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AP Physics C: E&M Unit 10 Review: Conductors and Capacitors

Q: What topics are covered in AP Physics E&M Unit 10?

AP Physics E&M Unit 10 covers four topics: Electrostatics with Conductors (10.1), Redistribution of Charge Between Conductors (10.2), Capacitors (10.3), and Dielectrics (10.4). Together these topics explain how charge is stored and redistributed in materials, building the foundation for understanding electric circuits. See the full topic breakdown at AP Physics E&M Unit 10 .

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

Unit 10 makes up 10-15% of the AP Physics E&M exam, making it one of the more heavily tested units. It covers conductors, capacitors, and dielectrics, including how charge is stored, redistributed between conductors, and affected by insulating materials placed inside capacitors.

Q: What's on the AP Physics E&M Unit 10 progress check (MCQ and FRQ)?

The AP Physics E&M Unit 10 progress check includes both MCQ and FRQ parts drawn from all four unit topics: Electrostatics with Conductors, Redistribution of Charge Between Conductors, Capacitors, and Dielectrics. MCQ questions test conceptual understanding of conductor behavior and capacitor properties, while the FRQ portion asks you to derive expressions, analyze charge redistribution, and explain how dielectrics change capacitance. Practice with matched questions at AP Physics E&M Unit 10 .

Q: How do I practice AP Physics E&M Unit 10 FRQs?

The best way to practice Unit 10 FRQs is to focus on the three topic areas that generate the most free-response questions: Electrostatics with Conductors, Capacitors, and Dielectrics. Typical FRQ prompts ask you to derive capacitance for a given geometry, calculate the effect of a dielectric on stored energy, or justify conductor behavior using Gauss's Law. Work through each derivation step-by-step and write out your reasoning clearly, since partial credit depends on your explanation. Find FRQ practice at AP Physics E&M Unit 10 .

Q: Where can I find AP Physics E&M Unit 10 practice questions?

You can find AP Physics E&M Unit 10 practice questions, including multiple-choice and practice test sets, at AP Physics E&M Unit 10 . The page includes MCQ questions covering conductors, capacitors, and dielectrics so you can test yourself on each topic individually before taking a full unit practice test.

Q: How should I study AP Physics E&M Unit 10?

Start Unit 10 by solidifying your understanding of conductors in electrostatic equilibrium from Topic 10.1, since that logic carries through the whole unit. Then work through charge redistribution problems in 10.2 using conservation of charge. For capacitors in 10.3, practice deriving capacitance from geometry using Gauss's Law rather than memorizing formulas. Finish with dielectrics in 10.4 and make sure you can explain how inserting a dielectric changes capacitance, electric field, and stored energy. Do at least one timed FRQ per topic so you get comfortable showing your reasoning under pressure. Get study resources at AP Physics E&M Unit 10 .

Review AP Physics C: E&M Unit 10 to understand how charge distributes on conductors, how capacitors store charge and energy, and how dielectrics change capacitor properties. This unit bridges the electrostatics of Units 8 and 9 with the circuit behavior of Unit 11.

Use the topic guides, practice questions, and FRQ practice available for this unit to work through conductor and capacitor problems systematically.

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AP exam review verified for 2027

What is AP Physics C: E&M unit 10?

Unit 10 asks you to reason about how charge behaves in and on conductors, how two conductors share charge when they interact, and how a capacitor stores energy in an electric field. Dielectrics add a final layer by showing how an insulating material between capacitor plates changes every quantity: capacitance, field, voltage, and stored energy.

A conductor in electrostatic equilibrium has zero internal field, all excess charge on its surface, and a uniform potential throughout. A capacitor stores charge Q at potential difference DeltaV with capacitance C = Q/DeltaV. Inserting a dielectric of constant kappa multiplies capacitance by kappa and divides the electric field by kappa.

Conductors in equilibrium

In electrostatic equilibrium, the electric field inside any conductor is zero, excess charge lives entirely on the outer surface, and the conductor is one equipotential. Just outside the surface, the field is perpendicular to the surface and equals sigma/epsilon_0, where sigma is the local surface charge density.

