Microscopic explanation of the Joule effect and dielectric on capacitors

Question

I understand intuitively that the resistor heats up when electrical energy travels through it and mathematically that the capacitor stores more charge when we fill the space between the conductive plates with a dielectric. However, I cannot see a microscopic explanation for these phenomena. Why is this, at the atom level, like this?

Superfast Jellyfish · Accepted Answer

When electrons zip around in a wire, they occasionally collide with the atoms of the wire. As a result, the electron can impart some (or all) of its kinetic energy to the atoms of the wire. This causes them to vibrate. And it’s this vibration that is macroscopically seen as the heating up.

Consider the atoms in the dielectric. They can be thought of as dipoles that can swivel around to align themselves to the external electric field. When an electric field is turned on, then all the negative ends point towards the positive plate, and all the positive ends towards the negative plate.

The field polarises the dielectric. Now, each of these dipoles attract the respective charges on the plates more so than if the dielectric was absent.

trula · Answer

Current trough a wire gets the electrons moving, ( I would not talk about traveling energy)  they are slowed by bumping in atoms and get them to move or vibrate faster, which is heat.
The stuff between the plates of the capacitor ist partly polarized, so the electric field has shorter ways, the same thing as getting the plates closer zigether.

Microscopic explanation of the Joule effect and dielectric on capacitors

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