AP Physics C: E&M · Topic 8.6

Topic 8.6: Gauss's Law

Q: What is Gauss's law in AP Physics C?

Gauss's law relates the electric flux through a closed surface to the charge enclosed by that surface. Essential knowledge 8.6.A.1 states it that way and the equation sheet prints the closed-surface integral of E dot dA equal to the enclosed charge divided by the permittivity of free space. The closed surface is called a Gaussian surface, defined in 8.6.A.2 as a three-dimensional closed surface, and it is imaginary: nothing physical has to be there. The law holds for any closed surface around any arrangement of charge, but it only lets you solve for the field when the charge has enough symmetry.

Q: How do you choose a Gaussian surface?

Let the symmetry of the charge choose it for you. Essential knowledge 8.6.A.4 says Gaussian surfaces are typically constructed so that the field is either perpendicular or parallel to different regions of the surface, which is what simplifies the integral. Where the field is perpendicular to the surface and constant in magnitude, the integral becomes the field times an area; where the field is parallel to the surface, that region contributes nothing. In practice that means a concentric sphere for spherical symmetry, a coaxial cylinder for cylindrical symmetry, and a pillbox straddling the plane for planar symmetry.

Q: When can you not use Gauss's law?

You can always use it, and you can only solve for a field with it when the charge distribution has spherical, cylindrical or planar symmetry, or is a point charge. That is exactly what the Topic 8.6 boundary statement says: AP Physics C: Electricity and Magnetism only expects students to quantitatively apply Gauss's law to point charges and charge distributions that have spherical, cylindrical, or planar symmetry. For a finite rod, a ring, an arc or a dipole, the law is still true but the field varies over any surface you draw, so it cannot be pulled out of the integral. Those cases are handled by direct integration in Topic 8.4.

Q: Is Gauss's law on AP Physics 2?

No. The word Gauss does not appear anywhere in the AP Physics 2 course and exam description, and neither does the phrase electric flux. Gauss's law is Topic 8.6 of AP Physics C: Electricity and Magnetism, and electric flux is Topic 8.5, and both exist only in the calculus-based course. AP Physics 2 covers electric charge, force, field, potential energy and potential in its Unit 10, and it does use the word flux nineteen times, but always for magnetic flux in its Unit 12 treatment of electromagnetic induction.

Q: Does the size of the Gaussian surface change the flux?

Not if the enclosed charge stays the same. Essential knowledge 8.6.A.3 states that 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. Growing a sphere around a fixed point charge weakens the field as one over the radius squared while growing the area as the radius squared, and the two effects cancel exactly. The condition matters, though: expanding a Gaussian sphere inside a uniformly charged ball encloses more charge, so both the flux and the field increase.

Q: How do you find the enclosed charge when the charge density is a function?

Integrate the density over the region inside the Gaussian surface. Essential knowledge 8.6.A.5 says the total charge can be determined by integrating the charge density over the length, area or volume of the distribution, and prints the volume version as Q total equals the integral of rho dV. The step that gets dropped is the volume element. For spherical symmetry a thin shell at radius s of thickness ds has volume four pi s squared ds, so you integrate rho of s times four pi s squared with respect to s. A units check catches the omission, since charge per volume times a length is not a charge.

Q: What is the difference between Topic 8.5 and Topic 8.6?

Topic 8.5 defines electric flux and asks you to compute it through an arbitrary surface, open or closed, with no charge necessarily involved. Topic 8.6 closes the surface and states that the net flux out of it equals the enclosed charge divided by epsilon zero, which turns a flux calculation into a statement about charge, and in symmetric cases into a way of finding the field. A useful marker: if a question gives you a field and asks what passes through a surface, that is 8.5. If it gives you charge and asks for a field, or gives you a flux and asks for charge, that is 8.6.

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

Gauss's law says the net electric flux out of any closed surface equals the charge enclosed divided by the permittivity of free space. It is always true, and it is useful when the charge has spherical, cylindrical or planar symmetry, because then the field is constant over the surface.

