AP Physics C: E&M · Topic 10.3

Topic 10.3: Capacitors

Q: What is capacitance in AP Physics C?

Capacitance relates the magnitude of the charge stored on each plate to the electric potential difference created by separating those charges, C = Q divided by delta V, which is CED statement 10.3.A.2. It is measured in farads, one farad being one coulomb per volt. The defining equation measures capacitance rather than determining it: statement 10.3.A.2.i says capacitance depends only on the physical properties of the capacitor, its shape and the material between the conductors, so putting more charge on raises the potential difference proportionally and leaves C unchanged.

Q: Is the electric field between capacitor plates on the AP Physics C equation sheet?

No. Statement 10.3.A.3.i gives E = Q divided by epsilon-zero A, but it is labelled a Derived equation rather than a Relevant equation, and it does not appear on the exam booklet's equation sheet. The CED's Required Equations note explains the label: derived equations demonstrate the final results of derivations expected of students on the exam. The AP Physics 2 sheet does print its version, E_C = Q divided by kappa epsilon-zero A, which is one of the sharpest differences between the two courses.

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

Three, named in the Topic 10.3 boundary statement: parallel-plate capacitors, concentric spherical capacitors, and coaxial cylindrical capacitors. The statement notes that while other shapes are also able to separate charges, the course only expects the quantitative analysis and description of those three. AP Physics 2's corresponding boundary statement requires only parallel-plate capacitors, and adds that edge effects will be ignored unless explicitly stated otherwise, a sentence the AP Physics C statement does not include.

Q: Why is there a factor of one half in the energy stored in a capacitor?

Because the potential difference rises as the capacitor charges rather than sitting at its final value throughout. Moving an increment dq when the charge already there is q costs q over C times dq, and integrating from zero to Q gives Q squared over 2C, which equals one half Q delta V. The one half is an average: the first increment crossed no potential difference at all and only the last one crossed the full delta V. The result is printed on the equation sheet as statement 10.3.A.5.

Q: What happens to a capacitor when the battery is disconnected?

The charge is trapped and becomes the fixed quantity, while the potential difference becomes dependent. Any change to the geometry or the material changes C, and delta V adjusts as Q over C. Use U equals Q squared over 2C for the energy, so a rise in capacitance lowers the stored energy. With the battery left connected, delta V is fixed instead, the charge adjusts as C times delta V, and U equals one half C delta V squared rises with capacitance. Deciding which quantity is pinned is the first step in any capacitor comparison.

Q: How is a charged particle between capacitor plates like a projectile?

Statement 10.3.A.3.iii says a charged particle between two oppositely charged parallel plates undergoes constant acceleration, so its motion shares characteristics with the projectile motion of a mass in Earth's gravitational field. The parallel motion is at constant velocity, the transverse motion is uniformly accelerated with a equal to qE over m, and time links them. The analogy holds because the field between plates is constant in magnitude and direction, which statement 10.3.A.3 grants except near the edges of the plates.

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

The equations are printed on the AP Physics C: Electricity and Magnetism equation sheet, but they do not belong to Unit 10 in the course framework. Unit 10 covers a single capacitor's physical properties: what capacitance is, what it depends on, the field between the plates, and the stored energy. Combining capacitors in series and parallel, and the RC time constant that is also on the sheet, belong to Unit 11 on electric circuits.

Unit 10: Conductors and Capacitors10-15% of the multiple-choice section

A capacitor is two conductors holding equal and opposite charge. Its capacitance is the charge on one conductor divided by the potential difference between them, and it depends only on the geometry and the material between them, not on how much charge you put there.

AP Physics: Unit 10 (topics 10.3 Capacitors). Topic 10.3 of the current AP Physics C: Electricity and Magnetism course and exam description, inside Unit 10, weighted 10 to 15% of the multiple-choice section at about 8 to 16 class periods. One learning objective, 10.3.A, using the task verb describe, with ten essential-knowledge statements, five at the top level and five sub-statements, covering the parallel-plate construction, capacitance as C = Q over delta V, its dependence on physical properties alone, C = kappa epsilon-zero A over d, the constant field between plates except near the edges, the derived field magnitude E = Q over epsilon-zero A from Gauss's law and superposition, proportionality of field to surface charge density, the projectile analogy for a charged particle in the gap, and the stored energy as external work with U_C = one half Q delta V. This topic carries Unit 10's only boundary statement: 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. Suggested skills are 1.C, 2.A, 2.D and 3.B, a set with no overlap at all with the skills AP Physics 2 lists for its Topic 10.6, which are 1.B, 2.B, 2.C, 3.A and 3.C. Three of the four equations are printed on the equation sheet; E = Q over epsilon-zero A is labelled a Derived equation and is not, which the CED's Required Equations note defines as demonstrating the final results of derivations expected of students on the exam. AP Physics 2's Topic 10.6 instead prints its version, E_C = Q over kappa epsilon-zero A, as a relevant equation and on its own sheet, and its boundary statement requires only parallel-plate capacitors while adding that edge effects will be ignored unless explicitly stated otherwise.

