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I have encountered something that is very confusing. My problem is this. I am assuming two infinite cubical Gaussian surfaces sharing a common side. One of the cubes contains a charge $q_1$ at a finite distance from the common surface and the second cube contains a charge $q_2$ at a finite distance from the common surface.

Now both the charges contribute flux to their respective Gaussian surfaces only at the common surface as the electric field at infinity goes to zero. So, the flux of first cube should be negative to that of second cube, which implies $q_1=-q_2$ which is certainly not true for all cases. Where is my argument flawed?

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  • $\begingroup$ If the charges are of different magnitudes then the electric fields due to each of them will be of different magnitude, thus the flux of the first is not simply the negative of the second. $\endgroup$
    – wgrenard
    Commented Feb 25, 2015 at 4:55
  • $\begingroup$ @wgrenard exactly ,but then where is my argument flawed $\endgroup$
    – avz2611
    Commented Feb 25, 2015 at 4:55
  • $\begingroup$ You used the assumption that the fluxes are equal and opposite to draw the conclusion that the charges are equal and opposite. But you cannot make that first assumption. $\endgroup$
    – wgrenard
    Commented Feb 25, 2015 at 4:59
  • $\begingroup$ i made some arguments to make that assumption that seemed right to me . maybe they are wrong but idk why $\endgroup$
    – avz2611
    Commented Feb 25, 2015 at 5:00
  • $\begingroup$ the flux through the common surface is equal and opposite for both cubes right ? $\endgroup$
    – avz2611
    Commented Feb 25, 2015 at 5:01

2 Answers 2

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You have two distinct errors. One is claiming that because the electric field goes to zero at infinity so does the flux. The flux is the integral of the electric field over the surface. The electric field goes down as $\frac 1{r^2}$, but the area goes up as $r^2$, s the flux constant. Ask Gauss about this. The second is to claim that because (from the first erroneous argument, but not supported by it) the flux though the five non-common sides is zero, the flux through the common side reflects the charge in each cube.

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  • $\begingroup$ yes you are right , i did not realize the surface area also increases so we should be dealing in limits , but if my first argument held true then my second does makes sense right ? i did not get what you wrote in the last part of your answer $\endgroup$
    – avz2611
    Commented Feb 25, 2015 at 5:16
  • $\begingroup$ You are claiming that if your first holds true there is no flux out the non-common faces because the total flux is zero. Logically, that is correct, as a false antecedent makes an implication true. I don't think that is useful. Then we know the flux integrated over a sphere is proportional to the charge within it. If the flux through the other five faces were zero, the flux through the common face would reflect the charge inside the cube, but it isn't so you can't make that argument. $\endgroup$
    – Ross Millikan
    Commented Feb 25, 2015 at 5:22
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Your problem is the assumption "both the charges contribute flux to their respective Gaussian surfaces only at the common surface."

When you take your surface out to infinity you are also increasing its area, so even though the electric field goes to zero, the integral of the electric field over your surface will be non-zero.

In fact, the total flux through your Gaussian cubes will be constant and equal to $q/\epsilon_0$ no matter the size of the cubes, which is just Gauss' Law.

Your instinct about the common surface is correct: it contributes equal and opposite fluxes to each of your integrals. However, the other (unshared) surfaces will have a non-zero contribution, and make up for the difference in the magnitude of the charges.

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  • $\begingroup$ 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. $\endgroup$
    – wgrenard
    Commented Feb 25, 2015 at 5:47
  • $\begingroup$ 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. $\endgroup$
    – Geoff Ryan
    Commented Feb 25, 2015 at 5:55
  • $\begingroup$ 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. $\endgroup$
    – wgrenard
    Commented Feb 25, 2015 at 6:01
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    $\begingroup$ 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. $\endgroup$
    – Geoff Ryan
    Commented Feb 25, 2015 at 6:11
  • $\begingroup$ 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. $\endgroup$
    – Geoff Ryan
    Commented Feb 25, 2015 at 6:11

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