Capacitance and energy storage

A parallel-plate capacitor has capacitance C = kappa*epsilon_0*A/d. The energy stored is U = (1/2)*Q*DeltaV = (1/2)*C*(DeltaV)^2 = Q^2/(2C). These three equivalent forms let you switch between charge-controlled and voltage-controlled scenarios.

Dielectrics and field reduction

A dielectric polarizes in an external field, creating an opposing internal field that reduces the net field between the plates by a factor of kappa. For an isolated capacitor (constant Q), voltage and field both drop by kappa. For a battery-connected capacitor (constant DeltaV), charge increases by kappa.

Charge, field, and energy are always connected

Every calculation in Unit 10 traces back to the relationship between charge distribution, electric field, and electric potential energy. Whether you are applying Gauss's law to find the field outside a charged sphere, using conservation of charge when two conductors touch, or tracking how stored energy changes when a dielectric is inserted, the same core principles from Units 8 and 9 drive the reasoning. Understanding this unit means being able to move fluidly between Q, E, DeltaV, C, and U in any configuration.

AP Physics C: E&M unit 10 topics

10.1

Electrostatics with Conductors

Covers charge distribution in conductors at electrostatic equilibrium: zero internal field, surface charge density sigma, E = sigma/epsilon_0 just outside, equipotential surfaces, and electrostatic shielding by conducting shells.

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10.2

Redistribution of Charge Between Conductors

Covers charge sharing when conductors contact each other, conservation of total charge, equalization of electric potential, grounding as an infinite charge reservoir, and charging by induction using a ground connection.

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10.3

Capacitors

Covers the parallel-plate capacitor, the defining equation C = Q/DeltaV, the geometry-dependent formula C = kappa*epsilon_0*A/d, the uniform field between plates, and the three equivalent expressions for stored energy.

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10.4

Dielectrics

Covers dielectric polarization, the dielectric constant kappa = epsilon/epsilon_0, the field reduction kappa = E_0/E, and how inserting a dielectric changes C, Q, E, DeltaV, and U differently depending on whether the capacitor is isolated or battery-connected.

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Hardest AP Physics C: E&M unit 10 topics

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61%average MCQ accuracy

Across 1.0k multiple-choice practice attempts for this unit.

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Hardest topics in unit 10

MCQ miss rate

10.2

Redistribution of Charge Between Conductors

Review Redistribution of Charge Between Conductors with attention to how the concept appears in AP-style source and evidence questions.

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Unit 10 review notes

10.1

Electrostatics with Conductors

An ideal conductor has electrons free to move. When charge is added or an external field is applied, electrons redistribute in a negligible time until electrostatic equilibrium is reached. At equilibrium, three rules always hold: the electric field inside the conductor is zero, all excess charge resides on the outer surface, and the entire conductor is at a single electric potential (an equipotential surface). Just outside the surface, the field is perpendicular to the surface with magnitude E = sigma/epsilon_0. Charge accumulates more densely at sharp points and edges, producing stronger local fields there. A closed conducting shell shields its interior from external fields entirely, a principle called electrostatic shielding.

E inside = 0: At electrostatic equilibrium, any Gaussian surface drawn inside the conductor encloses zero net field, so no net charge exists in the interior.
Surface field E = sigma/epsilon_0: The electric field just outside a conductor surface points perpendicular to the surface and has magnitude equal to the local surface charge density divided by epsilon_0.
Equipotential conductor: Because E = 0 inside, no work is done moving a charge through the conductor, so every point is at the same potential.
Electrostatic shielding: A closed conducting shell forces E = 0 in its interior regardless of external fields, protecting the enclosed region.
Charge at sharp points: Surface charge density sigma is higher at regions of small radius of curvature, producing stronger fields at tips and edges.

If a solid conducting sphere carries charge +Q, where does the charge sit and what is the field at the center? The charge sits on the outer surface; the field at the center is zero.

