AP Physics C: E&M · Topic 8.5

Topic 8.5: Electric Flux

Unit 8: Electric Charges, Fields, and Gauss's Law15-25% of the multiple-choice section

Electric flux measures how much electric field passes through a surface. In general it is the surface integral of the field over that area. When the field is constant across the area it collapses to the dot product of the field vector and the area vector, and the sign follows.

AP Physics: Unit 8 (topics 8.5 Electric Flux). AP Physics C: Electricity and Magnetism Unit 8, Topic 8.5. One learning objective, 8.5.A, describe the electric flux through an arbitrary area or geometric shape. Three essential-knowledge statements, one carrying two sub-statements: 8.5.A.1 (flux describes the amount of a given quantity that passes through a given area), 8.5.A.2 (for an electric field E that is constant across an area A, the electric flux through the area is defined as the dot product of E and A), 8.5.A.2.i (the direction of the area vector is defined as perpendicular to the plane of the surface and outward from a closed surface), 8.5.A.2.ii (the sign of flux is given by the dot product of the electric field vector and the area vector), and 8.5.A.3 (the total electric flux passing through a surface is defined by the surface integral of the electric field over the surface, with the relevant equation Phi sub E equals the integral of E dot dA). Topic 8.5 prints no boundary statement; Unit 8's three boundary statements sit under Topics 8.1, 8.4 and 8.6. Suggested skills are 1.A, 2.A, 2.C and 3.B. This topic has no AP Physics 2 counterpart: the phrase electric flux appears nowhere in the AP Physics 2 CED, and Gauss and Gaussian appear zero times in it, while the word flux appears nineteen times there and every occurrence is magnetic flux, mostly in Unit 12 Topic 12.4. The CED's sample multiple-choice Question 12 aligns to 8.5.A and essential knowledge 8.5.A.2 at skill 2.A: a cube of side a with one corner at the origin in a field E equals bx in the i-hat direction, total flux b a cubed, answer B. Unit 8 is weighted 15 to 25% of the multiple-choice section over about 12 to 24 class periods.

What Topic 8.5 requires

Topic 8.5 has one learning objective and three essential-knowledge statements, one of which carries two sub-statements. It prints no boundary statement, which puts it with Topics 8.2 and 8.3 in Unit 8; the unit's three boundary statements sit under 8.1, 8.4 and 8.6.

8.5.A, describe the electric flux through an arbitrary area or geometric shape.

8.5.A.1 states that flux describes the amount of a given quantity that passes through a given area.
8.5.A.2 states that for an electric field $\vec{E}$ that is constant across an area $\vec{A}$ , the electric flux through the area is defined as $\Phi_E = \vec{E} \cdot \vec{A}$ .
8.5.A.2.i the direction of the area vector is defined as perpendicular to the plane of the surface and outward from a closed surface.
8.5.A.2.ii the sign of flux is given by the dot product of the electric field vector and the area vector.
8.5.A.3 states that the total electric flux passing through a surface is defined by the surface integral of the electric field over the surface, with the relevant equation

\Phi_E = \int \vec{E} \cdot d\vec{A}

Two words in the learning objective are doing work. "Arbitrary" means the surface is not required to be a plane, a rectangle or anything else convenient. "Area or geometric shape" covers both a flat patch and a three-dimensional surface such as a hemisphere or a cylinder wall. Nothing in the objective says closed, and that omission is deliberate: the closed surface arrives one topic later, in Topic 8.6.

The four suggested skills are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Skill 1.A being listed first is not decoration. Most flux errors are drawing errors: the area vector was never drawn, or the angle was measured to the surface rather than to the normal.

Electric flux does not exist in AP Physics 2

This is one of the sixteen AP Physics C: Electricity and Magnetism topics with no algebra-based counterpart. There is no Physics 2 sibling page to link, because the phrase "electric flux" appears nowhere in the AP Physics 2 course and exam description.

