Can we extract energy out of a capacitor in the ocean?

[maial]

Hello, sorry about my english it's not my mother tongue. I hope this is the right section to place this.

1. Homework Statement
A cylindrical capacitor is placed in the sea so that when a wave comes (the water goes up), the water becomes the capacitors dielectric, when the wave has passed (the water goes down) the air becomes the dielectric.
So now when the water is at its highest, i attach a battery to the capacitor.

When the water goes down i detach the capacitor from the battery, this creates an isolated system in which the charge remains constant and can change the potential difference between the capacitors plates.

The force of gravity makes the water that was between the capacitors plates go down, and so it reduces the capacitors capacitance. The dielectric constant of water is quite high (about 81) and the change of capacitance is also quite high.

The charge remained the same but the potential difference between the capacitors plates has increased.
This means that the energy in the system has increased.
So i found a way to harvest energy from the waves movement without moving bodies... but...

1) Explain why this system in practice can not really work.
2) Think about how to make it work (even if with reduced performance).

Homework Equations

The Attempt at a Solution

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I think this does not work because sea water is salty which means it conducts current. So now there's a current flowing through the capacitors plates which i think would damage the capacitor itself.
Actually when i attach the battery to the capacitor with water inside it makes a short circuit?

Honestly I'm not sure what happens when there's salty water in between the plates...

To make it work i think one would need to put distilled water in the capacitor or another insulating material and make it go up and down the capacitor thanks to the waves movement.

What do you think?

 

[mfb]

You can isolate the plates to avoid contact to sea water (good to avoid corrosion as well).
How do you keep the capacitor at the same height?

You can also estimate the power you can get out of such a device.

 

[scottdave]

Not that this answers the question, but just something to note: 80 is for pure water. Adding salt will create ions and decrease the dielectric. See this https://chemistry.stackexchange.com...ntration-and-electrical-permittivity-of-water

 

[Tom.G]

1) Explain why this system in practice can not really work.

The charge remained the same

Congratulations, you answered your own question!

 

[willem2]

The charge remained the same

Congratulations, you answered your own question!

The charge is supposed to remain the same. If you reduce the capacity of a disconnected capacitor by removing the dielectric the voltage of the capacitor and also the energy will go up, while the charge remains the same. The energy from this comes from the potential energy of the water.

 

[collinsmark]

I have a couple of questions which you should ask yourself.

1. Are there any forces on the cylinder of water caused by the electric field in the capacitor? (i.e., besides the force of gravity. Remember, the water is a dielectric.) If so, what effect are these forces going to have on the column of water?

2. I was going to ask about the inefficiency losses (I^2R losses) of directly connecting a battery to the capacitor as described, but let's ignore that for now. One thing at a time. Concentrate on 1. for now. Perhaps we can come back to 2. later.