0
$\begingroup$

See for example this circuit

enter image description here

The resistor shown has a "potential difference" of 10V across it. Now we know that this refers to the work done per unit charge... The change in energy per unit charge. The name implies that the type of this energy is electric potential energy, but is this really the case? Electric potential energy is energy due to the position of a charge in the electric field so that means the only factor affecting it is the distance of the charge from the positive terminal, that means the drop in electrical potential energy for will be uniform as it moves through the length of the circuit regardless of whether or not passing through a resistor.

To further clarify my question, imagine an object falling in a uniform gravitational field. This rock will lose, say, 5J of gravitational potential energy for every 10m that it falls. Comparing this scenario with that of the electron... Shouldn't the electron lose some fixed quantity (E) of energy for every distance (x) that it moves through the circuit? Why is it that the greatest drop of electrical potential energy occurs in the resistor even if it doesn't cover that much distance?

$\endgroup$
0
$\begingroup$

Simply put, the electric field is not constant in all parts of this circuit: it is strong in the resistor and weak in the wires. A weak electric field means only a small amount of potential change per meter. To see why $\vec E$ must change, Ohms law relates the current to the electric field of the circuit: $$ \vec J = \sigma \vec E$$ Where $\vec J$ is the current density, $\sigma$ is the conductivity of the material, and $\vec E$ is the electric field. If the conductivity goes up (the wires) the electric field need not be as strong to produce the same current density.

To re-use your gravity analogy, $\vec g$ is the same as $\vec E$, and $\vec g$ would have to change during the masses' fall.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see! exactly what i was looking for. $\endgroup$ – user52347 Jan 17 '18 at 21:06
0
$\begingroup$

The electron loses energy when moving in the circuit towards the positive battery terminal according to the change of the local electrical potential. It loses most of its energy by traversing the resistor because there most (ideally all) of the potential drop happens. The energy lost by the electron in this process is delivered by microscopic scattering events as heat to the lattice of the resistor material.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.