How do i interpret the following electric potential diagrams?

Question

We recently started to learn about electricity in school and I'm trying to understand the concept of voltage(potential differance). From what i understand, electrons flow from the negative terminal to the positive. The electrons at the negative terminal will have 1.5 J of potential energy per Columb(see diagram), and when they reach the positive terminal, all the potential energy will have been transformed.

I have a few questions about this circuit:

1. Is the potential energy of a negatively charged particle in the negative terminal equal to the work needed to "push" it from the positive terminal to the negative terminal in the internal circuit, in other words is it equal to E * Q * d, where E is the electric field strength, Q is the charge of the particle and d is the distance between the terminals.

2. The diagram to the right shows how the voltage drops in the circuit. According to the illustration, the electrical potential remains constant from point A to point B. But is this the case? Isn't electric potential really energy due to position per columb? And since the position change, shouldn't the potential change aswell? Similary, after the charges pass the lightbulb the diagram shows that all the potential energy is transformed to light and thermal energy. But wouldn't there be some potential energy left, since a charge at position C has potential energy with respect to the positive terminal.

3. If the only cause of voltage drop in the circuit is when the charges encounters the lighbuld, then that suggests there would be no drop in potential from A to B if the circuit didn't have the lightbulb. This doesn't make sense to me.

   4. And lastly, how is the light energy produced in the lightbulb? Is it the kinetic energy from the electrons doing work on the filament in the lightbulb?

I would greatly appriciate it if you can answer/correct any or all of these questions.

Answer 1

1. Correct as far as it goes, and it works well in situations where the electric field is known. But it's not often used in circuits, because the electric field is not easily described.

2. It is due to its position in the electric field. But the shape of the field is not simple. Because of the (mobile) charges in the conducting wire, the field inside the wire is zero. So it takes (almost) no work to move a charge along the wire. The potential of a charge does not change from one end to the other.

3. That's correct. In an ideal conductor, there is no change in potential from one side to the other. If you leave the lightbulb out (open circuit), or if there is a reasonable resistance in the circuit, then there is no problem assuming the wire has zero resistance. If you just connect the two conductors (short circuit), then the differences between an ideal conductor and the actual conductor becomes significant. The real conductor does have some resistance. As the current rose in the circuit, this resistance (or the resistance in other parts of the circuit) would be sufficient to drop the voltage of the battery (or something would start breaking).

4. Yes, you can think of the energy transfer in a resistor as due to thermal heating from the charge collisions.