Be precise about that, because a nearby claim is false. The word "flux" does appear in the Physics 2 CED, nineteen times, and every one of them is magnetic flux. Nine sit in Unit 12 under Topic 12.4, Electromagnetic Induction and Faraday's Law; eight are in the scoring guidelines for that document's own sample free-response question on induction; one is in a suggested classroom activity for Unit 12; and one is the symbol legend on the equation sheet. So a Physics 2 student meets the idea of flux, defined for the magnetic field, in the second-to-last unit of the course. They never meet it for the electric field.

The words "Gauss" and "Gaussian" appear zero times in that document.

Two things follow for a student in the calculus-based course:

You are not revisiting anything here. Topics 8.1, 8.2 and much of 8.3 restate an algebra-based topic. Topic 8.5 does not, so there is nothing to skim.
If you have done magnetic flux, you already have the structure. Topic 13.1 in this same course defines magnetic flux with statements that mirror these ones almost exactly, including the same definition of the area vector. The definitions match; what differs is what the closed-surface version equals. For the magnetic field the sheet prints $\oint \vec{B} \cdot d\vec{A} = 0$ , always. For the electric field it prints the enclosed charge over $\varepsilon_0$ , which is not always zero, because isolated electric charges exist and isolated magnetic poles do not.

That contrast, in a single sentence, is a large fraction of what Maxwell's equations say.

The integral is the definition, and the dot product is the special case

The CED presents the constant-field case first (8.5.A.2) and the general one second (8.5.A.3). Physically the order is the other way round.

Start from the general statement. Chop the surface into patches small enough that the field does not vary across any one of them. Give each patch an area vector $d\vec{A}$ perpendicular to it, with magnitude equal to the patch's area. Take the dot product $\vec{E} \cdot d\vec{A}$ for each patch, which picks out only the part of the field that actually goes through rather than skimming along. Add them all up:

\Phi_E = \int \vec{E} \cdot d\vec{A}

Everything that makes a flux problem easy is a reason that sum turns into arithmetic.

The field is constant across the surface. Then $\vec{E}$ comes out of the integral, $\int d\vec{A} = \vec{A}$ , and $\Phi_E = \vec{E} \cdot \vec{A} = EA\cos\theta$ . This is statement 8.5.A.2, and it is the case most exam questions are built on. What licenses it is the constancy of the field, not the flatness of the surface.

The field is perpendicular to the surface everywhere and has constant magnitude. Then $\vec{E} \cdot d\vec{A} = E\,dA$ with $E$ constant, and $\Phi_E = EA$ with $A$ the total area. This is the case Gauss's law is engineered to produce, and statement 8.6.A.4 says so explicitly.

The field is parallel to the surface everywhere. Then $\vec{E} \cdot d\vec{A} = 0$ patch by patch, and the flux is zero with no work. A flat plate held edge-on to the field passes no flux.

The field varies across the surface. Then you integrate for real. Choose a strip or a band across which the field is constant, write its area as a differential, and put the limits at the edges of the surface. The second and third worked examples below are both this case, and it is the one the algebra-based course has no way to pose.

The surface is closed and the field is uniform. Then the flux is zero, whatever the shape. Every field line that enters leaves. That result belongs to Topic 8.5, not 8.6, because it needs no charge at all, and the first worked example uses it to make a hemisphere as easy as a disc.

The area vector, and the angle that is measured wrong

Statement 8.5.A.2.i gives the area vector two properties, and they apply in different situations.

Perpendicular to the plane of the surface. Always. The area vector never lies in the surface.

Outward from a closed surface. Only when the surface is closed. For a closed surface the direction is chosen for you, and that is what makes the sign of the flux in Gauss's law meaningful: positive flux means net field leaving, negative means net field entering.

For an open surface, the direction is your choice. Either perpendicular direction is allowed. Nothing in physics picks it, so you pick one, state it, and keep it. Statement 8.5.A.2.ii then makes that choice visible: the sign of the flux is given by the dot product of the field vector and the area vector.