AP Physics: Unit 8 (topics 8.6 Gauss's Law). AP Physics C: Electricity and Magnetism Unit 8, Topic 8.6. One learning objective, 8.6.A, describe the properties of a charge distribution by applying Gauss's law. Six essential-knowledge statements, none with sub-statements, which is more than any other single learning objective in Unit 8 carries: 8.6.A.1 (Gauss's law relates electric flux through a Gaussian surface to the charge enclosed by that surface, with two relevant equations printed, Phi sub E equals q enclosed over epsilon zero and the closed integral of E dot dA equals q enclosed over epsilon zero), 8.6.A.2 (a Gaussian surface is a three-dimensional, closed surface), 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), 8.6.A.4 (Gaussian surfaces are typically constructed such that the electric field generated by the enclosed charge is either perpendicular or parallel to different regions of the Gaussian surface, resulting in a simplified surface integral), 8.6.A.5 (if a function of charge density is given, the total charge can be determined by integrating the charge density over the length, area or volume of the distribution, for example Q total equals the integral of rho dV), and 8.6.A.6 (Maxwell's equations are the collection of equations that fully describe electromagnetism, and Gauss's law is Maxwell's first equation). The boundary statement reads: AP Physics C: Electricity & Magnetism only expects students to quantitatively apply Gauss's law to point charges and charge distributions that have spherical, cylindrical, or planar symmetry. The word quantitatively is the limit; qualitative reasoning about flux through arbitrary closed surfaces is not restricted. Suggested skills are 1.A, 2.A, 2.B and 2.D, all from science practices 1 and 2; Topic 8.6 is the only topic in Unit 8 with no science practice 3 skill listed. This topic has no AP Physics 2 counterpart, and the word Gauss appears zero times in the AP Physics 2 CED. The CED's sample multiple-choice Question 9 aligns to 8.6.A and essential knowledge 8.6.A.1 at skill 2.A, answer A: a uniform sphere of charge Q and radius R sub S with a point charge q at its centre, net field at R sub S over 2 equal to one over four pi epsilon zero times R sub S over 2 squared, times the quantity q plus Q over 8. Unit 8 is weighted 15 to 25% of the multiple-choice section over about 12 to 24 class periods, tied with Unit 11 for the largest of the six units.

What Topic 8.6 requires

Topic 8.6 has one learning objective carrying six essential-knowledge statements, none of them with sub-statements. No other single objective in Unit 8 carries six. It is also one of the sixteen topics in this course whose title has no counterpart in AP Physics 2.

8.6.A, describe the properties of a charge distribution by applying Gauss's law.

8.6.A.1 Gauss's law relates electric flux through a Gaussian surface to the charge enclosed by that surface. Two relevant equations are printed with it:

\Phi_E = \frac{q_{\text{enc}}}{\varepsilon_0} \qquad \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}

8.6.A.2 a Gaussian surface is a three-dimensional, closed surface.
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.
8.6.A.4 Gaussian surfaces are typically constructed such that the electric field generated by the enclosed charge is either perpendicular or parallel to different regions of the Gaussian surface, resulting in a simplified surface integral.
8.6.A.5 if a function of charge density is given for a charge distribution, the total charge can be determined by integrating the charge density over the length (one dimension), area (two dimensions), or volume (three dimensions) of the charge distribution. The example equation printed is $Q_{\text{total}} = \int \rho(\vec{r})\,dV$ .
8.6.A.6 Maxwell's equations are the collection of equations that fully describe electromagnetism, and Gauss's law is Maxwell's first equation. The closed-surface integral is printed again here.

The boundary statement, in full. "AP Physics C: Electricity & Magnetism only expects students to quantitatively apply Gauss's law to point charges and charge distributions that have spherical, cylindrical, or planar symmetry."

Note the word quantitatively. The law itself is stated without restriction, and qualitative reasoning about it, ranking fluxes through oddly shaped surfaces, arguing that external charge contributes nothing, is not fenced off. What is fenced is being asked to compute a field. That needs symmetry.

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.B, calculate or estimate an unknown quantity with units from known quantities by selecting and following a logical computational pathway; and 2.D, predict new values or factors of change of physical quantities using functional dependence between variables.

All four are from science practices 1 and 2. No science practice 3 skill is listed for Topic 8.6, and it is the only topic in Unit 8 for which that is true. The unit's other five topics each carry 3.B, 3.C or both. Take the hint: this topic is assessed by making you compute and derive, not by making you justify.

Choosing the surface is the method, not a step in it

Statement 8.6.A.4 is the most useful sentence in the topic and the one most often read past: Gaussian surfaces are typically constructed such that the electric field generated by the enclosed charge is either perpendicular or parallel to different regions of the Gaussian surface, resulting in a simplified surface integral.

Unpack why that makes the whole thing work. On the left of Gauss's law sits $\oint \vec{E} \cdot d\vec{A}$ , an integral you generally cannot do. There are exactly two ways to make it doable, and the statement names both.

Where the field is perpendicular to the surface, meaning parallel to $d\vec{A}$ , the dot product is $E\,dA$ . If in addition $E$ has the same magnitude everywhere on that region, it comes out of the integral and what is left is $E \times (\text{area})$ . Pure algebra.

Where the field is parallel to the surface, meaning perpendicular to $d\vec{A}$ , the dot product is zero and the whole region contributes nothing.

So a good Gaussian surface is one built entirely out of those two kinds of region. That is the design criterion, and it is why the choice of surface is the physics rather than a preliminary to it.

