AP Physics C: E&M · Unit 8 of 6

Unit 8: Electric Charges, Fields, and Gauss's Law

15-25% of the multiple-choice section6 topics

Topics in this unit

  1. 8.1Electric Charge and Electric Force
  2. 8.2Conservation of Electric Charge and the Process of Charging
  3. 8.3Electric Fields
  4. 8.4Electric Fields of Charge Distributions
  5. 8.5Electric Flux
  6. 8.6Gauss's Law

Electric Charges, Fields, and Gauss's Law is Unit 8 of AP Physics C: Electricity and Magnetism, worth 15 to 25 percent of the multiple-choice section over about 12 to 24 class periods. Six topics, nine learning objectives, and one new tool: the integral that replaces a sum over point charges.

AP Physics: Unit 8 (topics 8.1 Electric Charge and Electric Force, 8.2 Conservation of Electric Charge and the Process of Charging, 8.3 Electric Fields, 8.4 Electric Fields of Charge Distributions, 8.5 Electric Flux, 8.6 Gauss's Law). Unit 8 of the current AP Physics C: Electricity and Magnetism course and exam description, weighted 15 to 25% of the multiple-choice section at about 12 to 24 class periods, tied with Unit 11 for the largest in the course. Six topics carry nine learning objectives: 8.1.A, 8.1.B and 8.1.C in Topic 8.1, 8.2.A, 8.3.A and 8.3.B, 8.4.A, 8.5.A, and 8.6.A. All nine use the task verb describe. The unit prints three boundary statements. Under Topic 8.1: the course only expects calculations of the electric force between four or fewer interacting charged objects or systems, the analysis of the resulting electric force from more charges is allowed in situations of high symmetry, and students are expected to calculate the electric fields of charge distributions as described in Topics 8.4 and 8.6. Under Topic 8.4: calculus is expected for the electric field of an infinitely long uniformly charged wire or cylinder at a distance from its central axis, a thin ring of charge at a location along the axis of the ring, a semicircular arc or part of a semicircular arc at its center, and a finite wire or line charge at a point collinear with the line charge or at a location along its perpendicular bisector. Under Topic 8.6: Gauss's law is quantitatively applied only to point charges and to charge distributions with spherical, cylindrical, or planar symmetry. Topics 8.2, 8.3 and 8.5 print no boundary statement. Unit 8 contains no equations carrying the Derived Equation label. Six of its eight equations are printed on the equation sheet; the plain dot-product flux and the flux form of Gauss's law are not. Suggested skills by topic: 8.1 uses 1.A, 2.B, 2.D, 3.B; 8.2 uses 1.A, 2.C, 3.B, 3.C; 8.3 uses 1.B, 2.A, 2.D, 3.A, 3.B; 8.4 uses 1.C, 2.A, 2.C, 3.C; 8.5 uses 1.A, 2.A, 2.C, 3.B; 8.6 uses 1.A, 2.A, 2.B, 2.D. The unit opener names skills 1.A, 1.C and 2.A as the ones Unit 8 builds.

What the CED requires across Unit 8

Unit 8 of AP Physics C: Electricity and Magnetism is Electric Charges, Fields, and Gauss's Law. The course and exam description weights it at 15 to 25% of the multiple-choice section and suggests about 12 to 24 class periods. Unit 11 (Electric Circuits) carries the same 15 to 25% band and the same 12 to 24 periods. Units 9, 12 and 13 sit at 10 to 20%, and Unit 10 is the lightest at 10 to 15%. So Unit 8 and Unit 11 are the two heavyweight units of the course, and the sequence starts with one of them.

Six topics and nine learning objectives. Topic 8.1 carries three of the nine on its own, which is why it takes three printed pages of the framework while Topic 8.5 takes one.

TopicLearning objectivesSuggested skillsBoundary statement
8.1 Electric Charge and Electric Force8.1.A, 8.1.B, 8.1.C1.A, 2.B, 2.D, 3.Byes
8.2 Conservation of Electric Charge and the Process of Charging8.2.A1.A, 2.C, 3.B, 3.Cno
8.3 Electric Fields8.3.A, 8.3.B1.B, 2.A, 2.D, 3.A, 3.Bno
8.4 Electric Fields of Charge Distributions8.4.A1.C, 2.A, 2.C, 3.Cyes
8.5 Electric Flux8.5.A1.A, 2.A, 2.C, 3.Bno
8.6 Gauss's Law8.6.A1.A, 2.A, 2.B, 2.Dyes

Every one of those nine objectives opens with the task verb describe. The CED says that verb, used in nearly all learning objectives, "encompasses the range of possible graphical, mathematical, or verbal skill applications", and adds that within those multiple representations students should be able to describe a physical concept graphically, mathematically, and verbally. So describing an electric field can mean drawing a field map, writing the superposition integral, or saying in words what a test charge would feel.

The CED's own framing is that students will begin the study of electric force, which is exerted on all objects with a property called charge, and that the electric force, in contrast to gravitational force, is one of attraction or repulsion and therefore leads to different effects on objects. It says this knowledge will help students understand the role electrostatics plays in devices such as photocopiers, defibrillators, and printers, as well as television, radio, and radar industries.

