Topic 10.6: Capacitors

What Topic 10.6 requires

Topic 10.6 has a single learning objective:

• 10.6.A Describe the physical properties of a parallel-plate capacitor.

Six essential knowledge statements sit under it, 10.6.A.1 through 10.6.A.6, with four sub-statements: 10.6.A.2.i, 10.6.A.2.ii, 10.6.A.3.i and 10.6.A.3.ii. The CED's suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

That skill list is unusual for the unit, and it is a hint about how this topic gets tested. Three of the five, 1.B, 2.B and 3.A, are among the four skills the CED lists for the Experimental Design and Analysis free-response question, which is worth 10 points with a suggested time of 25 to 30 minutes. Topic 10.6 is laboratory-shaped material.

The boundary statement is two sentences and both matter:

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.

So no spherical or cylindrical capacitors, and the fringing field at the rim of the plates is off the table unless a question puts it back on. One more piece of exam scaffolding lives outside the topic pages: the conventions list printed with the AP Physics 2 Table of Information states that capacitors are air-filled with a dielectric constant of 1.0 unless otherwise stated. Every question that does not mention a dielectric has already told you $\kappa = 1.0$.

Unit 10 is weighted at 15 to 18 percent of the multiple-choice section and gets about 14 to 21 class periods. Topic 10.6 is where the unit's abstractions become an object: Topic 10.3 and Topic 10.5 describe fields and potentials in general, and a capacitor is the device built to hold a specific one.

Capacitance is a geometry number (10.6.A.1 and 10.6.A.2)

Statement 10.6.A.1 defines the object: a parallel-plate capacitor consists of two separated parallel conducting surfaces that can hold equal amounts of charge with opposite signs.

Read that carefully, because it fixes a piece of vocabulary that trips people up all year. A capacitor described as holding a charge Q has $+Q$ on one plate and $-Q$ on the other. The net charge on the device is zero. $Q$ in every capacitor equation means the magnitude on one plate, never a total.

Statement 10.6.A.2 defines the measure: capacitance relates the magnitude of the charge stored on each plate to the electric potential difference created by the separation of those charges. The equation, printed in the Electricity group of the AP Physics 2 equation sheet, is

The unit follows directly: a coulomb per volt, called the farad, symbol F. A farad is an enormous capacitance. The prefixes table printed with the Table of Information runs down to pico, and the numbers in worked example 1 land in the picofarad range, which is what a pair of hand-sized plates a millimeter or two apart actually gives you.

Then comes 10.6.A.2.i, which is the single most useful sentence in the topic: 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.

Only. Not on the charge you put on it. Not on the potential difference you apply. $C = Q/\Delta V$ looks like a formula in which $C$ depends on $Q$ and $\Delta V$, and it does not. Doubling $Q$ doubles $\Delta V$ and leaves the ratio untouched. The ratio is fixed by the metalwork, and $Q$ and $\Delta V$ are the two quantities free to move.

That distinction is a favorite target for suggested skill 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. If a question asks how to increase the capacitance of a capacitor, the only admissible answers change the geometry or the filling. Charging it harder is not one of them.

The parallel-plate equation, symbol by symbol (10.6.A.2.ii)

Statement 10.6.A.2.ii spells out the geometry: 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, and the constant of proportionality is the product of the dielectric constant of the material between the plates and the electric permittivity of free space. That is

with each symbol doing one job.

• $A$ is the area of one plate, not the pair. More area gives more room for charge at a given potential difference, so $C \propto A$.
• $d$ is the plate separation. Bringing the plates closer lets the opposite charges hold each other in place at a lower potential difference, so $C \propto 1/d$.
• $\varepsilon_0$ is the permittivity of free space, and the Constants group of the sheet prints it as $8.85 \times 10^{-12}$ in units of $\text{C}^2/(\text{N} \cdot \text{m}^2)$.
• $\kappa$ is the dielectric constant of whatever fills the gap. It is a pure number with no units, and it is 1.0 for the air-filled default the exam conventions assume.

Check the units and the equation earns its keep. Permittivity in $\text{C}^2/(\text{N} \cdot \text{m}^2)$ times an area in $\text{m}^2$ divided by a length in m leaves $\text{C}^2/(\text{N} \cdot \text{m})$, which is $\text{C}^2/\text{J}$, which is a coulomb per volt, which is a farad.

