Electric potential, conceptual problem

[_Andreas]

If you look at the diagram right below the headline "Electric potential diagrams" in http://www.physicsclassroom.com/Class/circuits/u9l1c.cfm" (about halfway down), you can see that in circuit A, the electric potential at point A is the same as that at point B, and likewise for points C and D. This makes little sense to me, perhaps because I think of electric potential as I think of gravitational potential: the + terminal is a point at some nonzero height above ground (if the charge carriers are positive), and thus the - terminal is at ground level. The wire running between the terminals is the vertical distance between this point and the ground, and thus there is an electric potential difference between every point on the wire. Viewing it like this, you can probably see why I don't understand how the electric potential at point A can be the same as at point B.

Could anyone help clear things up for me?

 

[Doc Al]

What matters is not the physical length of the wires, but the voltage drop across each element. Compared to the light bulbs, the wires have essentially no resistance so the voltage drop across them is zero.

 

[_Andreas]

What matters is not the physical length of the wires, but the voltage drop across each element. Compared to the light bulbs, the wires have essentially no resistance so the voltage drop across them is zero.

Thanks, but it still doesn't make sense. As I understand it, a positive test charge which is placed in an electrostatic field will lose potential energy if it flows along the field, regardless of any resistance. I think of it as a falling ball, which loses potential energy and gains kinetic energy, even if there is no air resistance. Thus, in my head, even if there is no resistance in the wire, the charge should lose potential energy as it moves from the + terminal to the - terminal.

 

[Doc Al]

Sure, there's got to be some loss of potential energy as the charges move across the wire. But not much! Especially compared to what it loses when moving across the light bulb, which has much more resistance. For all practical purposes it loses all its energy in the bulbs and not the wires.

 

[_Andreas]

Sure, there's got to be some loss of potential energy as the charges move across the wire. But not much! Especially compared to what it loses when moving across the light bulb, which has much more resistance. For all practical purposes it loses all its energy in the bulbs and not the wires.

But why is

ball falling in vacuum -> ball losing gravitational potential energy as it gets closer to the ground (and gaining kinetic energy)

true, but not

positive electric charge carrier moving along the field in a completely resistance-free wire -> electric charge carrier losing electric potential energy as it gets closer to the the - terminal (and gaining kinetic energy)?

 

[NascentOxygen]

positive electric charge carrier moving along the field in a completely resistance-free wire -> electric charge carrier losing electric potential energy as it gets closer to the the - terminal (and gaining kinetic energy)?

You are worrying needlessly over this because (a) there are no positive electric charge carriers in metal, and (b) you will never be using a completely resistance-free wire.

All wire conductors have some resistance per unit length, so there exists a potential gradient per unit length and this causes electrons to flow accordingly. As electrons move through the voltage gradient their electrical potential energy is converted it into heat energy (kinetic energy of vibration of the conductor atoms).

 

[_Andreas]

You are worrying needlessly over this because (a) there are no positive electric charge carriers in metal, and (b) you will never be using a completely resistance-free wire.

All wire conductors have some resistance per unit length, so there exists a potential gradient per unit length and this causes electrons to flow accordingly. As electrons move through the voltage gradient their electrical potential energy is converted it into heat energy (kinetic energy of vibration of the conductor atoms).

Yes, I do get that there aren't any resistance-free wires, but what I'm basically asking is this: why are gravitational potential gradients not dependent on resistance, while electric potential gradients are? Even in a vacuum (no resistance) there's a difference in gravitational potential between two points at different heights above ground, while in a resistance free-wire (again, I know such wires don't exist in the real world), there's apparently no potential difference between two points on the wire. There's obviously a fundamental misunderstanding of electric potential on my part here, but I need help to find out what it is.

 

[NascentOxygen]

Gravitational potential is determined by g and m and is independent of the viscosity or density of the medium. Electric potential energy is determined by the charge on the body and the voltage difference that exists, and is independent of the resistivity of the medium.

The two seem very analogous, to me.

I thought you were edging towards asking how zero-resistivity superconductivity manages to get current through a wire when there is zero voltage gradient across it. I'm not sure. But I believe superconductors exclude magnetic field lines, so clearly they are fascinating curiousities. And if we are really fortunate, we may some day have room temperature superconductors to experiment with, instead of, as now, having to chill most of them to at least the temperature of liquid N₂.

 

[_Andreas]

I thought you were edging towards asking how zero-resistivity superconductivity manages to get current through a wire when there is zero voltage gradient across it.

No, my question is much dumber than that.

I think I may have spotted what's wrong with my mental model, though (please tell me if I have!). The answer is probably spelled "electric field lines". This is the way I've been thinking about it until now: the electric field lines representing the electrostatic field generated by the battery follow the inside of the wire perfectly from the + terminal to the - terminal, i.e, where the wire is straight, the field lines are straight, and where the wire is curved, the field lines are, too. Since the equipotential surfaces are everywhere perpendicular to the field lines, then there'd be a voltage difference between any two points on the wire. But this is wrong, isn't it? The field doesn't behave this way, right?

 

[Doc Al]

Since the equipotential surfaces are everywhere perpendicular to the field lines, then there'd be a voltage difference between any two points on the wire. But this is wrong, isn't it? The field doesn't behave this way, right?

There's nothing particularly wrong about this picture. It's just that the voltage difference across the wire is very small compared to that across the light bulbs. For many purposes, you can ignore it.

 

[NascentOxygen]

This is the way I've been thinking about it until now: the electric field lines representing the electrostatic field generated by the battery follow the inside of the wire perfectly from the + terminal to the - terminal,

I associate electrostatic fields with static charges. There are no static charges in a battery, so I would not use the word. The correct term would be "electric field". I suppose you could speak of field lines in a conductor, but that is not usually done. (A field line is defined as mapping the path a single ( positive ) charge would follow if placed on that line. In a conductor, I think we can see what path the electrons will take.)

Equipotential lines is probably a better concept, as these highlight the voltage gradients.