AP Physics C: E&M · Unit 10 of 6

Unit 10: Conductors and Capacitors

10-15% of the multiple-choice section4 topics

Topics in this unit

  1. 10.1Electrostatics with Conductors
  2. 10.2Redistribution of Charge between Conductors
  3. 10.3Capacitors
  4. 10.4Dielectrics

Conductors and Capacitors is Unit 10 of AP Physics C: Electricity and Magnetism, worth 10 to 15 percent of the multiple-choice section over about 8 to 16 class periods. Four topics, four learning objectives, and three equations you are expected to derive rather than look up.

AP Physics: Unit 10 (topics 10.1 Electrostatics with Conductors, 10.2 Redistribution of Charge between Conductors, 10.3 Capacitors, 10.4 Dielectrics). Unit 10 of the current AP Physics C: Electricity and Magnetism course and exam description, weighted 10 to 15% of the multiple-choice section at about 8 to 16 class periods, the smallest figures of any unit in the course. Four topics with one learning objective each: 10.1.A, 10.2.A, 10.3.A and 10.4.A, all using the task verb describe. Topics 10.1 and 10.2 print no equations. The unit prints exactly one boundary statement, at the end of Topic 10.3: while other shapes are also able to separate charges, the course only expects the quantitative analysis and description of parallel-plate capacitors, concentric spherical capacitors, and coaxial cylindrical capacitors. Topics 10.1, 10.2 and 10.4 print no boundary statement. Three of the unit's seven equations carry the Derived Equation label, which the framework's Required Equations page defines as the final results of derivations expected of students on the exam, and none of those three is printed on the equation sheet: the field between parallel plates E = Q/(epsilon-zero A) at 10.3.A.3.i, the dielectric constant as kappa = E-zero/E for an isolated capacitor at 10.4.A.4, and C = kappa C-zero at 10.4.A.5. The four printed equations are C = Q/deltaV, C = kappa epsilon-zero A/d, U-sub-C = half Q deltaV, and kappa = epsilon/epsilon-zero. The series and parallel capacitance rules and the RC time constant appear on the equation sheet but belong to Unit 11 in the framework, at Topics 11.5 and 11.8. Suggested skills by topic: 10.1 uses 1.A, 2.C, 3.B, 3.C; 10.2 uses 1.A, 2.A, 2.C, 3.B; 10.3 uses 1.C, 2.A, 2.D, 3.B; 10.4 uses 1.B, 2.B, 2.D, 3.A, 3.B. Skill 3.B is listed for all four topics. The unit opener names skills 2.A, 3.B and 3.C as the ones Unit 10 builds, and points at the fourth free-response question, the Qualitative/Quantitative Translation question, worth 8 points. The exam conventions printed with the equation sheet include that capacitors are air-filled with kappa = 1.0.

What the CED requires across Unit 10

Unit 10 of AP Physics C: Electricity and Magnetism is Conductors and Capacitors. The course and exam description weights it at 10 to 15% of the multiple-choice section and suggests about 8 to 16 class periods. Both figures are the lowest in the course: every other unit starts at 10 to 20% and gets at least 10 periods. Units 8 and 11, which sit either side of it, are each weighted 15 to 25% over about 12 to 24 periods.

Four topics, and exactly one learning objective each.

TopicLearning objectiveSuggested skillsEquations
10.1 Electrostatics with Conductors10.1.A1.A, 2.C, 3.B, 3.Cnone
10.2 Redistribution of Charge between Conductors10.2.A1.A, 2.A, 2.C, 3.Bnone
10.3 Capacitors10.3.A1.C, 2.A, 2.D, 3.B4
10.4 Dielectrics10.4.A1.B, 2.B, 2.D, 3.A, 3.B3

Half the unit prints no equations at all. Topics 10.1 and 10.2 are fifteen numbered essential-knowledge statements of pure reasoning about where charge sits and where it goes, twelve in Topic 10.1 and three in Topic 10.2, and they are the half most likely to appear as a justification question rather than a calculation.

All four objectives open with the task verb describe, which the CED says "encompasses the range of possible graphical, mathematical, or verbal skill applications", adding that students should be able to describe a physical concept graphically, mathematically, and verbally.

The CED's framing is that in Unit 8 students investigated why all objects have an electric charge, and in Unit 10 they will examine how that charge can be stored. It says conductors, capacitors, and dielectrics are presented to demonstrate that the ability of charge to move is dependent on the material composition of an object, that each of these is important based on the type of movement or desired object behavior, and that this unit also examines how the behavior of charges is impacted by electric fields. It closes by pointing forward: knowledge of conductors, capacitors, and dielectrics will prepare students for understanding how electric circuits work in Unit 11 and how they behave when one or more electrical element is altered or modified.

The essential questions on the unit opener are the shortest set in the course: why the little red light on the TV stays on for a second after the TV is turned off, how a camera flash used for photography works, and how energy can be stored for later use. All three are the same question, and the answer is a charged capacitor discharging.

One boundary statement, and it names three capacitor shapes

Unit 10 prints exactly one boundary statement. It sits at the end of Topic 10.3, and Topics 10.1, 10.2 and 10.4 print none. Quoted whole, including the opening clause that most summaries drop:

"While other shapes are also able to separate charges, AP Physics C: Electricity & Magnetism only expects the quantitative analysis and description of parallel-plate capacitors, concentric spherical capacitors, and coaxial cylindrical capacitors."