Suggested skill 2.D is not on this topic's list, but the proportional reasoning it describes is exactly what 10.6.A.2.ii sets up, and skill 2.C, comparing two scenarios, asks the same questions in comparison form. Double the area and the capacitance doubles. Halve the separation and the capacitance doubles. Fill the gap with a material of dielectric constant 4 and the capacitance quadruples. None of those require you to know $A$, $d$, or $\varepsilon_0$, only the structure of the equation.

One warning about the geometry. The equation is derived for plates whose separation is small compared with their size, which is the same condition 10.6.A.3.i attaches to the field equation below. Pull the plates far apart and both statements stop being good approximations, which is another way of saying that edge effects have grown until they matter. The boundary statement's promise to ignore edge effects unless explicitly stated is what keeps you out of that territory.

The field between the plates (10.6.A.3)

Statement 10.6.A.3 is the reason parallel plates are worth building: 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.

A uniform field is a rare and useful thing. Everywhere else in the unit the field falls off with distance and changes direction from point to point. Between the plates it does neither, which is what makes the region a clean laboratory for energy problems.

Statement 10.6.A.3.i gives the magnitude, with its condition attached: 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 described with the equation

That condition is part of the statement, not a footnote to it. Notice what is missing from the right-hand side: $d$. Holding the charge fixed and pulling the plates apart does not change the field between them. It changes the potential difference, because the same field now acts over a longer distance.

You now have two independent routes to the same field, and a question can hand you either set of inputs.

They agree, and it is worth seeing why once. Substitute $Q = C \Delta V$ into the first, then $C = \kappa \varepsilon_0 A / d$, and the permittivity, the dielectric constant and the area all cancel, leaving $\Delta V / d$. Deriving that in three lines is suggested skill 2.A territory, and it is the cheapest way to check a numerical answer: compute the field twice and see if the two agree.

Statement 10.6.A.3.ii cashes the uniform field in: 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.

That is an explicit invitation to reuse AP Physics 1 machinery. A charge fired sideways into the gap traces a parabola for the same reason a thrown ball does: constant acceleration along one axis, constant velocity along the perpendicular one. The two-axis method in the projectile motion guide transfers with $a = qE/m$ in place of $g$, and the field between the plates plays the role gravity plays in AP Physics 1 Topic 1.5.

Stored energy, and where the one half comes from (10.6.A.4 and 10.6.A.5)

Statement 10.6.A.4 says what the stored energy is: 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. Statement 10.6.A.5 gives the equation, which the sheet also prints:

The one half is the part everybody queries, and the answer is in 10.6.A.4's phrasing. Charging a capacitor is not a single transfer of $Q$ across a fixed potential difference. The first bit of charge crosses a gap with almost no potential difference across it and costs almost nothing. The last bit crosses the full $\Delta V$. Since $C$ is a constant, the potential difference rises in direct proportion to the charge already moved, so the average potential difference over the whole process is $\frac{1}{2}\Delta V$, and the work is $Q$ times that average.

There is a graphical version of the same argument, and it is worth drawing because suggested skill 1.B asks for quantitative graphs. Plot $Q$ on the vertical axis against $\Delta V$ on the horizontal axis for a capacitor being charged. Because $Q = C \Delta V$ with $C$ constant, the graph is a straight line through the origin with slope $C$. The area under that line from 0 to $\Delta V$ is the area of a triangle, $\frac{1}{2} Q \Delta V$, and that area is the stored energy. Slope gives you capacitance, area gives you energy, from one graph.

The sheet prints exactly one form of the stored energy, the one above. The two variants you will want are not printed, and you get them by substituting $C = Q/\Delta V$ in one direction or the other:

Both are single-line derivations from printed material, which is suggested skill 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway. Knowing which one to reach for is most of the work: use $\frac{1}{2} C (\Delta V)^2$ when the potential difference is held fixed, as it is by an attached battery, and $Q^2/(2C)$ when the charge is held fixed, as it is on an isolated capacitor. Picking the form whose fixed quantity is actually fixed turns the next section's comparison into arithmetic you can do in your head.

Dielectrics, and the one question that settles everything (10.6.A.6)

Statement 10.6.A.6: adding a dielectric between two plates of a capacitor changes the capacitance of the capacitor and induces an electric field in the dielectric in the opposite direction to the field between the plates.

