Timeline for Why isn't electric field due to outside charges taken into account when calculating the "total" field in some Gauss law problems?
Current License: CC BY-SA 4.0
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| Oct 3, 2020 at 11:28 | comment | added | Rishab Navaneet | I get your point now. I now realise I've made a mistake... Let me think over it and I'll edit my answer. But to correct you a bit, I think in case of griffiths, it's not only $\int{E_o \cdot da} $ that is zero but $E_o$ itself is zero because of the concentric cylinder argument mentioned. | |
| Oct 3, 2020 at 11:17 | comment | added | Brain Stroke Patient | @RishabNavaneet But isn't $\mathbf{E_1}$ constant over $S_1$ ? So why can't we take it out of the integral? | |
| Oct 3, 2020 at 11:08 | comment | added | Rishab Navaneet | @BrainStrokePatient : you are absolutely right upto the vanishing of $\int_{S_1}{\vec{E_2}\cdot ds}$. But how can we find $\vec E_1$ from the left over integral? We cannot continue from there on because you cannot take $E_1$ out of the integral. | |
| Oct 3, 2020 at 10:52 | comment | added | jacob1729 | @BrainStrokePatient the answer's point is that what enters Gauss' law is the total field. The issue is that Gauss' law really is a formula for the charge enclosed by a surface. It's non-trivial to invert that for the field on the surface and not always possible. | |
| Oct 3, 2020 at 10:51 | comment | added | Brian | @BrainStrokePatient I think you should add this specific point in the answer so people know that you got that point already | |
| Oct 3, 2020 at 10:47 | comment | added | Brain Stroke Patient | I'm having trouble with your statement that Gauss law always finds the net electric field. If $\mathbf{E_1}$ and $\mathbf{E_2}$ are the electric fields due to $q_1$ and $q_2$, then it is true that $\mathbf{E} = \mathbf{E_1} + \mathbf{E_2}$ is not constant over the surface of, say $S_1$. But I can always write $\int_{S_1} \mathbf{E} \dot \mathbf{da}$ as $\int\mathbf{E_1} \dot \mathbf{da} + \int_{S_1} \mathbf{E_2} \dot \mathbf{da}$ and notice that the second integral vanishes so that I can pull $\mathbf{E_1}$ out of the integral. I find $E_1$ at P but the net field is $E_1 + E_2$ at P. | |
| Oct 3, 2020 at 10:42 | comment | added | Brian | I still think your answer doesn't answer his question | |
| Oct 3, 2020 at 10:03 | comment | added | Rishab Navaneet | Sorry that was a misinterpretation. I've edited the answer. | |
| Oct 3, 2020 at 10:02 | history | edited | Rishab Navaneet | CC BY-SA 4.0 |
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| Oct 3, 2020 at 9:45 | history | edited | Rishab Navaneet | CC BY-SA 4.0 |
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| Oct 3, 2020 at 9:43 | comment | added | Brain Stroke Patient | "when two spheres are kept as shown". I'm sorry if my explanation wasn't clear, English isn't my first language. $S_1$ and $S_2$ aren't spheres, they're (imaginary) Gaussian surfaces around point charges $q_1$ and $q_2$ respectively. | |
| Oct 3, 2020 at 9:38 | history | edited | Rishab Navaneet | CC BY-SA 4.0 |
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| Oct 3, 2020 at 9:37 | comment | added | Brian | Your answer is what I also had thought but doesn't really catch at the heart of what OP is really asking | |
| Oct 3, 2020 at 9:36 | history | edited | Brian | CC BY-SA 4.0 |
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| Oct 3, 2020 at 9:35 | history | answered | Rishab Navaneet | CC BY-SA 4.0 |