That first clause matters. It is not saying other shapes fail to store charge; it is saying you will not be asked to compute for them. So a question can show you an odd-shaped pair of conductors and ask you to reason about it, and the boundary statement does not rule that out. What it rules out is being asked for its capacitance.

The three shapes named are the three where symmetry lets Gauss's law give the field in closed form, which is not a coincidence. Compare the Topic 8.6 boundary statement, which restricts the quantitative use of Gauss's law to point charges and to distributions with spherical, cylindrical, or planar symmetry. Planar gives the parallel-plate capacitor, spherical gives the concentric spherical capacitor, cylindrical gives the coaxial cable. The two boundary statements are the same restriction seen from two ends of the course.

One more restriction is printed with the equation sheet rather than in the framework, and it changes what a bare capacitance number means. Among the conventions used on the exam unless otherwise stated: "Capacitors are air-filled (κ=1.0)(\kappa = 1.0)." So a capacitor in a problem has no dielectric until the problem says so, and the κ\kappa in the sheet's C=κε0A/dC = \kappa\varepsilon_0 A/d is 1 by default. The same conventions list also fixes that strings, springs, batteries, wires, and meters are ideal, which is why a capacitor problem can quietly assume zero wire resistance.

Notice what Unit 10 does not contain. The equation sheet prints the series and parallel combination rules for capacitors and prints the RCRC time constant, but none of those appears anywhere in this unit's required content. The framework puts equivalent capacitance in Topic 11.5 and RCRC circuits in Topic 11.8, both in the next unit. Unit 10 is about one capacitor.

How the four topics build

10.1 Electrostatics with Conductors is where the charge goes. Statement 10.1.A.1 defines an ideal conductor as a material in which electrons are able to move freely. Statement 10.1.A.2 says that in electrostatic equilibrium, mutual repulsion of excess charge carriers results in those carriers residing entirely on the surface, with 10.1.A.2.i covering a negative net charge as excess electrons on the surface and 10.1.A.2.ii covering a positive net charge as a surface deficient in electrons that can be modelled as if positive carriers reside on the surface.

Three of this topic's statements do most of the work later. Statement 10.1.A.3.i says the time interval over which charges reach electrostatic equilibrium is so short as to be negligible. Statement 10.1.A.3.ii says that at equilibrium all points on the surface have the same electric potential and the conductor becomes an equipotential surface. Statement 10.1.A.3.iii says the charge density is greater where there are points or edges compared to planar areas, which is why lightning rods are pointed. Then 10.1.A.4 says all excess charges reside on the surface, so there is no net charge in the interior and the field inside is zero; 10.1.A.5 says the field is perpendicular to the outer surface; 10.1.A.6 says a conductor can be polarized by an external field as a consequence of remaining an equipotential surface; and 10.1.A.7 defines electrostatic shielding as surrounding an area with a closed conducting shell to create a region inside the conductor free from external electric fields.

10.2 Redistribution of Charge between Conductors is three sentences long and worth memorising all three. Statement 10.2.A.1: when conductors are in electrical contact, charges will be redistributed such that the surfaces of each conductor are at the same electric potential. Statement 10.2.A.2: ground is an idealized reference point that has zero electric potential and can absorb or provide an infinite amount of charge without changing its electric potential. Statement 10.2.A.3: charge can be induced on a conductor by grounding the conductor in the presence of an external electric field. Note the criterion in the first one. Contact equalises potential, not charge and not charge density.

10.3 Capacitors is the calculation topic. Statement 10.3.A.1 describes a parallel-plate capacitor as two separated parallel conducting surfaces that can hold equal amounts of charge with opposite signs. Statement 10.3.A.2 defines capacitance as C=Q/ΔVC = Q/\Delta V, with 10.3.A.2.i insisting it depends only on the physical properties of the capacitor, such as shape and the material separating the plates, and 10.3.A.2.ii giving C=κε0A/dC = \kappa\varepsilon_0 A/d. Statement 10.3.A.3 says the field between two charged parallel plates with uniformly distributed charge is constant in both magnitude and direction except near the edges of the plates, and 10.3.A.3.i says its magnitude, where the plate separation is much smaller than the plate dimensions, can be determined by applying Gauss's law and the principle of superposition, giving the derived equation E=Q/(ε0A)E = Q/(\varepsilon_0 A). Statement 10.3.A.3.iii is the bridge back to mechanics: a charged particle between two oppositely charged parallel plates undergoes constant acceleration, so its motion shares characteristics with the projectile motion of an object with mass in the gravitational field near Earth's surface. Then 10.3.A.4 says the stored energy equals the work done by an external force to separate that amount of charge, and 10.3.A.5 gives UC=12QΔVU_C = \frac{1}{2}Q\,\Delta V.

10.4 Dielectrics finishes the unit and carries the most equations per statement. Statement 10.4.A.1 says that in a dielectric material, electric charges are not as free to move as in a conductor, and instead the material becomes polarized in the presence of an external field. Statement 10.4.A.2 defines the dielectric constant as κ=ε/ε0\kappa = \varepsilon/\varepsilon_0, tying it straight back to Unit 8's permittivity objective 8.1.C. Statement 10.4.A.3 says the field created by a polarized dielectric is opposite in direction to the external field, 10.4.A.4 gives the derived equation κ=E0/E\kappa = E_0/E for the field between the plates of an isolated parallel-plate capacitor, and 10.4.A.5 gives the derived equation C=κC0C = \kappa C_0.