Working the criterion backwards gives you the surface for each symmetry:

Symmetry of the charge	What the symmetry forces about $\vec{E}$	Surface that satisfies 8.6.A.4
Spherical	radial, and the same magnitude at every point at the same radius	a concentric sphere: field perpendicular everywhere, magnitude constant
Cylindrical, infinitely long	radially outward from the axis, magnitude depends only on distance from the axis	a coaxial cylinder: field perpendicular on the curved wall, parallel to the two flat ends so they contribute nothing
Planar, infinite sheet or slab	perpendicular to the plane, magnitude depends only on distance from it	a pillbox or box straddling the plane: field perpendicular on the two end faces, parallel to the sides

The symmetry argument comes before the surface, and both come before any algebra. You are not choosing a sphere because the charge is a sphere; you are choosing a sphere because the symmetry of the charge forces the field to be radial with constant magnitude at constant radius, and a concentric sphere is the surface on which that fact is useful.

The Gauss's law guide drills the five-line routine on standard cases. This page covers what the CED requires and where the method stops.

Where the symmetry limit really bites

The boundary statement allows point charges and spherical, cylindrical or planar symmetry, quantitatively. Three practical consequences.

Gauss's law is always true and often useless. Draw a closed surface around a charged dumbbell and the law holds exactly: the net flux is still $q_{\text{enc}}/\varepsilon_0$ . But $E$ varies in magnitude and direction over every surface you can draw, so you cannot pull it out of the integral and you learn nothing about the field. The law being true and the law being useful are different questions, and only symmetry connects them.

When symmetry fails, integrate instead. Topic 8.4 has its own boundary statement listing rings, arcs and finite lines, and none of those has the symmetry Gauss's law needs. The two boundary statements are complementary rather than overlapping: between them they cover the field problems the course expects you to solve, and each names the geometries the other cannot handle.

"Infinitely long" and "infinite sheet" are the actual conditions, not stylistic flourishes. Cylindrical symmetry requires that the distribution look the same from every point along the axis, which a finite rod does not. Planar symmetry requires the same in two directions. When a problem says "very long" or "large plate, near the middle", it is licensing the idealisation. Without it there is no symmetry and no Gauss's law solution.

What does not need symmetry. Three results in this topic hold for any closed surface at all:

The net flux depends only on the enclosed charge (8.6.A.1).
Charge outside contributes exactly zero net flux, because every field line from an external source that enters must leave.
The flux is independent of the size of the surface, as long as the enclosed charge is unchanged (8.6.A.3).

Those three are what get tested qualitatively, and they are the part of the topic the boundary statement does not restrict.

Size independence, and what 8.6.A.3 does not say

Statement 8.6.A.3 says the total electric flux through a Gaussian surface is independent of the size of the surface if the amount of enclosed charge remains constant. Both halves matter.

Grow a sphere around a fixed point charge and two things change in a way that exactly cancels: the field at the surface falls as $1/r^2$ , and the area grows as $r^2$ . Their product, the flux, is constant. That cancellation is not a coincidence; it is the inverse-square law and three-dimensional space agreeing, and it is why Gauss's law takes the form it does.

What the statement does not say:

It does not say the field is independent of the size of the surface. The field at radius $2r$ is a quarter of the field at $r$ . Only the flux is fixed.
It does not apply when growing the surface swallows more charge. Inside a uniformly charged ball, a bigger Gaussian sphere encloses more charge, so the flux grows and so does the field. The condition in 8.6.A.3 is exactly what excludes that case.

The consequence worth carrying into an exam. For any distribution, sketching $E$ against $r$ is really two questions: how does $q_{\text{enc}}(r)$ behave, and how does the Gaussian area behave. Outside all the charge, $q_{\text{enc}}$ is constant and the field falls as $1/r^2$ for spherical symmetry, as $1/r$ for cylindrical, and not at all for planar. Inside, $q_{\text{enc}}$ grows and the competition can go either way.

Getting q enclosed when the density is a function

Statement 8.6.A.5 is where the calculus lives on the right-hand side of the law. If a charge density is given as a function, the total charge follows from an integral over length, area or volume, with the printed example

Q_{\text{total}} = \int \rho(\vec{r})\,dV

Three cases, and the volume element is the part that gets dropped.

Uniform density, spherical. $q_{\text{enc}} = \rho \cdot \frac{4}{3}\pi r^3$ for a Gaussian sphere of radius $r$ inside the distribution. Equivalently, $q_{\text{enc}} = Q(r/R)^3$ : the fraction of the charge enclosed is the fraction of the volume enclosed. That ratio form is faster and less error-prone when a total charge is given rather than a density.

Uniform density, cylindrical. $q_{\text{enc}} = \rho \cdot \pi r^2 \ell$ for a coaxial cylinder of radius $r$ and length $\ell$ inside the distribution, using $V = \pi r^2 \ell$ from the sheet's geometry table. For a line charge, $q_{\text{enc}} = \lambda \ell$ and the arbitrary length always cancels.

Density varying with radius, spherical. Now the shell element matters. A thin spherical shell at radius $s$ of thickness $ds$ has volume $dV = 4\pi s^2\,ds$ , so

q_{\text{enc}}(r) = \int_0^{r} \rho(s)\, 4\pi s^2\, ds

The $4\pi s^2$ is the surface area of a sphere, printed as $S = 4\pi r^2$ in the Geometry and Trigonometry table of the appendix. Forgetting it, and integrating $\rho(s)\,ds$ alone, is the standard error, and a units check catches it: charge per volume times length is not a charge.

