Timeline for Is there a physical explanation for why increasing a capacitor's capacitance and the circuit's resistance decrease the capacitor's rate of discharge?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 12, 2020 at 22:35 | vote | accept | Bøbby Leung | ||
| Aug 12, 2020 at 22:35 | vote | accept | Bøbby Leung | ||
| Aug 12, 2020 at 22:35 | |||||
| Oct 1, 2018 at 11:35 | comment | added | Alfred Centauri | @BøbbyLeung, to charge a capacitor, the external circuit, e.g, a battery essentially 'pumps' electrons from one plate of he capacitor to the other. This takes work (moving the electrons against the increasing electric field) and the work done is stored in the energy of the electric field between the two plates. | |
| Oct 1, 2018 at 5:05 | comment | added | Bøbby Leung | @AlfredCentauri Thanks for the answer and comment. You mentioned capacitor stores energy. May I ask to confirm, is it because the charge Q on one plate and the charge -Q on the other plate are energy carriers or carried energy from e.g. a battery? | |
| Oct 1, 2018 at 1:02 | comment | added | David White | @AlfredCentauri, I certainly realize that. I was trying to give the OP a physical representation that was a good analogy, so he could hopefully "connect the dots". | |
| Sep 30, 2018 at 18:49 | comment | added | Alfred Centauri | To be sure, capacitors do not store electric charge; with charge $Q$ on one plate and charge $-Q$ on the other plate, the net charge stored in a capacitor is zero. A charged capacitor stores energy and not electric charge. | |
| Sep 30, 2018 at 18:24 | history | answered | David White | CC BY-SA 4.0 |