The Unit 10 equations, and the three you derive

Seven distinct equations appear in Unit 10's required content, all of them in Topics 10.3 and 10.4. Four are printed on the AP Physics C: E&M formula sheet and three are not, and the three that are not are exactly the three the CED labels Derived equation.

EquationWhere the CED puts itLabelPrinted on the sheet
C=QΔVC = \dfrac{Q}{\Delta V}10.3.A.2relevantyes
C=κε0AdC = \dfrac{\kappa \varepsilon_0 A}{d}10.3.A.2.iirelevantyes
E=Qε0AE = \dfrac{Q}{\varepsilon_0 A}10.3.A.3.iderivedno
UC=12QΔVU_C = \dfrac{1}{2}Q\,\Delta V10.3.A.5unlabelledyes
κ=εε0\kappa = \dfrac{\varepsilon}{\varepsilon_0}10.4.A.2relevantyes
κ=E0E\kappa = \dfrac{E_0}{E}10.4.A.4derivedno
C=κC0C = \kappa C_010.4.A.5derivedno

That pattern is the most useful fact about this unit, so it is worth saying precisely what the label means. The CED's Required Equations page states 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". So all three of Unit 10's derived equations are results you are expected to be able to produce, and none of them will be sitting on the sheet when you need it.

Each one has a stated route.

  • E=Q/(ε0A)E = Q/(\varepsilon_0 A) comes from Gauss's law plus superposition, and 10.3.A.3.i says so in the statement itself. It also carries a condition in the same sentence: the plate separation must be much smaller than the dimensions of the plates.
  • κ=E0/E\kappa = E_0/E follows from 10.4.A.3, that the polarized dielectric's field opposes the external field, and it is stated for an isolated capacitor, meaning the charge is fixed rather than the potential difference.
  • C=κC0C = \kappa C_0 follows from C=κε0A/dC = \kappa\varepsilon_0 A / d compared with the air-filled case, and the sheet's convention that capacitors are air-filled with κ=1.0\kappa = 1.0 is what makes C0C_0 well defined.

One thing the sheet gives you that the framework does not: the series and parallel capacitance rules and the RCRC time constant are printed, even though the framework places them in Unit 11. If a Unit 10 question hands you two capacitors, the combination rules are available on the sheet; what is not available is the field between the plates.

And because the C: E&M sheet reprints the whole C: Mechanics table, statement 10.3.A.3.iii can be taken literally. A charged particle launched between parallel plates is a projectile problem, and x=x0+vx0t+12axt2x = x_0 + v_{x0}t + \frac{1}{2}a_x t^2 is printed on the same page as C=Q/ΔVC = Q/\Delta V.

What calculus changes, and where AP Physics 2 stops

AP Physics C: Electricity and Magnetism is equivalent to the second course in an introductory college sequence in calculus-based physics, and its prerequisites say students should have taken or be concurrently taking calculus. Unit 10 is the unit where the calculus is least visible in the equations and most decisive in the reasoning.

  • Capacitance for a curved geometry is a Gauss-plus-integral calculation. The boundary statement names concentric spherical and coaxial cylindrical capacitors. Neither has a C=κε0A/dC = \kappa\varepsilon_0 A/d shortcut. You get the field from Gauss's law, then the potential difference from ΔV=abEdr\Delta V = -\int_a^b \vec{E}\cdot d\vec{r}, then divide. AP Physics 2's capacitor topic is parallel plates only.
  • The derived equations are derivations, not formulas. AP Physics 2 prints E=ΔV/dE = \Delta V/d and UC=12QΔVU_C = \frac{1}{2}Q\Delta V and expects them used. This course withholds E=Q/(ε0A)E = Q/(\varepsilon_0 A), κ=E0/E\kappa = E_0/E and C=κC0C = \kappa C_0 from the sheet and expects them produced.
  • The conductor results come from Unit 8. "The field inside a conductor is zero" is a Gauss's law argument, and 10.1.A.4 states the field and the interior charge in one breath because they are the same statement.
  • Charge distribution can be non-uniform. Statement 10.1.A.3.iii says the charge density is greater at points and edges. Once density varies with position, finding a total charge is the Qtotal=ρdVQ_{\text{total}} = \int \rho\, dV integral of Unit 8.

The nearest algebra-based pages on this site sit inside AP Physics 2 Unit 10, Electric Force, Field, and Potential: Topic 10.6, Capacitors is the direct counterpart to Topic 10.3 here, and its Topic 10.5 on electric potential carries the contact-equalises-potential statement that this course puts in Topic 10.2. Those pages are for AP Physics 2 students; this unit is for AP Physics C students. If you are in the algebra-based course you want parallel plates, a printed field formula and one capacitor topic inside a larger unit. If you are in Physics C you want a whole unit, three shapes of capacitor, and three equations you have to build.

One structural difference is worth flagging for anyone switching courses. AP Physics 2 keeps capacitors inside its electric-force-and-field unit and reaches RCRC circuits in its Unit 11. Physics C gives conductors and capacitors a unit of their own, then does compound circuits and RCRC circuits in Unit 11. The material is not reordered so much as unpacked.

Fifteen statements with no equations, and what they are for

Topics 10.1 and 10.2 print no equations between them, which makes them easy to skim and expensive to skip. Skill 3.B, applying an appropriate law, definition, theoretical relationship, or model to make a claim, is listed for all four topics in this unit, and 3.B is the only skill that is. These two topics are where it lives.

The reasoning chain runs like this.