A note on notation. The CED writes the density as $\rho(\vec{r})$ with a vector argument, allowing a density that depends on direction as well as distance. The equation sheet prints the same line as $\rho(r)$ . Every case the boundary statement permits has a density depending on distance alone, since a direction-dependent density would break the symmetry the boundary statement requires.

And a warning about the letter. The sheet's symbol list defines $\rho$ as "resistivity or charge density". Both appear in this course. Context decides, and it costs one clause to say which you mean.

Gauss's law as Maxwell's first equation

Statement 8.6.A.6 is the only place in Unit 8 where Maxwell's equations are named: they are the collection of equations that fully describe electromagnetism, and Gauss's law is the first of them.

The CED numbers all four, one at a time, in four different topics, and all four are printed on the equation sheet even though the sheet never groups or labels them as a set:

Maxwell equation	As printed on the sheet	Where the CED numbers it
First: Gauss's law	$\oint \vec{E} \cdot d\vec{A} = \dfrac{q_{\text{enc}}}{\varepsilon_0}$	8.6.A.6, here
Second: Gauss's law for magnetism	$\oint \vec{B} \cdot d\vec{A} = 0$	12.1.A.3.i, in Unit 12
Third: Faraday's law of induction	$\mathcal{E} = \oint \vec{E} \cdot d\vec{\ell} = -\dfrac{d\Phi_B}{dt}$	13.2.A.3, in Topic 13.2
Fourth: Ampere's law with Maxwell's addition	$\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$	12.4.A.4, in Unit 12

Two structural points worth noticing early, because they make the later units easier.

The first two rows are the same statement asked of two fields, and they get different answers. Both integrate a field over a closed surface. The electric one equals the enclosed charge over $\varepsilon_0$ , because isolated electric charges exist. The magnetic one equals zero, always, because isolated magnetic poles do not. Every field line of $\vec{B}$ closes on itself, so every line entering a closed surface leaves it.

The last two rows are the same shape as each other, but around a loop rather than over a surface. Ampere's law is to the magnetic field what Gauss's law is to the electric one in the sense of relating a field integral to its source, and the parallel between choosing a Gaussian surface by symmetry and choosing an Amperian loop by symmetry is exact. Learning the surface-selection habit here pays for itself in Unit 12.

One honest limitation, and the CED states it twice. The version printed on the sheet is Ampere's law without Maxwell's addition, and the Topic 12.4 boundary statement says the course does not expect students to use Maxwell's fourth equation with a changing electric field, while adding that students should understand that a changing electric field generates a magnetic field. The Topic 13.2 boundary statement then says the course does not expect students to mathematically derive the speed of light in free space from Maxwell's equations. Statement 8.6.A.6 names the set; nothing asks you to work with it as a set.

How Topic 8.6 is tested

Of the fifteen sample multiple-choice questions the CED prints for this course, four align to Unit 8, and one of those is Topic 8.6: sample Question 9, aligned to 8.6.A and essential knowledge 8.6.A.1, at skill 2.A.

The item is built to test exactly the thing this topic is about. A sphere of uniform charge density has net charge $+Q$ and radius $R_S$ . A small particle of charge $+q$ is held at the centre of the sphere, and the question asks for the magnitude of the net field at $R_S/2$ from the centre. The answer key gives choice A:

E = \frac{1}{4\pi\varepsilon_0 \left(R_S/2\right)^2}\left(q + \frac{Q}{8}\right)

Everything hard about the topic is in that expression. The point charge is fully enclosed, so it contributes $q$ . The uniform sphere contributes only the fraction inside the Gaussian sphere, which is the volume fraction $(1/2)^3 = 1/8$ . And the two sources are combined by adding charges inside a single application of the law, not by computing two fields and adding them, although either route gives the same answer.

Its partner, sample Question 8, uses the same figure to ask about the change in electric potential energy and aligns to learning objective 9.3.A in Unit 9, which is a reminder that Unit 8 setups get reused for later units.

Patterns worth rehearsing:

Derive $E(r)$ in every region of a layered spherically symmetric arrangement (skills 2.A, 2.B). Inside a conductor the field is zero, and 8.3.B.1 tells you so before you draw anything.
Find $q_{\text{enc}}$ from a density function by integrating with the right volume element (8.6.A.5).
Rank or compare fluxes through different closed surfaces around the same charges (8.6.A.1, 8.6.A.3).
Predict a factor of change when the radius, charge or enclosed fraction changes (skill 2.D). Doubling the radius of a Gaussian sphere outside all the charge leaves the flux alone and quarters the field.
Find induced surface charges on a conductor from the requirement that the field inside the metal is zero, which forces $q_{\text{enc}} = 0$ for any Gaussian surface drawn inside it. The third worked example.
Sketch $E$ against $r$ across the regions of a layered distribution.

