i understand that capacitance of capacitors is based around known factors, plate size, distance etc trying to figure out what is happening in the pic below. A is of a known size suspended in an electrolyte. B are of unknown size. from what i understand, the size of A has a direct bearing of the capacitance, but what happens if i move A in either direction. will the capacitance change with getting closer to one side than the other ? or does the surface area dictate the figure and it remains constant ?
Thanks in advance

 

First, you are not showing any insulators anywhere, so your picture looks more like a battery than a capacitor.

FYI, the electrolyte of an electrolytic capacitor forms a thin insulating oxide layer on one of the electrodes. It is this layer that forms the capacitance. The electrolyte itself is just a conductor of electricity. It doesn't play a part in the overall capacitance, and therefore, neither would the position of your 'A' electrode.

All IMHO, of course.

 

Think about the orbits of electrons around molecules of the dielectric. As you apply voltage to the capacitor, the orbits change from round to elliptical. The higher the voltage, the more elliptical. Now, if you disconnect the voltage supply and move the plates further apart, the orbits will stretch out further. Like a rubber band pulled further apart, it gives more potential energy. (Potential energy?? Sound familiar?) Of course this is what I remember from tech school 50 years ago and sometimes I have trouble remembering what I had for lunch yesterday.

ed. After reading joeyd999's post, you need to change your term "electrolyte" to dielectric for my thoughts to apply.

 

Think about the orbits of electrons around molecules of the dielectric. As you apply voltage to the capacitor, the orbits change from round to elliptical. The higher the voltage, the more elliptical. Now, if you disconnect the voltage supply and move the plates further apart, the orbits will stretch out further. Like a rubber band pulled further apart, it gives more potential energy. (Potential energy?? Sound familiar?) Of course this is what I remember from tech school 50 years ago and sometimes I have trouble remembering what I had for lunch yesterday.

Never heard this explanation before....sounds kinda fishy.

Assuming 'dielectric' as opposed to 'electrolyte', IIRC Maxwell's equations provide a quantitative solution to the problem.

 

Never heard this explanation before....sounds kinda fishy.

 

Check this out...... http://webphysics.davidson.edu/physlet_resources/bu_potential/Potential_Capacitor_Text.html

and this.... http://answers.yahoo.com/question/index?qid=20090613161637AAZyk8a

and this.....http://books.google.com/books?id=KgNhk-HcI4oC&pg=PA226&lpg=PA226&dq=elliptical+electron+orbit+in+capacitor&source=bl&ots=ttXdD3nwD8&sig=gWFqMoBw-bxjTrI9WOxIK1ENxd4&hl=en&sa=X&ei=0WAGT4qkIMivsALZ2ISRCg&ved=0CDcQ6AEwAA#v=onepage&q=elliptical%20electron%20orbit%20in%20capacitor&f=false

and this one shows charge /discharge animation of elliptical orbits......http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE11805

 

LOL
so Ok , its a battery not a cap i'm trying to find a way of (after getting a base figure) of replacing part A with an object of unknown size and getting a surface area. obviously resistance is out, as trying to measure to 5 degimal points of an ohm is tricky, and was trying to see if it could be done by measuring the capacitance (if there was any), as surface area does have a direct bearing. maybe im barking up the wrong tree and need to find a different aproach

 

Time to whip out one of my macros.

Please explain the entire problem in detail, we may find an overall simpler solution to your problem than a solution for your solution.

 

I do small scale electroplating and anodising, the process involves using the correct current through the electrolyte for the surface area size to acheive the required thickness of coating. so the surface area measurement is critical. at the moment i draw it in cad and press the relevant calc button, but some shapes (like sculptures etc.) can take me a week to draw, so got to thinking that as surface area is directly proportional to current draw within a solution, this must be measurable by an easier method. to use the existing electrolyte tank would be a bonus. For example, resistance of 25cm sq surface gives 310 ohms, double the size, halve the resistance. very quickly you are into very small resistance measurements . when youve got upto say a car wheel, you are down to 0.000162 ohms. hence my inquiry on possibly using natural capacitance.