The essential questions printed on the unit opener are the everyday versions: why hair stands up after brushing it with a plastic comb, how a charged rubber rod bends a stream of water, how a balloon can be made to stick to the wall, why cell phones do not work in concrete buildings, and why a bird can land safely on a high voltage wire. The last two are answered in Unit 10 rather than here, by electrostatic shielding and by a conductor sitting at a single electric potential.

Three boundary statements, quoted with their exception clauses

Boundary statements are how the CED fences off a treatment, and they are the only authority on how far the calculus goes. Unit 8 prints three of them, under Topics 8.1, 8.4 and 8.6. Topics 8.2, 8.3 and 8.5 print none. Here they are whole, because in the first one the sentences after the first reverse what the first one appears to say.

Topic 8.1, on how many charges you can be handed:

"AP Physics C: Electricity & Magnetism only expects students to make calculations of the electric force between four or fewer interacting charged objects or systems. The analysis of the resulting electric force from more charges is allowed in situations of high symmetry. Note that students are expected to calculate the electric fields of charge distributions, as described in Topics 8.4 and 8.6."

Stop after the first sentence and you would conclude that a five-charge arrangement is out of bounds. The second sentence allows it whenever there is high symmetry, and the third says outright that continuous charge distributions, which are infinitely many charges, are expected. The limit is on brute-force vector addition, not on the number of charges in the picture.

Topic 8.4, on which distributions you integrate:

"AP Physics C: Electricity & Magnetism only expects students to use calculus to find the electric field resulting from the following charge distributions and locations: an infinitely long, uniformly charged wire or cylinder at a distance from its central axis, a thin ring of charge at a location along the axis of the ring, a semicircular arc or part of a semicircular arc at its center, and a finite wire or line charge at a point collinear with the line charge or at a location along its perpendicular bisector."

That is a closed list, and every geometry on it comes with a location clause that is part of the boundary rather than decoration. A ring is fair game on its axis, not off it. An arc is fair game at its center, not elsewhere. A finite line is fair game at a point collinear with it or on its perpendicular bisector, not at an arbitrary point in the plane.

Topic 8.6, on Gauss's law:

"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."

Three symmetries, plus the point charge. Notice that the 8.4 and 8.6 lists interlock rather than duplicate. The infinitely long charged wire or cylinder appears on both, so its field is reachable either by integrating or by drawing a Gaussian surface. The ring and the arc appear only under 8.4, because neither has a symmetry that makes a Gaussian surface do any work. The plane appears only under 8.6, because a uniformly charged infinite sheet is one of the cleanest Gauss problems available and one of the nastiest integrals. Statement 8.6.A.4, quoted further down, explains why that list and no other.

How the six topics build

8.1 Electric Charge and Electric Force sets up the objects. Statement 8.1.A.1 makes charge a fundamental property of all matter, 8.1.A.1.i makes it a scalar, 8.1.A.1.ii calls the elementary charge ee the smallest indivisible amount of charge, and 8.1.A.1.iv defines the point-charge model as one in which the physical size of the object is negligible in the context being analyzed. Then 8.1.A.2 gives Coulomb's law and 8.1.A.3 puts its direction along the line of separation between the objects. Objective 8.1.B compares the two long-range forces: 8.1.B.2 says that for any two objects with both mass and charge the gravitational force is usually much smaller in magnitude, and 8.1.B.3 explains why gravity nonetheless dominates at large scales, because large systems tend to be electrically neutral. Objective 8.1.C is permittivity: 8.1.C.1 defines it as a measurement of the degree to which a material or medium is polarized in the presence of an electric field, and 8.1.C.4.ii distinguishes a conductor from an insulator by how easily charge carriers move. That is the seed of Unit 10's dielectric constant.

8.2 Conservation of Electric Charge and the Process of Charging is the only topic in the unit that prints no equation at all. Statement 8.2.A.1.ii defines induced charge separation as what happens when the electrostatic force between two systems alters the distribution of charges within them, and 8.2.A.1.iii adds that it can occur in neutral systems, which is the balloon on the wall. Statement 8.2.A.2.ii is conservation stated plainly: the net charge of a system will be constant unless there is a transfer of charge to or from the system. Statement 8.2.A.3 defines grounding as electrically connecting a charged object to a much larger and approximately neutral system, giving Earth as the example.

8.3 Electric Fields introduces the field as a ratio. Statement 8.3.A.2 defines the electric field at a point as the ratio of the electric force on a test charge at that point to the charge of the test charge, with 8.3.A.2.i requiring the test charge to be small enough that its presence does not significantly affect the field in its vicinity. Objective 8.3.B then does charged conductors and insulators: 8.3.B.1 puts the excess charge of a conductor in electrostatic equilibrium on its surface with zero field inside, 8.3.B.1.ii says the field outside an isolated sphere with a spherically symmetric charge distribution matches that of a point charge of the same net charge at the center, and 8.3.B.2 says an insulator's excess charge is spread through its interior as well as its surface and the field inside it may be nonzero.

8.4 Electric Fields of Charge Distributions is the calculus topic. Statement 8.4.A.1 says expressions for the field of specified charge distributions can be found using integration and the principle of superposition:

E=14πε0dqr2r^\vec{E} = \frac{1}{4\pi\varepsilon_0} \int \frac{dq}{r^2} \hat{r}

Statement 8.4.A.2 adds that symmetry considerations can simplify the analysis. That sentence carries more than it looks: on every geometry in the 8.4 boundary list, symmetry is what kills one or two components of the integral so that only one survives.