Property	Inside conductor	Just outside conductor surface
Electric field E	0	sigma/epsilon_0, perpendicular to surface
Electric potential V	Constant (same as surface)	Continuous across surface
Charge density	0	sigma (nonzero if charged)

10.2

Redistribution of Charge Between Conductors

When two conductors are brought into electrical contact, charge flows until both reach the same electric potential. Total charge is conserved throughout this process. For two identical spheres sharing charge Q_total, each ends up with Q_total/2. Ground acts as an infinite charge reservoir at V = 0; connecting a conductor to ground either drains its charge or, if an external charged object is nearby, allows an induced charge of opposite sign to remain when the ground connection is removed before the external object. This sequence, bring charged object near, ground the conductor, remove ground, then remove the object, is the standard charging by induction procedure.

Equal potential on contact: Charge redistributes until both conductors share the same electric potential, not necessarily the same charge.
Conservation of charge: The total charge on the system of conductors is unchanged by contact or redistribution.
Ground as reservoir: Ground is defined as V = 0 and can supply or absorb any amount of charge without changing its potential.
Charging by induction: A conductor can acquire a net charge without direct contact by grounding it in the presence of an external charge, then removing the ground before removing the external charge.
Charge partitioning: For non-identical conductors in contact, charge distributes in proportion to their individual capacitances so that both reach the same potential.

Two identical conducting spheres, one with charge +6Q and one neutral, are touched together and then separated. What charge does each carry? Each carries +3Q.

Method	Contact required?	Sign of induced charge	Net charge on conductor after
Charging by conduction	Yes	Same as source	Same sign as source
Charging by induction (with grounding)	No	Opposite to external charge	Opposite sign to external charge

10.3

Capacitors

A parallel-plate capacitor consists of two conducting plates of area A separated by distance d. The defining relationship is C = Q/DeltaV. For a parallel-plate capacitor, C = kappa*epsilon_0*A/d, showing that capacitance increases with plate area and decreases with plate separation. The uniform electric field between the plates is E = sigma/epsilon_0 = Q/(epsilon_0*A), derived using a Gaussian pillbox. The potential difference is DeltaV = E*d. Energy stored in the capacitor is U = (1/2)*C*(DeltaV)^2 = (1/2)*Q*DeltaV = Q^2/(2C). The energy density in the electric field between the plates is u = (1/2)*epsilon_0*E^2.

C = Q/DeltaV: Capacitance is the ratio of charge stored on one plate to the potential difference between the plates; units are farads (F).
C = kappa*epsilon_0*A/d: For a parallel-plate capacitor, capacitance depends only on geometry (A and d) and the material between the plates (kappa).
E = Q/(epsilon_0*A): The uniform field between the plates follows from Gauss's law applied to a pillbox surface straddling one plate.
U = (1/2)*C*(DeltaV)^2: Energy stored in a capacitor; equivalent forms are (1/2)*Q*DeltaV and Q^2/(2C), useful depending on which quantities are held constant.
Energy density u = (1/2)*epsilon_0*E^2: Energy stored per unit volume in the electric field between the plates.

A parallel-plate capacitor has plate area A and separation d. If d is doubled while the charge Q is held constant, how does the stored energy change? U = Q^2/(2C) and C halves, so U doubles.

10.4

Dielectrics

A dielectric is an insulating material whose molecules polarize in an external electric field. The polarized dielectric creates an internal field opposing the external field, reducing the net field between the plates. The dielectric constant kappa = epsilon/epsilon_0 quantifies this effect; kappa is always greater than or equal to 1. Inserting a dielectric multiplies capacitance by kappa: C = kappa*C_0. The key distinction for exam problems is whether the capacitor is isolated (constant Q) or connected to a battery (constant DeltaV). For isolated: E and DeltaV each decrease by kappa, U decreases by kappa. For battery-connected: Q increases by kappa, E stays the same, U increases by kappa.

Polarization: In a dielectric, bound charges shift slightly in response to an external field, creating dipoles aligned with the field and an opposing internal field.
kappa = E_0/E: The dielectric constant equals the ratio of the original field to the reduced field after the dielectric is inserted.
C = kappa*C_0: Inserting a dielectric increases capacitance by the factor kappa regardless of whether the capacitor is isolated or connected to a battery.
Isolated capacitor (constant Q): With Q fixed, inserting a dielectric reduces E by kappa, reduces DeltaV by kappa, and reduces stored energy U by kappa.
Battery-connected capacitor (constant DeltaV): With DeltaV fixed, inserting a dielectric increases Q by kappa, keeps E unchanged, and increases stored energy U by kappa.