Electrons move freely in a conductor (10.1.A.1), so if any field remained inside, they would still be moving. Equilibrium therefore requires zero interior field (10.1.A.4), and the excess charge has nowhere to sit but the surface (10.1.A.2). The surface must then be an equipotential (10.1.A.3.ii), because a potential difference along the surface would be a field along the surface and the charges would move again. That forces the field just outside to be perpendicular to the surface (10.1.A.5), since any parallel component would be a field along an equipotential, which statement 9.2.B.3.iv rules out. And all of this happens in a time so short as to be negligible (10.1.A.3.i), which is why "electrostatic equilibrium" can be assumed the moment a switch closes.

Each step of that chain answers a specific question type:

  • Why does a coaxial shield work? 10.1.A.7, electrostatic shielding, which is the concrete-buildings essential question from Unit 8's opener.
  • Why is a lightning rod pointed? 10.1.A.3.iii, charge density greater at points and edges than on planar areas.
  • Why can a bird sit on a high-voltage wire? 10.1.A.3.ii, the whole conductor at one potential, so no potential difference across the bird.
  • Why does a neutral conductor get attracted to a charged rod? 10.1.A.6, polarization as a consequence of the conductor remaining an equipotential surface.
  • What happens when two charged spheres touch? 10.2.A.1, charge moves until the surfaces are at the same potential. Equal potentials, not equal charges.
  • What does grounding actually do? 10.2.A.2, ground is an idealized reference at zero potential that can absorb or provide unlimited charge without changing its own potential, and 10.2.A.3, which is how a charge is induced.

The unit's first optional sample activity is 10.1.A.7 made physical: put a neon gas discharge tube inside an enclosed metal wire mesh container, touch the tube to a Tesla coil so it lights, then put it in the mesh and have students predict what happens and justify the claim with evidence.

Traps that span more than one topic

Contact equalises potential, not charge. Statement 10.2.A.1 says the surfaces end at the same potential. Two conducting spheres of different radii joined by a wire end with different charges, different surface charge densities and different surface fields. Splitting the charge equally is the single most common Topic 10.2 error.

The smaller sphere ends with the higher surface field. That follows from the same statement and is the quantitative face of 10.1.A.3.iii. Equal potentials with unequal radii means the surface field goes as 1/R1/R, so a sharp region is the one that breaks down first.

"Isolated" and "connected to a battery" give opposite answers. Statement 10.4.A.4 states κ=E0/E\kappa = E_0/E for an isolated capacitor, where the charge is fixed and inserting a dielectric lowers the field and the potential difference. Leave the battery attached and the potential difference is fixed instead, so the field between the plates does not change and the charge rises. Which quantity is held constant is the whole question, and the word isolated in the CED is doing that job.

Capacitance does not depend on the charge or the voltage on it. Statement 10.3.A.2.i says it depends only on the physical properties of the capacitor, such as shape and the material between the plates. C=Q/ΔVC = Q/\Delta V is a definition and a measurement recipe, not a statement that CC grows with QQ.

The uniform field between plates has an exception written into the statement. Statement 10.3.A.3 says the field is constant in magnitude and direction "except near the edges of the plates", and 10.3.A.3.i adds the condition that the plate separation be much smaller than the plate dimensions. Both clauses are part of the required content, not caveats to drop.

UC=12QΔVU_C = \frac{1}{2}Q\Delta V has a factor of one half for a reason. Statement 10.3.A.4 says the stored energy equals the work an external force does to separate that amount of charge, and the potential difference climbs from zero to its final value as the charge accumulates, so the work is the average rather than the product. Writing QΔVQ\Delta V doubles the answer.

A dielectric is not a conductor. Statement 10.4.A.1 says charges in a dielectric are not as free to move as in a conductor and the material polarizes instead. The field inside a dielectric is reduced by a factor κ\kappa; the field inside a conductor is zero. Filling a capacitor with metal is a short circuit, not a κ=\kappa = \infty dielectric.

Capacitors are air-filled unless told otherwise. The exam conventions printed with the equation sheet say so explicitly, with κ=1.0\kappa = 1.0. So C=κε0A/dC = \kappa\varepsilon_0 A/d reduces to ε0A/d\varepsilon_0 A/d by default, and a problem that mentions no material has not omitted anything.

How Unit 10 is assessed

The exam is 3 hours long: 42 multiple-choice questions in 85 minutes for 50%, then 4 free-response questions in 95 minutes for the other 50%, one of each type in a fixed order, with a calculator allowed throughout. Unit 10 contributes 10 to 15%, the smallest share of any unit, which works out to roughly four to six of the 42 multiple-choice questions.

The unit's Preparing for the AP Exam note points at the fourth free-response question, the Qualitative/Quantitative Translation question, which is worth 8 points with a suggested time of 15 to 20 minutes. The note says the QQT requires students to re-express key elements of physical phenomena across multiple representations of the domain, that students first make a claim and provide evidence and reasoning to support it without reference to equations, then derive an equation or set of equations to represent the same scenario, and finally connect the claim in the first part to the equations derived in the second. It then makes an observation about who finds this hard: students exposed primarily to numerical problem solving often struggle with the QQT because it requires them to communicate a conceptual understanding of course content.