8.5 Electric Flux defines the quantity Gauss's law counts. Statement 8.5.A.1 says flux describes the amount of a given quantity that passes through a given area. Statement 8.5.A.2 gives ΦE=EA\Phi_E = \vec{E} \cdot \vec{A} for a field constant across the area, with 8.5.A.2.i fixing the area vector as perpendicular to the surface and outward from a closed surface, and 8.5.A.2.ii pointing out that the sign of the flux is just the sign of that dot product. Statement 8.5.A.3 gives the general case as a surface integral.

8.6 Gauss's Law ties the two together. Statement 8.6.A.1 relates the flux through a Gaussian surface to the charge enclosed by that surface, and 8.6.A.6 names Gauss's law as Maxwell's first equation. The four statements in between are procedure rather than physics, and they are worth reading as a method, which the next section does.

Read in order, the unit is one sentence with six clauses: name the charge, conserve it, turn force into field, integrate the field over a distribution, count the field crossing a surface, then use the count to get the field back.

The Unit 8 equations, and which ones are printed

Eight distinct equations appear in Unit 8's required content. Six of the eight are printed on the AP Physics C: E&M formula sheet, and the two that are missing are the two simple ones.

EquationWhere the CED puts itPrinted on the sheet
FE=14πε0q1q2r2=kq1q2r2\lvert \vec{F}_E \rvert = \dfrac{1}{4\pi\varepsilon_0} \dfrac{\lvert q_1 q_2 \rvert}{r^2} = k \dfrac{\lvert q_1 q_2 \rvert}{r^2}8.1.A.2, relevantyes, both forms
E=FEq\vec{E} = \dfrac{\vec{F}_E}{q}8.3.A.2, relevantyes
E=14πε0dqr2r^\vec{E} = \dfrac{1}{4\pi\varepsilon_0} \int \dfrac{dq}{r^2} \hat{r}8.4.A.1, relevantyes
ΦE=EA\Phi_E = \vec{E} \cdot \vec{A}8.5.A.2, unlabelledno
ΦE=EdA\Phi_E = \int \vec{E} \cdot d\vec{A}8.5.A.3, relevantyes
ΦE=qencε0\Phi_E = \dfrac{q_{\text{enc}}}{\varepsilon_0}8.6.A.1, relevantno
EdA=qencε0\oint \vec{E} \cdot d\vec{A} = \dfrac{q_{\text{enc}}}{\varepsilon_0}8.6.A.1 and 8.6.A.6, relevantyes
Qtotal=ρdVQ_{\text{total}} = \int \rho\, dV8.6.A.5, unlabelledyes

Four things in that table are worth carrying into the exam.

Unit 8 has no Derived Equations. The CED's Required Equations page explains that not all equations in the framework appear on the equation sheet, that many are provided for reference and guidance or to demonstrate the final results of derivations expected of students on the exam, and that those are denoted "Derived Equations". Unit 10 labels three of its equations that way and Unit 13 labels four. Unit 8 labels none: every labelled equation in it reads "Relevant equation". Nothing in this unit is a canned result you are expected to reproduce, because in this unit the setting up of the integral is the work.

Gauss's law is printed in one form and stated in two. Statement 8.6.A.1 gives both ΦE=qenc/ε0\Phi_E = q_{\text{enc}}/\varepsilon_0 and the closed-surface integral as relevant equations. Only the integral form is on the sheet. If a question hands you a flux and asks for the enclosed charge, you are using the first form, and you got there yourself.

The plain dot-product flux is not printed either. The sheet gives you the surface integral and nothing simpler, exactly as it does for magnetic flux in Unit 13. Collapsing EdA\int \vec{E} \cdot d\vec{A} to EAcosθEA\cos\theta for a uniform field is a step you take, and on a free-response question it is a step you justify.

Coulomb's law is printed with kk as well as ε0\varepsilon_0. The sheet gives k=1/(4πε0)=9.0×109 Nm2/C2k = 1/(4\pi\varepsilon_0) = 9.0 \times 10^9\ \mathrm{N \cdot m^2/C^2} and prints ε0=8.85×1012 C2/(Nm2)\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/(N \cdot m^2)} separately. The two are rounded independently, since 1/(4π×8.85×1012)1/(4\pi \times 8.85 \times 10^{-12}) is closer to 8.99×1098.99 \times 10^9, so the same problem worked both ways can differ in the third significant figure. Pick one and stay on it. The scoring guidelines for the CED's own sample Question 1 make this explicit, noting that the answer can be in terms of either ε0\varepsilon_0 or kk.

One resource this site's formula page does not carry but the official Table of Information does: alongside the geometry and trigonometry tables, the appendix prints a calculus box with the power rule, the chain rule, and the standard derivatives and integrals, including xndx\int x^n dx and dx/(x+a)=lnx+a\int dx/(x+a) = \ln\lvert x + a\rvert. For a unit built on integrals, knowing the integral table travels with you is worth as much as knowing any single formula on it.

