Timeline for Potential Drop across a Capacitor
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2019 at 7:58 | comment | added | Bob D | @Aditya Ahuja Are you aware there is an “Accept” button? | |
| Nov 25, 2019 at 7:54 | comment | added | Aditya Ahuja | @BobD Yes.Thank You very much. | |
| Nov 25, 2019 at 7:26 | comment | added | Bob D | @Aditya Ahuja Yup that’s about it. The battery converts chemical energy to electric energy to do the necessary work to move the charge between plates. After all this do you now find my answer acceptable? | |
| Nov 25, 2019 at 6:43 | comment | added | Aditya Ahuja | @BobD So what forces the electrons to leave one plate and arrive at the other is the electric field created by the cell which attracts the electrons towards the negative plate,give them potential energy through some chemical reactions and send them on their way to the positive terminal. Am I wrong? | |
| Nov 25, 2019 at 6:40 | comment | added | Aditya Ahuja | @Bob D Thank you I got it. | |
| Nov 25, 2019 at 3:56 | comment | added | Bob D | So can you see that displacement of the electron by another makes no sense? | |
| Nov 25, 2019 at 3:55 | comment | added | Bob D | @Aditya Ahuja No that’s not what happens. Start with two plates each with an equal number of electrons and protons, i.e. two electrically neutral plates. Take one electron away from one plate and move it to the other plate. The plate it came from now has a net charge of +1. The plate it went to now has a net charge of -1. It doesn’t cause another electron to leave the plate it went to. Where would the displaced electron go? And if it left, the plate would be neutral again. If this happened for each electron you wind up with a positive plate and neutral plate, not a positive and negative plate. | |
| Nov 24, 2019 at 6:57 | comment | added | Aditya Ahuja | @BobD what I mean by displacing a charge is that the electron provides a repulsive force to an electron on the other plate (which then leaves that plate). Also , yes I believe we're on the same page about the work done in the circuit. | |
| Nov 22, 2019 at 13:25 | comment | added | Bob D | @AdityaAhuja What you said was the charge is "only displacing another charge". I don't see an electron leaving one plate and being deposited on another as "displacing another charge". Perhaps I see the meaning of a charge "displacing another charge" different from you. Bottom line, if you agree that all that is happening is that work is done to remove electrons from one plate and deposit them on another in addition to the work to push the electrons through the resistor, then we are on the same page and I will attempt an answer to your question. If not, I'll take a pass on it. | |
| Nov 22, 2019 at 7:10 | comment | added | Aditya Ahuja | @Bob D The charge displaced is on the other plate. Also I know that the charge doesn't go across the dielectric. | |
| Nov 21, 2019 at 17:52 | comment | added | Bob D | @AdityaAhuja First of all, I assume you understand that no charge "goes across" the capacitor if you mean it goes through the dielectric to get from one plate to the other. I also assume you also understand that when a battery moves charge (electrons) from one plate to the other those electrons arriving at the plate don't "displace" charges on that plate. The result is a net negative charge on the plate they are deposited on leaving an equal net positive charge, or a deficit of electrons, on the plate they were taken from. Is that how you understand it? | |
| Nov 21, 2019 at 16:43 | comment | added | Aditya Ahuja | @Bob D I'll rephrase.If the charge is not actually going across the capacitor and only displacing another charge why do we say that the potential difference is q/C. Isn't q/C the potential difference between the PLATES of the capacitor. Let me give an example of what I mean. Let us say a charge goes through a battery of 5 V (anode is the reference) and arrives at a capacitor plate. This charge displaces another charge . If the potential drop across the capacitor is 3 V at this time. Is it right to say that the displaced charge has a potential of 2 V? | |
| Nov 20, 2019 at 19:14 | comment | added | Bob D | @AdityaAhuja whether you call it a potential drop or potential rise depends on which way you go around the circuit when applying KVL. In your circuit, if go around clockwise it is a potential drop across the capacitor. If you go counter-clockwise it's a potential rise. | |
| Nov 11, 2019 at 13:13 | comment | added | Aditya Ahuja | But when we talk about potential drop we usually take in terms of charges right ? For example, A charge with some potential goes across a resistor and loses some potential energy. Is that not the case here? | |
| Nov 10, 2019 at 8:50 | history | answered | Farcher | CC BY-SA 4.0 |