Now the angle. In $\Phi_E = EA\cos\theta$ , the angle $\theta$ is between the field and the area vector, which is the normal to the surface, not the surface itself. Two phrasings that sound alike and mean opposite things:

The problem says	The area vector is	$\theta$	Flux
"the plane of the surface is perpendicular to the field"	along the field	$0^\circ$	$EA$ , the maximum
"the surface is tilted at $60^\circ$ to the field"	at $30^\circ$ to the field	$30^\circ$	$0.866\,EA$
"the area vector is at $60^\circ$ to the field"	at $60^\circ$ to the field	$60^\circ$	$0.5\,EA$
"the plane of the surface contains the field"	perpendicular to the field	$90^\circ$	zero
"the field points straight in through the surface"	opposite to the field	$180^\circ$	$-EA$

Read the sentence, draw the surface, then draw the arrow sticking out of it. Skill 1.A, first on this topic's list.

Flux is a scalar. It is produced by a dot product of two vectors, so it carries a sign but no direction. Fluxes through different parts of one surface add as ordinary signed numbers, which is why breaking a closed surface into faces and adding is legitimate.

The units. Flux is field times area, so newton metres squared per coulomb, $\mathrm{N \cdot m^2 / C}$ . Since a newton per coulomb is also a volt per metre, the same unit is often written $\mathrm{V \cdot m}$ . Neither form is printed in the CED; both follow from the definitions.

What flux is not

Four confusions account for most lost marks here, and three of them come from importing a picture that is only a metaphor.

Flux is not a count of field lines. Field lines are a representation, and statement 8.3.A.3.iii calls them a simplified model that gives relative magnitudes only. The count depends on how many lines whoever drew the diagram decided to draw. Flux is a number with units, computed from the field. The line picture is a good intuition for the sign and for the cancellation, and useless for a value.

Flux is not "field times area" in general. It is field times area times the cosine of the angle to the normal, and only when the field is constant across the surface. Skip either qualifier and you will get a wrong answer on a tilted surface or a varying field.

Flux is not a property of a closed surface. Nothing in learning objective 8.5.A mentions closure. A flat square in a uniform field has a perfectly well-defined flux. The closed surface is a device introduced in Topic 8.6 because closing the surface is what lets you relate flux to the charge inside.

Flux is not zero just because the net charge inside is zero. That statement is about the net flux through a closed surface. The flux through one face of a box can be large while the total over all six faces is zero.

One more, worth its own line because it separates Topic 8.5 from Topic 8.6 cleanly. Flux does not require any charge to be present. A uniform field in an empty region has flux through any surface you draw in it. Charge enters the story only when you close the surface and ask what the total comes to.

How Topic 8.5 is tested

Of the fifteen sample multiple-choice questions the CED prints for this course, four align to Unit 8 objectives, and one of those is Topic 8.5: sample Question 12, aligned to 8.5.A and essential knowledge 8.5.A.2, at skill 2.A, derive a symbolic expression.

The item is a good model of what this topic looks like on an exam. A cube of side $a$ sits with one corner at the origin and its edges along the axes, in a region where the field is $\vec{E} = (bx)\,\hat{i}$ with $b$ a positive constant. The question asks for the total electric flux through the surface of the cube, and the answer key gives choice B, $ba^3$ .

Three observations about it.

The field is not uniform, so statement 8.5.A.2 does not apply directly and 8.5.A.3 does. But the field is uniform across each face, because the field depends only on $x$ and each face perpendicular to the $x$ -axis is at a single value of $x$ . So the surface integral collapses face by face.
Four of the six faces contribute nothing, because on them the field lies in the plane of the face and the dot product vanishes.
The question calls the cube a Gaussian surface but does not need Gauss's law. It is a flux calculation, which is why the alignment is to 8.5.A and not 8.6.A. Gauss's law would then tell you what charge is inside, and the second worked example below finishes that thought.