On the multiple-choice section, skill 2.A carries 25 to 30 percent and skill 2.B carries 20 to 25 percent, the two largest, and both are listed for this topic. Unit 8 itself is weighted 15 to 25 percent of that section, tied with Unit 11 for the largest of the six units.

Where Topic 8.6 goes wrong

Using charge outside the surface in $q_{\text{enc}}$ . Only enclosed charge appears. External charge changes the field at points on the surface and contributes zero to the net flux.

Concluding that zero flux means zero field. A closed surface with no charge inside has zero net flux, and the field on it can be large everywhere. The two statements are about different quantities.

Forgetting the volume element in a density integral. $dV = 4\pi s^2\,ds$ for spherical shells, $dV = 2\pi s\,\ell\,ds$ for cylindrical shells. Integrating $\rho\,ds$ gives the wrong units.

Using the total charge instead of the enclosed charge inside a distribution. At $r < R$ inside a uniformly charged ball, $q_{\text{enc}} = Q(r/R)^3$ , not $Q$ .

Reaching for Gauss's law when there is no symmetry. A finite rod, a ring, a dipole. The law is true and gives you nothing. Integrate instead, under Topic 8.4.

Forgetting that a conductor's interior forces $q_{\text{enc}} = 0$ . Any Gaussian surface drawn entirely within conducting material has zero field on it, so zero flux, so zero enclosed charge. That is how you find induced charge on the inner wall of a cavity, and it is the single most productive use of the law in Unit 10.

Dropping the end caps of a cylinder without saying why. They contribute zero because the field is parallel to them, which is statement 8.6.A.4 in action. Say it; on free response it is often a scored line.

A point charge at the centre of a uniformly charged ball

A solid nonconducting sphere of radius $R_S = 0.12 \ \mathrm{m}$ carries a total charge $+6.0 \ \mathrm{nC}$ spread uniformly through its volume. A small particle of charge $+2.0 \ \mathrm{nC}$ is held at the centre. Find the magnitude of the net electric field at (a) $r = 0.060 \ \mathrm{m}$ and (b) $r = 0.24 \ \mathrm{m}$ from the centre. This is the CED's sample multiple-choice Question 9 with numbers attached.

Check the symmetry first, because the boundary statement only licenses a quantitative Gauss's law treatment when it holds. A uniform ball is spherically symmetric, and a point charge at its centre preserves that symmetry exactly. So the field is radial and has the same magnitude at every point at a given radius.
That fact chooses the surface. Take a concentric sphere of radius $r$ . On it, the field is perpendicular to the surface everywhere (parallel to $d\vec{A}$ ) and constant in magnitude, so $\oint \vec{E} \cdot d\vec{A} = E \oint dA = E(4\pi r^2)$ , using $S = 4\pi r^2$ from the sheet's geometry table. That is statement 8.6.A.4 doing its job.
Gauss's law then reads $E(4\pi r^2) = q_{\text{enc}}/\varepsilon_0$ , so $E = \dfrac{q_{\text{enc}}}{4\pi\varepsilon_0 r^2} = \dfrac{k\,q_{\text{enc}}}{r^2}$ . Every part of both answers is now just a question of what $q_{\text{enc}}$ is.
(a) At $r = 0.060 \ \mathrm{m} = R_S/2$ , the Gaussian sphere is inside the ball. It encloses the whole point charge, $+2.0 \ \mathrm{nC}$ , plus the part of the ball inside radius $r$ . Because the density is uniform, the enclosed fraction is the volume fraction: $(r/R_S)^3 = (1/2)^3 = 1/8$ .
So the ball contributes $(6.0 \ \mathrm{nC})/8 = 0.75 \ \mathrm{nC}$ , and $q_{\text{enc}} = 2.0 + 0.75 = 2.75 \ \mathrm{nC}$ . This is the CED's answer choice A written out: $q$ plus $Q/8$ .
$E = \dfrac{(9.0 \times 10^9)(2.75 \times 10^{-9})}{(0.060)^2} = \dfrac{24.75}{3.6 \times 10^{-3}} = 6875 \ \mathrm{N/C}$ , so $6.9 \times 10^3 \ \mathrm{N/C}$ , radially outward.
(b) At $r = 0.24 \ \mathrm{m} = 2R_S$ , the Gaussian sphere is outside everything, so $q_{\text{enc}} = 2.0 + 6.0 = 8.0 \ \mathrm{nC}$ , the full total.
$E = \dfrac{(9.0 \times 10^9)(8.0 \times 10^{-9})}{(0.24)^2} = \dfrac{72}{5.76 \times 10^{-2}} = 1250 \ \mathrm{N/C}$ , so $1.3 \times 10^3 \ \mathrm{N/C}$ .
Two checks. From outside, the whole arrangement behaves as a point charge of $8.0 \ \mathrm{nC}$ at the centre, which is statement 8.3.B.1.ii arrived at independently. And the outside answer would be unchanged if the $6.0 \ \mathrm{nC}$ were redistributed into a shell, a hollow ball or a thin surface layer, as long as the symmetry survived.
A consistency note on the two printed constants. Using $1/(4\pi\varepsilon_0)$ with the sheet's $\varepsilon_0 = 8.85 \times 10^{-12}$ instead of $k = 9.0 \times 10^9$ gives $6869 \ \mathrm{N/C}$ for part (a) rather than $6875$ , a difference of 0.1 percent that vanishes at two significant figures. The CED's own scoring notes accept either.

