AP Physics C: Electricity & Magnetism Study Guide 2026

Unit 0Official AP Questions

Practice Quiz

10 official AP Physics C: E&M multiple-choice questions spanning electrostatics, Gauss's law, electric potential, and capacitance. Select an answer to see immediate feedback.

AP Physics C: E&M Practice Quiz

Questions sourced from College Board released AP Physics C: E&M exams. Select an answer to reveal correct/incorrect feedback, then use ExamGPT to get a full explanation.

10 Questions

Coulomb's Law

1A conducting sphere with a radius of 0.10 meter has

1.0 \times 1 0^{- 9}

coulomb of charge deposited on it. The electric field just outside the surface of the sphere is

Choose 1 answer:

Coulomb's Law

2Two small spheres have equal charges

q

and are separated by a distance

d

. The force exerted on each sphere by the other has magnitude

F

. If the charge on each sphere is doubled and

d

is halved, the force on each sphere has magnitude

Choose 1 answer:

Gauss's Law

3The electric field

E

just outside the surface of a charged conductor is

Choose 1 answer:

Gauss's Law

4A closed surface in the shape of a cube of side

a

is placed in a region where there is a constant electric field of magnitude

E

parallel to the

x

-axis. The total electric flux through the cubical surface is

Choose 1 answer:

Gauss's Law

5A solid nonconducting sphere of radius

R

has a charge

Q

uniformly distributed throughout its volume. A Gaussian surface of radius

r

with

r < R

is used to find the magnitude of the electric field

E

at distance

r

from the center. Which equation results from a correct application of Gauss's law?

Choose 1 answer:

Electric Potential

6A positive charge of

3.0 \times 1 0^{- 8}

coulomb is placed in an upward-directed uniform electric field of

4.0 \times 1 0^{4}

N/C. When the charge is moved 0.5 meter upward, the work done by the electric force on the charge is

Choose 1 answer:

Electric Potential

7A sphere of radius

R

has positive charge

Q

uniformly distributed on its surface. For a point at distance

r < R

from the center, which of the following correctly gives the electric field

E

and the electric potential

V

?

Choose 1 answer:

Capacitance

8Three 6-microfarad capacitors are connected in series with a 6-volt battery. The equivalent capacitance of the set of capacitors is

Choose 1 answer:

Capacitance

9An isolated capacitor with air between its plates has a potential difference

V_{0}

and a charge

Q_{0}

. After the space between the plates is completely filled with oil, the new difference in potential is

V

and the charge is

Q

. Which of the following pairs of relationships is correct?

Choose 1 answer:

Capacitance

10A capacitor is constructed of two identical conducting plates parallel to each other and separated by a distance

d

. The capacitor is charged to a potential difference of

V_{0}

by a battery, which is then disconnected. Assuming edge effects are negligible, what is the magnitude of the electric field between the plates?

Choose 1 answer:

Unit 126–34% of exam

Electrostatics

Electric charge, Coulomb's law, electric fields from discrete and continuous charge distributions, and Gauss's law — the most heavily weighted unit on the exam.

Electrostatics is the study of stationary electric charges and the fields they produce. AP Physics C: E&M is calculus-based — unlike Physics 1, you will integrate over continuous charge distributions and use Gauss's law to find fields with symmetry.

Electric Force and Field

Two point charges exert forces on each other described by Coulomb's law. The electric field is the force per unit positive test charge — it exists at every point in space around a source charge, independent of whether a test charge is there.

The electric field points away from positive sources and toward negative sources.
Fields from multiple charges add as vectors (superposition principle).
Continuous distributions require integrating dE from each infinitesimal charge element dq.

Gauss's Law

Gauss's law states that the total electric flux through any closed surface equals the enclosed charge divided by ε₀. It is always true, but only simplifies the field calculation when the charge distribution has spherical, cylindrical, or planar symmetry. Recognizing which symmetry applies is the key skill for Gauss's law FRQs.