What Topic 10.3 requires

Topic 10.3 has one learning objective, ten essential-knowledge statements, five at the top level and five sub-statements, and the only boundary statement in Unit 10.

10.3.A, describe the physical properties of a parallel-plate capacitor.

10.3.A.1 states that a parallel-plate capacitor consists of two separated parallel conducting surfaces that can hold equal amounts of charge with opposite signs.
10.3.A.2 states that capacitance relates the magnitude of the charge stored on each plate to the electric potential difference created by the separation of those charges, with the relevant equation $C = \dfrac{Q}{\Delta V}$ .
10.3.A.2.i states that the capacitance of a capacitor depends only on the physical properties of the capacitor, such as the capacitor's shape and the material used to separate the plates.
10.3.A.2.ii states that the capacitance of a parallel-plate capacitor is proportional to the area of one of its plates and inversely proportional to the distance between its plates, with the constant of proportionality being the product of the dielectric constant $\kappa$ of the material between the plates and the electric permittivity of free space $\varepsilon_0$ . Relevant equation: $C = \dfrac{\kappa\varepsilon_0 A}{d}$ .
10.3.A.3 states that the electric field between two charged parallel plates with uniformly distributed electric charge, such as in a parallel-plate capacitor, is constant in both magnitude and direction, except near the edges of the plates.
10.3.A.3.i states that the magnitude of the electric field between two charged parallel plates, where the plate separation is much smaller than the dimensions of the plates, can be determined by applying Gauss's law and the principle of superposition. Its derived equation: $E = \dfrac{Q}{\varepsilon_0 A}$ .
10.3.A.3.ii states that the electric field is proportional to the surface charge density on either plate of the capacitor.
10.3.A.3.iii states that a charged particle between two oppositely charged parallel plates undergoes constant acceleration, and therefore its motion shares characteristics with the projectile motion of an object with mass in the gravitational field near Earth's surface.
10.3.A.4 states that the electric potential energy stored in a capacitor is equal to the work done by an external force to separate that amount of charge on the capacitor.
10.3.A.5 states that the electric potential energy stored in a capacitor is described by the equation $U_C = \dfrac{1}{2}Q\Delta V$ .

The boundary statement, in full: "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."

Suggested skills: 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Unit 10 is weighted 10 to 15% of the multiple-choice section over a suggested 8 to 16 class periods.

The Derived Equation label, and the one equation here that carries it

Statement 10.3.A.3.i is labelled Derived equation rather than Relevant equation, and the framework defines that distinction rather than leaving it to guesswork. Under the heading "Required Equations", the CED says:

"Not all equations in this course framework appear on the equation sheet provided to students while taking the AP Physics C: Electricity and Magnetism Exam. Many of the equations in this document are provided for reference and guidance, or to demonstrate the final results of derivations expected of students on the exam. These equations are denoted as 'Derived Equations.'"

So the label is a promise about your workings. $E = \dfrac{Q}{\varepsilon_0 A}$ for the field between parallel plates is a result you are expected to derive, and it is deliberately absent from the equation sheet. Quoting it on a free-response derivation is quoting the answer.

Check the sheet against this topic line by line:

Statement	Equation	On the sheet?
10.3.A.2	$C = Q/\Delta V$	Yes
10.3.A.2.ii	$C = \kappa\varepsilon_0 A / d$	Yes
10.3.A.3.i	$E = Q/(\varepsilon_0 A)$	No, labelled Derived
10.3.A.5	$U_C = \frac{1}{2}Q\Delta V$	Yes

Three printed, one to derive. Two more capacitor lines appear on the sheet that do not belong to this topic: the combination rules $\frac{1}{C_{\text{eq,s}}} = \sum_i \frac{1}{C_i}$ and $C_{\text{eq,p}} = \sum_i C_i$ , and the time constant $\tau = R_{\text{eq}}C_{\text{eq}}$ . Those are Unit 11 content, so do not go looking for them in the Unit 10 framework.