Patterns worth rehearsing:

Flux through a flat surface at an angle to a uniform field (skill 2.C). Check the angle is measured to the normal.
Flux through a curved surface in a uniform field by projecting it onto a plane. The hemisphere case in the first worked example is standard.
Face-by-face flux through a box in a position-dependent field (skill 2.A). The CED's sample Question 12 exactly.
A genuine surface integral over an open surface, where the field varies across it. The third worked example.
Rank or compare fluxes through several surfaces in one scenario (skill 2.C).
Argue a sign from the dot product (skill 3.B, statement 8.5.A.2.ii).

On the free-response section, skill 2.A carries a 25 to 30 percent weighting on the multiple-choice section and science practice 2 as a whole carries 40 to 45 percent on free response. Symbolic setup is where the marks are.

Where Topic 8.5 goes wrong

Measuring the angle to the surface instead of to the normal. The single most common error in the topic, and the reason the table above exists.

Using $EA$ when the field varies across the surface. Check first whether $\vec{E}$ is the same at every point of the surface. If it is not, statement 8.5.A.2 does not apply.

Forgetting that the outward convention is only for closed surfaces. For an open surface you choose, and then you must keep the choice for the rest of the problem. A normal that flips halfway through a multi-part question is a sign error waiting to happen.

Treating flux as a vector. It is a scalar with a sign. There is no such thing as flux "in the $x$ direction", only flux through a surface whose area vector points along $x$ .

Assuming a closed surface means Gauss's law. Closing the surface is a geometric act. Whether you need the enclosed charge depends on what is being asked.

Adding face contributions without their signs. On a closed box in a uniform field the six face fluxes cancel exactly, and they only cancel if you keep the signs the outward convention gives them.

Confusing electric flux with magnetic flux. They are defined identically and their closed-surface integrals are completely different. See Topic 13.1 for the magnetic version and the $\oint \vec{B} \cdot d\vec{A} = 0$ result that has no electric analogue.

A disc, a tilt, and a hemisphere that is no harder

A uniform electric field of magnitude $450 \ \mathrm{N/C}$ points in the $+x$ direction. A flat circular disc of radius $0.20 \ \mathrm{m}$ is placed in the field. Find the flux through it when (a) its area vector points along $+x$ , (b) its area vector is at $60^\circ$ to the field, (c) the plane of the disc contains the $x$ -axis. (d) The disc is then replaced by a hemispherical cap of the same radius, bulging in the $+x$ direction, with the same circular rim. Find the flux through the cap, with its area vector taken outward.

Area of the disc, from the sheet's geometry table: $A = \pi r^2 = \pi(0.20)^2 = 0.12566 \ \mathrm{m^2}$ .
The field is uniform across the disc, so statement 8.5.A.2 applies and $\Phi_E = \vec{E} \cdot \vec{A} = EA\cos\theta$ with $\theta$ measured to the area vector, per 8.5.A.2.i.
(a) $\theta = 0$ : $\Phi_E = (450)(0.12566)(1) = 56.5 \ \mathrm{N \cdot m^2/C}$ . This is the maximum for this disc in this field.
(b) $\theta = 60^\circ$ , $\cos 60^\circ = 0.500$ : $\Phi_E = (450)(0.12566)(0.500) = 28.3 \ \mathrm{N \cdot m^2/C}$ .
(c) "The plane of the disc contains the $x$ -axis" means the field lies in the plane of the disc, so the area vector is perpendicular to the field: $\theta = 90^\circ$ , $\cos 90^\circ = 0$ , and $\Phi_E = 0$ . Nothing passes through; the field skims across.
Watch the wording against part (a). "The plane of the disc is perpendicular to the field" would mean the opposite of (c): area vector along the field, $\theta = 0$ , maximum flux.
(d) The hemisphere. You could set up $\int \vec{E} \cdot d\vec{A}$ over a curved surface, with the angle between $\vec{E}$ and $d\vec{A}$ changing from $0$ at the pole to $90^\circ$ at the rim. Do not.
Instead close the surface. The cap plus the flat disc that caps its rim form a closed surface. The field is uniform, so every field line that enters through the flat disc leaves through the curved cap, and the net flux out of the closed surface is zero.
With outward normals, the flat disc's normal points in $-x$ and gives flux $-56.5 \ \mathrm{N \cdot m^2/C}$ . For the total to be zero, the cap must give $+56.5 \ \mathrm{N \cdot m^2/C}$ .
So the flux through the hemispherical cap equals the flux through the flat disc with the same rim: $56.5 \ \mathrm{N \cdot m^2/C}$ . The general statement is that in a uniform field, the flux through any surface depends only on the projected area perpendicular to the field, which for both shapes here is $\pi r^2$ .
This is what the learning objective's phrase "arbitrary area or geometric shape" is pointing at. A cone, a dented sheet or a paraboloid with the same rim all give the same answer, and none of them needs an integral.