Spherical symmetry (e.g., uniformly charged sphere): use a concentric spherical Gaussian surface → E is constant on the surface → E · 4πr² = Q_enc/ε₀
Cylindrical symmetry (e.g., long line charge): use a coaxial cylindrical Gaussian surface → E · 2πrL = Q_enc/ε₀
Planar symmetry (e.g., infinite sheet of charge): use a pillbox Gaussian surface → E · 2A = Q_enc/ε₀ → E = σ/(2ε₀)

\text{Coulomb's Law: } \vec{F} = k\frac{q_1 q_2}{r^2}\hat{r} \qquad k = \frac{1}{4\pi\varepsilon_0} \approx 8.99\times10^9 \;\text{N·m}^2/\text{C}^2 \\[14pt] \text{Electric Field: } \vec{E} = \frac{\vec{F}}{q_0} = k\frac{Q}{r^2}\hat{r} \qquad \text{Superposition: } \vec{E}_{\text{total}} = \sum_i \vec{E}_i \\[14pt] \text{Gauss's Law: } \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \\[14pt] \text{Sphere (outside, } r>R\text{): } E = \frac{Q}{4\pi\varepsilon_0 r^2} \qquad \text{Line charge: } E = \frac{\lambda}{2\pi\varepsilon_0 r} \qquad \text{Plane: } E = \frac{\sigma}{2\varepsilon_0}

Electric Field Visualizer — Point Charges

Interactive · Custom

Select a charge configuration to see the electric field arrows at every point in space. Arrows point in the direction of E; brightness encodes field strength (∝ 1/r²). Observe how a dipole's field lines form closed loops and how parallel plates create a nearly uniform field between them — the basis of capacitors.

Field lines radiate outward from a positive charge — intensity falls off as 1/r².

+ charge

− charge

Arrow direction = E · Brightness = field strength

Electric Potential

Electric potential V is the electric potential energy per unit charge. It is a scalar — much easier to work with than the vector E field. The relationship between V and E is:

E = −dV/dr (the E-field points from high potential to low potential)
The work done moving charge q from A to B: W = q(V_A − V_B)
Equipotential surfaces (V = constant) are always perpendicular to E field lines.

For a point charge: V = kQ/r. For a continuous distribution, integrate dV = k dq/r over the distribution — this is simpler than integrating the vector field dE.

V = k\frac{Q}{r} \qquad (\text{point charge; }V=0\text{ at }r\to\infty) \\[14pt] V_{\text{total}} = \sum_i k\frac{q_i}{r_i} \qquad \text{(scalar sum — no vector components)} \\[14pt] E = -\frac{dV}{dr} \qquad \text{(in 3D: } \vec{E} = -\nabla V\text{)} \\[14pt] W_{A\to B} = q(V_A - V_B) \qquad U = qV = k\frac{q_1 q_2}{r}

Charges and Fields

Interactive · PhET

Charges and Fields

Place positive and negative charges on the canvas. See the electric field (blue arrows) and equipotential lines (yellow) update in real time. Confirm that equipotentials are always perpendicular to field lines — a key fact for FRQ part answers.

Open simulation

Opens in a new tab · Powered by PhET · University of Colorado Boulder

Key Concepts

Electric field EForce per unit positive test charge; vector quantity in N/C or V/m.

Electric potential VPotential energy per unit charge; scalar in volts (J/C).

Gauss's Law∮E·dA = Q_enc/ε₀ — total flux through closed surface equals enclosed charge/ε₀.

Gaussian surfaceImaginary closed surface chosen to exploit symmetry; field must be uniform on it.

SuperpositionNet field or potential is the sum of contributions from each source charge.

Equipotential surfaceSurface where V is constant; always perpendicular to E field lines.