Where AP Physics 2 stops, and one equation that changes shape

Topic 10.6 in AP Physics 2 is called Capacitors too, and most of its statements correspond one to one with this topic's in near-identical wording: 10.6.A.1, 10.6.A.2 with $C = Q/\Delta V$ , both of its sub-statements, 10.6.A.3, the projectile analogy and the two energy statements. The projectile statement differs only by a comma, "undergoes constant acceleration, and therefore" in Physics C against "undergoes constant acceleration and therefore" in Physics 2.

Three things differ, and one of them is the sharpest single illustration in Unit 10 of what the calculus-based course does differently.

The plate field is given in Physics 2 and derived in Physics C. Physics 2's statement 10.6.A.3.i says the field "can be described with the equation" $E_C = \dfrac{Q}{\kappa\varepsilon_0 A}$ , printed as a relevant equation and printed again on the AP Physics 2 equation sheet. Physics C's statement 10.3.A.3.i says the same field "can be determined by applying Gauss's law and the principle of superposition", labels the result $E = \dfrac{Q}{\varepsilon_0 A}$ as derived, and leaves it off the sheet. Same physics, opposite pedagogy: one course hands you the formula, the other hands you the method and expects the formula out the other end.

Physics C adds statement 10.3.A.3.ii, that the electric field is proportional to the surface charge density on either plate. There is no Physics 2 counterpart. It is the statement that connects the plate field to Topic 10.1's conductor surface, and it is what makes the Gauss's law derivation feel inevitable rather than clever.

The boundary statements name different shapes. Physics 2: "While other shapes are also able to separate charges, only the analysis and descriptions of parallel-plate capacitors are required for AP Physics 2. Edge effects will be ignored unless explicitly stated otherwise." Physics C: three shapes, parallel-plate, concentric spherical and coaxial cylindrical, with no sentence about edge effects. Two extra geometries, and no blanket permission to ignore fringing, though statement 10.3.A.3 does say the field is constant "except near the edges of the plates".

There is one more difference and it is unusual. The two courses list completely disjoint sets of suggested skills for this topic. Physics 2 lists 1.B, 2.B, 2.C, 3.A and 3.C. Physics C lists 1.C, 2.A, 2.D and 3.B. Not one code appears on both lists. Physics 2 aims at plotting data, calculating and designing an experiment; Physics C aims at sketching model behaviour, deriving symbolically, predicting factors of change, and making a claim from a law.

Physics 2's statement 10.6.A.6, on adding a dielectric, is not in Physics C's Topic 10.3 either. Physics C gives dielectrics a topic of their own, 10.4, with five statements. If you want the algebra-based treatment of capacitors, read the AP Physics 2 topic.

Deriving the plate field and the parallel-plate capacitance

Since the field equation is one you are expected to produce, here is the production, in the two steps statement 10.3.A.3.i names.

Step one, Gauss's law for one plate. Take one plate with surface charge density $\sigma = Q/A$ , wide enough to treat as an infinite sheet. A Gaussian pillbox straddling it has flux out of both faces, so $2EA_{\text{box}} = \dfrac{\sigma A_{\text{box}}}{\varepsilon_0}$ , giving a field of magnitude $\dfrac{\sigma}{2\varepsilon_0}$ on each side, pointing away from a positive sheet.

Step two, superposition. The second plate carries $-\sigma$ and produces a field of the same magnitude pointing toward it. Between the plates the two fields point the same way and add; outside, they point in opposite directions and cancel. So between the plates

E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}

and outside, in the ideal case, zero. Statement 10.3.A.3.ii is the middle expression: the field is proportional to the surface charge density on either plate. This is also the $E = \sigma/\varepsilon_0$ you get just outside any conductor's surface, which is no coincidence, since each plate is a conductor.

The capacitance then follows without new physics. The field is uniform, so the line integral of Topic 9.2 collapses to a product:

\Delta V = Ed = \frac{Qd}{\varepsilon_0 A}

C = \frac{Q}{\Delta V} = \frac{\varepsilon_0 A}{d}

which is the printed $C = \dfrac{\kappa\varepsilon_0 A}{d}$ with $\kappa = 1$ . The conventions box on the equation sheet fixes that for you: "Capacitors are air-filled $(\kappa = 1.0)$ " unless a question says otherwise.