(a) $56.5 \ \mathrm{N \cdot m^2/C}$ . (b) $28.3 \ \mathrm{N \cdot m^2/C}$ . (c) zero, because the field lies in the plane of the disc and so has no component along the area vector. (d) $56.5 \ \mathrm{N \cdot m^2/C}$ , identical to the flat disc, because in a uniform field the flux through any surface depends only on the area of its projection perpendicular to the field.

A cube in a field that grows with position

A cube of side $0.30 \ \mathrm{m}$ sits with one corner at the origin and its edges along the coordinate axes, occupying $0 \le x,y,z \le 0.30 \ \mathrm{m}$ . The electric field in the region is $\vec{E} = (bx)\,\hat{i}$ with $b = 240 \ \mathrm{N}/(\mathrm{C \cdot m})$ . Find the total electric flux out of the cube, first symbolically and then numerically.

Note first that the field is not uniform, so 8.5.A.2 cannot be applied to the whole cube in one line. Use 8.5.A.3 and go face by face, with every area vector pointing outward, as 8.5.A.2.i requires for a closed surface.
The four faces parallel to the $x$ -axis are the ones at $y = 0$ , $y = a$ , $z = 0$ and $z = a$ . Their outward normals point along $\pm\hat{j}$ and $\pm\hat{k}$ , and the field points along $\hat{i}$ . The dot product is zero at every point of each of those faces, so each contributes exactly zero. Four of six faces gone, with no arithmetic.
The face at $x = 0$ . Its outward normal is $-\hat{i}$ . The field there is $\vec{E} = (b \cdot 0)\hat{i} = 0$ . Zero field, zero flux.
The face at $x = a$ . Its outward normal is $+\hat{i}$ . The field is the same at every point of this face, because $\vec{E}$ depends only on $x$ and this whole face sits at $x = a$ . So the field is constant across this face even though it is not constant across the cube, and 8.5.A.2 applies to the face: $\Phi = (ba)(a^2) = ba^3$ .
Total: $\Phi_E = ba^3$ . That is the CED's own sample multiple-choice Question 12, answer B.
Numbers. $a^3 = (0.30)^3 = 0.027 \ \mathrm{m^3}$ , so $\Phi_E = (240)(0.027) = 6.48 \ \mathrm{N \cdot m^2/C}$ .
Check it the long way on the one face that matters. At $x = 0.30 \ \mathrm{m}$ the field is $E = (240)(0.30) = 72 \ \mathrm{N/C}$ . The face area is $(0.30)^2 = 0.090 \ \mathrm{m^2}$ . Product: $(72)(0.090) = 6.48 \ \mathrm{N \cdot m^2/C}$ . Agrees.
Why the answer is not zero. More field leaves the far face than enters the near one, because the field grows with $x$ . A net outward flux is the signature of source charge inside, which is where Topic 8.6 picks the story up: Gauss's law converts this $6.48 \ \mathrm{N \cdot m^2/C}$ into an enclosed charge of $\varepsilon_0 \Phi_E = (8.85 \times 10^{-12})(6.48) = 5.7 \times 10^{-11} \ \mathrm{C}$ .
Why the answer is not $6ba^3$ . That would be the result of applying $EA$ to all six faces with the field evaluated at $x = a$ , which is wrong twice over: four faces have the field parallel to their surface, and the fifth sits where the field is zero.

