# Relationship between potential in batteries and zero potential at infinity

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#1
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!
0
3 years ago
#2
(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!
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 E-field, 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 E-field 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.
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#3
(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 E-field, 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 E-field 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.
0
3 years ago
#4
(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.
Don't think you can, charges outside a circuit don't acquire a potential / feel force from a circuit's E-field(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.
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#5
(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 E-field(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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3 years ago
#6
(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.....
Still doesn't make sense to talk about infinity when any interesting Physics here only occurs at a micro level, and you need a radial field to integ from infinity to r. There's a reason you've not seen it done on circuits.

Doubt I have anything more to add on this 0
#7
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.
0
3 years ago
#8
(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.
Again, that definition applies to radial fields, which a point charge provides. A battery or capacitor connected in a circuit doesn't. You know a capacitor has a uniform E-field with its own formula for potential. In a simple resistor circuit, V is defined using a closed loop integral, only connected components have a measurable V across them.

(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.
Difference is type of E-field, one is open & radial, the other isn't. In a radial field with inverse square law, field lines extends to infinity. Work is done on a particle to move against the field all the way to/from infinity. Unlike the battery scenario. RE charges inside batteries/wires 1) insulation 2) fields cancel out.
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#9
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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3 years ago
#10
(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 anymore
Uniform E-field 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 E-field. Work is always done to move a particle, here it wouldn't be against an E-field. I'm unsure about infinite area capacitors.
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#11
(Original post by Physics Enemy)
Uniform E-field 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 E-field. Work is always done to move a particle, here it wouldn't be against an E-field. I'm unsure about infinite area capacitors.
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
0
3 years ago
#12
(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
I fail to see how finite capacitors can interact at infinite distance to produce an E-field and a V, let alone infinite V. I don't think those formulae hold for d -> infinity, or even large d. They're conditional.
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