Electric Flux vs Magnetic Flux: The Difference

The two are defined identically: a field dotted with an area vector over a surface. They part only on a closed surface. The electric flux out of one is the enclosed charge divided by epsilon nought. The magnetic flux out of one is always exactly zero, because there is no magnetic charge.

AP Physics: Unit 8 (topics 8.5 Electric Flux, 8.6 Gauss's Law, 13.1 Magnetic Flux). This pair is AP Physics C: Electricity and Magnetism material. Electric flux is Topic 8.5 and Gauss's law is Topic 8.6, both in Unit 8, Electric Charges, Fields, and Gauss's Law, weighted at 15 to 25 percent of the multiple-choice section over a suggested 12 to 24 class periods. Magnetic flux is Topic 13.1, opening Unit 13, Electromagnetic Induction, at 10 to 20 percent. The definitions are deliberately parallel: EK 8.5.A.2 and EK 13.1.A.1 both define the flux of a constant field across an area as the dot product of the field and the area vector; EK 8.5.A.2.i and EK 13.1.A.1.i both define the area vector as perpendicular to the plane of the surface and outward from a closed surface; EK 8.5.A.2.ii and EK 13.1.A.1.ii both give the sign of the flux by that dot product; and EK 8.5.A.3 and EK 13.1.A.2 both give the general surface integral. EK 8.5.A.1 adds that flux describes the amount of a given quantity that passes through a given area. The asymmetry is in the closed-surface laws. EK 8.6.A.1 relates the electric flux through a Gaussian surface to the charge enclosed, printing both the ratio form and the integral form, and EK 8.6.A.6 names Gauss's law as Maxwell's first equation. On the magnetic side, EK 12.1.A.3 states that magnetic field lines must form closed loops, as described by Gauss's law for magnetism, with the printed equation setting the closed-surface integral to zero, and EK 12.1.A.3.i names it as Maxwell's second equation. A boundary statement on Topic 8.6 restricts quantitative application of Gauss's law to point charges and to charge distributions with spherical, cylindrical or planar symmetry. AP Physics 2 has magnetic flux, inside Topic 12.4, and no electric flux; its equation sheet prints no Gauss's law of either kind.

The distinction, stated once

Start with what is the same, because almost everything is.

AP Physics C: Electricity and Magnetism defines the two in matching sentences. EK 8.5.A.2: for an electric field $\vec{E}$ that is constant across an area $\vec{A}$ , the electric flux through the area is defined as $\Phi_E = \vec{E} \cdot \vec{A}$ . EK 13.1.A.1: for a magnetic field $\vec{B}$ that is constant across an area $\vec{A}$ , the magnetic flux through the area is defined as $\Phi_B = \vec{B} \cdot \vec{A}$ .

The sub-points match too. Both courses define the area vector as perpendicular to the plane of the surface and outward from a closed surface (EK 8.5.A.2.i and EK 13.1.A.1.i). Both give the sign of the flux as the dot product of the field vector with the area vector (EK 8.5.A.2.ii and EK 13.1.A.1.ii). Both generalise to a surface integral (EK 8.5.A.3 and EK 13.1.A.2). One definition, applied twice.

So the difference is not in the definition. It is entirely in what happens when the surface is closed. The sheet prints both results:

\oint \vec{E} \cdot d\vec{A} = \frac{q_{\mathrm{enc}}}{\varepsilon_0} \qquad \oint \vec{B} \cdot d\vec{A} = 0

The left one is Gauss's law and its right-hand side is whatever charge is inside. The right one is Gauss's law for magnetism and its right-hand side is zero, on every closed surface, in every situation, with no exceptions and no conditions.

That zero is not an approximation and not a special case. It is the whole content of the equation, and the reason for it is that there is no such thing as magnetic charge.

