I have been wondering really hard about what is the relationship between the defined zero potential at infinity as we do in physics, and the potential differences in circuits using batteries or other voltage sources.
Suppose we have a circuit with a battery and a resistor, and we measure the voltage across the resistor, what is the relationship between this potential and the zero defined at infinity?
As we know, the potential at a point due to an electric field is the integral from infinity to that point of E.dl, so how do you use this to calculate the potential across the resistor?
I know the zero of the circuit is in the circuit not at infinity as that would be dumb, but still this question hurts my brain. I want to know how do you calculate the potential across the resistor using the zero at infinity.
Can I define the potential across the resistor in relation to the zero at infinity or this doesn't exist ?
Can I calculate the potential across the resistor by taking a path integral from one end of it to the other ? Can I do it by taking the integral from one end of it to infinity, and then from infinity to the other end, and will the answer be the same as the voltage source reading?
Afterall, all the battery does is to split charges, so it creates an electric field, but I find it hard to visualize this electric field. I've heard that the electric field is zero across a conductor, but non zero in the resistor, so if I find the electric field inside the resistor to be constant, then the integral from one end of it to infinity and then back to the other end will be zero everywhere except inside the resistor if the electric field points directly from one end of the resistor to the other, but not if it doesn't.
What direction does the electric field inside this resistor point to?
Thanks a million!

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 15072017 02:53

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 15072017 04:28
(Original post by PAULOCONSTANTINO)
I have been wondering really hard about what is the relationship between the defined zero potential at infinity as we do in physics, and the potential differences in circuits using batteries or other voltage sources.
Suppose we have a circuit with a battery and a resistor, and we measure the voltage across the resistor, what is the relationship between this potential and the zero defined at infinity?
As we know, the potential at a point due to an electric field is the integral from infinity to that point of E.dl, so how do you use this to calculate the potential across the resistor?
I know the zero of the circuit is in the circuit not at infinity as that would be dumb, but still this question hurts my brain. I want to know how do you calculate the potential across the resistor using the zero at infinity.
Can I define the potential across the resistor in relation to the zero at infinity or this doesn't exist ?
Can I calculate the potential across the resistor by taking a path integral from one end of it to the other ? Can I do it by taking the integral from one end of it to infinity, and then from infinity to the other end, and will the answer be the same as the voltage source reading?
Afterall, all the battery does is to split charges, so it creates an electric field, but I find it hard to visualize this electric field. I've heard that the electric field is zero across a conductor, but non zero in the resistor, so if I find the electric field inside the resistor to be constant, then the integral from one end of it to infinity and then back to the other end will be zero everywhere except inside the resistor if the electric field points directly from one end of the resistor to the other, but not if it doesn't.
What direction does the electric field inside this resistor point to?
Thanks a million!
In a circuit, system is a closed loop between battery's ends, so integ for V is done over a closed loop. Hence Kirschoff laws. Between capacitor plates, you have a uniform Efield, straight & parallel equipotential lines. Note in uniform and radial fields, equipotential lines are perpendic to field lines.
In a circuit you can replace the battery with solenoid & magnet, to generate a varying Efield with potential lines, via EM induction. RE moving electrons, yes they carry their own field but these are small (vectors) and cancel out.
The electric field lines across a resistor point in the direction of conventional current, as the direction is defined by how they'd affect +ve charge.Last edited by Physics Enemy; 15072017 at 16:46. 
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 15072017 04:46
(Original post by Physics Enemy)
Integ from infinity applies to radial fields from a point source, be it grav, electric, magnetic. Has concentric, equipotential rings and can calc change in potential between them. You can calc potential on a ring by integ from infinity to it.
In a circuit, system is a closed loop between battery's ends, so integ for V is done over a closed loop. Hence Kirschoff laws. Between capacitor plates, you have a uniform Efield, straight & parallel equipotential lines. Note in uniform and radial fields, equipotential lines are perpendic to field lines.
What you can do in a circuit is replace the battery with a solenoid/magnet, to generate a varying Efield with potential lines, via EM induction. On moving electrons, yes they carry their own field but these are small (vectors) and cancel out.
The electric field lines across a resistor point in the direction of conventional current, as the direction is defined by how they'd affect +ve charge.
I think you should be able to integrate from infinity even in a closed loop battery system, because all the battery does is to split charges and create the electric fields which are identical to your point charge ones.
I understand Kirchoff and Faraday for closed loops, but I need to know if the answer matches with the infinity reference. 
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 15072017 14:25
(Original post by PAULOCONSTANTINO)
I think you should be able to integrate from infinity even in a closed loop battery system, because all the battery does is to split charges and create the electric fields which are identical to your point charge ones.
I understand Kirchoff and Faraday for closed loops, but I need to know if the answer matches with the infinity reference.
I've never heard of 2 simple circuits in close proximity interact in a discernable way, or people be affected.Last edited by Physics Enemy; 15072017 at 14:35. 
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 15072017 14:40
(Original post by Physics Enemy)
Don't think you can, charges outside a circuit don't acquire a potential / feel force from a circuit's Efield(s). Unless maybe at v small distances.
I've never heard of 2 simple circuits in close proximity interact in a discernable way, or people be affected.
I'm confused. All a battery does is to split charges, so what if we open circuit the battery and so there will be positive charges on the +ve end, and negative on the ve end, then going from infinity to the +ve end will give positive work because there's a small difference in distance between +ve and ve..... 
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 15072017 14:58
(Original post by PAULOCONSTANTINO)
I'm confused. All a battery does is to split charges, so what if we open circuit the battery and so there will be positive charges on the +ve end, and negative on the ve end, then going from infinity to the +ve end will give positive work because there's a small difference in distance between +ve and ve.....
Doubt I have anything more to add on thisLast edited by Physics Enemy; 15072017 at 16:50. 
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 15072017 17:52
You could say the same thing about static charges then. If you have q charge at point p, then the potential at a point R is defined as the work from infinity to R. So all I want is to know if the same can be defined for a battery. Why do I want to know this? Because it seems there's something inherently special about batteries, and different from static charges as in basic physics. I want to know if there is any difference between a static charge Q suspended in space, and the charges located at ve or ve of a battery. I get bugged by these things. I understand that in a battery it's a closed loop and all voltages that matter are differentials, but still, there must be a finite amount of work to go from infinity to the battery's ve and so on.