$\Phi_E = ba^3 = 6.48 \ \mathrm{N \cdot m^2/C}$ . Only the face at $x = a$ contributes: four faces have the field lying in their plane, and the face at $x = 0$ sits where the field vanishes. The net outward flux is nonzero, which by Gauss's law means the cube encloses about $5.7 \times 10^{-11} \ \mathrm{C}$ of charge.

A genuine surface integral over an open surface

A square flat surface of side $d = 0.25 \ \mathrm{m}$ lies in the $yz$ -plane, occupying $0 \le y \le d$ and $0 \le z \le d$ . The electric field in the region is $\vec{E} = E_0\left(1 + \dfrac{y}{d}\right)\hat{i}$ with $E_0 = 200 \ \mathrm{N/C}$ . Take the area vector along $+\hat{i}$ and find the flux through the square.

Declare the normal, since the surface is open and 8.5.A.2.i leaves the choice to you: $\hat{n} = +\hat{i}$ . Every flux below is measured with respect to that choice.
Check whether 8.5.A.2 applies. The field varies with $y$ , and $y$ runs across the surface from $0$ to $d$ . So it does not: this needs the surface integral of 8.5.A.3.
The field is along $\hat{i}$ and the normal is along $\hat{i}$ , so the dot product is just $E\,dA$ with no cosine. That simplification comes from the geometry, not from the field being constant.
Choose an element over which the field is constant. The field depends on $y$ only, so a horizontal strip at height $y$ , of width $dy$ and full length $d$ in the $z$ direction, has the same field at every point. Its area is $dA = d\,dy$ .
Set up: $\Phi_E = \displaystyle\int_0^{d} E_0\left(1 + \frac{y}{d}\right) d\,dy = E_0 d \int_0^{d}\left(1 + \frac{y}{d}\right)dy$ .
Integrate with the power rule from the sheet's calculus table: $\displaystyle\int_0^d \left(1 + \frac{y}{d}\right) dy = \left[y + \frac{y^2}{2d}\right]_0^d = d + \frac{d}{2} = \frac{3d}{2}$ .
So $\Phi_E = E_0 d \cdot \dfrac{3d}{2} = \dfrac{3}{2}E_0 d^2$ . Stop here if the question wants a symbolic answer, which is what skill 2.A is asking for.
Numbers: $d^2 = (0.25)^2 = 0.0625 \ \mathrm{m^2}$ , so $\Phi_E = 1.5(200)(0.0625) = 18.75 \ \mathrm{N \cdot m^2/C}$ , which is $18.8 \ \mathrm{N \cdot m^2/C}$ to three figures.
Check it against the average. The field runs from $E_0 = 200 \ \mathrm{N/C}$ at $y = 0$ to $2E_0 = 400 \ \mathrm{N/C}$ at $y = d$ , linearly. A linear variation has an average equal to its midpoint value, $300 \ \mathrm{N/C}$ , and $(300)(0.0625) = 18.75 \ \mathrm{N \cdot m^2/C}$ . The integral and the average agree, as they must for a linear field.
Why that check is not a shortcut in general. It works only because the field happens to be linear in $y$ . Make the field go as $y^2$ and the midpoint value is no longer the average, and only the integral is right. Setting up the integral is the transferable skill; recognising when the average trick is legitimate is a bonus.