Notice what dropped out. $Q$ appears in both numerator and denominator and cancels. That is statement 10.3.A.2.i in algebraic form: capacitance depends only on the physical properties of the capacitor, its shape and the material between the conductors. Doubling the charge doubles the potential difference and leaves $C$ alone. A student who says "the capacitance goes up because we put more charge on it" has read the defining equation as a formula for $C$ rather than as a measurement of it.

Three shapes, one recipe

The boundary statement names parallel-plate, concentric spherical and coaxial cylindrical capacitors. All three yield to the same four steps, and knowing that is worth more than memorising three results.

Put $+Q$ on one conductor and $-Q$ on the other. Statement 10.3.A.1 guarantees the equal-and-opposite arrangement.
Get the field in the gap from Gauss's law. Each shape has one of the three symmetries the Topic 8.6 boundary statement allows: planar, spherical, cylindrical. The Gaussian surface is a pillbox, a sphere, or a coaxial cylinder respectively.
Integrate the field across the gap to get $\Delta V$ , taking the path straight across so the dot product is just $E\,dr$ .
Divide. $C = Q/\Delta V$ , and $Q$ cancels every time.

Run it and you get:

Shape	Field in the gap	Capacitance
Parallel plates, area $A$ , gap $d$	$\dfrac{Q}{\varepsilon_0 A}$ , uniform	$\dfrac{\varepsilon_0 A}{d}$
Concentric spheres, radii $R_a < R_b$	$\dfrac{kQ}{r^2}$	$\dfrac{4\pi\varepsilon_0 R_a R_b}{R_b - R_a}$
Coaxial cylinders, radii $a < b$ , length $L$	$\dfrac{Q}{2\pi\varepsilon_0 L r}$	$\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$

Only the first is printed on the sheet. The other two are exactly what skill 2.A is for, and the equation sheet supplies every ingredient: Gauss's law, the line integral, and $C = Q/\Delta V$ .

Two checks worth applying to any capacitance you derive. The units must come out as farads, and since $\varepsilon_0$ has units of $\mathrm{C^2/(N \cdot m^2)}$ , every one of these is $\varepsilon_0$ times a length. The limits must behave: shrink the spherical capacitor's gap, so $R_b \to R_a$ , and the capacitance grows without bound, which is right, because a tiny gap holds a great deal of charge per volt. Pull the plates far apart and $C \to 0$ .

One structural point the CED does not spell out but the boundary statement implies. All three shapes are two conductors facing each other across a gap, which is why the recipe is identical: the geometry only ever changes step 2.

Energy stored, and where the one half comes from

Statement 10.3.A.4 defines the stored energy exactly as Topic 9.1 defines potential energy: as work done by an external force, here to separate that amount of charge on the capacitor. Statement 10.3.A.5 gives the result:

U_C = \frac{1}{2}Q\Delta V

The factor of one half is the part students distrust, so it is worth seeing where it comes from. The potential difference is not constant during charging: it starts at zero and rises with the charge already moved, since $\Delta V' = q/C$ at every stage. Moving the next increment $dq$ against that costs $dW = \dfrac{q}{C}dq$ , and

W = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C} = \frac{1}{2}Q\Delta V

So the half is an average: you paid the full $\Delta V$ only for the last increment, and nothing for the first. That derivation also produces the two rewrites the sheet does not print but the algebra hands you free, using $C = Q/\Delta V$ :

U_C = \frac{1}{2}Q\Delta V = \frac{1}{2}C(\Delta V)^2 = \frac{Q^2}{2C}

Choose the form whose two quantities are held fixed in your problem, which is usually the fastest route through a comparison question:

Charge fixed, meaning the capacitor is isolated and the battery is disconnected: use $Q^2/2C$ , so anything that raises $C$ lowers the energy.
Potential difference fixed, meaning a battery stays connected: use $\frac{1}{2}C(\Delta V)^2$ , so anything that raises $C$ raises the energy.

That single choice resolves most of the traps in this unit, and it is the whole of the dielectric comparison in Topic 10.4.

Constant charge or constant voltage, and the projectile analogy

Which quantity is held fixed is the first thing to establish in any capacitor comparison. The CED does not name this distinction anywhere, and it decides the answer to most questions in the unit.