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 15072017 19:58
(Original post by PAULOCONSTANTINO)
You could say the same thing about static charges then. If you have q charge at point p, then the potential at a point R is defined as the work from infinity to R. So all I want is to know if the same can be defined for a battery.
(Original post by PAULOCONSTANTINO)
Why do I want to know this? Because it seems there's something inherently special about batteries, and different from static charges as in basic physics. I want to know if there is any difference between a static charge Q suspended in space, and the charges located at ve or ve of a battery. I get bugged by these things. I understand that in a battery it's a closed loop and all voltages that matter are differentials, but still, there must be a finite amount of work to go from infinity to the battery's ve and so on.Last edited by Physics Enemy; 15072017 at 20:04. 
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 15072017 20:34
Well a capacitor doesn't have a uniform E field, it bends on the edges, so a +ve charge put exactly in between the plates, at light years away distance from the plates will have work done on it as it returns to the capacitor. So the integral from inf to the capacitor is non zero... Unless the cap is infinitely large in area. This stuff is really confusing. I don't even know what to think anymore

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 15072017 22:00
(Original post by PAULOCONSTANTINO)
Well a capacitor doesn't have a uniform E field, it bends on the edges, so a +ve charge put exactly in between the plates, at light years away distance from the plates will have work done on it as it returns to the capacitor. So the integral from inf to the capacitor is non zero... Unless the cap is infinitely large in area. This stuff is really confusing. I don't even know what to think anymoreLast edited by Physics Enemy; 15072017 at 22:05. 
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 15072017 22:05
(Original post by Physics Enemy)
Uniform Efield apart from at the edges then. If plates are infinitely far apart there'd be no circuit and even if hypoth there was, there'd be no Efield. Work is always done to move a particle, here it wouldn't be against an Efield. I'm unsure about infinite area capacitors.
This is hardcore confusing 
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 15072017 22:20
(Original post by PAULOCONSTANTINO)
Sir, as far as I know if you move the plates of a cap further away, the electric field remains constant as it is just sigma/epsilon. The voltage between the plates increases as it is V = Q/C, and C = epsilon0 * A / d, so V = Q*d/(epsilon0*A), so as d > inf, V > inf.
This is hardcore confusing
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