$\Phi_E = \dfrac{3}{2}E_0 d^2 = 18.8 \ \mathrm{N \cdot m^2/C}$ , with respect to an area vector along $+\hat{i}$ . Reversing that choice reverses the sign and changes nothing physical. This is the kind of flux the algebra-based course cannot pose, because it has no way to add up a field that changes across the surface.

Frequently asked questions

What is electric flux in AP Physics C?

Electric flux measures how much electric field passes through a surface. Essential knowledge 8.5.A.3 defines the total electric flux through a surface as the surface integral of the electric field over the surface, and the equation sheet prints that integral form. Essential knowledge 8.5.A.2 gives the special case in which the field is constant across the area, where the flux collapses to the dot product of the electric field vector and the area vector. Flux is a scalar with a sign but no direction, and its units are newton metres squared per coulomb, which is the same as volt metres.

Is electric flux on AP Physics 2?

No. The phrase electric flux appears nowhere in the AP Physics 2 course and exam description, and neither Gauss nor Gaussian appears in it at all. The word flux does appear there nineteen times, but every one of those refers to magnetic flux, mostly under Unit 12 Topic 12.4 on electromagnetic induction and Faraday's law. So an AP Physics 2 student meets flux only for the magnetic field, late in the course. Electric flux and Gauss's law are AP Physics C: Electricity and Magnetism topics 8.5 and 8.6 and exist only in the calculus-based course.

Does a surface have to be closed to have electric flux through it?

No. Learning objective 8.5.A asks students to describe the electric flux through an arbitrary area or geometric shape, and says nothing about closure. A flat square, a hemispherical cap or a tilted disc all have a well-defined flux. The closed surface enters in the next topic, 8.6, because closing the surface is what allows the flux to be related to the enclosed charge. The one thing that changes when a surface is closed is that the direction of the area vector stops being your choice: essential knowledge 8.5.A.2.i defines it as outward for a closed surface.

How do you find the direction of the area vector?

Essential knowledge 8.5.A.2.i says the area vector is perpendicular to the plane of the surface, and outward from a closed surface. For a closed surface the direction is therefore fixed for you, which is what makes the sign of the flux in Gauss's law meaningful. For an open surface such as a flat plate, either perpendicular direction is allowed, so you choose one, state it, and keep it for the whole problem. The choice fixes the sign of the flux through the dot product, per 8.5.A.2.ii, but never changes a physical result.

Why is the flux zero when the surface is parallel to the field?

Because nothing passes through it. If the field lies in the plane of the surface then it is perpendicular to the area vector, the dot product of the two vectors is zero, and so is the flux. Picture the field skimming along the surface rather than crossing it. This is also the reason four of the six faces of a cube contribute nothing when the field points along one coordinate axis: on those four faces the field lies in the plane of the face. Watch the wording, though, since a surface perpendicular to the field has the maximum flux, not zero.

What are the units of electric flux?

Newton metres squared per coulomb. Flux is an electric field multiplied by an area, and the electric field is measured in newtons per coulomb because it is defined as force per unit charge in essential knowledge 8.3.A.2. Since a newton per coulomb is also a volt per metre, the same unit is often written as a volt metre. Neither form is printed in the CED or on the equation sheet, and both follow from the definitions. Flux has no direction, only a sign, so it never carries a unit vector.

Does AP Physics C Topic 8.5 have a boundary statement?

No. Topic 8.5 prints no boundary statement, and neither do Topics 8.2 and 8.3. Unit 8 carries exactly three boundary statements, under Topics 8.1, 8.4 and 8.6, and none of them restricts what shape of surface you may be asked to find the flux through. The learning objective points the other way, since it asks about the flux through an arbitrary area or geometric shape. What does bound the work in practice is that a flux integral is only tractable when the field is constant across the surface, constant in magnitude and perpendicular to it, or varies in a single coordinate.