Do electric fields in a capacitor add to its weight?

Question

So this article got me thinking: https://www.scientificamerican.com/article/do-electric-charges-and-m/

It states that according to GR, the energy in an electric field should curve spacetime and therefore produce a gravitational effect. And so I started wondering if a charged capacitor would weigh more than an uncharged capacitor, meaning the electric field would be adding gravitational pull to the capacitor.

It seems that it must since it is curving spacetime itself. However, to my understanding electric fields would follow the null geodesic, like light, and be bent by curved spacetime, not accelerated by it.

So to be clear, I'm not talking about the gravitational effect FROM the capacitor. The article above makes it clear the charged capacitor would exert more gravitational pull on objects around it. I'm concerned with the charge of the capacitor causing the capacitor to get pulled harder by the gravity of other objects, like earth.

Answer 1

I'm concerned with the charge of the capacitor causing the capacitor to get pulled harder by the gravity of other objects, like earth.

General relativity is not required to answer this question. Consider (for simplicity) a parallel plate capacitor where the field is constrained within the parallel plates. In this case the field is uniform (let it be $E_0$) and thus the energy stored by the capacitor is $\frac{1}{2}\epsilon_0 E_0^2$ and by $E=mc^2$ (assuming the capacitor to be at rest) the mass of the capacitor will increase to $m_{new}=m_{initial}+\epsilon_0\frac{E_0^2}{2c^2}$ meaning by Newtonian gravity it will have a greater weight.

However, to my understanding electric fields would follow the null geodesic, like light, and be bent by curved spacetime, not accelerated by it.

First of all in the realm of General Relativity, any object solely under the influence of gravity follows geodesics and is unaccelerated irrespective of whether the geodesic is null or timelike. Light follows null geodesics in the geometric optics approximation. In case of electric/magnetic fields no path as such is followed we are just concerned with evolution of the field values at any given point.

Answer 2

Yes, the electric fields in a capacitor add to its weight. But not so that you’d notice with anything so crude as a balance.

Suppose you had a one-farad “supercapacitor” that you could charge up to one kilovolt. The energy stored in the electric field would be

$$ U = \frac12 C V^2 = \frac12\times10^6 \rm\,J $$

This is an awful lot of energy for a capacitor, but the gravitational field it creates will correspond to a mass

$$ m_\text{effective} = U/c^2 \approx 10^6\,\mathrm{J}/c^2 \approx 10^{-11}\rm\,kg \approx 10\,ng $$

Supercapacitors are pretty amazing these days, but a farad-scale capacitor starts out at a mass of a few grams, not a few nanograms. And a small farad-scale capacitor that can actually hold a kilovolt? That’s a nontrivial challenge. The nanogram correction to the mass of our hypothetical supercapacitor, due to its electric field, is a part-per-trillion correction at the most. (This is an order-of-magnitude estimate; see the comments below for one factor-of-two correction.)

For what it’s worth, the electric-field correction to the mass of the hydrogen atom, a binding energy of $-13.6\rm\,eV$ on a GeV-scale mass, is a part-per-billion correction.

Why have I written that the electric fields in a capacitor add to its weight, when in the case of the hydrogen atom the binding energy is negative? The relativistic mass of the charged capacitor is actually less than the relativistic mass of the two charged plates: you would have to apply work to the system to pull the plates apart. But the relativistic mass of the charged capacitor is more than the combined mass of the two neutral plates. You can reduce the system’s effective mass by shorting the capacitor and allowing the charges to recombine, confining the strong electric fields to the atomic scale instead of the capacitor-gap length scale.

Answer 3

This was tested experimentally by Kreuzer, Phys. Rev. 169 (1968) 1007. The "capacitors" in this experiment were actually atomic nuclei. The electric field of a heavy nucleus makes a fairly significant contribution to its own weight, which is easy to measure. Kreuzer used a Cavendish balance to test whether this was also true for the active gravitational mass, as required by the equivalence principle.

Answer 4

Energy and mass are equivalent. If the electric field stores energy, then it stores mass. Weight is proportional to mass, other things being equal, so the electric field will contribute to the weight of the capacitor.

Likewise a wound up mechanical watch has a greater weight than the same one unwound, other things (e.g. temperature) being equal.

It's really very simple, as explained in "Relativity Visualized" by Lewis Carroll Epstein.

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N.B. It seems that professional relativists have been using a new convention regarding the terms "mass" and "energy". The physical laws are the same, so Epstein was not wrong, but it seems that his use of "mass" and "energy" (which matches Albert Einstein's use which is also out of date, it seems) is now frowned on as dated by professional relativists, or at least the majority of English speaking ones.

I will let my answer above stand, because it's not wrong, but allegedly uses "mass" and "energy" in a dated way. However I will add another version that uses terms that cannot be faulted by anyone.

In the old convention (Einstein's) mass and energy were the same thing. But in the new convention used by professional relativists since about 1970, "mass" means "rest mass" and "energy" means "total energy including rest mass".

If the electric field stores energy, or if there is some extra energy in or attached to the capacitor or its field or charge in any way, that extra energy means extra weight. The weight is given my W = mg and in this case the part called "m" is replaced by "E" for "energy", where E is the total energy including all forms of energy including KE and mechanical energy and including the rest mass. So we have W = Eg where E is the total energy in kilograms.

Likewise a wound up mechanical watch has a greater weight than the same one unwound, other things (e.g. temperature) being equal.

A simple and crystal clear explanation of this can be found it Lewis Epstein's wonderful book, "Relativity Visualized", but be warned that he uses the terms "mass" and "energy" in the way relativists including Einstein used them up until about 1970, it seems. It's not a problem, an in fact it might even be a strength as the new conventions are highly confusing, to me for one. One reason it is so confusing is that instead of using terms of art that are clear to all physicists and intelligent nonphysicists, such as "rest mass" (professional relativists call it just, "mass") and "total energy including rest mass" (they just say, "energy") they use in effect a private language, AKA "jargon" that not the majority of fellow physics graduates understand. In fact, the majority of them misunderstand it.

HardlyCurious raised a great point in his comment, which is that it is far from clear how the weight of the energy of the electric field pushes down on the capacitor. I don't know how, but I've read that it's figurative to say the energy of a capacitor is in the field.

Answer 5

IMO No. The electromagnetic energy from the source is used to coherently rearrange the existing charges in the circuit. It does not add more matter or energy in this open system. The electric potential energy of the source is transformed in kinetic energy of the moving charges and all of this energy is expelled out of the system as heat. After the capacitor is fully charged there is no more work done in the system. To prove conclusively that the capacitor gained mass after it was fully charged you must prove that there is an excess energy given by the source minus the expelled heat from the electrons motion that is stored in the capacitor. Notice the stored electromagnetic potential energy is not given by the source but sources from the electrons which where there in the capacitor before the source was applied. No extra energy was stored in the capacitor.

Part of the capacitor's own matter energy was made more coherent (less entropy). You use the energy given by the source to make the capacitor with less entropy on its matter field (polarization) and this energy given by the source is then all transformed into heat and expelled out of the system.

The confined energy of the capacitor remains the same no extra energy is stored in the capacitor given from the source. The extra energy is all expelled out as heat.