That is the clearest signal the CED gives about how to study this unit. The words-first, equations-second order of the QQT is the same order as this unit's own topics: 10.1 and 10.2 print no equations at all, then 10.3 and 10.4 print seven. The unit opener names skills 2.A, 3.B and 3.C as the ones Unit 10 develops, and says the unit encourages students to derive new expressions from fundamental principles, to predict using functional dependence, and to make claims and support them with evidence, giving as its example having students predict the net charge on a capacitor from its geometry and then justify those predictions with appropriate physics principles. The unit's AP Classroom Progress Check runs about 18 multiple-choice questions and 4 free-response questions, one of each type.

On the CED's sample exam, one of the fifteen sample multiple-choice questions aligns to Unit 10, and it is a Topic 10.4 question aligned to essential knowledge 10.4.A.5 with skill 2.A. Two air-filled capacitors of capacitance CC and C/3C/3 sit in parallel; a dielectric is inserted into the C/3C/3 one, and the new equivalent capacitance is twice the original. The dielectric constant is then found from C+κC/3=2(C+C/3)C + \kappa C/3 = 2(C + C/3), giving κ=5\kappa = 5. Notice how much of that question comes from outside Unit 10: the parallel combination rule is Topic 11.5 material and is on the sheet, while C=κC0C = \kappa C_0 is Unit 10's derived equation and is not.

The sample free-response Question 4, the Qualitative/Quantitative Translation question, aligns to learning objectives 10.3.A, 11.1.A, 11.7.A, 11.3.B and 11.8.B, with skills 2.A, 2.D, 3.B and 3.C. So the one sample free-response question that touches this unit reaches it through a circuit, which is a fair picture of how a capacitor usually arrives on this exam. One instruction from that question is worth carrying into every RCRC-adjacent problem: it asks students to "Derive, but do not solve, a differential equation". Setting up correctly is what earns the points.

The five optional sample instructional activities are the most lab-heavy of any unit in this part of the course, and they skip Topic 10.2 entirely: one on 10.1 (the neon tube in a wire mesh), two on 10.3 (researching the ionosphere and Earth's surface as a capacitor, and building a capacitor from aluminium foil and waxed paper then measuring it with a capacitance meter), and two on 10.4 (arguing from molecular polarity why pure water has a high dielectric constant while impure water makes a leaky capacitor, and building a foil-and-paper parallel-plate capacitor, then adding sheets of paper to find the dielectric constant of the paper). Topic 10.4 is also the only topic in the unit listing skill 3.A, creating experimental procedures, which those last two activities are built to practise.

A parallel-plate capacitor, then the same one with a dielectric and no battery

A parallel-plate capacitor has plates of area A=0.025 m2A = 0.025\ \mathrm{m^2} separated by d=1.5d = 1.5 mm of air, and is charged by a 12 V battery. Find (a) its capacitance, the charge stored, the field between the plates and the stored energy. The battery is then disconnected and a slab of dielectric constant κ=2.5\kappa = 2.5 is slid in to fill the gap. Find (b) the new capacitance, field, potential difference and stored energy, and (c) account for the change in energy. Use ε0=8.85×1012 C2/(Nm2)\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}.

  1. Declare the setup: the air-filled state is the printed exam convention, capacitors air-filled with κ=1.0\kappa = 1.0, and the plate separation is far smaller than the plate dimensions, which is the condition statement 10.3.A.3.i attaches to the uniform-field result. Call the initial quantities C0C_0, QQ, E0E_0, V0V_0.

  2. (a) From 10.3.A.2.ii with κ=1\kappa = 1: C0=ε0Ad=(8.85×1012)(0.025)1.5×103=1.475×1010C_0 = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85 \times 10^{-12})(0.025)}{1.5 \times 10^{-3}} = 1.475 \times 10^{-10} F, which is 147.5147.5 pF.

  3. From 10.3.A.2, Q=C0V0=(1.475×1010)(12)=1.77×109Q = C_0 V_0 = (1.475 \times 10^{-10})(12) = 1.77 \times 10^{-9} C, or 1.771.77 nC.

  4. The field comes from the derived equation of 10.3.A.3.i, which is not on the sheet: E0=Qε0A=1.77×109(8.85×1012)(0.025)=8000E_0 = \dfrac{Q}{\varepsilon_0 A} = \dfrac{1.77 \times 10^{-9}}{(8.85 \times 10^{-12})(0.025)} = 8000 V/m. Cross-check it against the uniform-field version of ΔV=Edr\Delta V = -\int \vec{E}\cdot d\vec{r}, which for a constant field gives E=V0/d=12/(1.5×103)=8000E = V_0/d = 12/(1.5 \times 10^{-3}) = 8000 V/m. The two agree, which is a check that the derived equation was assembled correctly.

  5. From 10.3.A.5, U0=12QΔV=12(1.77×109)(12)=1.062×108U_0 = \frac{1}{2}Q\,\Delta V = \frac{1}{2}(1.77 \times 10^{-9})(12) = 1.062 \times 10^{-8} J, or 10.610.6 nJ.

  6. (b) The battery is disconnected, so QQ is now the fixed quantity, not VV. This is the condition the word isolated is doing in statement 10.4.A.4, and it decides every answer below.

  7. From the derived equation of 10.4.A.5, C=κC0=2.5(147.5 pF)=368.75C = \kappa C_0 = 2.5(147.5\ \mathrm{pF}) = 368.75 pF.

  8. From the derived equation of 10.4.A.4, κ=E0/E\kappa = E_0/E, so E=E0/κ=8000/2.5=3200E = E_0/\kappa = 8000/2.5 = 3200 V/m. The polarized dielectric's own field opposes the external one, per 10.4.A.3, which is why the net field falls.