What calculus changes, and where AP Physics 2 stops

AP Physics C: Electricity and Magnetism is a calculus-based, college-level course, equivalent to the second course in an introductory college sequence in calculus-based physics. Its prerequisites say students should have taken or be concurrently taking calculus. Unit 8 is where that prerequisite starts earning its place, and the difference from the algebra-based treatment is not cosmetic.

  • The field of a distribution is an integral, not a sum. AP Physics 2 adds the fields of a handful of point charges as vectors. This course integrates dqdq over a wire, a ring or an arc, which is the only way to get a field when the charge is spread out rather than lumped.
  • Flux is a surface integral, not a product. A field that varies across the surface, or a surface that curves through a varying field, gives a flux you have to integrate for.
  • Gauss's law does not exist in AP Physics 2. The phrases electric flux, Gaussian surface and Gauss's law appear nowhere in that course's framework, which reaches flux only as magnetic flux, in its Unit 12. Topics 8.5 and 8.6, a third of this unit, have no algebra-based counterpart at all.
  • The charge itself can be a function of position. Statement 8.6.A.5 puts a charge density ρ\rho inside an integral, so "find the total charge" becomes a calculus question before the field question even starts.

The nearest algebra-based page on this site is the AP Physics 2 Unit 10 hub, Electric Force, Field, and Potential. Its Topics 10.1, 10.2 and 10.3 carry the same three titles as this unit's Topics 8.1, 8.2 and 8.3, word for word. Those pages are for AP Physics 2 students; this unit is for AP Physics C students, and they are not the same material at two reading levels. If you are in the algebra-based course you want Topic 10.1, Electric Charge and Electric Force and Topic 10.3, Electric Fields, and you can stop before the integral. If you are in Physics C you want those three titles plus three topics the other course never reaches.

The split also runs the other way. AP Physics 2 packs force, field, potential energy, potential and capacitors into one seven-topic unit. Physics C spreads the same ground over three units: Unit 8 for charge and field, Unit 9 for energy and potential, Unit 10 for conductors and capacitors. If you are coming from Physics 2, the reorganisation is as much of an adjustment as the calculus.

Flux and Gaussian surfaces: the sentences that do the work

Four of Topic 8.6's six essential-knowledge statements are procedure rather than physics. Read them as a method and the unit's hardest problems get shorter.

8.6.A.2: a Gaussian surface is a three-dimensional, closed surface. So a disc is not one, a hemisphere is not one, and a cube face is not one. If a question asks for the flux through an open surface, Gauss's law does not apply to it directly; you either integrate, or you close the surface yourself and subtract what you added.

8.6.A.3: the total flux is independent of the size of the surface if the enclosed charge is unchanged. This is what lets you shrink or grow a sphere around a point charge at will, and it is why a flux answer usually contains no radius at all. It is also the statement most naturally tested as a comparison question, which fits skill 2.C being listed for Topic 8.5.

8.6.A.4: Gaussian surfaces are typically constructed such that the field is either perpendicular or parallel to different regions of the surface, resulting in a simplified surface integral. That is the whole trick in one sentence. Perpendicular regions contribute EdAE \, dA with the cosine equal to one; parallel regions contribute nothing; nothing in between is ever needed. Which is why the boundary statement restricts you to spherical, cylindrical and planar symmetry: those are the three shapes where a closed surface can be built out of only those two kinds of region.

8.6.A.5: a charge density given as a function of position is integrated over length, area or volume. The CED prints the volume case as Qtotal=ρ(r)dVQ_{\text{total}} = \int \rho(\vec{r})\, dV with a vector argument; the equation sheet prints the same line with a plain rr. Either way, the enclosed charge in Gauss's law is itself sometimes an integral, and for a spherically symmetric ρ(r)\rho(r) the volume element you want is a shell, dV=4πr2drdV = 4\pi r^2 dr. The Gauss's law guide walks the routine end to end.

One sentence from Topic 8.3 belongs in this list even though it sits elsewhere. Statement 8.3.B.1 says the field inside a conductor in electrostatic equilibrium is zero. Combine that with Gauss's law and you get the result that carries most of Unit 10: any Gaussian surface drawn wholly inside conducting material encloses zero net charge, because the flux through it is zero.

Traps that span more than one topic

A charge distribution is not a point charge at its center, except when it is. Statement 8.3.B.1.ii grants that licence to one shape only: an isolated sphere with a spherically symmetric charge distribution, viewed from outside. A ring, a rod, a disc or a cube gets no such licence, and treating a semicircular arc of total charge QQ as a point charge at distance RR overstates the field at the center by a factor of π/2\pi/2.

Zero field does not mean zero enclosed charge, and zero enclosed charge does not mean zero field. Gauss's law equates enclosed charge to the flux through the whole closed surface, not to the field at any one point on it. A surface enclosing a dipole has zero net enclosed charge and zero net flux, while the field is nonzero everywhere on it.

Charge outside the surface changes the field on the surface and contributes nothing to the flux. Both halves of that sentence are true at once, and mixing them up is the most common Gauss error. Statement 8.6.A.1 says the flux is set by qencq_{\text{enc}}. It does not say the field is.

The area vector points outward from a closed surface, always. Statement 8.5.A.2.i fixes it, and 8.5.A.2.ii makes the sign of the flux the sign of the dot product. Field entering the surface gives negative flux; field leaving gives positive. On an open surface the CED gives you no convention, so declare the direction you chose before you use it.

