Max voltage of both air and water capacitors

[HelloCthulhu]

Homework Statement

Theses questions have been modified to solve my own inquiries. The originals can be found in the attached file.

A parallel plate capacitor has area A = 1 cm², a plate separation of d = 0.01m, and is filled with air.(a) If the breakdown field is E₀ = 3 × 10⁶V/m, calculate the maximum voltage and charge the capacitor can hold.(b) Water at room temp (20°C) is poured between the plates filling the volume of 1cm³. Find the new capacitance and maximum charge if the breakdown field is 21.6 times larger than air.(c) Water heated to 100°C (dielectric constant k = 55.3) is poured between the plates, filling the volume. Find the new capacitance and maximum charge:

Homework Equations

Q=CVV_max = E₀ x dC_new = k x C_old

The Attempt at a Solution

(a) The breakdown field is the minimum electric field at which a material ionizes; in air, this means a spark

forms.

Maximum voltage:

$$V{max}=\varepsilon_0\times d =(3\times10^6\times0.01) =(3\times10^4)V = 30kV$$

To find the maximum charge corresponding to V = 30kV, we need the capacitance of the parallel plate

capacitor,

$$\frac{(8.85×10^{-12})\times(0.0001)}{0.01}=\frac{(8.85×10^-16)}{0.01}=8.85×10^{-14}F=0.0885pF$$

$$V{max}=(3\times10^4 )\times(0.0885\times10^{-12}) =(3\times10^4)V=30kV$$

The charge is determined from C = Q/V :

$$Q=CV=(3\times10^4 )\times(0.0885\times10^{-12}) =2.655\times10^9C=2.655nC$$

(b) Water at room temp (20°C) is poured between the plates filling the volume of 1cm3. Find the new capacitance and maximum charge if the breakdown field is 21.6 times larger than air.

$$80.1\times0.0885pF=7.08885pF$$

The breakdown field, and hence the maximum voltage, increase by a factor of 21.6, so the new maximum voltage is Vnew = 21.6 x 30kV = 650kV. The new maximum charge,

Q_new = C_new * V_new, is a factor of 80.1 x 21.6 greater than the old charge,

(c) Water heated to 100°C (dielectric constant k = 55.3) is poured between the plates, filling the volume. Find the new capacitance and maximum charge:

The dielectric increases the capacitance by a factor k,

$$C_{new}=k\times C_{old}=55.3 \times0.0885pF=4.894pF=4.894\times10^{-12}F$$

(d) With a new max capacitance of 4.894 x 10^-12C and a charge of 4.593 x 10^-6C what is the new max voltage inside of the capacitor?

$$V\times(4.894\times10^{-12}F)=4.593\times10^{-6}C$$

$$\frac{4.593\times10^{-6}C}{(4.894\times10^{-12}F)}=938.5kV{max}$$

 

[Baluncore]

Teflon does not disassociate because it is solid and the PTFE fluorine bonds are very strong. Water is an unusually good dielectric, but at a couple of volts it will hydrolyse and so become the gases 2H₂ + O₂. Metal electrode surfaces in contact with water will be stripped and enter solution, that will increase the conductance of the water in the capacitor cell.
Avoid water, use instead an oil or some other insulator.

 

[HelloCthulhu]

Thank you for the response! Did you check any of the math though? I want to make sure I calculated everything correctly.

 

[Baluncore]

Did you check any of the math though?

No.
There was no point me trying to check the mathematics of irreality. Any agreement with the math might be interpreted as agreement with the false premise that water is a dielectric like Teflon with a breakdown voltage greater than air.

 

[HelloCthulhu]

Ok that's understandable. Can you explain a little further what is happening in the capacitor under these circumstances?

 

[Baluncore]

https://en.wikipedia.org/wiki/Electrolysis_of_water

 

[Borek]

To make thing even more complicated, there is the https://en.wikipedia.org/wiki/Double_layer_(surface_science) - and when voltages get high, there are additional effects changing properties of the whole system (like https://en.wikipedia.org/wiki/Wien_effect ).

 

[Baluncore]

More complicated can also be more beautiful, and very expensive. Good insulators in the presence of water may develop a surface imperfection and then do interesting things. Water and DC can make trees grow.
https://en.wikipedia.org/wiki/Electrical_treeing

 

[Borek]

High voltage and treeing?