Side by side

	Electric flux $\Phi_E$	Magnetic flux $\Phi_B$
Definition, constant field	$\Phi_E = \vec{E} \cdot \vec{A}$	$\Phi_B = \vec{B} \cdot \vec{A}$
General form	$\Phi_E = \int \vec{E} \cdot d\vec{A}$	$\Phi_B = \int \vec{B} \cdot d\vec{A}$
Unit	$\mathrm{N \cdot m^2/C}$	Weber (Wb), which is $\mathrm{T \cdot m^2}$
Through a closed surface	$q_{\mathrm{enc}}/\varepsilon_0$ , zero only if no net charge is inside	Zero, always
Named law for that result	Gauss's law, Maxwell's first equation	Gauss's law for magnetism, Maxwell's second equation
Why	Isolated electric charge exists	Isolated magnetic charge does not
Field lines	Begin and end on charges	Close on themselves
What it is used for	Finding the field, under symmetry	Feeding Faraday's law, on an open surface
Which AP course	C: E&M only, Topic 8.5	AP Physics 2 (inside Topic 12.4) and C: E&M Topic 13.1
On the AP Physics 2 sheet	Not printed	Printed twice

The through a closed surface row is the page. Everything else either sets it up or follows from it.

The what it is used for row is the practical consequence and it surprises people. Electric flux is a means to an end: you compute it on a cleverly chosen closed surface in order to extract the field. Magnetic flux is never used that way, because the answer on a closed surface is known in advance and carries no information. Magnetic flux earns its keep on open surfaces, where it is the quantity Faraday's law differentiates.

The which AP course row is worth checking before you revise. AP Physics 2 has magnetic flux and no electric flux, and prints no Gauss's law of either kind.

The case that separates them: a closed surface with something inside

Take a cube of side $0.20 \ \mathrm{m}$ and use it as a closed surface. Put it first in a uniform electric field of $300 \ \mathrm{N/C}$ pointing along $+x$ , then in a uniform magnetic field of $0.50 \ \mathrm{T}$ pointing along $+x$ .

Each face has area $0.040 \ \mathrm{m^2}$ . The four faces parallel to the field contribute nothing, since their area vectors are perpendicular to it. The face the field enters has an outward area vector pointing $-x$ , so it contributes a negative flux; the face the field leaves has an outward area vector along $+x$ , so it contributes an equal positive one.

	Uniform $\vec{E}$ of $300 \ \mathrm{N/C}$	Uniform $\vec{B}$ of $0.50 \ \mathrm{T}$
Entry face	$-12 \ \mathrm{N \cdot m^2/C}$	$-0.020 \ \mathrm{Wb}$
Exit face	$+12 \ \mathrm{N \cdot m^2/C}$	$+0.020 \ \mathrm{Wb}$
Four side faces	$0$	$0$
Total through the closed cube	$0$	$0$

So far the two behave identically, and that is the trap: a uniform field gives zero net flux through any closed surface, electric or magnetic, because whatever enters leaves.

Now put a charge of $+4.0 \ \mathrm{nC}$ inside the cube.

	Electric	Magnetic
Total flux through the closed cube	$+452 \ \mathrm{N \cdot m^2/C}$	$0$

The electric answer changed because there is now something inside for field lines to start on. The magnetic answer cannot change, because there is nothing you could put inside a surface that would make it change. Bar magnets, current loops, spinning charges, whole electric motors: enclose any of them and the net magnetic flux out of the surface is still zero, because every line that leaves comes back.

Why the magnetic answer is always zero

The reason is one sentence long and the CED states it. EK 12.1.A.1.i: magnetic fields can be produced by magnetic dipoles or combinations of dipoles, but never by monopoles.

A field line has to start somewhere and finish somewhere, or else close on itself. Electric field lines start on positive charge and finish on negative charge, so enclosing a lone positive charge traps a set of lines that leave and never come back, and the surface registers a net outflow. There is no magnetic equivalent of a lone positive charge. EK 12.1.A.3 puts the consequence directly: magnetic field lines must form closed loops, as described by Gauss's law for magnetism.

