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I think the point is that the electric field contribution from the patch (which is the surface charge inside the Gaussian box) is the significant contributing one in the formula for the following reason:

In the proof after we form our Gaussian box we let the sides of the box (of length say $\epsilon$) tend to zero. Then we arrive at the result that the difference in the electric field produced by charges in the box are $\vec{E}_{\text{above box}} - \vec{E}_{\text{below box}} = \frac{\sigma}{\epsilon_0} \hat{n}$. Your concern is about why the other surface charge contribution outside the Gaussian box does not seem to be present in the derivation of the general formula. The outside contribution simply cancel, since $$\vec{E}_{above} = \vec{E}_{other} + \vec{E}_{\text{above box}}$$ and similarly $$\vec{E}_{below} = \vec{E}_{other} + \vec{E}_{\text{below box}}$$ hence $$\vec{E}_{above} - \vec{E}_{below} = \frac{\sigma}{\epsilon_0} \hat{n}.$$ This is expected since the discontinuity is only from the patch covered by the Gaussian surface, if we removed that patch we would have a continuous electric field in this space. It is also clear to see for cases where we have symmetry, like an infinite charged plane (with uniform surface charge), that $\vec{E}_{above} = \vec{E}_{\text{above box}}$ and $\vec{E}_{below} = \vec{E}_{\text{ below box}}$ and also the formula from the proof still clearly holds.

I think the point is that the electric field contribution from the patch (which is the surface charge inside the Gaussian box) is the significant contributing one in the formula for the following reason:

In the proof after we form our Gaussian box we let the sides of the box (of length say $\epsilon$) tend to zero. Then we arrive at the result that the difference in the electric field produced by charges in the box are $\vec{E}_{\text{above box}} - \vec{E}_{\text{below box}} = \frac{\sigma}{\epsilon_0} \hat{n}$. Your concern is about why the other surface charge contribution outside the Gaussian box does not seem to be present in the derivation of the general formula. The outside contribution simply cancel, since $$\vec{E}_{above} = \vec{E}_{other} + \vec{E}_{\text{above box}}$$ and similarly $$\vec{E}_{below} = \vec{E}_{other} + \vec{E}_{\text{below box}}$$ hence $$\vec{E}_{above} - \vec{E}_{below} = \frac{\sigma}{\epsilon_0} \hat{n}.$$ This is expected since the discontinuity is only from the patch covered by the Gaussian surface, if we removed that patch we would have a continuous electric field in this space. It is also clear to see for cases where we have symmetry, like an infinite charged plane (with uniform surface charge), that $\vec{E}_{above} = \vec{E}_{\text{above box}}$ and $\vec{E}_{below} = \vec{E}_{\text{ below box}}$ (also the formula from the proof clearly holds).

I think the point is that the electric field contribution from the patch (which is the surface charge inside the Gaussian box) is the significant contributing one in the formula for the following reason:

In the proof after we form our Gaussian box we let the sides of the box (of length say $\epsilon$ tend to zero). Then we arrive at the result that the difference in the electric field produced by charges in the box are $\vec{E}_{\text{above box}} - \vec{E}_{\text{below box}} = \frac{\sigma}{\epsilon_0} \hat{n}$. Your concern is about why the other surface charge contribution outside the Gaussian box does not seem to be present in the derivation of the general formula. The outside contribution simply cancel, since $$\vec{E}_{above} = \vec{E}_{other} + \vec{E}_{\text{above box}}$$ and similarly $$\vec{E}_{below} = \vec{E}_{other} + \vec{E}_{\text{below box}}$$ hence $$\vec{E}_{above} - \vec{E}_{below} = \frac{\sigma}{\epsilon_0} \hat{n}.$$ This is expected since the discontinuity is only from the patch covered by the Gaussian surface, if we removed that patch we would have a continuous electric field in this space. It is also clear to see for cases where we have symmetry, like an infinite charged plane (with uniform surface charge), that $\vec{E}_{above} = \vec{E}_{\text{above box}}$ and $\vec{E}_{below} = \vec{E}_{\text{ below box}}$ and also the formula from the proof still clearly holds.