 

[HelloCthulhu]

Are you saying the breakdown voltage in reality isn't going to go higher than 650kV regardless of the dielectric constant? I know the dielectric is breaking down I'm just trying to understand at what voltage.

 

[Baluncore]

Are you saying the breakdown voltage in reality isn't going to go higher than 650kV regardless of the dielectric constant?

Who is saying what? Please quote statement and member, or post number.
A liquid water path across a capacitor will give a breakdown voltage of less than 2 volts total.
Dielectric constant and breakdown voltage are NOT directly related.
Air will also breakdown, so HV circuits such as EHT transformers and contactors are immersed in oil or SF₆ to displace air and water.

 

[HelloCthulhu]

I hope you'll forgive my ignorance. I'm just trying to understand these concepts and I'm really confused. I posted this in a thread back in December:

https://www.physicsforums.com/threa...tric-strength-per-volume.935008/#post-5906962

In that post I proposed (I typed "purpose" in the original) a scenario: A 1 cubic centimeter container is filled with water and electrodes are placed on either side. With roughly a distance of 1 centimeter (0.01m) between the electrodes, would the dielectric breakdown of the water equal 650kV (corrected from 600kV)? But if a capacitor is fashioned in the same way, you're saying that because of electrolysis it will have a breakdown voltage of less than 2V. Without an electrolyte, I don't understand how this is possible. I have a very limited understanding of these concepts and I greatly appreciate your patience.

 

[Borek]

What will happen around 2V is that the water will start to electrolyze. I am not sure it counts as a dielectric breakdown, I was always under impression DB is a rather dramatic, fast event. Electrolysis will be quite slow.

Pure water is a very good insulator (ultra pure water is sometimes called an 18 MΩ water, as its specific resistance is 18 MΩ/cm). Still, because of autodissociation it always contains minute amounts of H⁺ and OH^- and these are quite happy to react on the electrodes when the voltage is right (and 2V is perfectly enough). This electrolysis will be limited by the water resistance - electrolysis needs a closed circuit so that oxidation and reduction can take place simultaneously on opposite electrodes. Still, even if it is slow it will eventually create gas bubbles, changing capacitor properties.

Try to search literature, I find it hard to believe nobody checked experimentally how ultra pure water behaves as a dielectric in capacitors.

Note that ultra pure water is notoriously difficult to work with, even short exposition to air makes it no longer pure, as it quickly absorbs carbon dioxide.

 

[HelloCthulhu]

What will happen around 2V is that the water will start to electrolyze. I am not sure it counts as a dielectric breakdown, I was always under impression DB is a rather dramatic, fast event. Electrolysis will be quite slow...

Try to search literature, I find it hard to believe nobody checked experimentally how ultra pure water behaves as a dielectric in capacitors.

Thank you for this clarification. I've found some research on using electric fields to split water, but I'm not savvy on most of the terminology. I'll create a new post for it.

 

[Baluncore]

1. Here we discuss only liquid water, the electrical properties of water as ice are quite different.

2. How do you define breakdown voltage? A DC voltage or an AC voltage at what frequency?
Dielectric breakdown involves tearing the insulator structure apart.

3. Pure water has a pH of 7. That means 1/10⁷ of the molecules are dissociated into ions. Water is always a resistive conductor.

4. If the water is pure, then given a few volts DC across the cell, there will be some small DC current flowing with an exchange reaction at each electrode where hydrogen or oxygen gas will be released.

5. If the water is contaminated with any dissolved salt or electrode corrosion products, a higher DC current will flow, with the release of gas, or electroplating of the electrode surfaces. The velocity of ions in water is much slower than electrons on the surface. An AC voltage will not polarise the reversible reactions so there should be no net gas released. The water will then be more like a resistor and significant heat may be generated.

6. I believe the water bridge experiment is a trick that uses deionised water and a fast, maybe a 10μs step pulse. Such transients prefer to propagate through the air, Er=1, over the external surface of the water, rather than through the water, Er=80, at a velocity factor of 1/Sqrt(80), which is about one ninth the speed of light in air. For a fast pulse, the external arc will form first and so reduce the voltage across the bulk of the water.

7. Pure water between the plates of a capacitor must have a boundary. If the boundary is a gas or air then ionisation on that boundary is the probable mode of breakdown.

8. In a very few situations water can be used as a dielectric in a capacitor. It must be subjected to AC voltages only, while being continuously filtered, deionised and cooled.
https://en.wikipedia.org/wiki/Water_capacitor