Change made	Battery connected	Battery disconnected
What is fixed	$\Delta V$	$Q$
What adjusts	$Q = C\Delta V$	$\Delta V = Q/C$
Energy	$U = \frac{1}{2}C(\Delta V)^2$ , follows $C$	$U = Q^2/2C$ , opposes $C$
Field between plates	$E = \Delta V/d$ , unchanged if $d$ is	$E = Q/(\varepsilon_0 A)$ , unchanged if $A$ is

Read the last row carefully, because it is the least intuitive. With the battery connected and the plate separation held, changing what sits between the plates does not change the field, because the potential difference and the gap both stay put. With the battery disconnected, changing the separation does not change the field, because the charge and the area both stay put. Two different quantities are pinned in the two cases.

The other half of this section is statement 10.3.A.3.iii, and it is a gift. A charged particle between two oppositely charged parallel plates undergoes constant acceleration, so its motion shares characteristics with projectile motion near Earth's surface. Every technique from projectile motion transfers directly: the motion parallel to the plates is at constant velocity, the motion across the gap is uniformly accelerated, and time is the shared variable.

The translation table is short:

$g$ becomes $a = \dfrac{qE}{m} = \dfrac{q\Delta V}{md}$ .
"Down" becomes "toward the plate of opposite sign to the particle", which for an electron is the positive plate.
Range becomes the plate length divided by the entry speed, giving the time in the gap.
"Does it clear the table?" becomes "does it exit before hitting a plate?", which compares the deflection with half the gap.

This works only because the field is uniform, which statement 10.3.A.3 grants except near the edges of the plates. Physics C's boundary statement, unlike Physics 2's, carries no blanket instruction to ignore edge effects, so if a question puts the particle near an edge, read it carefully.

Traps, and how Topic 10.3 is assessed

Believing capacitance depends on charge or voltage. Statement 10.3.A.2.i says it depends only on the physical properties. $C = Q/\Delta V$ measures capacitance; it does not determine it.

Quoting $E = Q/(\varepsilon_0 A)$ on a derivation. It is a Derived Equation and is not on the sheet. Produce it, from Gauss's law and superposition, in the two steps the statement names.

Forgetting the factor of one half. $U_C = \frac{1}{2}Q\Delta V$ , not $Q\Delta V$ . The full product is the work a battery does moving that charge through a fixed potential difference, which is a different quantity.

Mixing up which quantity is fixed. Establish battery connected or disconnected before comparing anything.

Using $\Delta V = Ed$ for the spherical or cylindrical capacitor. Only the parallel-plate field is uniform. The other two need the integral.

Treating $\kappa$ as optional. The sheet prints $C = \kappa\varepsilon_0 A/d$ with $\kappa$ in it, and the conventions box says capacitors are air-filled with $\kappa = 1.0$ unless stated. The $\kappa$ is always there; it is usually one.

Assuming zero field outside a real capacitor. Zero is the ideal-plate result from perfect cancellation. Statement 10.3.A.3 restricts the constant-field claim to the region away from the edges.

On assessment: the exam is 3 hours, with 42 multiple-choice questions worth 50% in 85 minutes and 4 free-response questions worth 50% in 95 minutes. The Unit 10 opener names skills 2.A, 3.B and 3.C as the ones this unit builds, saying that alongside gaining proficiency with the specific equations for capacitance, the unit encourages students to derive new expressions from fundamental principles. It points at the fourth free-response question, the Qualitative/Quantitative Translation, worth 8 points with a suggested time of 15 to 20 minutes, and gives as its unit example having students predict the net charge on a capacitor from its geometry and then justify those predictions with appropriate physics principles. The Unit 10 Progress Check is about 18 multiple-choice questions and 4 free-response questions.

The CED also suggests two classroom activities for this topic: researching the electrical properties of the ionosphere and Earth's surface, then treating the pair as a spherical capacitor to find its capacitance, charge and potential difference; and building a capacitor from sheets of aluminium foil and waxed paper, predicting its capacitance, then measuring it with a capacitance meter.

A coaxial cylindrical capacitor, derived and then evaluated

A coaxial cable has an inner conductor of radius $a = 2.0$ mm carrying charge $+Q$ and an outer conducting shell of radius $b = 6.0$ mm carrying $-Q$ , with air between them, over a length $L = 0.50$ m. Derive an expression for the capacitance, evaluate it, then find the charge stored, the energy stored and the field at $r = 4.0$ mm when the capacitor is charged to $150$ V. Use $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$ and neglect end effects.

