before the electrons can be liberated from the metal surface. Commented Aug 15, 2015 at 23:00

Electrons (and other charge carriers, e.g., ions) in vacuum travel without resistance. However, as pointed out, correctly, in the other answers, there are no charge carriers in vacuum.

Nevertheless, electrons can escape from the terminals if they have a kinetic energy which is bigger than the potential barrier of the terminal surface, i.e., the work function $W$ of the material. If the kinetic energy of the electrons is bigger than the work function, the battery can discharge in a finite time. This condition can be obtained either if the potential difference between the two terminals or their temperature is high enough. In the first case one obtains an electric arc (a small-scale lightening), while the second case one has a steady charge transfer (thermionic effect).

The reason I give is simply that attraction from some opposite charge induces motion in an electron. The resistance to its flow is caused by the medium. If the medium was empty, there would be no resistance, hence empty space should conduct electricity better than copper wires!

A metal wire is conductive because metals have lots of energy levels near the Fermi Level $E_F$. This means that lots of electrons are available to be moved around which is why the electrical conductivity of metals is so high.

You mentioned:

If the medium was empty, there would be no resistance

However, in a vacuum there is no medium to provide a conductive path in the first place. The electrical resistance $R$ is defined as the inverse of electrical conductance $G$. In a vacuum, the conductance is $G = 0$, and therefore the resistance is $$\lim_{G \to 0} R = \frac{1}{G} = \infty$$

Instead of thinking about resistance as:

The resistance to its flow is caused by the medium.

Resistance is not caused because there is a medium, rather, it is caused by the medium not behaving as an ideal conductor. The only reason that electrons can flow in the first place is because there is a conductive medium.

As other posters have pointed out, there are more exotic ways to get electrons to flow in a vacuum (thermionic emission, extremely large electric fields, etc), but I hope this answer clarifies why electrons would not flow faster through vacuum than through a wire.

Lines of electrostatic force exist between the positive and negative poles of the battery, even though they're separated by a vacuum. Vacuum permittivity is ε0 = 8.854 * 10^-12 farads per meter. By convention, this is called the dielectric constant of 1, a baseline against which the dielectric permittivities of other materials are compared. (https://en.wikipedia.org/wiki/Vacuum_permittivity).

But even though a vacuum is highly permittive of electrostatic field lines, it contains no atoms to ionize, nor any electron cloud such as exists in copper through which the electrons can transfer their charge, so a greater potential difference is needed. For example, high voltage capacitors use vacuum as the dielectric to maximize the charge stored on the capacitor plates. They charge and discharge through the vacuum, but at very high voltages and high temperatures.

You could augment the potential difference by shaping the negative terminal into a point in order to concentrate electrons there, or you could roughen the surface of the negative terminal to create many points of concentrated charge, or you could heat the terminal to give kinetic energy to the electrons concentrated there. If the concentrated charge exceeds the dielectric property of the vacuum, electrons will leave the negative terminal. But an ordinary battery will not discharge very well through a vacuum unless the potential difference is augmented by temperature, geometric shape, and/or material properties.

• These statements from you contradict each other: 'it (vacuum) contains no atoms to ionize, nor any electron cloud such as exists in copper through which the electrons can transfer their charge' and 'To get electrons to flow in a vacuum, you would need an enormous potential difference'. Suppose we use enormous potential difference, there is still no atom in vacuum to ionize, nor any electron cloud. There is an electron cloud though- the current, but the current can't be the reason why the current is at high Potential difference.
– Prem
Commented Aug 16, 2015 at 18:50
• No more work is required to "push" an electron through vacuum than is require to push a rock through vacuum. I.e., no work at all is required. The problem is, how do you get the electrons to want to leave the battery's metal "-" terminal? If the "-" terminal isn't glowing red hot, then it's going to take a lot more than 1.5 V to do that. Commented Aug 16, 2015 at 19:02
• @jameslarge: Work = force * displacement. Pushing a rock through a vacuum requires force to overcome the rock's inertia. Are you sure an electric field doesn't do work? The electric field is force per unit charge. An electric field is defined even in a vacuum: Commented Aug 16, 2015 at 20:53
• @Raja: There is no contradiction. Ionization of air, or existence of an electron cloud in copper, make it easier for current to flow, but current will flow even in a vacuum. I pointed out that electrons flow more easily through other materials. Potential difference is not caused by the current. It causes the current to flow. Because vacuum is an insulator, current flows slowly through the vacuum unless the potential difference is great, and/or the negative terminal surface has been prepared to concentrate electric charges, and/or it has been heated to impart kinetic energy. Commented Aug 16, 2015 at 21:06
• @Raja: Vacuum has extremely high dielectric strength: 10^12 MV/m. Compare that to 3 MV/m for air. Dielectric strength measures the maximum electric field something can withstand without breaking down and losing its dielectric property. Although vacuum has high permittivity of field lines, its dielectric strength requires high voltage and high heat to get electrons to leave the cathode and follow the field lines. Here's a link that may help resolve this: qdl.scs-inc.us/2ndParty/Pages/8819.html. Basically, the electrons flow freely, but first you have to get them off the cathode! Commented Aug 18, 2015 at 23:55