I think the point is that the electric field contribution from the patch (which is the surface charge inside the Gaussian box) is the significant contributing one in the formula for the following reason:

In the proof after we form our Gaussian box we let the sides of the box (of length say $\epsilon$) tend to zero. Then we arrive at the result that the difference in the electric field produced by charges in the box are $\vec{E}_{\text{above box}} - \vec{E}_{\text{below box}} = \frac{\sigma}{\epsilon_0} \hat{n}$. Your concern is about why the other surface charge contribution outside the Gaussian box does not seem to be present in the derivation of the general formula. The outside contribution simply cancel, since $$\vec{E}_{above} = \vec{E}_{other} + \vec{E}_{\text{above box}}$$ and similarly $$\vec{E}_{below} = \vec{E}_{other} + \vec{E}_{\text{below box}}$$ hence $$\vec{E}_{above} - \vec{E}_{below} = \frac{\sigma}{\epsilon_0} \hat{n}.$$ This is expected since the discontinuity is only from the patch covered by the Gaussian surface, if we removed that patch we would have a continuous electric field in this space. It is also clear to see for cases where we have symmetry, like an infinite charged plane (with uniform surface charge), that $\vec{E}_{above} = \vec{E}_{\text{above box}}$ and $\vec{E}_{below} = \vec{E}_{\text{ below box}}$ and also the formula from the proof still clearly holds.

I think the point is that the electric field contribution from the patch (which is the surface charge inside the Gaussian box) is the significant contributing one in the formula for the following reason:

In the proof after we form our Gaussian box we let the sides of the box (of length say $\epsilon$ tend to zero). Then we arrive at the result that the difference in the electric field produced by charges in the box are $\vec{E}_{\text{above box}} - \vec{E}_{\text{below box}} = \frac{\sigma}{\epsilon_0} \hat{n}$. Your concern is what about the other surface charge contribution outside the box and therefore how do we get the general formula. The outside contribution simply cancel, since $$\vec{E}_{above} = \vec{E}_{other} + \vec{E}_{\text{above box}}$$ and similarly $$\vec{E}_{below} = \vec{E}_{other} + \vec{E}_{\text{below box}}$$ hence $$\vec{E}_{above} - \vec{E}_{below} = \frac{\sigma}{\epsilon_0} \hat{n}.$$ It is then clear to see for cases where we have symmetry, like an infinite charged plane (with uniform surface charge) that $\vec{E}_{above} = \vec{E}_{\text{above box}}$ and $\vec{E}_{below} = \vec{E}_{\text{ below box}}$ and also the formula from the proof still clearly holds.

I think the point is that the electric field contribution from the patch (which is the surface charge inside the Gaussian box) is the significant contributing one in the formula for the following reason:

In the proof after we form our Gaussian box we let the sides of the box (of length say $\epsilon$ tend to zero). Then we arrive at the result that the difference in the electric field produced by charges in the box are $\vec{E}_{\text{above box}} - \vec{E}_{\text{below box}} = \frac{\sigma}{\epsilon_0} \hat{n}$. Your concern is about why the other surface charge contribution outside the Gaussian box does not seem to be present in the derivation of the general formula. The outside contribution simply cancel, since $$\vec{E}_{above} = \vec{E}_{other} + \vec{E}_{\text{above box}}$$ and similarly $$\vec{E}_{below} = \vec{E}_{other} + \vec{E}_{\text{below box}}$$ hence $$\vec{E}_{above} - \vec{E}_{below} = \frac{\sigma}{\epsilon_0} \hat{n}.$$ This is expected since the discontinuity is only from the patch covered by the Gaussian surface, if we removed that patch we would have a continuous electric field in this space. It is also clear to see for cases where we have symmetry, like an infinite charged plane (with uniform surface charge), that $\vec{E}_{above} = \vec{E}_{\text{above box}}$ and $\vec{E}_{below} = \vec{E}_{\text{ below box}}$ and also the formula from the proof still clearly holds.