A capacitor connected to a 12 V battery has capacitance C_0. A dielectric with kappa = 3 is inserted. What is the new charge on the plates? Q = kappa*C_0*DeltaV = 3*C_0*12 = 36*C_0 coulombs.

Quantity	Isolated capacitor (Q constant)	Battery-connected (DeltaV constant)
Capacitance C	kappa * C_0	kappa * C_0
Charge Q	Q_0 (unchanged)	kappa * Q_0
Electric field E	E_0 / kappa	E_0 (unchanged)
Voltage DeltaV	DeltaV_0 / kappa	DeltaV_0 (unchanged)
Stored energy U	U_0 / kappa	kappa * U_0

Practice AP Physics C: E&M unit 10 questions

Try AP-style multiple-choice questions and written prompts after you review the notes.

Example AP-style MCQs

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MCQ

AP-style practice question

Question

A positive point charge is placed off-center inside a spherical cavity within a neutral, spherical conducting shell. Which of the following correctly compares the distribution of the induced surface charge density on the cavity wall ( $\sigma_{cav}$ ) to the charge density on the outer surface of the shell ( $\sigma_{out}$ )?

$\sigma_{cav}$ is non-uniform, while $\sigma_{out}$ is uniform

$\sigma_{cav}$ is uniform, while $\sigma_{out}$ is non-uniform

Both $\sigma_{cav}$ and $\sigma_{out}$ are distributed uniformly

Both $\sigma_{cav}$ and $\sigma_{out}$ are distributed non-uniformly

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MCQ

AP-style practice question

Question

An isolated parallel-plate capacitor contains a dielectric with $\kappa > 1$ . A student removes the dielectric. Which statement correctly justifies the change in the potential difference $\Delta V$ ?

$\Delta V$ increases because the capacitance decreases while the charge remains constant.

$\Delta V$ decreases because the electric field strength decreases as the dielectric is removed.

$\Delta V$ remains constant because the battery is disconnected and cannot supply potential.

$\Delta V$ increases because the stored energy decreases as the dielectric is removed.

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Example FRQs

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FRQ

Charge distribution between identical conducting spheres

4. Two identical conducting spheres, Sphere A and Sphere B, each have radius 2.0 cm and are mounted on insulating stands, as shown in Figure 1. Sphere A has an initial net charge of +6.0 nC, and Sphere B is initially uncharged. The spheres are initially far apart so that their electric influence on each other is negligible. A parallel-plate capacitor is also available. The capacitor has two large conducting plates, each with area 0.020 m^2, separated by 1.0 mm. The capacitor is connected to a 12 V ideal battery, and a dielectric slab with dielectric constant $$

\kappa = 4.0

Figure 1. Two identical conducting spheres on insulating stands, initially far apart, with a removable conducting wire used to touch the spheres together.

Sphere A and Sphere B are brought into contact with each other and then separated so they are far apart again.

Let $Q_A$ be the net charge on Sphere A after they are separated, and let $Q_B$ be the net charge on Sphere B after they are separated.

Indicate whether $Q_B$ is greater than, less than, or equal to $Q_A$ by writing one of the following.

$Q_B > Q_A$
$Q_B < Q_A$
$Q_B = Q_A$

Justify your answer.

Derive an expression for the magnitude of the electric field $E$ at a point on the surface of Sphere A after the spheres have been separated, in terms of $Q_A$ , the sphere radius $R = 2.0\ \text{cm}$ , and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Figure 2. Parallel-plate capacitor (A = 0.020 m^2, plate separation d = 1.0 mm) connected to a 12 V ideal battery, with dielectric slab (κ = 4.0) fully inserted between the plates while still connected.