(a) $E = 6.9 \times 10^3 \ \mathrm{N/C}$ , from an enclosed charge of $2.75 \ \mathrm{nC}$ : the whole point charge plus one eighth of the ball, because half the radius encloses one eighth of the volume. (b) $E = 1.3 \times 10^3 \ \mathrm{N/C}$ , from the full $8.0 \ \mathrm{nC}$ . From outside, the arrangement is indistinguishable from a single point charge at the centre.

Planar symmetry: inside and outside a charged slab

An infinite nonconducting slab of thickness $0.040 \ \mathrm{m}$ carries a uniform volume charge density $\rho = 2.0 \times 10^{-6} \ \mathrm{C/m^3}$ . Take the mid-plane of the slab as $x = 0$ , so the slab occupies $-d \le x \le d$ with $d = 0.020 \ \mathrm{m}$ . Find the field magnitude (a) at $x = 0.010 \ \mathrm{m}$ , inside the slab, and (b) anywhere outside it. (c) Check the outside result against the field of an infinite sheet.

Symmetry first. The slab looks the same from every point in the $y$ and $z$ directions and is symmetric about $x = 0$ , so the field must point along $\pm x$ , away from the mid-plane for positive charge, with a magnitude that depends only on $|x|$ . Planar symmetry, which the boundary statement allows.
Choose a pillbox: a box of cross-sectional area $A$ centred on the mid-plane, with its two flat faces at $x = -a$ and $x = +a$ , perpendicular to the $x$ -axis. On the flat faces the field is perpendicular to the surface; on the four sides it is parallel to the surface and contributes nothing. Statement 8.6.A.4 again.
The flux is therefore $\oint \vec{E} \cdot d\vec{A} = EA + EA = 2EA$ . Both faces contribute positively, because the field points outward through each, away from the mid-plane on both sides.
(a) Inside, take $a = 0.010 \ \mathrm{m} < d$ . The pillbox encloses the slab material between $-a$ and $+a$ , a volume $2aA$ , so $q_{\text{enc}} = \rho(2aA)$ .
Gauss's law: $2EA = \dfrac{\rho(2aA)}{\varepsilon_0}$ . The area $A$ cancels, which it must, since $A$ was never specified. $E = \dfrac{\rho a}{\varepsilon_0}$ .
Numbers: $E = \dfrac{(2.0 \times 10^{-6})(0.010)}{8.85 \times 10^{-12}} = \dfrac{2.0 \times 10^{-8}}{8.85 \times 10^{-12}} = 2260 \ \mathrm{N/C}$ , so $2.3 \times 10^3 \ \mathrm{N/C}$ , pointing away from the mid-plane.
Note the functional dependence, which is skill 2.D: inside the slab the field grows in direct proportion to the distance from the mid-plane, and it is exactly zero at the centre by symmetry.
(b) Outside, take $a > d$ . Now growing the pillbox stops adding charge, because there is no more slab. The enclosed volume is $2dA$ and $q_{\text{enc}} = \rho(2dA)$ , independent of $a$ .
$2EA = \dfrac{\rho(2dA)}{\varepsilon_0}$ , so $E = \dfrac{\rho d}{\varepsilon_0} = \dfrac{(2.0 \times 10^{-6})(0.020)}{8.85 \times 10^{-12}} = \dfrac{4.0 \times 10^{-8}}{8.85 \times 10^{-12}} = 4520 \ \mathrm{N/C}$ , so $4.5 \times 10^3 \ \mathrm{N/C}$ .
The field outside does not depend on how far out you go. That is the planar signature: spherical symmetry gives $1/r^2$ , cylindrical gives $1/r$ , planar gives a constant. Growing the pillbox adds area and adds enclosed charge in exactly the same proportion, so 8.6.A.3 does not apply here, and yet the field still comes out constant.
(c) Collapse the slab to a sheet. Its charge per unit area is $\sigma = \rho(2d) = (2.0 \times 10^{-6})(0.040) = 8.0 \times 10^{-8} \ \mathrm{C/m^2}$ . The standard infinite-sheet result is $E = \sigma/(2\varepsilon_0) = (8.0 \times 10^{-8})/(2 \times 8.85 \times 10^{-12}) = 4520 \ \mathrm{N/C}$ . Identical, as it must be: from outside, a slab of finite thickness is indistinguishable from a sheet carrying the same charge per unit area.
The sketch. $E$ against $x$ rises linearly from zero at the mid-plane to $4.5 \times 10^3 \ \mathrm{N/C}$ at each face, then stays flat forever. The curve is continuous, with a kink at each face, and it is antisymmetric in direction about the centre.