Step 2 of the recipe, the field from Gauss's law. Draw a coaxial Gaussian cylinder of radius $r$ with $a < r < b$ and length $L$ . By cylindrical symmetry $\vec{E}$ is radial and constant on the curved face, and the flat ends contribute no flux. So $E(2\pi r L) = \dfrac{Q}{\varepsilon_0}$ , giving $E(r) = \dfrac{Q}{2\pi\varepsilon_0 L r}$ .
Step 3, integrate across the gap. Take the path straight out along a radius, so $\vec{E}\cdot d\vec{r} = E\,dr$ : the magnitude of the potential difference is $\Delta V = \displaystyle\int_a^b \dfrac{Q}{2\pi\varepsilon_0 L r}dr = \dfrac{Q}{2\pi\varepsilon_0 L}\ln\!\left(\dfrac{b}{a}\right)$ .
Step 4, divide. $C = \dfrac{Q}{\Delta V} = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$ . The charge cancels, as statement 10.3.A.2.i requires: what is left is $\varepsilon_0$ times a length divided by a dimensionless logarithm.
Evaluate. $\ln(6.0/2.0) = \ln 3 = 1.0986$ . And $2\pi\varepsilon_0 L = 2\pi(8.85 \times 10^{-12})(0.50) = 2.780 \times 10^{-11}$ , so $C = \dfrac{2.780 \times 10^{-11}}{1.0986} = 2.53 \times 10^{-11}$ F, that is $25.3$ pF.
Charge at $150$ V. $Q = C\Delta V = (2.53 \times 10^{-11})(150) = 3.80 \times 10^{-9}$ C, that is $3.80$ nC.
Energy stored. $U_C = \frac{1}{2}Q\Delta V = \frac{1}{2}(3.80 \times 10^{-9})(150) = 2.85 \times 10^{-7}$ J, that is $285$ nJ. Cross-check with $\frac{1}{2}C(\Delta V)^2 = \frac{1}{2}(2.53 \times 10^{-11})(2.25 \times 10^4) = 2.85 \times 10^{-7}$ J.
Field at $r = 4.0$ mm. $E = \dfrac{Q}{2\pi\varepsilon_0 L r} = \dfrac{3.80 \times 10^{-9}}{(2.780 \times 10^{-11})(4.0 \times 10^{-3})} = 3.41 \times 10^4$ V/m, radially outward from the inner conductor.
Check the field limits. At $r = a$ it is $6.83 \times 10^4$ V/m and at $r = b$ it is $2.28 \times 10^4$ V/m, a factor of three apart, matching the $1/r$ dependence and $b/a = 3$ . The field is strongest at the inner conductor, which is where a real cable breaks down first.

$C = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)} = 25.3$ pF. At $150$ V: $Q = 3.80$ nC, $U_C = 285$ nJ, and $E = 3.41 \times 10^4$ V/m at the midpoint radius. The derivation is Gauss's law, then the radial line integral, then $C = Q/\Delta V$ , and the charge cancels at the last step.

Pulling the plates apart with the battery disconnected

A parallel-plate air capacitor has plates of area $A = 0.040\ \mathrm{m^2}$ separated by $d = 0.50$ mm. It is charged to $24$ V and the battery is then disconnected. The plates are pulled apart to $1.5$ mm. Find the capacitance, charge, field and stored energy before and after, and the work done pulling the plates apart.

