How does a salt water capacitor work?

AI Thread Summary
A saltwater capacitor operates by utilizing two conductors separated by a dielectric, where the dielectric consists of saltwater and the plastic bottle wall. The high capacitance arises from the electrolyte's ability to allow ions to migrate, forming a double layer at the electrodes, which enhances charge density. Increasing the salt concentration can further boost capacitance due to reduced Debye screening length. While the saltwater acts as a conductor, it also provides a high polarizability, allowing the capacitor to function effectively. Ultimately, the saltwater is crucial for achieving measurable capacitance compared to using just the plastic dielectric alone.
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What do you mean? It's a capacitor ... you have two conductors separated by a dielectric.

If you mean why does it have such a high capacitance, that is because the dielectric medium chosen for this capacitor includes an electrolyte (the total dielectric is the salt-water plus the thin plastic wall of the bottle ... the plastic is important to avoid electrochemical reactions). Some of the ions in the solution will migrate to the oppositely charged electrode and form a double layer. This is analogous to the polarization effect in a normal dielectric, but MUCH more effective at shielding the charge, so much higher charge densities can be achieved. In fact, you should be able to increase the capacitance (at least up to a point), by increasing the concentration of the electrolyte. This is because as the concentration goes up, the Debye screening length drops, allowing higher charge densities to be stabilized at a given voltage.
 
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SpectraCat said:
What do you mean? It's a capacitor ... you have two conductors separated by a dielectric.

If you mean why does it have such a high capacitance, that is because the dielectric medium chosen for this capacitor includes an electrolyte (the total dielectric is the salt-water plus the thin plastic wall of the bottle ... the plastic is important to avoid electrochemical reactions). Some of the ions in the solution will migrate to the oppositely charged electrode and form a double layer. This is analogous to the polarization effect in a normal dielectric, but MUCH more effective at shielding the charge, so much higher charge densities can be achieved. In fact, you should be able to increase the capacitance (at least up to a point), by increasing the concentration of the electrolyte. This is because as the concentration goes up, the Debye screening length drops, allowing higher charge densities to be stabilized at a given voltage.

1.8 nF isn't a lot of capacitance for something as big as a water bottle. I think the salt water just works as a conductor and forms one of the plates.

A polyethylene capacitor of 10cmx10cm area and 0.1 mm thickness gives a capacitance
of \epsilon_0 \epsilon_r A/d = 8.9 * 10^(-12) * 2.2 * 0.01 / 10^(-4) = 1.9 nF
 
willem2 said:
1.8 nF isn't a lot of capacitance for something as big as a water bottle. I think the salt water just works as a conductor and forms one of the plates.

A polyethylene capacitor of 10cmx10cm area and 0.1 mm thickness gives a capacitance
of \epsilon_0 \epsilon_r A/d = 8.9 * 10^(-12) * 2.2 * 0.01 / 10^(-4) = 1.9 nF

Well, "a lot" is a relative concept ... :wink: If you leave out the salt water, you still have a capacitor ... but its capacitance will be (almost) immeasurably small ... the key to observing the capacitance that is seen is the salt water.

Still you make a good point ... you are likely correct that the magnitude of the capacitance is dominated by the low-dielectric constant plastic material separating the foil electrode and the salt water. However, the only reason that you see any measurable capacitance at all with such a construction is the salt water, and the ion mobility that I described. I am not sure what you mean when you said the salt-water "forms one of the plates" .. if you mean that the free ions (of the appropriate charge) become polarized so that they form a layer on the inside of the plastic, and thus the voltage drop happens almost entirely across the low-dielectric plastic, then I agree with that.
 
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I also think that the main function of the salt water here is to act like a conductor rather than dielectric. It's just a simple way to put an electrode on the inside of the plastic bottle. An alternative, usually described in school experiment books, would be to fill the bottle with small pieces of aluminum foil and squeeze them inside until they form a compact block so the contact area with the wall is large enough.

For a "Leyda" bottle from PET (DC dielectric constant 2.6) with 0.5 mm wall I've got around 1nF. (dimensions used: 5 cm diameter, 14 cm height).
 
nasu said:
I also think that the main function of the salt water here is to act like a conductor rather than dielectric. It's just a simple way to put an electrode on the inside of the plastic bottle. An alternative, usually described in school experiment books, would be to fill the bottle with small pieces of aluminum foil and squeeze them inside until they form a compact block so the contact area with the wall is large enough.

For a "Leyda" bottle from PET (DC dielectric constant 2.6) with 0.5 mm wall I've got around 1nF. (dimensions used: 5 cm diameter, 14 cm height).

Well ... salt water really isn't a good conductor of electrons, although it is an excellent ion conductor. Since it doesn't conduct electrons, when you put a voltage across it, it acts like a dielectric with an EXTREMELY high polarizability (due to the high mobility of the solvated ions). Thus I think my description captured the essence of the mechanism.
 
SpectraCat said:
Well ... salt water really isn't a good conductor of electrons, although it is an excellent ion conductor. Since it doesn't conduct electrons, when you put a voltage across it, it acts like a dielectric with an EXTREMELY high polarizability (due to the high mobility of the solvated ions). Thus I think my description captured the essence of the mechanism.

Even if the salt water would make a capacitor, it will be in series with a capacitor with the bottle as dielictric. What the polarizability of the water is, is unimportant, because you can't do better than the capacitance of just the bottle with two sides of it made of metal.
 
willem2 said:
Even if the salt water would make a capacitor, it will be in series with a capacitor with the bottle as dielictric.

I agree with that .. I said so in post#4.

What the polarizability of the water is, is unimportant, because you can't do better than the capacitance of just the bottle with two sides of it made of metal.

Really? Try taking the water out then, or better yet, just try making it without the salt. The capacitance will be much lower, if it is even measurable at all. It is the high mobility of the ions that allows the capacitor to behave essentially *as if* it is a double plate capacitor with metal on the inside of the bottle. In fact, I would expect the capacitance to be proportional to the salt concentration, at least over some range. It may be that the protocol for building the capacitor already has the salt concentration well over the point where the effect is basically saturated.
 
SpectraCat said:
Well ... salt water really isn't a good conductor of electrons, although it is an excellent ion conductor. Since it doesn't conduct electrons, when you put a voltage across it, it acts like a dielectric with an EXTREMELY high polarizability (due to the high mobility of the solvated ions). Thus I think my description captured the essence of the mechanism.

I think you make an interesting point, especially in your last post. Pure water is a dielectric with a quite high dielectric constant, due to its permanent dipole. When you add enough ions in it to make it an electrolyte, it becomes conductor due to the free ions. These ions are not "polarized" in electric field but rather undergo drift motion. As you increase concentration of dissolved ions, it would be interesting to study how the properties change from "dielectric" to "conductor" (there is no sharp cut).
For pure water the dielectric will be very thick, about 2-3 cm assuming a cylindrical symmetry), including both plastic and water (with dielectric constant about 80). For conductive electrolyte, the dielectric is limited to the thin plastic layer (0.5- 1mm and dielectric constant 2-3).
It's not obvious to me which case produces higher capacitance. However really pure water is really difficult to obtain so it's more of an ideal case.

For the problem with the capacitor I think that the solution is concentrated enough to consider it a conductor. You can fill it with mercury to pretty much the same effect but it's more inconvenient from several points of view (expensive, poison controlled in some countries)

A semiconductor is another example of material that have both mobile charge (of both signs) and also lattice fixed charges that can be polarized in electric field so you can find both dielectric constant and conductivity for the same material.
 
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