(a) $E = \rho a/\varepsilon_0 = 2.3 \times 10^3 \ \mathrm{N/C}$ at $0.010 \ \mathrm{m}$ from the mid-plane, growing linearly with distance from the centre. (b) $E = \rho d/\varepsilon_0 = 4.5 \times 10^3 \ \mathrm{N/C}$ everywhere outside, independent of distance. (c) The same $4.5 \times 10^3 \ \mathrm{N/C}$ comes from the infinite-sheet result with $\sigma = 2\rho d = 8.0 \times 10^{-8} \ \mathrm{C/m^2}$ .

A coaxial cable, and the charge the conductor is forced to hold

A very long straight wire carrying linear charge density $\lambda = +3.0 \times 10^{-8} \ \mathrm{C/m}$ runs along the axis of a long conducting cylindrical shell of inner radius $a = 0.010 \ \mathrm{m}$ and outer radius $b = 0.014 \ \mathrm{m}$ . The shell itself carries a net linear charge density of $-2\lambda$ . The whole arrangement is in electrostatic equilibrium. Find the field magnitude and direction at (a) $r = 0.0050 \ \mathrm{m}$ , (b) $r = 0.012 \ \mathrm{m}$ and (c) $r = 0.020 \ \mathrm{m}$ , and (d) find the linear charge density induced on each surface of the shell.

Symmetry. Everything is infinitely long and rotationally symmetric about the axis, so the field is radial and depends only on $r$ : cylindrical symmetry, which the boundary statement allows.
The surface follows. Take a coaxial cylinder of radius $r$ and length $\ell$ . On the curved wall the field is perpendicular to the surface and constant in magnitude, giving $E(2\pi r\ell)$ from the sheet's geometry table entry $S = 2\pi r\ell + 2\pi r^2$ , where the $2\pi r\ell$ is the curved part. On the two flat ends the field is parallel to the surface, so they contribute exactly zero. Say that out loud; it is a scored line.
So the general form is $E(2\pi r \ell) = q_{\text{enc}}/\varepsilon_0$ , that is $E = \dfrac{q_{\text{enc}}/\ell}{2\pi\varepsilon_0 r}$ . Only the enclosed charge per unit length changes between the three regions, and $\ell$ always cancels.
(a) At $r = 0.0050 \ \mathrm{m}$ , inside the cavity, the surface encloses only the central wire: $q_{\text{enc}}/\ell = +\lambda$ . Using $1/(2\pi\varepsilon_0) = 2k$ : $E = \dfrac{2k\lambda}{r} = \dfrac{2(9.0 \times 10^9)(3.0 \times 10^{-8})}{0.0050} = \dfrac{540}{0.0050} = 1.08 \times 10^5 \ \mathrm{N/C}$ , radially outward.
(b) At $r = 0.012 \ \mathrm{m}$ you are inside the metal of the shell, between $a$ and $b$ . The shell is a conductor in electrostatic equilibrium, so by statement 8.3.B.1 the field there is zero. $E = 0$ , with no calculation.
(d, first half) That zero is what determines the induced charge. Draw a Gaussian cylinder of radius $r$ with $a < r < b$ , entirely inside the metal. The field is zero on it, so the flux is zero, so by Gauss's law the enclosed charge is zero. The surface encloses the wire's $+\lambda$ per unit length plus whatever sits on the shell's inner wall, so the inner wall must carry $-\lambda$ per unit length, that is $-3.0 \times 10^{-8} \ \mathrm{C/m}$ .
(d, second half) The shell's total is $-2\lambda$ per unit length, and $-\lambda$ of that is now committed to the inner wall. So the outer wall carries $-2\lambda - (-\lambda) = -\lambda$ per unit length, again $-3.0 \times 10^{-8} \ \mathrm{C/m}$ . Charge conservation on the shell, and nothing else.
If a surface charge density is wanted rather than a linear one, divide by the circumference: inner wall $\sigma_a = -\lambda/(2\pi a) = -(3.0 \times 10^{-8})/(2\pi \times 0.010) = -4.8 \times 10^{-7} \ \mathrm{C/m^2}$ ; outer wall $\sigma_b = -\lambda/(2\pi b) = -(3.0 \times 10^{-8})/(2\pi \times 0.014) = -3.4 \times 10^{-7} \ \mathrm{C/m^2}$ .
(c) At $r = 0.020 \ \mathrm{m}$ , outside everything, the enclosed charge per unit length is $+\lambda - 2\lambda = -\lambda$ . So $E = \dfrac{2k\lambda}{r} = \dfrac{540}{0.020} = 2.7 \times 10^4 \ \mathrm{N/C}$ , and the sign of $q_{\text{enc}}$ makes it point radially inward.
Check the pattern before finishing. Between $0.0050$ and $0.020 \ \mathrm{m}$ the radius grew by a factor of four and the enclosed charge per unit length kept the same magnitude, so the field magnitude fell by a factor of four: $1.08 \times 10^5$ to $2.7 \times 10^4$ . That is the $1/r$ fall of cylindrical symmetry, not $1/r^2$ , and it is the fastest way to catch a slip.
One more check on the direction. Far outside, the whole cable looks like a line of net density $-\lambda$ , and the field of a negative line points toward it. Agrees.