Before, capacitance. $C_0 = \dfrac{\kappa\varepsilon_0 A}{d}$ with $\kappa = 1.0$ for air, per the printed conventions: $C_0 = \dfrac{(8.85 \times 10^{-12})(0.040)}{5.0 \times 10^{-4}} = 7.08 \times 10^{-10}$ F, that is $708$ pF.
Before, charge. $Q = C_0\Delta V = (7.08 \times 10^{-10})(24) = 1.70 \times 10^{-8}$ C, that is $17.0$ nC.
Before, field and energy. $E = \dfrac{\Delta V}{d} = \dfrac{24}{5.0 \times 10^{-4}} = 4.8 \times 10^4$ V/m. Check against the derived equation: $\dfrac{Q}{\varepsilon_0 A} = \dfrac{1.70 \times 10^{-8}}{(8.85 \times 10^{-12})(0.040)} = 4.8 \times 10^4$ V/m, agreeing. Energy $U_0 = \frac{1}{2}Q\Delta V = \frac{1}{2}(1.70 \times 10^{-8})(24) = 2.04 \times 10^{-7}$ J, that is $204$ nJ.
Identify what is fixed. The battery is disconnected, so the charge is pinned at $17.0$ nC and the potential difference becomes the dependent quantity.
After, capacitance. Tripling $d$ divides $C$ by three: $C' = 236$ pF.
After, potential difference and field. $\Delta V' = \dfrac{Q}{C'} = \dfrac{1.70 \times 10^{-8}}{2.36 \times 10^{-10}} = 72$ V, three times as much. But the field is unchanged: $E = \dfrac{Q}{\varepsilon_0 A} = 4.8 \times 10^4$ V/m, since neither $Q$ nor $A$ changed. Confirm with $\Delta V'/d' = 72/(1.5 \times 10^{-3}) = 4.8 \times 10^4$ V/m.
After, energy. With $Q$ fixed, use $U = \dfrac{Q^2}{2C}$ : dividing $C$ by three multiplies $U$ by three, so $U' = 612$ nJ. Check directly: $\frac{1}{2}(1.70 \times 10^{-8})(72) = 6.12 \times 10^{-7}$ J.
Work done. Nothing else supplied energy, so the work you did equals the increase in stored energy: $W = U' - U_0 = 612 - 204 = 408$ nJ. Spread over the $1.0$ mm you pulled, that is an average force of $\dfrac{4.08 \times 10^{-7}}{1.0 \times 10^{-3}} = 4.1 \times 10^{-4}$ N, the attraction between the oppositely charged plates you had to overcome.

Before: $C_0 = 708$ pF, $Q = 17.0$ nC, $E = 4.8 \times 10^4$ V/m, $U_0 = 204$ nJ. After: $C' = 236$ pF, $Q$ unchanged at $17.0$ nC, $\Delta V' = 72$ V, $E$ unchanged at $4.8 \times 10^4$ V/m, $U' = 612$ nJ, and you did $408$ nJ of work. With the battery left connected instead, the potential difference would have been pinned at $24$ V, the charge would have fallen to one third, and the stored energy would have fallen rather than risen.

An electron crossing the gap: the projectile analogy in numbers

Two parallel plates of length $L = 0.050$ m are separated by $d = 0.020$ m and held at a potential difference of $100$ V. An electron enters midway between them, moving parallel to the plates at $v_0 = 2.0 \times 10^7$ m/s. Find the field, the electron's acceleration, its deflection on exit, and whether it clears the plates. Take $e = 1.60 \times 10^{-19}$ C and $m_e = 9.11 \times 10^{-31}$ kg, and neglect gravity and edge effects.

Field between the plates. Uniform by statement 10.3.A.3, so $E = \dfrac{\Delta V}{d} = \dfrac{100}{0.020} = 5.0 \times 10^3$ V/m, directed from the positive plate toward the negative plate.
Acceleration. $a = \dfrac{eE}{m_e} = \dfrac{(1.60 \times 10^{-19})(5.0 \times 10^3)}{9.11 \times 10^{-31}} = \dfrac{8.0 \times 10^{-16}}{9.11 \times 10^{-31}} = 8.78 \times 10^{14}\ \mathrm{m/s^2}$ , directed toward the positive plate, because the electron's charge is negative so the force opposes the field.
Set up the projectile analogy, which statement 10.3.A.3.iii licenses. Along the plates: constant velocity $v_0$ . Across the gap: constant acceleration $a$ from zero transverse velocity. Time is shared.
Time in the gap. $t = \dfrac{L}{v_0} = \dfrac{0.050}{2.0 \times 10^7} = 2.5 \times 10^{-9}$ s.
Deflection. $y = \frac{1}{2}at^2 = \frac{1}{2}(8.78 \times 10^{14})(2.5 \times 10^{-9})^2 = \frac{1}{2}(8.78 \times 10^{14})(6.25 \times 10^{-18}) = 2.74 \times 10^{-3}$ m, that is $2.74$ mm.
Does it clear the plates? It entered midway, so it has $d/2 = 10.0$ mm before it strikes a plate. A $2.74$ mm deflection is comfortably inside that, so it exits.
Exit angle. $v_y = at = (8.78 \times 10^{14})(2.5 \times 10^{-9}) = 2.20 \times 10^6$ m/s, so the deflection angle is $\arctan\!\left(\dfrac{2.20 \times 10^6}{2.0 \times 10^7}\right) = 6.3^\circ$ .
Energy cross-check, using Topic 9.3. Crossing $2.74$ mm of a $5.0 \times 10^3$ V/m field means crossing $13.7$ V, so the kinetic energy gained is $(1.60 \times 10^{-19})(13.7) = 2.20 \times 10^{-18}$ J. And $\frac{1}{2}m_e v_y^2 = \frac{1}{2}(9.11 \times 10^{-31})(2.20 \times 10^6)^2 = 2.20 \times 10^{-18}$ J. The two methods agree, which is the kind of consistency check the Qualitative/Quantitative Translation question asks you to make explicitly.