  9. The potential difference follows from either ΔV=Q/C\Delta V = Q/C or ΔV=Ed\Delta V = Ed. Both give 1.77×1093.6875×1010=4.80\dfrac{1.77 \times 10^{-9}}{3.6875 \times 10^{-10}} = 4.80 V and (3200)(1.5×103)=4.80(3200)(1.5 \times 10^{-3}) = 4.80 V.

  10. U=12QΔV=12(1.77×109)(4.80)=4.248×109U = \frac{1}{2}Q\,\Delta V = \frac{1}{2}(1.77 \times 10^{-9})(4.80) = 4.248 \times 10^{-9} J, or 4.254.25 nJ.

  11. (c) The energy fell from 10.6210.62 nJ to 4.254.25 nJ, a drop of 6.376.37 nJ. The ratio is 4.248/10.62=0.4004.248/10.62 = 0.400, exactly 1/κ1/\kappa, which is what U=12Q2/CU = \frac{1}{2}Q^2/C with fixed QQ and CκCC \to \kappa C predicts. The energy did not vanish: with the charge fixed, the field pulls the dielectric slab into the gap, so the slab does positive work on whatever was holding it and the field gives up energy in exchange.

  12. Contrast the other case in one line, because the exam asks for it. Had the battery stayed connected, ΔV\Delta V would have stayed at 12 V, the field would have stayed at 8000 V/m, the charge would have risen to κQ=4.43\kappa Q = 4.43 nC, and the stored energy would have risen to κU0=26.6\kappa U_0 = 26.6 nJ, supplied by the battery.

(a) C0=147.5C_0 = 147.5 pF, Q=1.77Q = 1.77 nC, E0=8000E_0 = 8000 V/m, U0=10.6U_0 = 10.6 nJ. (b) With the battery disconnected, C=369C = 369 pF, E=3200E = 3200 V/m, ΔV=4.80\Delta V = 4.80 V, U=4.25U = 4.25 nJ. (c) The energy falls by a factor of exactly κ\kappa, to 4.254.25 nJ, because the field does work pulling the slab in. With the battery left connected, ΔV\Delta V and EE would have held and the energy would have risen by a factor of κ\kappa instead.

Two conducting spheres in contact: what equalises and what does not

An isolated conducting sphere of radius a=0.060a = 0.060 m carries a charge of +9.0+9.0 nC. A second, uncharged conducting sphere of radius b=0.030b = 0.030 m sits far away. The two are joined by a long, thin conducting wire whose own capacitance is negligible. Find (a) the final charge on each sphere, (b) the common potential, (c) the ratio of their surface charge densities and of their surface field magnitudes, and (d) which way electrons moved.

  1. Declare the model: both spheres are far enough apart that each behaves as an isolated sphere, so by 8.3.B.1.ii the potential at the surface of each is that of a point charge with the same net charge at its center, V=kQ/RV = kQ/R. Take k=9.0×109 Nm2/C2k = 9.0 \times 10^9\ \mathrm{N \cdot m^2/C^2}.

  2. (a) Statement 10.2.A.1 gives the condition: charges redistribute so that the surfaces of each conductor are at the same electric potential. So kQaa=kQbb\dfrac{kQ_a}{a} = \dfrac{kQ_b}{b}, which means Qaa=Qbb\dfrac{Q_a}{a} = \dfrac{Q_b}{b}.

  3. Charge is conserved by 8.2.A.2.ii, so Qa+Qb=9.0Q_a + Q_b = 9.0 nC. Solving, Qa=Qtotalaa+b=9.0(0.0600.090)=6.0Q_a = Q_{\text{total}}\dfrac{a}{a+b} = 9.0\left(\dfrac{0.060}{0.090}\right) = 6.0 nC and Qb=3.0Q_b = 3.0 nC.

  4. The charge split 2:1, in the ratio of the radii, not equally. Equal charges would have given the larger sphere the lower potential and the electrons would have kept moving.

  5. (b) V=kQaa=(9.0×109)(6.0×109)0.060=900V = \dfrac{kQ_a}{a} = \dfrac{(9.0 \times 10^9)(6.0 \times 10^{-9})}{0.060} = 900 V. Check on the other sphere: (9.0×109)(3.0×109)0.030=900\dfrac{(9.0 \times 10^9)(3.0 \times 10^{-9})}{0.030} = 900 V. Equal, as required.

  6. (c) Surface charge density is charge over surface area, σ=Q/(4πR2)\sigma = Q/(4\pi R^2). Taking the ratio, σbσa=QbQa(ab)2=12(2)2=2\dfrac{\sigma_b}{\sigma_a} = \dfrac{Q_b}{Q_a}\left(\dfrac{a}{b}\right)^2 = \dfrac{1}{2}(2)^2 = 2. The smaller sphere ends up with twice the surface charge density.

  7. The surface fields go the same way: E=kQ/R2E = kQ/R^2, so EbEa=QbQa(ab)2=2\dfrac{E_b}{E_a} = \dfrac{Q_b}{Q_a}\left(\dfrac{a}{b}\right)^2 = 2. Numerically Ea=(9.0×109)(6.0×109)(0.060)2=1.5×104E_a = \dfrac{(9.0\times10^9)(6.0\times10^{-9})}{(0.060)^2} = 1.5 \times 10^4 V/m and Eb=3.0×104E_b = 3.0 \times 10^4 V/m.

