# How do i interpret the following electric potential diagrams?

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. 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.

1. 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.