Timeline for Paradox in electrostatics in relation to Gaussian surfaces?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 25, 2015 at 6:16 | comment | added | wgrenard | Okay, agreed. It seems I misunderstood your original point :) | |
| Feb 25, 2015 at 6:11 | comment | added | Geoff Ryan | Sorry, I maybe wrote that poorly. All I'm saying is that if we have charges $q_1$, $q_2$ and gaussian surfaces $S1$ and $S2$ which share a face $A$, then: $\frac{q_1}{\epsilon_0} = \Phi_{S1} = \Phi_{S1-A} + \Phi_{A}$ and $\frac{q_2}{\epsilon_0} = \Phi_{S2} = \Phi_{S2-A} - \Phi_{A}$. The minus sign there just fixes the orientation. | |
| Feb 25, 2015 at 6:11 | comment | added | Geoff Ryan | The issue is you're comparing two completely different systems. If you have a single charge $q$ in isolation I 100% agree the flux through each face is $\frac{q}{6\epsilon_0}$. If you have two charges $q_1$ and $q_2$ then the flux through the common face (call it $A$) assuming each cube is centred on its charge will be $\frac{q_1-q_2}{6\epsilon_0}$. In general $q_1$ and $q_2$ will add different amounts to the flux through $A$, but the contribution of that flux to the total integral will be the same for each surface. | |
| Feb 25, 2015 at 6:01 | comment | added | wgrenard | Right, I relied on symmetry in my argument, but if the fluxes are not equal in one case it cannot be said that they are equal in the general case. In addition, about the more general case in which you integrate. Different charges will create different electric fields. Sure, the total field at the faces can be integrated to get the net flux, but the contribution to this total flux from each charge is not equal. | |
| Feb 25, 2015 at 5:55 | comment | added | Geoff Ryan | The flux through a face is $\int \vec{E} \cdot \hat{n} dA$. Since the faces coincide in space, this integral will give the same result for both surface 1 and 2 (modulo a minus sign for orientation). Gauss' Law only tells you what the total flux through a closed surface is, not how its distributed. In your example, the flux through the common surface is only $\frac{q}{6\epsilon_0}$ if there are no other charges present, otherwise you destroy the symmetry you relied on to calculate it in the first place. | |
| Feb 25, 2015 at 5:47 | comment | added | wgrenard | Honest question here, aside from flawed part of the OP's argument, how can you say that the fluxes through the common face are equal and opposite? Even if you consider two cubes with finite length, 1 with charge $q_1$ in its center and the other with charge $q_2$ in its center, then the flux through the face from the first charge is ${q_1 \over 6 \epsilon_o}$ and from the second is ${q_2 \over 6 \epsilon_o}$ Clearly these are only equal if $q_1=q_2$; they aren't equal in general. | |
| Feb 25, 2015 at 5:22 | history | answered | Geoff Ryan | CC BY-SA 3.0 |