Indicate whether $Q_\text{new}$ is greater than, less than, or equal to $Q_0$ by writing one of the following. The parallel-plate capacitor (plate area $A = 0.020\ \text{m}^2$ , separation $d = 1.0\ \text{mm}$ ) is connected to a $12\ \text{V}$ ideal battery. The capacitor is initially filled with air. The dielectric slab with dielectric constant $\kappa = 4.0$ is then inserted completely between the plates while the capacitor remains connected to the battery, as shown in Figure 2. Let $Q_\text{new}$ be the new magnitude of the charge on each plate after insertion and let $Q_0$ be the original magnitude of the charge on each plate before insertion.

$Q_\text{new} > Q_0$
$Q_\text{new} < Q_0$
$Q_\text{new} = Q_0$

Briefly justify your answer by referencing your derivation in part B.

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FRQ

Charge distribution in parallel-plate capacitors

3. Students investigate charge distribution in conductors, charge transfer by contact, and the properties of a parallel-plate capacitor. The students first create a known potential difference using two identical conducting spheres. The spheres are then connected to the plates of a parallel-plate capacitor, as shown in Figure 1. The capacitor plates are square with side length 0.120 m, and the plate separation is adjustable. The students have access to an electrometer (very high input resistance) for measuring potential difference between the plates, a set of insulating gloves and handles, a meterstick, a micrometer, a switch, and dielectric sheets of known dielectric constant.

Figure 1. Two identical conducting spheres used to charge a parallel-plate capacitor; an electrometer measures the potential difference ΔV between the plates. The capacitor plates are squares of side length 0.120 m with adjustable separation d.

Describe a procedure the students could use to create a reproducible fixed amount of charge on the capacitor plates using the two identical conducting spheres and then collect data needed to determine the capacitor’s capacitance. Include in your procedure how charge is distributed in a conductor at electrostatic equilibrium, how charge moves when conductors contact, and at least one step that reduces experimental uncertainty.

Describe how the students could use their measurements to graphically determine the capacitance $C$ of the parallel-plate capacitor and how the graph would be analyzed. In your description, specify what would be plotted on each axis and how a feature of the graph (slope or intercept) is related to $C$ through an equation.

Figure 2. Blank grid for graphing two quantities to determine the capacitance C from a straight-line relationship.

Plate separation, d (m)	Measured potential difference, \|ΔV\| (V)
0.0020	14.8
0.0030	22.9
0.0040	30.6
0.0050	39.2
0.0060	46.3

The students keep the charging method the same so that the magnitude of charge $|Q|$ placed on the capacitor plates is approximately constant from trial to trial. They vary the plate separation $d$ and measure the magnitude of the potential difference $|\Delta V|$ between the plates. The data are recorded in Table 1.

Indicate two quantities, either measured quantities from Table 1 or additional calculated quantities, that could be graphed to produce a straight line that could be used to determine $C$ .

Vertical axis: Horizontal axis:

ii.

On the grid provided in Figure 2, create a graph of the quantities indicated in part C(i).

•

Use Table 2 to record the measured or calculated quantities that you will plot.

•

Clearly label the axes, including units as appropriate.

•

Plot the points you recorded in Table 2.

iii.

Draw a best-fit line for the data graphed in part C(ii).

Using the best-fit line from part C(iii) (or an equivalent relationship consistent with your graph), calculate an experimental value for the capacitance $C$ in air. Then calculate the expected potential difference $|\Delta V|$ when the dielectric is inserted and the charge is unchanged. For one run, the students set $d = 0.0040\ \text{m}$ and measure $|\Delta V| = 30.6\ \text{V}$ . The plate area is $A = (0.120\ \text{m})^2$ . Assume the capacitor is in air so $\kappa = 1.00$ and $\varepsilon_0 = 8.85× 10^{-12}\ \text{F/m}$ . In a second run with the same plate separation, the students insert a dielectric sheet that completely fills the space between the plates. The dielectric has $\kappa = 3.2$ , and the capacitor is isolated so the charge on the plates remains the same.

