How much field gets through

For a flat surface in a uniform field,

$$\Phi = \vec{E}\cdot\vec{A} = EA\cos\theta,$$

where $\vec{A}$ is the area vector: magnitude $A$, direction along the surface NORMAL. So $\theta$ is the angle between the field and the perpendicular to the surface, never the angle between the field and the surface itself. Getting that wrong swaps cosine for sine and inverts every answer.

1. Field along the normal, $\theta = 0$: $\Phi = EA$, the maximum. The surface is broadside to the field.
2. Field in the plane of the surface, $\theta = 90^\circ$: $\Phi = 0$. The surface is edge-on and nothing passes through it.
3. Anything between: project. A surface whose PLANE makes $30^\circ$ with $\vec{E}$ has a normal making $60^\circ$ with it, so $\Phi = EA\cos 60^\circ = EA/2$.

The words in a problem usually describe the surface, and the formula wants the normal, so build the habit of converting once, in writing, before substituting. A sheet described as "parallel to the field" is the case that catches nothing, even though "parallel" sounds like alignment.

Units are N$\cdot$m$^2$/C, and flux is a scalar. It has a sign, set by which way the normal is chosen, but no direction.

$\Phi = EA\cos\theta$ assumes $\vec{E}$ is the same everywhere on the surface. When it is not, chop the surface into patches small enough that it is:

$$\Phi = \int \vec{E}\cdot d\vec{A}, \qquad \Phi_{\text{closed}} = \oint \vec{E}\cdot d\vec{A}.$$

On a closed surface the convention is fixed for you: $d\vec{A}$ points OUTWARD everywhere. That makes the sign meaningful. Field leaving the volume contributes positive flux, field entering contributes negative, and the net is what is left over.

Worked case. Take $\vec{E} = (3x)\,\hat{x}$ N/C and a cube of side $2$ m with one corner at the origin and edges along the axes. The four faces whose normals are $\pm\hat{y}$ or $\pm\hat{z}$ catch nothing, because $\vec{E}$ lies in their planes. On the face at $x = 0$ the field is zero. On the face at $x = 2$ m the field is $6$ N/C and points straight out:

$$\Phi_{\text{net}} = (6\ \text{N/C})(4\ \text{m}^2) - 0 = 24 \ \text{N}\cdot\text{m}^2/\text{C}.$$

Notice that the answer came from two faces out of six, and which two was decided entirely by orientation.

Put a closed box anywhere in a uniform field. Field pours in one side and out the other in exactly equal amounts, so the net flux is zero. The field on that box is large everywhere, and it is zero nowhere.

Keep the two statements separate:

1. Zero net flux says inflow balanced outflow, which says the net charge inside is zero. It says nothing whatsoever about the value of $\vec{E}$ at any point on the surface.
2. Zero field on the surface is a much stronger claim, and it would give zero flux, but the reverse implication does not run.

The mirror error is just as common: seeing a strong field on a surface and concluding there must be charge inside. A point charge sitting just outside a box makes a strong field on the near face, and every field line that enters the box leaves it again, so the net flux is zero.

A useful reading of the flux integral is a count of net field lines crossing the surface, with lines out counted positive and lines in counted negative. Lines that pass straight through cancel themselves. Only lines that BEGIN or END inside, which is to say charge inside, leave a net count.

Flux through an open surface depends on the surface: a bigger loop in a uniform field catches proportionally more, and $\Phi = EA\cos\theta$ scales with $A$. Flux through a closed surface does not care about the shape at all, only about what is inside.

That gives a shortcut worth knowing before Gauss's law formalizes it. Cap a hemisphere of radius $R$ with its flat disk to make a closed surface, and put it in a uniform field $\vec{E}$ aligned with the axis. No charge inside means zero net flux, so

$$\Phi_{\text{curved}} = -\Phi_{\text{disk}} = E\pi R^2.$$

The curved surface has area $2\pi R^2$, but it catches the same flux as the flat disk of area $\pi R^2$, because the projection factor eats the difference. Flux depends on the surface's shadow across the field, not its area.

Two related facts that get conflated:

1. Double the radius of a sphere drawn around a point charge and the flux is unchanged: $E$ falls by four while $A$ grows by four.
2. Double the area of a flat loop held at a fixed orientation in a uniform field and the flux doubles, because nothing here weakens $E$.