  8. Both ratios reduce to a/ba/b, the inverse radius ratio, which is 10.1.A.3.iii in numbers: charge density is greater where the surface curves more tightly. A sphere of radius zero is a point, and that is a lightning rod.

  9. (d) The sphere that gained positive charge lost electrons. Charge flowed from A to B in the conventional-current sense, so electrons moved from B to A, from the small sphere to the large one. Statement 8.2.A.2.i is the reminder that the charging of a system typically involves the transfer of electrons.

  10. Two checks before finishing. The final potentials are equal, which was the condition. And 6.0+3.0=9.06.0 + 3.0 = 9.0 nC, so charge is conserved. Note also 10.1.A.3.i: this whole redistribution takes a time so short as to be negligible.

(a) 6.06.0 nC on the larger sphere and 3.03.0 nC on the smaller, splitting in the ratio of the radii. (b) V=900V = 900 V on both. (c) The smaller sphere has twice the surface charge density and twice the surface field. (d) Electrons moved from the small sphere to the large one. Contact equalises potential, not charge.

A concentric spherical capacitor, derived from Gauss and a line integral

A concentric spherical capacitor consists of an inner conducting sphere of radius Ra=0.050R_a = 0.050 m carrying charge +Q+Q and an outer concentric conducting shell of inner radius Rb=0.080R_b = 0.080 m carrying Q-Q, with air between them. Derive an expression for the capacitance, then evaluate it, and find the charge, the stored energy and the field just outside the inner sphere when the capacitor is charged to 200200 V.

  1. The concentric spherical capacitor is one of the three shapes the Topic 10.3 boundary statement names for quantitative analysis, so this derivation is inside the course. There is no shortcut equation for it: C=κε0A/dC = \kappa\varepsilon_0 A/d is the parallel-plate result and does not apply.

  2. Step 1, field from Gauss's law. Draw a spherical Gaussian surface of radius rr with Ra<r<RbR_a < r < R_b. The enclosed charge is +Q+Q, and by 8.6.A.4 the field is radial and constant over the surface, so EdA=E(4πr2)=Q/ε0\oint \vec{E}\cdot d\vec{A} = E(4\pi r^2) = Q/\varepsilon_0 and E(r)=Q4πε0r2E(r) = \dfrac{Q}{4\pi\varepsilon_0 r^2}.

  3. Step 2, potential difference from the line integral. Apply 9.2.B.2 radially inward from the outer shell to the inner sphere. Taking magnitudes, ΔV=RaRbQ4πε0r2dr=Q4πε0[1r]RaRb=Q4πε0(1Ra1Rb)\Delta V = \displaystyle\int_{R_a}^{R_b} \frac{Q}{4\pi\varepsilon_0 r^2}\,dr = \frac{Q}{4\pi\varepsilon_0}\left[-\frac{1}{r}\right]_{R_a}^{R_b} = \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{R_a} - \frac{1}{R_b}\right).

  4. Step 3, divide. From 10.3.A.2, C=QΔV=4πε01Ra1Rb=4πε0RaRbRbRaC = \dfrac{Q}{\Delta V} = \dfrac{4\pi\varepsilon_0}{\dfrac{1}{R_a} - \dfrac{1}{R_b}} = \dfrac{4\pi\varepsilon_0 R_a R_b}{R_b - R_a}. The charge cancelled, which it had to: statement 10.3.A.2.i says the capacitance depends only on the physical properties of the capacitor.

  5. Evaluate. 4πε0=1.1121×10104\pi\varepsilon_0 = 1.1121 \times 10^{-10}, RaRb=4.0×103 m2R_aR_b = 4.0 \times 10^{-3}\ \mathrm{m^2} and RbRa=0.030R_b - R_a = 0.030 m, so C=(1.1121×1010)4.0×1030.030=1.483×1011C = (1.1121 \times 10^{-10})\dfrac{4.0 \times 10^{-3}}{0.030} = 1.483 \times 10^{-11} F, which is 14.814.8 pF.

  6. Q=CΔV=(1.483×1011)(200)=2.966×109Q = C\,\Delta V = (1.483 \times 10^{-11})(200) = 2.966 \times 10^{-9} C, so 2.972.97 nC.

  7. UC=12QΔV=12(2.966×109)(200)=2.97×107U_C = \frac{1}{2}Q\,\Delta V = \frac{1}{2}(2.966 \times 10^{-9})(200) = 2.97 \times 10^{-7} J, or 297297 nJ.

  8. Field just outside the inner sphere, from the Step 1 expression at r=Rar = R_a: E=2.966×1094π(8.85×1012)(0.050)2=1.07×104E = \dfrac{2.966 \times 10^{-9}}{4\pi(8.85 \times 10^{-12})(0.050)^2} = 1.07 \times 10^4 V/m. It is perpendicular to the surface, as 10.1.A.5 requires, and points radially outward from the positive inner sphere.

  9. Check the derivation by running it backwards: Q4πε0(10.05010.080)=2.9657×1091.1121×1010(2012.5)=200\dfrac{Q}{4\pi\varepsilon_0}\left(\dfrac{1}{0.050} - \dfrac{1}{0.080}\right) = \dfrac{2.9657 \times 10^{-9}}{1.1121 \times 10^{-10}}(20 - 12.5) = 200 V, the potential difference we started from.