Symmetry kills components; it does not kill the integral. On the arc, the ring and the finite line, the boundary statement's location clause is exactly the position where one component cancels. Off that position nothing cancels and the problem leaves the course, so if you find yourself integrating two components at once, check whether you are still at the location the boundary statement named.

Permittivity is a property of the material, not a universal constant. Statement 8.1.C.3 gives free space one constant value ε0\varepsilon_0, and 8.1.C.4 says the permittivity of matter differs from it. Coulomb's law with ε0\varepsilon_0 in it is the free-space law, and that distinction is what makes Unit 10's dielectric constant mean anything.

Charge is a scalar; force and field are vectors. Statement 8.1.A.1.i says so about charge. That is why potentials add arithmetically and fields do not, which is the most productive single idea in Unit 9.

How Unit 8 is assessed

The exam is 3 hours long. Section I is 42 multiple-choice questions in 85 minutes for 50% of the score. Section II is 4 free-response questions in 95 minutes for the other 50%, always one of each type in a fixed order: Mathematical Routines (10 points), Translation Between Representations (12 points), Experimental Design and Analysis (10 points), and Qualitative/Quantitative Translation (8 points). A four-function, scientific, or graphing calculator is allowed on both sections. The course also requires that 25 percent of instructional time be spent on hands-on laboratory work.

The unit's own Preparing for the AP Exam note points at the first free-response question. It says the Mathematical Routines question focuses on assessing students' ability to create and use mathematical models and representations, that students will be required to derive an expression for a physical quantity which may culminate in a numerical calculation, and that while Unit 8 offers content perfect for practicing the MR question, the MR question can pull content from any of the six units of the course. Read the second half of that as carefully as the first: nothing guarantees Unit 8 appears in a given year's free-response section. The reliable figure is the multiple-choice weighting, 15 to 25%. The unit's AP Classroom Progress Check runs about 18 multiple-choice questions and 4 free-response questions, one of each type.

On the science practices, the unit opener names skills 1.A, 1.C and 2.A as the ones Unit 8 develops. Across the exam, skill 2.A, deriving a symbolic expression by following a logical mathematical pathway, carries a 25 to 30% weighting on the multiple-choice section, the largest of any single skill there, and Practice 2 as a whole carries 40 to 45% of the free-response section. Practice 1 is not assessed on the multiple-choice section at all. Four of the six topics list 2.A among their suggested skills; the two that do not are 8.1 and 8.2.

The CED prints a sample exam with an alignment table, and four of its fifteen sample multiple-choice questions align to Unit 8 objectives: 8.1.A on the signs of the work done by an external force and by the electrostatic forces as a test charge moves between two points near two fixed charges, 8.3.A on where along a line the field of two spheres carrying 3Q-3Q and +4Q+4Q is greatest, 8.5.A on the total flux through a cube sitting in a field E=(bx)i^\vec{E} = (bx)\hat{i}, and 8.6.A on the net field at radius RS/2R_S/2 inside a uniformly charged sphere with a point charge at its center. That last one needs enclosed charge scaling as r3r^3, superposition, and no integration at all.

The sample free-response Question 1, the Mathematical Routines question, is worth 10 points and aligns to learning objectives 8.4.A, 9.2.A, 9.2.B and 9.3.A, using skills 2.A, 3.B and 3.C. It puts a thin nonconducting ring of radius RR and charge +Q+Q in a plane with a small sphere of charge Q-Q on the axis, then asks where the net potential can be zero, for a derivation of the sphere's speed from conservation of energy, and for a derivation of the field due to the ring at a point on the axis. Its scoring guidelines award one point for a multistep derivation that starts with an equation for the electric field of a charge distribution, and a separate point for indicating that only the axial component need be considered, for instance by including a cosθ\cos\theta term in the integral. They then print an alternate solution that gets the field by differentiating the potential, and award it the same three points. Method earns points on its own, and more than one method is accepted.

The five optional sample instructional activities sit at the front of the unit, two on Topic 8.1 and one each on 8.2, 8.3 and 8.6. Topics 8.4 and 8.5 get none, which is a fair signal that the integral and the flux are taught with a pencil rather than an experiment.

A semicircular arc of charge, at the one place the CED allows

A thin insulating rod is bent into a semicircle of radius R=0.050R = 0.050 m and carries a uniformly distributed total charge Q=+6.0Q = +6.0 nC. Find (a) the linear charge density, (b) the magnitude and direction of the electric field at the center of the semicircle, and (c) what happens to that field if the rod is bent into a full circle instead.

  1. Declare the geometry first. Put the center of the semicircle at the origin, with the arc occupying the upper half plane from angle 00 to π\pi. The symmetry axis is the yy-axis, so call up positive yy. Hold that for the whole problem.

  2. (a) The arc length of a semicircle of radius RR is πR\pi R, so λ=QπR=6.0×109π(0.050)=3.82×108\lambda = \dfrac{Q}{\pi R} = \dfrac{6.0 \times 10^{-9}}{\pi (0.050)} = 3.82 \times 10^{-8} C/m, which is 38.238.2 nC/m.