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FRQ

Charge redistribution between conducting spheres and capacitor

2. Two isolated conducting spheres, sphere 1 and sphere 2, have radii $R_1 = 2.0\ \text{cm}$ and $R_2 = 6.0\ \text{cm}$ , respectively. Sphere 1 is given a net charge $+Q_0 = +12\ \text{nC}$ , and sphere 2 is initially uncharged. The spheres are far apart compared to their radii so that each sphere can be treated as an isolated conductor with negligible influence from the other except when they are physically connected. The spheres are then connected by a thin conducting wire of negligible resistance, as shown in Figure 1. After electrostatic equilibrium is reached, the wire is removed. Sphere 2 is then used as one plate of a parallel-plate capacitor. The capacitor consists of two large, parallel conducting plates of area $A = 0.020\ \text{m}^2$ separated by a distance $d = 1.0\ \text{mm}$ . Sphere 2 is connected to the top plate so that all charge on sphere 2 is transferred to the top plate, and the bottom plate is connected to ground. Assume edge effects are negligible, and the region between the plates is initially vacuum.

Figure 1. Two isolated conducting spheres connected by a thin wire to allow charge redistribution.

Figure 2. Bar-chart template for comparing charge magnitude |Q| and potential magnitude |V| for initial and final states of the two-sphere system (with one reference bar provided).

In Figure 2, draw bars to represent the relative magnitudes of (i) the charge |Q| on each sphere and (ii) the electric potential |V| of each sphere for the initial and final states. Use the given bar as a reference. If a quantity is zero, write a "0" in that column. After the wire is connected and electrostatic equilibrium is reached, the wire is removed. The partially completed bar chart in Figure 2 shows a bar that represents the magnitude of the electric potential |V| of sphere 2 in the final state.

Derive an expression for the final charge on sphere 2, $Q_{2f}$ , after the wire is removed, in terms of $Q_0$ , $R_1$ , $R_2$ , and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Figure 3. Axes for sketching potential difference V across the capacitor versus dielectric insertion depth x (0 to d).

Sphere 2 is then touched to (and thereby charges) the isolated top plate of the parallel-plate capacitor, and the bottom plate remains grounded, so the capacitor has charge magnitude $|Q| = |Q_{2f}|$ on its plates. A dielectric slab of thickness $d = 1.0\ \text{mm}$ and dielectric constant $\kappa = 3.0$ is slowly inserted between the plates from one side so that it fills a fraction $x/d$ of the plate area, where $x$ is the insertion depth measured in meters and $0 ≤ x ≤ d$ . On the axes shown in Figure 3, sketch a graph of the potential difference $V$ across the capacitor as a function of $x$ .

Indicate whether the graph you sketched in part C is or is not consistent with your bars in part A. Briefly justify your answer by referencing the functional dependence between $V$ and the capacitance $C$ for a capacitor with fixed charge, and how the effective capacitance changes as the dielectric is inserted. Use $\varepsilon_0 = 8.85× 10^{-12}\ \text{F/m}$ . The capacitor has $A = 0.020\ \text{m}^2$ and $d = 1.0\ \text{mm}$ . The dielectric constant is $\kappa = 3.0$ . The initial charge on sphere 1 is $Q_0 = 12\ \text{nC}$ , and the sphere radii are $R_1 = 2.0\ \text{cm}$ and $R_2 = 6.0\ \text{cm}$ . Assume the capacitor remains isolated after being charged, so the charge magnitude on the plates stays equal to $|Q_{2f}|$ while the dielectric is inserted.

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Key terms

Term	Definition
electrostatic equilibrium	The state in which excess charge in a conductor has fully redistributed to the surface, leaving zero electric field inside and a uniform potential throughout the conductor.
surface charge density	Charge per unit area (sigma) on a conductor surface; the electric field just outside the surface equals sigma/epsilon_0 and points perpendicular to the surface.
equipotential surface	A surface where every point is at the same electric potential; a conductor in electrostatic equilibrium is entirely one equipotential surface.
electrostatic shielding	The elimination of external electric fields inside a closed conducting shell; the interior field is zero regardless of charges or fields outside the shell.
conducting shell	A hollow conductor whose inner and outer surfaces carry induced or excess charge; the field inside the conductor material is zero, and charge distributes on both surfaces depending on internal and external charges.
parallel-plate capacitor	Two parallel conducting plates of area A separated by distance d, storing equal and opposite charges; capacitance is C = kappaepsilon_0A/d.
dielectric material	An insulating material that polarizes in an electric field, reducing the net field between capacitor plates and increasing capacitance by the factor kappa.
superposition principle	The total electric field at any point is the vector sum of fields from each charge distribution individually; used to find the field between capacitor plates by adding contributions from each plate.