Exam tip: Gauss's law FRQs always ask you to: (1) draw the Gaussian surface, (2) state why E is uniform on the surface, (3) evaluate the flux integral as E·A, and (4) set equal to Q_enc/ε₀. If the problem has a solid sphere with uniform charge density ρ, you'll need Q_enc inside radius r: Q_enc = ρ·(4/3)πr³. Show every step — the rubric awards points at each stage.

Common mistake: Gauss's law gives the E field only when the field is constant on the Gaussian surface — which requires symmetry. Do not try to use Gauss's law for an asymmetric charge distribution (e.g., two nearby point charges). For those, use superposition of Coulomb's law instead.

Unit 214–17% of exam

Conductors, Capacitors & Dielectrics

Electrostatic equilibrium of conductors, the capacitance of various geometries, energy stored in capacitors, and the effect of dielectric materials.

Conductors in electrostatic equilibrium obey a set of rules that follow directly from Gauss's law and the definition of E inside a conductor:

The electric field inside a conductor is zero in equilibrium — any field would accelerate charges until they redistribute.
Any net charge resides on the outer surface of a conductor.
The electric field just outside a conductor is perpendicular to the surface with magnitude σ/ε₀.
The interior of a conductor is an equipotential; the surface is an equipotential.

Capacitance

A capacitor stores separated charge. The capacitance C is defined as the charge Q stored per unit voltage V across the plates: C = Q/V. Larger C means more charge per volt. For a parallel-plate capacitor, C depends only on geometry — not on Q or V.

Energy in Capacitors

A charged capacitor stores electric potential energy. Three equivalent forms are useful:

U = ½QV (charge × voltage)
U = ½CV² (useful when V is given)
U = Q²/(2C) (useful when Q is given, e.g., isolated capacitor)

Dielectrics

Inserting a dielectric (insulating material) between the plates multiplies the capacitance by the dielectric constant κ: C_new = κC₀. The molecules in the dielectric polarize and partially cancel the E-field inside, allowing more charge for the same voltage.

C = \frac{Q}{V} \quad \text{(definition)} \qquad C_{\text{parallel plate}} = \frac{\varepsilon_0 A}{d} \qquad C_{\text{with dielectric}} = \kappa C_0 = \frac{\kappa\varepsilon_0 A}{d} \\[14pt] U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C} \\[14pt] \text{Series: } \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \qquad \text{Parallel: } C_{\text{eq}} = C_1 + C_2 + \cdots \\[10pt] \text{Energy density: } u = \frac{1}{2}\varepsilon_0 E^2 \;\text{(energy per unit volume)}

Parallel-Plate Capacitor — Capacitance vs. Plate Separation

Interactive · Desmos

Parallel-Plate Capacitor — Capacitance vs. Plate Separation

x = plate separation in mm; y = capacitance in pF (area = 100 cm²). C = ε₀A/d — as plate separation decreases, capacitance rises hyperbolically. Doubling d halves C. This shape appears when dielectrics are removed (d effectively increases) or when analyzing parallel vs. series combinations.

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Key Concepts

Capacitance (C)C = Q/V; ability to store charge per unit voltage. Units: farads (F = C/V).

Electrostatic shieldingE = 0 inside a conductor — the interior is shielded from external fields.

Dielectric constant (κ)Factor by which capacitance increases when dielectric fills the gap; κ ≥ 1.

Series capacitorsSame charge Q on each; voltages add: 1/C_eq = Σ1/Cᵢ.

Parallel capacitorsSame voltage V across each; charges add: C_eq = ΣCᵢ.

Energy densityu = ½ε₀E² — energy stored per unit volume of electric field.

Exam tip: When a fully charged capacitor is disconnected from the battery and a dielectric is inserted: Q is fixed, C increases by κ, so V = Q/C decreases, and U = Q²/(2C) decreases. When the battery remains connected: V is fixed, C increases, so Q = CV increases, and U = ½CV² increases. The FRQ will often ask you to compare before/after — nail down which quantity is held fixed.

