Timeline for Charged capacitors in series -- but connected at same polarity plates?
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15 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2019 at 14:17 | comment | added | Bob D | @Farcher I realized that there was negative charge equal to the positive charge, and was just not shown. Everything you have said is exactly as I thought it should be, but was looking for confirmation. Thank you! | |
| Jan 24, 2019 at 13:59 | comment | added | Farcher | @BobD They are at the same potential relative to any point that you wish to choose. If I connected another conducting wire between those two top plates no charge would move because the potential difference across the wire is zero. | |
| Jan 24, 2019 at 13:58 | comment | added | Farcher | @BobD Please note that just below my diagrams I wrote "I have not labelled the equal magnitude negative charges which reside on the other capacitor plates." So those negative charges which cannot move keep the positive charges locked in position. | |
| Jan 24, 2019 at 13:57 | comment | added | Bob D | @Farcher And if so, at the same potential with respect to what? | |
| Jan 24, 2019 at 13:47 | comment | added | Bob D | @Farcher So even though the charges on the positive plates are different, the plates are still at the same potential. Correct? | |
| Jan 24, 2019 at 13:06 | comment | added | Farcher | @BobD The top plates are at the same potential, it is the bottom plates that are at different potentials. | |
| Jan 24, 2019 at 12:57 | comment | added | Bob D | @Farcher Question about Diagram 1. The positive plates are connected together. One capacitor has a +2uC charge and the other a +3uC charge. I don't disagree with the diagram, but if someone were to question the differences in charge by saying since the two plates are connected together, they should be "at the same potential" how would you respond. | |
| Oct 18, 2016 at 11:52 | comment | added | Farcher | That is why I made the comment about being careful with the signs. What you have found is that the voltage across the $10 \mu \rm F$ capacitor is less than $10 \rm V$ showing that voltage across the $5 \mu \rm F$ was reversed with a plate initially having positive charges on it now has negative charges. | |
| Oct 18, 2016 at 11:41 | comment | added | NeedForMath | Thank you very much. Got to the answer, but one of the final charges was of negative sign. I guess that has to do with the current direction, so adjusting for that we'd get both to be positive? | |
| Oct 18, 2016 at 11:25 | vote | accept | NeedForMath | ||
| Oct 18, 2016 at 9:35 | comment | added | Massimo Ortolano | Your edit if fine ;-) | |
| Oct 18, 2016 at 9:14 | history | edited | Farcher | CC BY-SA 3.0 |
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| Oct 18, 2016 at 9:13 | comment | added | Farcher | @MassimoOrtolano Thank you for your comment. Do you think that I should amend my answer as the use of the formula directly in this case can cause a lot of problems as the OP found? I have changed the first paragraph. | |
| Oct 18, 2016 at 9:09 | comment | added | Massimo Ortolano | "The reason for that is that the capacitor in series formula was derived on the assumption that both capacitors were initially uncharged.": The capacitance of two capacitor in series is a parameter of the circuit, and it is independent on any assumption on the initial charges. | |
| Oct 18, 2016 at 7:40 | history | answered | Farcher | CC BY-SA 3.0 |