The bar magnet is the case worth picturing. Draw a closed surface around just the north end of a bar magnet, cutting the magnet in half so that the surface passes through the middle of the iron. Field lines stream out of the north pole into the air outside, and every one of them curves back around, re-enters at the south end, and travels back up through the inside of the magnet, crossing your surface in the other direction. The outward flux through the air and the inward flux through the metal cancel exactly.

Cut the magnet at that plane and you do not get a north pole in your hand. You get two shorter magnets, each with its own north and south. That is the experiment, and it has never once come out otherwise.

So the two Gauss's laws are the same equation written for two fields with different source structures, and the difference between $q_{\mathrm{enc}}/\varepsilon_0$ and $0$ is the difference between a field with monopoles and a field without. The Gauss's law for magnetism entry states the law on its own; the Gauss's law guide works the electric one.

Where these two sit among Maxwell's equations

The AP Physics C: Electricity and Magnetism CED numbers all four of Maxwell's equations, one at a time, in four different units, and it uses the same opening sentence each time: Maxwell's equations are the collection of equations that fully describe electromagnetism.

	Law	CED statement
First	Gauss's law	EK 8.6.A.6
Second	Gauss's law for magnetism	EK 12.1.A.3.i
Third	Faraday's law of induction	EK 13.2.A.3
Fourth	Ampere's law with Maxwell's addition	EK 12.4.A.4

The first two are the pair on this page, and they are the two flux laws: what a closed surface says about each field. The third and fourth are the circulation laws, about closed loops rather than closed surfaces, and they say how each field is generated by changes in the other.

That ordering is worth carrying, because it makes the asymmetry structural rather than incidental. Equations one and two ask the same question of the two fields and get different answers. Equations three and four ask a different question and also get different answers, since a changing magnetic flux produces an electric field with no current involved, while a magnetic field needs either a current or a changing electric field.

EK 13.2.A.4 adds the payoff: Maxwell's equations can be used to show that electric and magnetic fields obey wave equations and that electromagnetic waves travel at a constant speed in free space. A boundary statement then says students are not expected to derive the speed of light from them, which tells you the naming is examinable and the derivation is not.

One is a calculating tool, the other is a constraint

This is the difference that decides how you use each on a free response.

Electric flux computes. Pick a closed surface that matches the symmetry of the charge distribution, so that the field has a constant magnitude over it and is either perpendicular or parallel to each region. Then $\oint \vec{E} \cdot d\vec{A}$ collapses to $EA$ , and setting that equal to $q_{\mathrm{enc}}/\varepsilon_0$ hands you the field. EK 8.6.A.4 describes exactly this construction, and a boundary statement restricts quantitative applications to point charges and to charge distributions with spherical, cylindrical or planar symmetry. Those three symmetries are the whole examinable list.

One more property makes it powerful. EK 8.6.A.3: the total electric flux through a Gaussian surface is independent of the size of the Gaussian surface if the amount of enclosed charge remains constant. Grow the sphere and the field falls as $1/r^2$ while the area grows as $r^2$ , and the product does not move. The worked example below checks that digit by digit.

Magnetic flux on a closed surface computes nothing. The equation reads $0 = 0$ wherever you put the surface, so there is no field to extract and no unknown to solve for. What it does instead is rule things out: any proposed magnetic field configuration with a net outflow through some closed surface is impossible, and no amount of ingenuity will produce one. Its work is done when it tells you an answer cannot be right.

The calculating tool on the magnetic side is Ampere's law, $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\mathrm{enc}}$ , which uses a closed loop rather than a closed surface and a current rather than a charge. A boundary statement limits its quantitative use to symmetrical magnetic fields, naming long straight wires, long solenoids, and conductive slabs or cylindrical conductors carrying a current density.