(a) $1.1 \times 10^5 \ \mathrm{N/C}$ radially outward. (b) zero, because that point is inside conducting material in electrostatic equilibrium. (c) $2.7 \times 10^4 \ \mathrm{N/C}$ radially inward, since the surface now encloses a net $-\lambda$ per unit length. (d) The inner wall carries $-3.0 \times 10^{-8} \ \mathrm{C/m}$ , forced by the requirement that a Gaussian surface inside the metal enclose zero charge, and the outer wall carries the remaining $-3.0 \times 10^{-8} \ \mathrm{C/m}$ .

Frequently asked questions

What is Gauss's law in AP Physics C?

Gauss's law relates the electric flux through a closed surface to the charge enclosed by that surface. Essential knowledge 8.6.A.1 states it that way and the equation sheet prints the closed-surface integral of E dot dA equal to the enclosed charge divided by the permittivity of free space. The closed surface is called a Gaussian surface, defined in 8.6.A.2 as a three-dimensional closed surface, and it is imaginary: nothing physical has to be there. The law holds for any closed surface around any arrangement of charge, but it only lets you solve for the field when the charge has enough symmetry.

How do you choose a Gaussian surface?

Let the symmetry of the charge choose it for you. Essential knowledge 8.6.A.4 says Gaussian surfaces are typically constructed so that the field is either perpendicular or parallel to different regions of the surface, which is what simplifies the integral. Where the field is perpendicular to the surface and constant in magnitude, the integral becomes the field times an area; where the field is parallel to the surface, that region contributes nothing. In practice that means a concentric sphere for spherical symmetry, a coaxial cylinder for cylindrical symmetry, and a pillbox straddling the plane for planar symmetry.

When can you not use Gauss's law?

You can always use it, and you can only solve for a field with it when the charge distribution has spherical, cylindrical or planar symmetry, or is a point charge. That is exactly what the Topic 8.6 boundary statement says: AP Physics C: Electricity and Magnetism only expects students to quantitatively apply Gauss's law to point charges and charge distributions that have spherical, cylindrical, or planar symmetry. For a finite rod, a ring, an arc or a dipole, the law is still true but the field varies over any surface you draw, so it cannot be pulled out of the integral. Those cases are handled by direct integration in Topic 8.4.

Is Gauss's law on AP Physics 2?

No. The word Gauss does not appear anywhere in the AP Physics 2 course and exam description, and neither does the phrase electric flux. Gauss's law is Topic 8.6 of AP Physics C: Electricity and Magnetism, and electric flux is Topic 8.5, and both exist only in the calculus-based course. AP Physics 2 covers electric charge, force, field, potential energy and potential in its Unit 10, and it does use the word flux nineteen times, but always for magnetic flux in its Unit 12 treatment of electromagnetic induction.

Does the size of the Gaussian surface change the flux?

Not if the enclosed charge stays the same. Essential knowledge 8.6.A.3 states that 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. Growing a sphere around a fixed point charge weakens the field as one over the radius squared while growing the area as the radius squared, and the two effects cancel exactly. The condition matters, though: expanding a Gaussian sphere inside a uniformly charged ball encloses more charge, so both the flux and the field increase.

How do you find the enclosed charge when the charge density is a function?

Integrate the density over the region inside the Gaussian surface. Essential knowledge 8.6.A.5 says the total charge can be determined by integrating the charge density over the length, area or volume of the distribution, and prints the volume version as Q total equals the integral of rho dV. The step that gets dropped is the volume element. For spherical symmetry a thin shell at radius s of thickness ds has volume four pi s squared ds, so you integrate rho of s times four pi s squared with respect to s. A units check catches the omission, since charge per volume times a length is not a charge.

What is the difference between Topic 8.5 and Topic 8.6?

Topic 8.5 defines electric flux and asks you to compute it through an arbitrary surface, open or closed, with no charge necessarily involved. Topic 8.6 closes the surface and states that the net flux out of it equals the enclosed charge divided by epsilon zero, which turns a flux calculation into a statement about charge, and in symmetric cases into a way of finding the field. A useful marker: if a question gives you a field and asks what passes through a surface, that is 8.5. If it gives you charge and asks for a field, or gives you a flux and asks for charge, that is 8.6.