Timeline for Different result when deriving electric force as $dW/dr$ and $-dU/dr$
Current License: CC BY-SA 4.0
13 events
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| Apr 2, 2022 at 8:36 | history | edited | Marko Gulin | CC BY-SA 4.0 |
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| Apr 1, 2022 at 14:43 | vote | accept | radastro | ||
| Apr 1, 2022 at 13:58 | comment | added | Billy Istiak | I can't find any question in your last edit.... | |
| Apr 1, 2022 at 13:42 | history | edited | radastro | CC BY-SA 4.0 |
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| Apr 1, 2022 at 13:36 | history | edited | radastro | CC BY-SA 4.0 |
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| Apr 1, 2022 at 13:18 | comment | added | kricheli | Who says that $W$ is the Coulomb potential? Anyways: When working with work, always make sure to inspect who performs work on whom before discussing the sign! | |
| Apr 1, 2022 at 13:13 | comment | added | Marko Gulin | Am I correct to assume that you are confused how come $W$ and $dW$ mean the same thing? If yes, then see the post below. | |
| Apr 1, 2022 at 13:07 | answer | added | Marko Gulin | timeline score: 1 | |
| Apr 1, 2022 at 12:56 | comment | added | radastro | This might be a possible reason: $W=-\Delta U_r$, only $U_r$ can be express as a change, and we can then relate this to the derivative, whereas $W$ already means a change, so we don't write $dW$ | |
| Apr 1, 2022 at 12:45 | comment | added | radastro | $W$ and $U_r$ equals $\frac{kQq}{r}$, and by differentiating this, we get different equations of $F_e$, as the sign is different which is mathematically obvious as there is a missing negative sign in the derivative of work done. | |
| Apr 1, 2022 at 12:44 | history | edited | Qmechanic♦ |
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| Apr 1, 2022 at 12:34 | comment | added | Marko Gulin | From $-dU_r / dr = F_e$ and $dW/dr = F_e$ it follows $-dU_r/dr = dW/dr$ which equals $dW = -dU_r$. This is exactly what you wrote in the first line $W = -\Delta U$. What do you mean by "this contradicts with the first derivation as $W$ and $U_r$ both have the same equation"? | |
| Apr 1, 2022 at 12:28 | history | asked | radastro | CC BY-SA 4.0 |