And magnetic flux itself is genuinely useful, just not on closed surfaces. On an open surface, such as the area bounded by a loop of wire, its rate of change is the induced emf. That is Faraday's law, and it is covered in magnetic flux vs magnetic field.

Units, and what a course does and does not print

The two fluxes have different units, because the two fields do.

Electric flux is a field in N/C times an area in $\mathrm{m^2}$ , so $\mathrm{N \cdot m^2 / C}$ . Because N/C is also V/m, this can equally be written as volt metres, and both forms are correct. Dividing it by $\varepsilon_0$ , in $\mathrm{C^2/(N \cdot m^2)}$ , does return coulombs, which is a useful check on a Gauss's law answer.

Magnetic flux is a field in tesla times an area in $\mathrm{m^2}$ , so $\mathrm{T \cdot m^2}$ , which is the weber. It is also a volt second, because dividing a flux by a time has to give an emf in volts.

What each course prints matters more than usual here, because the two courses are lopsided in opposite directions.

AP Physics 2 prints magnetic flux twice on its sheet, as $\Phi_B = \vec{B} \cdot \vec{A}$ and as $\Phi_B = \lvert \vec{B} \rvert \cos\theta \lvert \vec{A} \rvert$ , and prints no electric flux at all. It has no Gauss's law of either kind anywhere on the sheet, and no flux topic. Flux appears only as the input to Faraday's law in Topic 12.4.

AP Physics C: Electricity and Magnetism prints all four: both flux integrals and both closed-surface laws, in the Electricity and Magnetism block. Electric flux gets its own topic, 8.5, immediately before Gauss's law in Topic 8.6, and magnetic flux gets Topic 13.1, immediately before induction in Topic 13.2. The placement in each case tells you what the flux is for.

So a Physics 2 student who has met flux has met only the magnetic one, and a question about electric flux is outside that course.

When it costs a mark

Writing a nonzero closed-surface magnetic flux. The most direct way to lose a point on this material. If the surface is closed, the answer is zero, and no configuration inside it changes that.

Reading the magnetic zero as "there is no magnetic field here". It says the net flux is zero, not that the field is. A cube sitting in a uniform $0.50 \ \mathrm{T}$ field has zero net flux through it and a strong field at every one of its points.

Assuming the electric flux is zero because the field is nonzero everywhere on the surface. Also common, and the same error in reverse. A uniform electric field through a closed surface gives zero net flux, because the entry and exit contributions cancel. Zero net flux means no net charge inside, and nothing more.

Putting the size of the surface into a Gauss's law answer. Grow or shrink a Gaussian surface around a fixed charge and the flux does not change, per EK 8.6.A.3. If your flux answer depends on the radius, you have made an arithmetic error.

Counting charge that is outside the surface. It is $q_{\mathrm{enc}}$ , the charge enclosed. External charges do contribute to the field at every point of the surface, and their contributions to the flux cancel out exactly. They belong in the field and not in the flux.

Using Gauss's law where the symmetry is not there. The boundary statement limits quantitative use to point charges and to spherical, cylindrical or planar symmetry. Without one of those, $\oint \vec{E} \cdot d\vec{A}$ does not collapse to $EA$ and the law, while still true, will not give you a field.

Measuring the angle from the surface rather than the area vector. Shared with every flux calculation. The area vector is perpendicular to the surface, and both CED definitions say so.

Where they behave the same, and what to read next

On an open surface the two are the same calculation with different letters. A flat loop tilted in a uniform field gives $EA\cos\theta$ or $BA\cos\theta$ , with the same area vector convention, the same sign rule and the same trap about the angle. Nothing on an open surface distinguishes them.

They also agree on the uniform-field closed surface, as the cube example shows, and this is the case that makes the asymmetry easy to miss. Both come out zero, for the same reason, and a student who has only seen that example has seen no difference at all.

The split appears at exactly one moment: when there is something inside the closed surface. On the electric side that something can be a net charge, and then the flux is not zero. On the magnetic side there is nothing that could play that role.