  3. (b) Start from 8.4.A.1: E=14πε0dqr2r^\vec{E} = \dfrac{1}{4\pi\varepsilon_0} \int \dfrac{dq}{r^2}\hat{r}. Every element of the arc sits at the same distance RR from the center, so r2=R2r^2 = R^2 is a constant and comes out of the integral. That is what makes the center the permitted location.

  4. Use 8.4.A.2 and let symmetry kill a component. For each element at angle θ\theta there is a mirror element at πθ\pi - \theta whose horizontal contribution is equal and opposite, so Ex=0E_x = 0 and only EyE_y survives. Writing dq=λRdθdq = \lambda R \, d\theta and taking the component along the axis gives a sinθ\sin\theta factor.

  5. Ey=14πε00πλRdθR2sinθ=kλR[cosθ]0π=2kλRE_y = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\int_0^{\pi} \dfrac{\lambda R \, d\theta}{R^2}\sin\theta = \dfrac{k\lambda}{R}\Big[-\cos\theta\Big]_0^{\pi} = \dfrac{2k\lambda}{R}.

  6. Substituting λ=Q/(πR)\lambda = Q/(\pi R) gives the symbolic answer E=2kQπR2E = \dfrac{2kQ}{\pi R^2}. Numerically, πR2=π(0.050)2=7.854×103 m2\pi R^2 = \pi (0.050)^2 = 7.854 \times 10^{-3}\ \mathrm{m^2}, so E=2(9.0×109)(6.0×109)7.854×103=1.375×104E = \dfrac{2(9.0 \times 10^9)(6.0 \times 10^{-9})}{7.854 \times 10^{-3}} = 1.375 \times 10^4 N/C, which is 1.4×1041.4 \times 10^4 N/C at two significant figures.

  7. Direction: the charge is positive, so the field at the center points away from the arc, along the negative yy-axis in the convention declared above.

  8. Compare that with the wrong answer. Treating the whole arc as a point charge at distance RR gives kQ/R2=2.16×104kQ/R^2 = 2.16 \times 10^4 N/C, too large by a factor of π/21.571\pi/2 \approx 1.571. The factor is exactly π/2\pi/2 for any semicircular arc, whatever the numbers, which makes it a good check that you did the integral rather than skipped it.

  9. (c) Bend the rod into a full circle and every element acquires a mirror element directly opposite it. Both components now cancel and E=0E = 0 at the center. The field is zero even though the potential there is not, which is the distinction Unit 9 is built on.

(a) λ=38.2\lambda = 38.2 nC/m. (b) E=2kQπR2=1.4×104E = \dfrac{2kQ}{\pi R^2} = 1.4 \times 10^4 N/C, pointing away from the arc along the symmetry axis. (c) Zero, by symmetry, for a complete ring. Treating the arc as a point charge at distance RR overstates the field by exactly π/2\pi/2.

Gauss's law with a charge density that varies

A solid nonconducting sphere of radius R=0.080R = 0.080 m carries a volume charge density that grows linearly with distance from the center, ρ(r)=ρ0r/R\rho(r) = \rho_0 \, r/R, with ρ0=3.0×105 C/m3\rho_0 = 3.0 \times 10^{-5}\ \mathrm{C/m^3}. Find (a) the total charge on the sphere, (b) the field magnitude at r=R/2r = R/2, and (c) the field magnitude at r=2Rr = 2R. Use ε0=8.85×1012 C2/(Nm2)\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/(N \cdot m^2)} throughout.

  1. The distribution is spherically symmetric, which is one of the three symmetries the Topic 8.6 boundary statement allows, so a spherical Gaussian surface concentric with the sphere is legal and useful.

  2. (a) Statement 8.6.A.5 says to integrate the charge density over the volume. For a spherically symmetric density the volume element is a thin shell, dV=4πr2drdV = 4\pi r^2 dr: Qtotal=0Rρ0rR4πr2dr=4πρ0RR44=πρ0R3Q_{\text{total}} = \displaystyle\int_0^R \rho_0 \frac{r}{R} 4\pi r^2 \, dr = \frac{4\pi\rho_0}{R}\cdot\frac{R^4}{4} = \pi \rho_0 R^3.

  3. Qtotal=π(3.0×105)(0.080)3=4.83×108Q_{\text{total}} = \pi (3.0 \times 10^{-5})(0.080)^3 = 4.83 \times 10^{-8} C, which is 4848 nC.

  4. (b) For a Gaussian sphere of radius r<Rr < R, the same integral run to rr instead of RR gives qenc=πρ0r4/Rq_{\text{enc}} = \pi\rho_0 r^4 / R. Note the fourth power: this is not the r3r^3 scaling of a uniformly charged sphere, and reaching for r3r^3 here is the whole trap.

  5. By 8.6.A.4 the field is radial and constant in magnitude over the surface, so EdA=E(4πr2)\oint \vec{E} \cdot d\vec{A} = E (4\pi r^2) and Gauss's law gives E=qenc4πε0r2=πρ0r4/R4πε0r2=ρ0r24ε0RE = \dfrac{q_{\text{enc}}}{4\pi\varepsilon_0 r^2} = \dfrac{\pi \rho_0 r^4 / R}{4\pi\varepsilon_0 r^2} = \dfrac{\rho_0 r^2}{4\varepsilon_0 R}.

