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1:35 PM
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Q: Is the 0 potential of the -ve terminal fo battery only an assumption?

Aditya MukherjeeI have just recently started studying electric circuits, and as far as I know if a cell has a potential difference of V, we can assume the negative terminal of the battery to have 0V potential and positive terminal to have V volts potential. I have solved a number of problems in this manner, and ...

 
Imagine your (b) circuit is floating the great voids of outer space between super-clusters of galaxies. It's totally alone. What's the voltage at any node in the circuit? (It still works just fine.) It should be impossible to say. All that you can say about any of it is that the nodes have relative voltages to each other, but that the values themselves at the nodes are arbitrary. You can add one million to all of them and nothing changes. Or subtract that much and nothing changes. Still works, too. The absolute numbers are meaningless. So you get to pick any one of them and call it zero.
 
@periblepsis , ok, so it is just a convenient assumption, so that we can solve the problem, right?
 
You can solve the problem by assuming one of the nodes is one, too. Or, if you like it abstract, then x and just leave it to be filled out later when you feel like it. (Keep in mind that a positive end of a voltage source can be assigned 0, or 1, or x, just as easily as any other.) On a more technical note, there's something called the Rank-Nullity Theorem that, applied in graph theory, informs the whole question, rigorously and clearly. But then I'd have to write a lot more. (See here for a sample.)
 
@periblepsis , can the reference point for measuring potentials be taken at infinity?
 
Do you mean in the way that those in orbital mechanics and discussing orbital energy do?
 
1:35 PM
@periblepsis , no like we do in measuring potential for say like a charged sphere in electrostatics
 
Okay. Then yes, similar idea. But no, I haven't seen it done. And thinking about it I'm not sure how you would define things to make that work right. Conservation laws and symmetries don't seem to apply. I wouldn't want to stop you from trying, though.
 
@periblepsis , yeah it is kind of useless, I can't make it work either
@periblepsis, I have another doubt, it is kind of off topic of this question, but I wanted to ask that can we always say that potential decreases when we move from +ve plate to -ve plate of a capacitor? I think it is true, because the electric field is from the +ve plate to -ve plate.
 
I need to hit the bed (was just going there.) But I'll take a short moment, anyway.
I'm not sure what that question entails...
Hmm. I think the best book for that topic (one that is both readable and accessible as well as getting into the necessary details) is Matter & Interactions. They cover the entire analysis of the field surrounding a capacitor, in fact.
But yes, in the region between the idealized plates, the field is directed. What's interesting, moreso, is what the field does around the rest of the regions surrounding that volume.
(I think the authors are Chabay and Sherwood. Memory serving.)
 
Ok, but can you please tell if I was right to say that the potential of the negatively charged plate of a capacitor is less than the potential of the +vely charged plate?
Since delta V=-E×d and E is directed from +ve plate to -ve plate
I will definitely try to read the book you recommend, but I have an upcoming exam, so for that is it safe to state this?
 
By convention, yes. But it's a difference. There are only potential differences.
 
1:52 PM
Yes but this potential difference (-ve plate - +ve plate) will always come out to be -ve though?
 
If you take two plates and apply a one volt difference between them and release a coulomb of charge near the plate similarly charged then they will impact the opposite plate with exactly one Joule of kinetic energy. And that fact is independent of the distance between the plates.
 
Ok
Thanks for your time
 
No problem. :)
 

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