On the exam, electric flux is Topic 8.5 and Gauss's law is Topic 8.6, inside Unit 8, which the CED weights at 15 to 25 percent of the multiple-choice section over a suggested 12 to 24 class periods. Magnetic flux is Topic 13.1, opening Unit 13 at 10 to 20 percent. Gauss's law for magnetism is not in either of those units; it is stated in Unit 12, at EK 12.1.A.3, as a property of the magnetic field rather than as a flux technique, which is itself a clue about how it is meant to be used.

From here, the Gauss's law guide runs the electric calculation end to end, electric flux and magnetic flux give the definitions on their own, and magnetic flux vs magnetic field covers the open-surface side, where magnetic flux does its real work. For choosing between the two ways of getting an electric field, see Coulomb's law vs Gauss's law, and for the closed-surface against closed-loop contrast, see Gauss's law vs Ampere's law.

A cube in a uniform field, then a charge inside it

A cube of side $0.20 \ \mathrm{m}$ is used as a closed surface. (a) Find the net electric flux through it in a uniform electric field of $300 \ \mathrm{N/C}$ directed along $+x$ , face by face. (b) Find the net magnetic flux through it in a uniform magnetic field of $0.50 \ \mathrm{T}$ along $+x$ . (c) Now place a charge of $+4.0 \times 10^{-9} \ \mathrm{C}$ inside the cube and find both net fluxes again. Use $\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C^2/(N \cdot m^2)}$ .

Each face has area $A = (0.20)^2 = 0.040 \ \mathrm{m^2}$ .
(a) Entry face: its outward area vector points along $-x$ , opposite the field, so $\theta = 180^\circ$ and $\Phi = EA\cos 180^\circ = -(300)(0.040) = -12 \ \mathrm{N \cdot m^2/C}$ .
Exit face: outward area vector along $+x$ , $\theta = 0$ , so $\Phi = +(300)(0.040) = +12 \ \mathrm{N \cdot m^2/C}$ .
The four remaining faces have outward area vectors perpendicular to the field, so $\cos 90^\circ = 0$ and each contributes nothing.
Net: $-12 + 12 + 0 + 0 + 0 + 0 = 0$ . Consistent with Gauss's law, since no charge is inside.
(b) The magnetic case is the identical arithmetic with a different field: $\mp(0.50)(0.040) = \mp 0.020 \ \mathrm{Wb}$ on the two end faces and zero on the others, for a net of $0$ .
(c) With the charge inside, Gauss's law gives $\Phi_E = q_{\mathrm{enc}}/\varepsilon_0 = (4.0 \times 10^{-9})/(8.85 \times 10^{-12})$ .
$4.0/8.85 = 0.452$ , and $10^{-9}/10^{-12} = 10^3$ , so $\Phi_E = 452 \ \mathrm{N \cdot m^2/C}$ . The uniform external field still contributes nothing, because its own net flux was zero.
The net magnetic flux is still exactly $0$ . There is no object you could place inside the cube that would change this, which is what $\oint \vec{B} \cdot d\vec{A} = 0$ asserts.

Empty cube: both net fluxes are zero. With $+4.0 \ \mathrm{nC}$ inside: the electric flux is $452 \ \mathrm{N \cdot m^2/C}$ and the magnetic flux is still zero. The charge is the thing that has no magnetic counterpart.

Two spheres, one charge: the flux does not care about the radius

A point charge $q = +4.0 \times 10^{-9} \ \mathrm{C}$ sits at the origin. Find the electric flux through a spherical Gaussian surface of radius $0.10 \ \mathrm{m}$ centred on it, and through one of radius $0.20 \ \mathrm{m}$ . Check each answer by computing the field and multiplying by the area.