  6. At r=R/2=0.040r = R/2 = 0.040 m: E=(3.0×105)(0.040)24(8.85×1012)(0.080)=4.80×1082.832×1012=1.695×104E = \dfrac{(3.0 \times 10^{-5})(0.040)^2}{4(8.85 \times 10^{-12})(0.080)} = \dfrac{4.80 \times 10^{-8}}{2.832 \times 10^{-12}} = 1.695 \times 10^4 N/C, so 1.7×1041.7 \times 10^4 N/C.

  7. (c) Outside the sphere the whole charge is enclosed, so E=Qtotal4πε0r2E = \dfrac{Q_{\text{total}}}{4\pi\varepsilon_0 r^2}. At r=2R=0.16r = 2R = 0.16 m: E=4.8255×1084π(8.85×1012)(0.16)2=1.695×104E = \dfrac{4.8255 \times 10^{-8}}{4\pi (8.85 \times 10^{-12})(0.16)^2} = 1.695 \times 10^4 N/C, so 1.7×1041.7 \times 10^4 N/C again.

  8. That agreement is exact, not a rounding accident. Symbolically, E(R/2)=ρ0R/(16ε0)E(R/2) = \rho_0 R / (16\varepsilon_0) and E(2R)=πρ0R3/(4πε04R2)=ρ0R/(16ε0)E(2R) = \pi\rho_0 R^3 / (4\pi\varepsilon_0 \cdot 4R^2) = \rho_0 R/(16 \varepsilon_0). For this particular density the field at half the radius equals the field at twice the radius, which is a useful reminder that the inside and outside expressions are two different functions that happen to agree at one pair of points.

  9. Consistency check on the constant: working part (c) with k=9.0×109k = 9.0 \times 10^9 instead of ε0\varepsilon_0 gives 1.696×1041.696 \times 10^4 N/C. The two routes differ in the fourth digit because the sheet's kk and ε0\varepsilon_0 are rounded independently. At two significant figures they agree, which is why declaring which constant you are using matters more than which one you pick.

(a) Qtotal=πρ0R3=48Q_{\text{total}} = \pi\rho_0 R^3 = 48 nC. (b) E=ρ0r2/(4ε0R)=1.7×104E = \rho_0 r^2/(4\varepsilon_0 R) = 1.7 \times 10^4 N/C at r=R/2r = R/2. (c) 1.7×1041.7 \times 10^4 N/C at r=2Rr = 2R, equal to (b) exactly for this density. The enclosed charge grows as r4r^4 here, not r3r^3.

When the charge sits on the boundary of the surface

A point charge q=+4.0q = +4.0 nC is held just outside a cube, at three positions in turn: at one corner of the cube, at the midpoint of one edge, and at the center of one face. In each case find the total electric flux out of that cube. For the corner case, also find the flux through each individual face. The Gauss's law guide handles the charge at the center; this example is about what changes when it is not.

  1. Set up the idea once and all three cases follow. Gauss's law gives the flux out of a closed surface that encloses the charge. A charge sitting exactly on the boundary is not enclosed by one cube, but it is enclosed by a group of identical cubes stacked around it, and by symmetry the total flux q/ε0q/\varepsilon_0 divides equally among them.

  2. First the total: qε0=4.0×1098.85×1012=452 Nm2/C\dfrac{q}{\varepsilon_0} = \dfrac{4.0 \times 10^{-9}}{8.85 \times 10^{-12}} = 452\ \mathrm{N \cdot m^2/C}. The cube's edge length never appears, which is statement 8.6.A.3: the total flux is independent of the size of the surface as long as the enclosed charge is unchanged.

  3. Corner. Eight identical cubes meet at a common vertex and together fill the space around it, so each takes one eighth: ΦE=q8ε0=451.988=56.5 Nm2/C\Phi_E = \dfrac{q}{8\varepsilon_0} = \dfrac{451.98}{8} = 56.5\ \mathrm{N \cdot m^2/C}.

  4. Edge midpoint. Four identical cubes share an edge and together fill the space around that line, so each takes one quarter: ΦE=q4ε0=451.984=113 Nm2/C\Phi_E = \dfrac{q}{4\varepsilon_0} = \dfrac{451.98}{4} = 113\ \mathrm{N \cdot m^2/C}.

  5. Face center. Two cubes share a face, so each takes one half: ΦE=q2ε0=451.982=226 Nm2/C\Phi_E = \dfrac{q}{2\varepsilon_0} = \dfrac{451.98}{2} = 226\ \mathrm{N \cdot m^2/C}.

  6. Now the face-by-face split in the corner case. The three faces that meet at the charge's corner lie in planes containing the charge, so the field along each of them is parallel to the surface and the dot product in EdA\int \vec{E} \cdot d\vec{A} vanishes. Those three faces carry zero flux, which is the parallel-region half of statement 8.6.A.4.

  7. The other three faces are equivalent to one another by symmetry, so they split the whole 56.5 Nm2/C56.5\ \mathrm{N \cdot m^2/C} three ways: 56.497/3=18.8 Nm2/C56.497/3 = 18.8\ \mathrm{N \cdot m^2/C} each.

  8. Check it closes: 3(0)+3(18.832)=56.497 Nm2/C=q/(8ε0)3(0) + 3(18.832) = 56.497\ \mathrm{N \cdot m^2/C} = q/(8\varepsilon_0).

