# How much electricity can a kilo of electrons produce?

I know how stupid this question sounds.

I heard my grandma say something close to it in a phrase "It's a 1000€ aircondition, it's not gonna consume 200 kilos of electricity !"

Made total no sence but i remenbered an article stating that a full memory stick actualy weighs more than an empty stick due to stored electrons. So technicaly it is possible to store a kilo of electrons.

So how many electrons whould that be and how much energy could they produce ?

• I question the statement that there more electrons in a full memory stick. – Ben51 Nov 26 '18 at 1:35
• The incredibly small fractional change in a memory sticks weight is more likely due to the higher energy density from the forced arrangement of magnetic domains – Triatticus Nov 26 '18 at 1:50
• – rob Nov 26 '18 at 2:42

So how many electrons would that be and how much energy could they produce?

As you note, this question makes no sense at all. This is like asking how much gravitational energy a kilo of rice will "produce": it is entirely dependent on what interactions it takes part of.

Moreover, as for the title of your question,

How much electricity can a kilo of electrons produce?

you just cannot quantify electricity like that - there are multiple inequivalent ways to measure it, and if you don't state which one, the question "how much electricity" is essentially meaningless.

Nevertheless, if you take the question text literally, as the two existing answers have pointed out, it is possible to calculate how much electric charge corresponds to one kilogram of electrons, via the ratio $$\frac{e}{m} = \frac{1.6\times 10^{-10}\:\mathrm{C}}{9.1\times 10^{-31}\:\mathrm{kg}} = 1.75\times 10^{11} \: \rm C/kg$$ between the electron's charge $$e$$ and its mass $$m_e$$. Thus, $$1\:\rm kg$$ of electrons corresponds to a total electric charge of $$Q=1.75\times 10^{11} \: \rm C$$.

However, what the existing answers don't mention is how much energy it costs to put that much excess charge together. Assuming for simplicity that you're putting it together into a spherical region with homogeneous charge distribution of radius $$r$$, the energy it will take to assemble that much charge is given by $$E_\mathrm{self} = K_\mathrm{geom} \frac{1}{4\pi\epsilon_0} \frac{Q^2}{r},$$ where $$\frac{1}{4\pi\epsilon_0}=9\times 10^9 \:\rm J\:m/C^2$$ is the Coulomb constant, and $$K_\mathrm{geom}$$ is a geometrical factor of order unity; for a homogeneous sphere of charge it's $$K_\mathrm{geom}=3/5$$. Assuming to start with that $$r=1\:\rm m$$ is of roughly human scale, that gives you \begin{align} E_\mathrm{self} & = K_\mathrm{geom} \frac{1}{4\pi\epsilon_0} \frac{Q^2}{r} \\ & = \frac35 (9\times 10^9 \:\rm J\:m/C^2) \frac{(1.75\times 10^{11} \: \rm C)^2}{1\:\rm m} \\ & = 1.7 \times 10^{32} \: \rm J. \tag{*} \end{align} It's hard to understate just how big that energy is. It's tempting to compare it to the energy in a nuclear bomb, but that's in the order of a kiloton of TNT equivalent, which is in the range of $$E_\mathrm{kiloton} = 4\times 10^{12} \: \rm J$$. (If you do want to reduce the energy to that level, then you'd need $$r$$ to be of the order of 1000 light years, i.e. some five times the radius of the Milky Way.) Wolfram Alpha is a bit more helpful - it points out that $$(*)$$ is on the order of 70% of the gravitational binding energy of the Earth. In other words, you're looking at Death-Star-like, planet-destroying amounts of energy.

However, this nonsensical result comes, basically, from the fact that you're either mis-remembering or mis-interpreting the article you read. This

a full memory stick actually weighs more than an empty stick due to stored electrons. So technically it is possible to store a kilo of electrons.

is vaguely close to a true statement, but it doesn't quite get there. A full memory stick probably does weigh a bit more than an empty one, but that is not (let me emphasize NOT) due to added electrons.

The argument you're likely referring to is that an empty USB stick will generally have a bunch of zeros in line, and those tend to have a smaller interaction energy than a neighbouring one and a zero. Thus, if your USB stick is full of information, then that means that there is some amount of potential energy (magnetic, or otherwise) stored in the interactions between neighbouring bits. And, since $$E=mc^2$$, that would conceivably raise the mass of the stick by some marginal amount.

Note, however, that the total number of electrons in the USB stick is exactly the same on both configurations.

And, moreover, that argument is mute as to how much interaction energy there is between neighbouring bits, which means that it is utterly impossible to produce a realistic estimate of how much information you'd need to store in such a way in order to get an additional kilogram of rest mass on the USB stick.

Still, it's kind of possible to give an upper bound, since it is extremely unlikely that the inter-bit interaction energy will be much larger than $$1\:\rm eV$$ (figures closer to a milli-eV are more likely, I think), and the mass equivalent of that is easy enough to calculate, at $$m_\text{upper bound} = \frac{1\:\rm eV}{c^2} \approx 10^{-36} \:\rm kg.$$ So if you want to have a $$1\:\rm kg$$ increase in rest mass, you're unlikely to need any less than $$10^{36}$$ inter-bit interaction pairs, and that's again in the astronomical regime of unfeasible numbers.

The invariant mass of an electron is approximately 9.109×10−31 kilograms so 1 kg is about 10^30 electrons.

The charge of an electron is 1.60217662 × 10^-19 coulombs, so 1kg has about 1.6x10^11 Coulombs.

1 ampere is 1 Coulomb/sec.

Therefore 1kg of electrons carry 1.6x10^11 Ampere x seconds.

All of this ignoring relativistic effects

The mass of an electron is $$9.1\times 10^{-31}$$ kg. So one kilogram of electrons is $$1.1\times 10^{30}$$ electrons.

An electron has a charge of $$1.6\times 10^{-19}$$ Coulombs. So a kilogram of electrons has a charge of $$1.8\times 10^{11}$$ Coulombs. This is a lot of charge.

The energy they could produce depends on what you do with them. If you let them move across a voltage difference of $$100$$ volts, then they will acquire $$1.8\times 10^{13}$$ Joules of energy.

This is all completely unrealistic and you are never going to have a kilogram of electrons.

• Mentioning the energy gained over a 100-volt potential is kinda disingenuous if you don't also mention the self-energy required to assemble that much excess charge in a volume smaller than the size of the solar system, which clocks in at some eight orders of magnitude larger than the energy you quote. If you make that a sphere with a 1m diameter, that's more like eighteen orders of magnitude - that's to 1 kiloton of TNT what that kiloton is to 1J. (all of these being rough back-of-the-envelope numbers.) – Emilio Pisanty Nov 26 '18 at 1:42