Common unit 10 mistakes

Placing excess charge inside a conductor

Excess charge always moves to the outer surface at equilibrium. The interior of a solid conductor has zero charge density and zero field. Only on the surface does sigma appear.

Confusing constant-Q and constant-DeltaV dielectric scenarios

Whether the capacitor is isolated or connected to a battery completely changes what happens to E, DeltaV, and U when a dielectric is inserted. Always identify which quantity is held fixed before calculating.

Forgetting that grounding sets V = 0, not Q = 0

Grounding a conductor forces its potential to zero by allowing charge to flow to or from ground. The conductor does not necessarily end up uncharged; it ends up at zero potential.

Using the wrong energy formula when a variable is held constant

U = (1/2)*C*(DeltaV)^2 is easiest when DeltaV is constant; U = Q^2/(2C) is easiest when Q is constant. Mixing these up leads to errors in dielectric and charge-redistribution problems.

Assuming the field inside a capacitor changes when a dielectric is inserted with a battery connected

For a battery-connected capacitor, DeltaV is fixed, so E = DeltaV/d is also fixed. The field does not change; instead, more charge flows onto the plates to maintain that field through the dielectric.

How this unit shows up on the AP exam

Derive-and-explain tasks combining Gauss's law with conductor rules

AP Physics C: E&M free-response problems frequently ask you to derive the electric field in different regions around a charged conductor or capacitor using Gauss's law, then explain why the field is zero inside the conductor. You must draw a clear Gaussian surface, state the symmetry argument, and carry the algebra through to a final expression with correct units.

Multi-part capacitor problems with changing geometry or dielectric insertion

A common task pattern presents a parallel-plate capacitor in an initial configuration, then changes one variable (plate separation, plate area, or dielectric insertion) and asks you to find new values of C, Q, E, DeltaV, and U. These problems test whether you can identify which quantity is held constant (charge or voltage) and apply the correct energy formula for that constraint.

Qualitative reasoning about charge redistribution and grounding

Exam questions often present a scenario involving two conductors in contact, a grounding event, or charging by induction, and ask you to predict the sign and relative magnitude of charge on each conductor. You are expected to justify your answer using conservation of charge, the equal-potential condition, and the definition of ground as a zero-potential reservoir.

Final unit 10 review checklist

Final Unit 10 review checklistUse this list to confirm you can handle every major task type in Unit 10 before exam day.
State and apply the three equilibrium rules for conductorsE = 0 inside, all excess charge on the surface, and the conductor is one equipotential. Be able to apply each rule to spheres, shells, and cavities.
Use Gauss's law to find E outside a charged conductorDraw a pillbox or spherical Gaussian surface, apply Gauss's law, and arrive at E = sigma/epsilon_0 just outside a flat surface or E = Q/(4*pi*epsilon_0*r^2) outside a spherical conductor.
Solve charge redistribution problemsApply conservation of charge and equal-potential conditions when conductors touch. Correctly identify what grounding does and trace the steps of charging by induction.
Calculate capacitance, field, voltage, and energy for a parallel-plate capacitorUse C = kappa*epsilon_0*A/d, E = Q/(epsilon_0*A), DeltaV = E*d, and all three forms of U. Know which form is most useful when Q or DeltaV is held constant.
Predict dielectric effects for both isolated and battery-connected capacitorsFor each quantity (C, Q, E, DeltaV, U), state whether it increases, decreases, or stays the same when kappa is inserted, and by what factor, for both constant-Q and constant-DeltaV cases.

How to study unit 10

Step 1: Conductor equilibrium rules (Topic 10.1)Read the Topic 10.1 guide and write out the three equilibrium rules from memory. Practice applying Gauss's law with a pillbox to derive E = sigma/epsilon_0 at a conductor surface. Sketch charge distributions for a solid sphere, a hollow shell, and a shell with an internal point charge.

