Resistance Between Two Rectangular Electrodes

[Vne147]

Hello everyone. I have what is probably a relatively simple question. I'm trying to calculate the resistance between two rectangular copper plates submerged in water. I found this thread that briefly discusses it:

https://www.physicsforums.com/threa...ween-two-electrodes-filled-with-water.147144/

The equation:

R = (ρ*L)/A

seems pretty straight forward but I have a question about the area, A. Is the area in the equation the area of a single copper plate, or the sum of both plates?

For instance, if I have two parallel rectangular copper plates each with a surface area facing the other of 1 in², should I use 1 in², or 2 in² in the equation? I'm thinking I should just use 1 in² but I'm not sure.

Perhaps if someone can share or point me to a derivation of the formula, that would clarify things.

Thanks in advance for any help you can provide.

 

[cnh1995]

I'm thinking I should just use 1 in2

Yes, provided it is same everywhere along the conducting path. It is the cross sectional area of the conducting path.

 

[Vne147]

Yes, provided it is same everywhere along the conducting path. It is the cross sectional area of the conducting path.

Thank you sir. That was my hunch. Just to satisfy my curiosity though, do you know where I can find a derivation?

 

[cnh1995]

Thank you sir. That was my hunch. Just to satisfy my curiosity though, do you know where I can find a derivation?

Resistance of a conductor is directly proportional to the length of the conductor and inversely proportional to its cross sectional area. These are experimentally observed facts (and are intuitive). So, resistivity ρ is simply the proportioality constant in the relation R∝L/A.
You can find this stuff in any college level physics book.

 

[Vne147]

Understood. Thanks.

 

[The Electrician]

The formula R∝L/A only applies if the water is confined to a rectangular cross section tube between the electrodes, the cross section of the tube being the same as the dimensions of the electrodes.

Otherwise, the current in the water will expand outward into a much larger volume of water than would be the case if the water were in the described tube.

In the non-confined case you will need to use advanced techniques to calculate the resistance.

See: http://www.bru.hlphys.jku.at/conf_map/index.html