The mechanism is in the second half. A dielectric is an insulator whose molecules polarize in an applied field, positive ends drifting toward the negative plate and negative ends toward the positive plate. That leaves a layer of bound charge on each face of the slab, and those layers produce their own field pointing opposite to the plates' field. The two partly cancel, the net field in the gap is weaker than it would be with vacuum, and a weaker field across the same gap means a smaller potential difference for the same charge, which by $C = Q/\Delta V$ means a larger capacitance. The factor by which capacitance goes up is $\kappa$, which is why $\kappa$ appears in the numerator of $C = \kappa \varepsilon_0 A/d$ and in the denominator of $E_C = Q/(\kappa \varepsilon_0 A)$.

Then comes the exam question, and it is always the same one. Is the battery still attached? Insert the same slab into the same capacitor and you get two different sets of answers depending on which quantity the circuit is holding fixed.

Only the first row is the same in both columns, because capacitance is a geometry number and 10.6.A.2.i already told you it does not care what the circuit is doing. Everything else splits.

The energy row is the one that looks paradoxical. With the battery attached the stored energy rises, and the extra energy comes out of the battery, which pushes more charge onto the plates. With the battery removed the stored energy falls, and the released energy does work on the slab: an isolated capacitor pulls a dielectric into the gap. Worked example 2 runs both columns on the same numbers so you can see the two answers side by side, which is suggested skill 2.C in its purest form.

The Physics 2 sheet does not print a definition of $\kappa$ in terms of permittivities. It appears only inside the two capacitor equations, so treat it as the multiplier the CED describes rather than as a ratio you have to justify.

The capacitor lab (skills 1.B and 3.A)

Skill 3.A, create experimental procedures that are appropriate for a given scientific question, appears on only some topics, and the CED's description of the Experimental Design and Analysis question tells you the standard it is held to. The procedure is expected to be scientifically sound: vary a single parameter, and measure how that change affects a single characteristic. Methods must be performable in a typical high school laboratory, and measurements must be made with realistically obtainable equipment or sensors.

A capacitor gives you three parameters you can vary one at a time, which is why this topic suits the format.

1. Vary the plate separation $d$, hold $A$ and $\kappa$ fixed. Prediction: $C \propto 1/d$.
2. Vary the overlapping plate area $A$, hold $d$ and $\kappa$ fixed. Slide one plate sideways so only part of it faces the other. Prediction: $C \propto A$.
3. Change the material in the gap, hold $A$ and $d$ fixed. Prediction: $C \propto \kappa$.

Skill 1.B then asks for a quantitative graph with appropriate scales and units, and the analysis step is always the same: choose axes that make the expected relationship a straight line, because a straight line is the only shape you can read a number off reliably. A plot of $C$ against $d$ is a curve and tells you little. A plot of $C$ against $1/d$ is a line through the origin whose slope is $\kappa \varepsilon_0 A$, and dividing that slope by $\kappa \varepsilon_0$ hands you the plate area. The CED's description of the question says as much: the slope or intercepts of the line may be used to determine a physical quantity, or the nature of the slope may itself answer the posed question. Worked example 3 does that from a table of data.

Two habits that earn points. State what you are holding constant, out loud, for every run. And when you report a quantity extracted from a slope, give its units, which come from the axis units divided into each other rather than from the symbol you were expecting.

What Topic 10.6 does not cover

Three things a capacitor page could plausibly contain sit outside this topic, and knowing where they live saves you from studying them in the wrong place.

Combinations of capacitors. The equation sheet prints both rules in its Electricity group, $1/C_{\text{eq,s}} = \sum_i 1/C_i$ for series and $C_{\text{eq,p}} = \sum_i C_i$ for parallel, but neither appears among Topic 10.6's essential knowledge statements. They belong to Topic 11.8, where 11.8.A.1.i and 11.8.A.1.iii state them and 11.8.A.2 adds that as a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate. Statement 11.8.A.1.ii is the sanity check worth remembering: the equivalent capacitance of a set of capacitors in series is less than the capacitance of the smallest capacitor.

Charging and discharging over time. The exponential approach to steady state, the time constant $\tau = R_{\text{eq}} C_{\text{eq}}$, and the 63 percent and 37 percent landmarks are all Topic 11.8 as well. Topic 10.6 is entirely about the settled state.

Shapes other than parallel plates, and edge effects. Both are excluded by the boundary statement quoted at the top of this page.

What Topic 10.6 does leave you with is a device that stores a known amount of energy in a known field across a known gap, and that is precisely the setup Topic 10.7 needs when it starts turning potential differences into speeds. Work through this topic until you can produce $C$, $Q$, $\Delta V$, $E$ and $U_C$ from any two of them plus the geometry, in either direction, and the rest of the unit's capacitor questions become bookkeeping.