By Gauss's law the answer does not depend on the radius at all: $\Phi_E = q/\varepsilon_0 = (4.0 \times 10^{-9})/(8.85 \times 10^{-12}) = 452 \ \mathrm{N \cdot m^2/C}$ for both spheres.
Check the small sphere the long way. Field at $r = 0.10 \ \mathrm{m}$ : $E = kq/r^2 = (9.0 \times 10^9)(4.0 \times 10^{-9})/(0.10)^2 = 36/0.010 = 3600 \ \mathrm{N/C}$ .
Area of that sphere: $A = 4\pi r^2 = 4\pi(0.010) = 0.12566 \ \mathrm{m^2}$ . The field is radial and the area vector is radial, so $\theta = 0$ everywhere and $\Phi_E = EA = (3600)(0.12566) = 452.4 \ \mathrm{N \cdot m^2/C}$ .
Check the large sphere. Field at $r = 0.20 \ \mathrm{m}$ : $E = 36/0.040 = 900 \ \mathrm{N/C}$ , one quarter of the previous value.
Area: $A = 4\pi(0.040) = 0.50265 \ \mathrm{m^2}$ , four times the previous value. Product: $(900)(0.50265) = 452.4 \ \mathrm{N \cdot m^2/C}$ , identical to the small sphere.
That is EK 8.6.A.3 made arithmetic. The field falls as $1/r^2$ and the area grows as $r^2$ , so the product is fixed by the enclosed charge and by nothing else.
A note on the last digit. The field-times-area route gives $452.4$ and the $q/\varepsilon_0$ route gives $452.0$ , a difference of one part in a thousand. It is there because the AP sheet rounds $k$ to $9.0 \times 10^9$ and $\varepsilon_0$ to $8.85 \times 10^{-12}$ independently, and $1/(4\pi\varepsilon_0)$ with that $\varepsilon_0$ comes to $8.99 \times 10^9$ rather than $9.0 \times 10^9$ . Report two significant figures, $4.5 \times 10^2 \ \mathrm{N \cdot m^2/C}$ , and the two routes agree.

$\Phi_E = 452 \ \mathrm{N \cdot m^2/C}$ through both spheres, confirmed by the field-times-area route in each case. Doubling the radius quartered the field and quadrupled the area, leaving the flux untouched.

One law hands you a field; the other hands you nothing

Using the same $+4.0 \times 10^{-9} \ \mathrm{C}$ point charge, (a) start from Gauss's law on a sphere of radius $0.20 \ \mathrm{m}$ and recover the electric field, without using Coulomb's law. (b) Attempt the same procedure with Gauss's law for magnetism and say what you get.

(a) Gauss's law: $\oint \vec{E} \cdot d\vec{A} = q_{\mathrm{enc}}/\varepsilon_0$ .
The charge distribution is spherically symmetric, which is one of the three symmetries the boundary statement allows, so the field magnitude is the same at every point of the sphere and is everywhere parallel to the outward area vector. That lets the integral collapse: $\oint \vec{E} \cdot d\vec{A} = E \times 4\pi r^2$ .
So $E \times 4\pi r^2 = q/\varepsilon_0$ , giving $E = q/(4\pi\varepsilon_0 r^2)$ . The whole of Coulomb's law has just come out of a flux statement.
Substitute: $E = (4.0 \times 10^{-9})/[(4\pi)(8.85 \times 10^{-12})(0.040)]$ . The denominator is $(1.112 \times 10^{-10})(0.040) = 4.449 \times 10^{-12}$ , so $E = 899 \ \mathrm{N/C}$ , which is $900 \ \mathrm{N/C}$ at two significant figures and matches $kq/r^2 = 36/0.040 = 900 \ \mathrm{N/C}$ .
(b) Now the magnetic version, on any closed surface you like: $\oint \vec{B} \cdot d\vec{A} = 0$ .
There is no enclosed magnetic charge on the right-hand side to be proportional to, so the equation is $0 = 0$ before you start. There is nothing to solve for and no field comes out. Repeating the step that worked in part (a) fails at the first line, because the symmetry argument needs an isolated point source and no isolated magnetic source exists.
So the magnetic law is a constraint, not a calculator. It tells you that a proposed field with a net outflow is impossible, and it stops there. The calculating tool on the magnetic side is Ampere's law, $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\mathrm{enc}}$ , which runs around a closed loop rather than over a closed surface and encloses a current rather than a charge.

