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Lets say we have a planar surface with uniform charge. I make a Gaussian pillbox box of area A on it.

We then have,

$$\oint \mathbf E \cdot d\mathbf a= q/\epsilon_0$$

For the surface integral I guess we can break it as,

$$\oint \mathbf E \cdot d\mathbf a = \int_{left side} \mathbf E \cdot d\mathbf a + \int_{right side} \mathbf E \cdot d\mathbf a + \int_{upward} \mathbf E \cdot d\mathbf a + \int_{downward} \mathbf E \cdot d\mathbf a + \int_{backward} \mathbf E \cdot d\mathbf a + \int_{forward} \mathbf E \cdot d\mathbf a$$

Where subscripts are the sides of the box.

I don't understand why all integrals on RHS except "upward" and "downward" are zero. The reason I find in books are in google is because "symmetry", which seems very vague to me. By symmetry they persumably mean "right side" integral and "left side" integral add up to zero, etc.

But why ?

Since in a closed integral the $da$ points normal to the surface at everywhere, the only way for "left side" and "right side" integrals to be zero would be if $E\cdot da_{left side} = -E\cdot da_{right side}$, which in turn - at least in my mind -mean that field lines are going from left in to the box and out from right sides ie field is making $180$ degree with area element on the left side of the box and $0$ degree with the right sides of the box. This assumption seems baseless for it to be true all the time.

So why does the other integrals cancel out?

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  • $\begingroup$ It needs not be true "all the time": each geometry has its own values of that integral. For the specific case at hand it does hold and the contributions do in fact cancel out (one can see that this is however almost always true when taking more or less regular domains, like cubes, spheres and the like). $\endgroup$
    – gented
    Jul 14 '17 at 15:30
  • $\begingroup$ @GennaroTedesco Why it is true for a plane ? I would to see an explanation. $\endgroup$ Jul 14 '17 at 15:32
  • $\begingroup$ In the example of the plane the scalar product is zero on the sides because the normal surface element is orthogonal to the direction of the electric field (if I have understood your depiction correctly). $\endgroup$
    – gented
    Jul 14 '17 at 16:23
  • $\begingroup$ It's not just a coincident that $E \cdot da_L = -E \cdot da_R$, it's that $E = 0$ in those directions. $\endgroup$
    – Señor O
    Jul 14 '17 at 17:23
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I'm not sure I get your notation or your argument so pardon me if my answer is off the mark.

In the case of a planar sheet with uniform charge density, the field by symmetry is everywhere normal to the sheet so the sides of your pillbox will have $0$ flux: geometrically this is because da is perpendicular to E on those sides. Indeed on the top and bottom part (upward and downward?) are the only sides of the box where there is a flux. Moreover, the direction of da for the bottom part is parallel to the field, and so is the direction of da for the top part. The result is then $$ \oint \vec E\cdot d\vec a = \vert E\vert A_{top} + \vert E\vert A_{bottom} = \frac{\sigma A_{cross\ section}}{\epsilon} $$ from which the usual result follows since $A_{top}=A_{bottom}=A_{cross\ section}$.


Edit: This is the best figure I can find to show what I have in mind: enter image description here

The Gaussian surface is cylindrical rather than Cartesian but the idea is the same. The vector da is everywhere perpendicular to the Gaussian surface. Thus, on the sides of the cylinder, the vector da is parallel to the plane of charge and thus perpendicular to the E-field. It is only on the front and back caps of the cylinder that the vector da is parallel to E, so the only contribution to the flux is through these caps.

Note also that the direction of da is by definition outward so that, for the near cap, da is exactly parallel to E there. On the back cap, da points in the opposite direction, but so does E. Thus, the scalar product E da is the same for both caps, and just E da, so that

$\int_{front\ cap}$E da =$\vert E\vert A$ = $\int_{back\ cap}$E da

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  • $\begingroup$ geometrically this is because $d\mathbf a$ is perpendicular to $\mathbf E$ on those sides. Why ? Your answer is fine with me if you can explain me the above quote. $\endgroup$ Jul 14 '17 at 16:51
  • $\begingroup$ @123 added comments $\endgroup$ Jul 14 '17 at 17:19
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Symmetry implies that $\bf{E}$ must be perpendicular to the planar surface, so $\bf{E}\cdot d\bf{a}$ is zero except on the upward and downward surfaces.

The symmetry argument is just that there is no preferred direction in which $\bf{E}$ could be "tilted."

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  • $\begingroup$ Why is $\mathbf E \perp d\mathbf a$ ? $\endgroup$ Jul 14 '17 at 17:01
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Answering the question "Why is the electric field normal to the surface of the plane at every point"?

In order to have a nonzero component parallel to the surface at some point, there must be more charge on one side of that point than the other. Since the plane is uniform (i.e. there is the same amount of charge in every direction), this is impossible.

If you don't like that explanation, here's another one: Any horizontal component created by a surface element on the plane at some point is cancelled out by an equal and opposite horizontal component from a charge element on the other side of that point.

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