  10. One comparison worth keeping. A lone conducting sphere of radius RaR_a has capacitance 4πε0Ra=5.564\pi\varepsilon_0 R_a = 5.56 pF. Adding the outer shell raises it to 14.814.8 pF, a factor of Rb/(RbRa)=2.67R_b/(R_b - R_a) = 2.67. Bringing the shell closer would raise it further without bound, which is the spherical version of why parallel-plate capacitance goes as 1/d1/d.

C=4πε0RaRbRbRa=14.8C = \dfrac{4\pi\varepsilon_0 R_a R_b}{R_b - R_a} = 14.8 pF. At 200200 V: Q=2.97Q = 2.97 nC, UC=297U_C = 297 nJ, and E=1.07×104E = 1.07 \times 10^4 V/m just outside the inner sphere. The derivation is Gauss's law for the field, then the line integral for the potential difference, then C=Q/ΔVC = Q/\Delta V, and the charge cancels on the way.

Frequently asked questions

How much of the AP Physics C E&M exam is Unit 10?

Unit 10, Conductors and Capacitors, is weighted at 10 to 15% of the multiple-choice section of the AP Physics C: Electricity and Magnetism exam, the smallest share of any unit in the course, and the course description suggests about 8 to 16 class periods, also the smallest estimate. Every other unit starts at 10 to 20% and gets at least 10 periods; Units 8 and 11 on either side of it are weighted 15 to 25% each. With 42 multiple-choice questions in total, 10 to 15% works out to roughly four to six questions.

What are the four topics in AP Physics C E&M Unit 10?

They are 10.1 Electrostatics with Conductors, 10.2 Redistribution of Charge between Conductors, 10.3 Capacitors, and 10.4 Dielectrics, each with exactly one learning objective. Topics 10.1 and 10.2 print no equations at all and are pure reasoning about where charge sits and where it moves; Topics 10.3 and 10.4 carry all seven of the unit's equations. Only Topic 10.3 has the title in common with an AP Physics 2 topic, its Topic 10.6 Capacitors. Electrostatics with conductors, redistribution of charge between conductors, and dielectrics are unique to the calculus-based course.

Which capacitor shapes does AP Physics C expect you to calculate?

Three. The Topic 10.3 boundary statement reads that while other shapes are also able to separate charges, AP Physics C: Electricity and Magnetism only expects the quantitative analysis and description of parallel-plate capacitors, concentric spherical capacitors, and coaxial cylindrical capacitors. The opening clause matters: other shapes still store charge and can still be reasoned about, they just will not be handed to you for a capacitance calculation. Those three are exactly the planar, spherical and cylindrical symmetries that the Topic 8.6 boundary statement allows Gauss's law to be applied to quantitatively, which is how the curved two get calculated at all: Gauss's law for the field, then the line integral for the potential difference, then capacitance as charge over potential difference.

Which Unit 10 equations are not on the AP Physics C E&M equation sheet?

Three, and they are exactly the three the course description labels Derived Equations: the field between parallel plates written as the charge divided by the permittivity of free space times the plate area, the dielectric constant written as the ratio of the field without the dielectric to the field with it, and the new capacitance written as the dielectric constant times the original capacitance. The course description reserves that label for equations that demonstrate the final results of derivations expected of students on the exam, so all three are results you should be able to produce. The four that are printed are capacitance as charge over potential difference, the parallel-plate capacitance with the dielectric constant and plate geometry, the stored energy as half the charge times the potential difference, and the dielectric constant as the ratio of permittivities.

What happens when a dielectric is inserted into a capacitor?

It depends on whether the capacitor is isolated or still connected to a battery, and that is the whole question. Essential knowledge 10.4.A.5 says the capacitance changes by a factor of the dielectric constant either way. Essential knowledge 10.4.A.4 covers the isolated case, where the charge is fixed: the field between the plates decreases by that same factor, because per 10.4.A.3 the field created by the polarized dielectric opposes the external field, so the potential difference and the stored energy both fall by a factor of the dielectric constant. If the battery stays connected, the potential difference is fixed instead, so the field between the plates is unchanged and the charge and the stored energy both rise by that factor, with the battery supplying the difference. The exam conventions printed with the equation sheet state that capacitors are air-filled with a dielectric constant of 1.0 unless a problem says otherwise.

What happens when two charged conducting spheres are connected by a wire?

Charge moves until the surfaces of both spheres are at the same electric potential, which is essential knowledge 10.2.A.1. It does not divide equally unless the spheres are the same size. For two well-separated spheres each behaving as an isolated sphere, equal potentials means the charges end up in the ratio of the radii, so the larger sphere takes more charge. The smaller sphere ends with the higher surface charge density and the higher surface field, in the inverse ratio of the radii, which is the quantitative form of essential knowledge 10.1.A.3.iii: the charge density on the surface of a conductor is greater where there are points or edges compared to planar areas. Essential knowledge 10.1.A.3.i adds that the redistribution takes a time so short as to be negligible.

Are series and parallel capacitor rules part of AP Physics C Unit 10?

No. The rules for combining capacitors in series and in parallel are printed on the AP Physics C: Electricity and Magnetism equation sheet, but the course framework places them in Unit 11, under Topic 11.5 Compound Direct Current Circuits, and the resistor-capacitor time constant in Topic 11.8. Nothing in Unit 10's required content covers more than one capacitor. That does not stop a question from using both at once: on the course description's own sample exam, the one multiple-choice question aligned to Unit 10 puts two capacitors in parallel, then inserts a dielectric into one of them and asks for the dielectric constant, so it needs the parallel rule from the sheet and the derived equation that is not on the sheet.