Step 2: Charge redistribution and grounding (Topic 10.2)Work through the Topic 10.2 guide focusing on the equal-potential condition and conservation of charge. Trace the four-step induction sequence in a diagram. Solve two or three charge-sharing problems with identical and non-identical spheres.

Step 3: Capacitor geometry and energy (Topic 10.3)Review the Topic 10.3 guide and derive C = epsilon_0*A/d from Gauss's law. Practice converting between all three energy expressions. Work problems where A, d, Q, or DeltaV changes and you must find the new C, E, or U.

Step 4: Dielectric effects (Topic 10.4)Use the Topic 10.4 guide and the comparison table for isolated vs. battery-connected capacitors. For each quantity (C, Q, E, DeltaV, U), practice stating the factor-of-kappa change and its direction for both scenarios. Then solve mixed problems that combine dielectric insertion with energy calculations.

Step 5: Full-unit FRQ practiceAttempt multi-part FRQ problems that combine conductor rules, capacitor geometry, and dielectric effects in a single setup. Use the AP score calculator to estimate your score and identify which topic areas need additional review.

More ways to review

Topic study guides

Open the individual guides for Unit 10 when you want a closer review of one topic.

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FRQ practice

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Cheatsheets

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Score calculator

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Frequently Asked Questions

What topics are covered in AP Physics E&M Unit 10?

AP Physics E&M Unit 10 covers four topics: Electrostatics with Conductors (10.1), Redistribution of Charge Between Conductors (10.2), Capacitors (10.3), and Dielectrics (10.4). Together these topics explain how charge is stored and redistributed in materials, building the foundation for understanding electric circuits. See the full topic breakdown at AP Physics E&M Unit 10.

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

Unit 10 makes up 10-15% of the AP Physics E&M exam, making it one of the more heavily tested units. It covers conductors, capacitors, and dielectrics, including how charge is stored, redistributed between conductors, and affected by insulating materials placed inside capacitors.

What's on the AP Physics E&M Unit 10 progress check (MCQ and FRQ)?

The AP Physics E&M Unit 10 progress check includes both MCQ and FRQ parts drawn from all four unit topics: Electrostatics with Conductors, Redistribution of Charge Between Conductors, Capacitors, and Dielectrics. MCQ questions test conceptual understanding of conductor behavior and capacitor properties, while the FRQ portion asks you to derive expressions, analyze charge redistribution, and explain how dielectrics change capacitance. Practice with matched questions at AP Physics E&M Unit 10.

How do I practice AP Physics E&M Unit 10 FRQs?

The best way to practice Unit 10 FRQs is to focus on the three topic areas that generate the most free-response questions: Electrostatics with Conductors, Capacitors, and Dielectrics. Typical FRQ prompts ask you to derive capacitance for a given geometry, calculate the effect of a dielectric on stored energy, or justify conductor behavior using Gauss's Law. Work through each derivation step-by-step and write out your reasoning clearly, since partial credit depends on your explanation. Find FRQ practice at AP Physics E&M Unit 10.

Where can I find AP Physics E&M Unit 10 practice questions?

You can find AP Physics E&M Unit 10 practice questions, including multiple-choice and practice test sets, at AP Physics E&M Unit 10. The page includes MCQ questions covering conductors, capacitors, and dielectrics so you can test yourself on each topic individually before taking a full unit practice test.

How should I study AP Physics E&M Unit 10?

Start Unit 10 by solidifying your understanding of conductors in electrostatic equilibrium from Topic 10.1, since that logic carries through the whole unit. Then work through charge redistribution problems in 10.2 using conservation of charge. For capacitors in 10.3, practice deriving capacitance from geometry using Gauss's Law rather than memorizing formulas. Finish with dielectrics in 10.4 and make sure you can explain how inserting a dielectric changes capacitance, electric field, and stored energy. Do at least one timed FRQ per topic so you get comfortable showing your reasoning under pressure. Get study resources at AP Physics E&M Unit 10.

Ready to review Unit 10?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.

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