Gauss's law gives $E = 900 \ \mathrm{N/C}$ at $0.20 \ \mathrm{m}$ , reproducing Coulomb's law from a flux argument. Gauss's law for magnetism gives $0 = 0$ and no field, which is why the magnetic side uses Ampere's law and a closed loop instead.

Frequently asked questions

What is the difference between electric flux and magnetic flux?

The definitions are the same: a field dotted with an area vector, summed over a surface, with the area vector perpendicular to the surface and outward from a closed one. The difference appears only on a closed surface. The net electric flux out of a closed surface equals the charge enclosed divided by epsilon nought, so it is nonzero whenever charge is inside. The net magnetic flux out of a closed surface is always exactly zero, because isolated magnetic charge does not exist. Their units also differ: newton square metres per coulomb for electric flux, and webers for magnetic flux.

Why is the magnetic flux through a closed surface always zero?

Because magnetic field lines close on themselves, so every line that enters a closed surface leaves it again. AP Physics C: E&M EK 12.1.A.1.i states that magnetic fields can be produced by magnetic dipoles or combinations of dipoles but never by monopoles, and EK 12.1.A.3 draws the consequence: magnetic field lines must form closed loops, as described by Gauss's law for magnetism. There is no magnetic equivalent of a lone charge for lines to start on, so there is nothing a closed surface could enclose that would give a net outflow.

Is Gauss's law for magnetism one of Maxwell's equations?

Yes, the second one. AP Physics C: E&M EK 12.1.A.3.i states that Maxwell's equations are the collection of equations that fully describe electromagnetism and that Gauss's law for magnetism is Maxwell's second equation. The CED names all four in the same form: EK 8.6.A.6 makes Gauss's law the first, EK 13.2.A.3 makes Faraday's law of induction the third, and EK 12.4.A.4 makes Ampere's law with Maxwell's addition the fourth.

Can the electric flux through a closed surface be zero if the field is not zero?

Yes. A closed surface sitting in a uniform electric field has zero net flux, because the flux entering one side exactly cancels the flux leaving the other, while the field at every point of the surface is whatever the uniform field is. Zero net flux means only that no net charge is enclosed. Charges outside the surface do contribute to the field on it, and their contributions to the flux always cancel.

Does the size of a Gaussian surface change the electric flux through it?

No, provided the enclosed charge stays the same. EK 8.6.A.3 says the total electric flux through a Gaussian surface is independent of its size if the amount of enclosed charge remains constant. For a point charge you can see why: the field falls off as one over r squared while the surface area grows as r squared, so the product is fixed. A 4.0 nC charge gives 452 newton square metres per coulomb through a sphere of any radius.

Does AP Physics 2 cover electric flux?

No. AP Physics 2 has magnetic flux, which appears inside Topic 12.4 as the input to Faraday's law, and its equation sheet prints magnetic flux twice. It prints no electric flux and no Gauss's law of either kind. Electric flux is AP Physics C: Electricity and Magnetism material, where it has its own topic, 8.5, feeding Gauss's law in Topic 8.6.

Why is Gauss's law useful but Gauss's law for magnetism is not used to calculate anything?

Because one has an unknown on the right-hand side and the other has zero. Choose a closed surface that matches the symmetry of a charge distribution and the electric flux integral collapses to field times area, which you then set equal to the enclosed charge over epsilon nought and solve for the field. The magnetic version reads zero equals zero on every surface, so there is nothing to solve. It rules configurations out rather than computing them, and the calculating tool on the magnetic side is Ampere's law, which uses a closed loop and an enclosed current.