$E = 5.0 \times 10^3$ V/m, $a = 8.78 \times 10^{14}\ \mathrm{m/s^2}$ toward the positive plate, a deflection of $2.74$ mm over a $2.5$ ns transit, exiting at $6.3^\circ$ . It clears the plates, since $2.74$ mm is less than the $10.0$ mm available. The energy method gives the same transverse kinetic energy, $2.20 \times 10^{-18}$ J.

Frequently asked questions

What is capacitance in AP Physics C?

Capacitance relates the magnitude of the charge stored on each plate to the electric potential difference created by separating those charges, C = Q divided by delta V, which is CED statement 10.3.A.2. It is measured in farads, one farad being one coulomb per volt. The defining equation measures capacitance rather than determining it: statement 10.3.A.2.i says capacitance depends only on the physical properties of the capacitor, its shape and the material between the conductors, so putting more charge on raises the potential difference proportionally and leaves C unchanged.

Is the electric field between capacitor plates on the AP Physics C equation sheet?

No. Statement 10.3.A.3.i gives E = Q divided by epsilon-zero A, but it is labelled a Derived equation rather than a Relevant equation, and it does not appear on the exam booklet's equation sheet. The CED's Required Equations note explains the label: derived equations demonstrate the final results of derivations expected of students on the exam. The AP Physics 2 sheet does print its version, E_C = Q divided by kappa epsilon-zero A, which is one of the sharpest differences between the two courses.

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

Three, named in the Topic 10.3 boundary statement: parallel-plate capacitors, concentric spherical capacitors, and coaxial cylindrical capacitors. The statement notes that while other shapes are also able to separate charges, the course only expects the quantitative analysis and description of those three. AP Physics 2's corresponding boundary statement requires only parallel-plate capacitors, and adds that edge effects will be ignored unless explicitly stated otherwise, a sentence the AP Physics C statement does not include.

Why is there a factor of one half in the energy stored in a capacitor?

Because the potential difference rises as the capacitor charges rather than sitting at its final value throughout. Moving an increment dq when the charge already there is q costs q over C times dq, and integrating from zero to Q gives Q squared over 2C, which equals one half Q delta V. The one half is an average: the first increment crossed no potential difference at all and only the last one crossed the full delta V. The result is printed on the equation sheet as statement 10.3.A.5.

What happens to a capacitor when the battery is disconnected?

The charge is trapped and becomes the fixed quantity, while the potential difference becomes dependent. Any change to the geometry or the material changes C, and delta V adjusts as Q over C. Use U equals Q squared over 2C for the energy, so a rise in capacitance lowers the stored energy. With the battery left connected, delta V is fixed instead, the charge adjusts as C times delta V, and U equals one half C delta V squared rises with capacitance. Deciding which quantity is pinned is the first step in any capacitor comparison.

How is a charged particle between capacitor plates like a projectile?

Statement 10.3.A.3.iii says a charged particle between two oppositely charged parallel plates undergoes constant acceleration, so its motion shares characteristics with the projectile motion of a mass in Earth's gravitational field. The parallel motion is at constant velocity, the transverse motion is uniformly accelerated with a equal to qE over m, and time links them. The analogy holds because the field between plates is constant in magnitude and direction, which statement 10.3.A.3 grants except near the edges of the plates.

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

The equations are printed on the AP Physics C: Electricity and Magnetism equation sheet, but they do not belong to Unit 10 in the course framework. Unit 10 covers a single capacitor's physical properties: what capacitance is, what it depends on, the field between the plates, and the stored energy. Combining capacitors in series and parallel, and the RC time constant that is also on the sheet, belong to Unit 11 on electric circuits.