  9. Note which step Gauss's law did and which step it did not. Gauss's law gave every total. Splitting a total between faces was always a symmetry argument, and it only works because the faces being compared really are equivalent. Tilt the cube and the totals are unchanged while the per-face answers become integrals.

452 Nm2/C452\ \mathrm{N \cdot m^2/C} would leave a surface that fully enclosed the charge. On the boundary, a cube receives q/(8ε0)=56.5 Nm2/Cq/(8\varepsilon_0) = 56.5\ \mathrm{N \cdot m^2/C} at a corner, q/(4ε0)=113 Nm2/Cq/(4\varepsilon_0) = 113\ \mathrm{N \cdot m^2/C} at an edge midpoint, and q/(2ε0)=226 Nm2/Cq/(2\varepsilon_0) = 226\ \mathrm{N \cdot m^2/C} at a face center. In the corner case the three faces touching the charge carry zero flux and the other three carry 18.8 Nm2/C18.8\ \mathrm{N \cdot m^2/C} each.

Frequently asked questions

How much of the AP Physics C E&M exam is Unit 8?

Unit 8, Electric Charges, Fields, and Gauss's Law, is weighted at 15 to 25% of the multiple-choice section of the AP Physics C: Electricity and Magnetism exam, and the course description suggests about 12 to 24 class periods for it. Unit 11, Electric Circuits, carries the same 15 to 25% band and the same class-period estimate, and the two are the heaviest units in the course. Units 9, 12 and 13 sit at 10 to 20% each, and Unit 10 is the lightest at 10 to 15%. The multiple-choice section is 42 questions in 85 minutes and counts for half the exam score.

What are the six topics in AP Physics C E&M Unit 8?

They are 8.1 Electric Charge and Electric Force, 8.2 Conservation of Electric Charge and the Process of Charging, 8.3 Electric Fields, 8.4 Electric Fields of Charge Distributions, 8.5 Electric Flux, and 8.6 Gauss's Law. Between them the six carry nine learning objectives, three of which sit in Topic 8.1 alone. The first three titles are shared word for word with AP Physics 2 Topics 10.1, 10.2 and 10.3. The last three are unique to the calculus-based course: the phrases electric flux, Gaussian surface and Gauss's law do not appear anywhere in the AP Physics 2 framework, which meets flux only as magnetic flux in its Unit 12.

Which charge distributions does AP Physics C expect you to integrate?

The Topic 8.4 boundary statement names four, each with a location: an infinitely long, uniformly charged wire or cylinder at a distance from its central axis; a thin ring of charge at a location along the axis of the ring; a semicircular arc or part of a semicircular arc at its center; and a finite wire or line charge at a point collinear with the line charge or at a location along its perpendicular bisector. The location clauses are part of the boundary, because each one is the position where symmetry cancels a component of the field. A ring off its axis, or an arc at a point other than its center, is outside the course.

How far does AP Physics C take Gauss's law?

The Topic 8.6 boundary statement says the course only expects students to quantitatively apply Gauss's law to point charges and charge distributions that have spherical, cylindrical, or planar symmetry. Those are the three shapes where a closed surface can be built so that the electric field is either perpendicular or parallel to every part of it, which is what essential knowledge 8.6.A.4 describes as producing a simplified surface integral. Gauss's law is also named as Maxwell's first equation in statement 8.6.A.6. Qualitative reasoning about flux is not restricted by that boundary statement: a question can ask about the flux through any closed surface, since the total flux depends only on the enclosed charge.

Which Unit 8 equations are on the AP Physics C E&M equation sheet?

Six of the eight equations in Unit 8's required content are printed. The sheet gives Coulomb's law in both the permittivity form and the k form, the electric field as force per unit charge, the superposition integral for the field of a charge distribution, electric flux as a surface integral, Gauss's law as a closed surface integral equal to the enclosed charge over the permittivity of free space, and the total charge as an integral of charge density over volume. Two are missing: the plain dot-product form of flux for a uniform field, and the form of Gauss's law written as flux equals enclosed charge over the permittivity. Unit 8 contains no Derived Equations, the label the course description uses for results students are expected to reproduce.

Does AP Physics C expect calculations with more than four charges?

Yes, under two conditions. The Topic 8.1 boundary statement says the course only expects calculations of the electric force between four or fewer interacting charged objects or systems, but the same boundary statement then allows the analysis of the force from more charges in situations of high symmetry, and adds that students are expected to calculate the electric fields of charge distributions as described in Topics 8.4 and 8.6. A continuous charge distribution is infinitely many charges. The four-charge limit is a limit on unstructured vector addition, not a cap on how much charge can appear in a problem.

What is the difference between AP Physics C Unit 8 and AP Physics 2 Unit 10?

AP Physics 2 Unit 10 covers electric force, field, potential energy, potential and capacitors in seven topics, with the field of a small number of point charges added as vectors. AP Physics C splits that same ground across three units and goes further: Unit 8 for charge and field, Unit 9 for potential energy and potential, Unit 10 for conductors and capacitors. Its Unit 8 replaces the vector sum with the superposition integral, then adds electric flux and Gauss's law, neither of which exists in AP Physics 2 at all. The first three topic titles are identical between the two courses. If you are in the algebra-based course, AP Physics 2 Unit 10 is your page. If you are in the calculus-based course, this unit is.