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When thinking of gauss laws, it states that the net outflow of the electric field through a closed surface is proportional to the enclosed charge.

This for the most surfaces is understandable, but not for the following:

enter image description here

The closed "volume" shall be the area within the green line, the green line shall be the enclsoing "area" By gauss law the net flux through the area must be zero, because there is no charge. But this does not seem to be the case.

What am I missing?>

Edit: I use this playground: https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_all.html

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  • $\begingroup$ Is there any particular reason why you are only looking on the two sides of your rectangle? There is flux flowing through the long sides of the area as well, as is also seen in your picture. $\endgroup$
    – mb28025
    Feb 11 at 19:56
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    $\begingroup$ To me it looks like they contribute the same (positive) amount. Recall that you should project the vector field onto the (outwards pointing) normal of the surface, so there is no extra minus sign from orientation like it would be the case if you were integrating along the curve. $\endgroup$
    – mb28025
    Feb 11 at 20:06
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    $\begingroup$ Electric field lines do not have any magnitude characteristics. Check Wikipedia article on Gauss law, - it has nice illustration about similar picture of zero flux where lines crosses enclosed empty sphere. $\endgroup$ Feb 11 at 20:16
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    $\begingroup$ Scalar product returns a scalar. The only negative contribution is the short side closest to the charge. All the rest contribute positively and exactly cancel the original negative contribution. $\endgroup$
    – mb28025
    Feb 11 at 20:16
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    $\begingroup$ You are right I must flip the normal vector for the lower side, but also the e field flips it values, thus the y-values have a net positive contribution. $\endgroup$
    – Niclas
    Feb 11 at 20:23

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enter image description here Very long thin Gaussian surfaces can be quite tricky. The answer will end up being that the small component of the flux going out the sides really does account for the decrease in magnitude coming through the ends. Let's say I have a point charge $Q$. It makes an electric field: $$ \mathbf{E}=\frac{Q}{4\pi\epsilon_0}\frac{\hat{r}}{r^2} $$ Now let me take a Gaussian surface - a long thin cylinder just like yours which starts a distance $a$ from the point charge and ends a distance $b$. The radius of the cylinder will be $R$. And to simplify the algebra let me assert $R\ll a,b$. Then the flux through the flat surface a distance $a$ from the point charge is: $$ \Phi_a=\pi R^2\frac{Q}{4\pi\epsilon_0}\frac{1}{a^2} $$ And the flux through the surface on the other side is: $$ \Phi_b=-\pi R^2\frac{Q}{4\pi\epsilon_0}\frac{1}{b^2} $$ So just as your question says... what went wrong? these two are clearly not equal to eachother. The answer is this: although the component of $\mathbf{E}$ pointing out of the curved surface we previously ignored is small, this surface is also much bigger than the two flat surfaces, and the total is the same order of magnitude as the flux going through the other surfaces. The flux through the curved surface of the cylinder is: $$ \Phi_c=-2\pi R\int_a^b dr\frac{Q}{4\pi\epsilon_0}\frac{1}{r^2}\frac{R}{r}=\pi R^2\frac{Q}{4\pi\epsilon_0}\left.\frac{1}{r^2}\right|_a^b=\pi R^2\frac{Q}{4\pi\epsilon_0}\frac{1}{b^2}-\pi R^2\frac{Q}{4\pi\epsilon_0}\frac{1}{a^2}$$ $\Phi_a+\Phi_b+\Phi_c=0$. Gauss' law is satisfied... of course it is. It always is.

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