Common mistake: The formula C = ε₀A/d applies only to parallel-plate capacitors. Do not use it for cylindrical or spherical capacitors — those require applying Gauss's law to find E between the conductors, then integrating E to find V, then using C = Q/V.

Unit 317–20% of exam

Electric Circuits

Ohm's law, Kirchhoff's rules, series and parallel resistors, power dissipation, and RC circuits analyzed with differential equations.

Electric circuits in Physics C extend beyond the steady-state analysis of Physics 1. The key addition is RC circuits — circuits where a capacitor charges or discharges through a resistor, described by a first-order differential equation. The exponential solutions are directly tested on the FRQ.

Kirchhoff's Rules

Junction rule (KCL): The sum of currents entering a junction equals the sum leaving — conservation of charge.
Loop rule (KVL): The sum of voltage changes around any closed loop is zero — conservation of energy.

For the loop rule: going through a battery from − to + is +ε. Going through a resistor in the direction of assumed current is −IR.

RC Circuits

Applying Kirchhoff's loop rule to a series RC circuit gives the differential equation: ε − IR − Q/C = 0. Substituting I = dQ/dt yields a separable ODE. The solution is exponential with time constant τ = RC.

Physical meaning of τ: after one time constant, the capacitor has reached 63.2% of its final charge (charging) or has lost 63.2% of its initial charge (discharging).

V = IR \quad \text{(Ohm's law)} \qquad P = IV = I^2R = \frac{V^2}{R} \quad \text{(power)} \\[14pt] \text{Series: } R_{\text{eq}} = R_1 + R_2 + \cdots \qquad \text{Parallel: } \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots \\[14pt] \text{RC charging: } Q(t) = C\varepsilon\!\left(1 - e^{-t/\tau}\right), \quad V_C(t) = \varepsilon\!\left(1 - e^{-t/\tau}\right), \quad I(t) = \frac{\varepsilon}{R}e^{-t/\tau} \\[10pt] \text{RC discharging: } Q(t) = Q_0 e^{-t/\tau}, \quad V_C = V_0 e^{-t/\tau}, \quad I = -\frac{V_0}{R}e^{-t/\tau} \\[10pt] \tau = RC \quad (\text{time constant — units: seconds})

RC Circuit — Charging & Discharging

Interactive · Custom

Adjust the resistance R and capacitance C using the sliders and observe how τ = RC changes the charging/discharging curves. The dashed line marks one time constant — at t = τ, the capacitor has charged to 63.2% of the supply voltage. Toggle between Charging and Discharging to see both solutions of the RC differential equation.

Switch closes at t = 0. V_C rises from 0 toward V_s; current decays from V_s/R toward 0.

R = 10 kΩC = 10 μF

V_C(t) — capacitor voltage

I(t) — circuit current

Dashed line = one time constant τ = RC

RC Circuits — Derivation and Intuition

Khan Academy · YouTube

Key Concepts

Ohm's Law (V = IR)Voltage across a resistor equals current × resistance. Applies to ohmic conductors at constant temperature.

Kirchhoff's Junction RuleSum of currents in = sum of currents out at any node; expresses charge conservation.

Kirchhoff's Loop RuleSum of all voltage drops around any closed loop = 0; expresses energy conservation.

Time constant τ = RCCharacteristic time for exponential charging/discharging; after 5τ the capacitor is ~99% charged.

Steady state (RC)Long after switching, capacitor is fully charged (charging) or discharged — current through capacitor is zero.

Power (P = I²R)Rate of energy dissipation in a resistor; a capacitor in steady state dissipates no power.

Exam tip: RC circuit FRQs almost always ask for three things: (1) the differential equation (show ε − IR − Q/C = 0 with I = dQ/dt substituted), (2) the solution Q(t) or V_C(t) with τ = RC identified, and (3) the behavior at t = 0 and t → ∞. At t = 0: the capacitor acts like a wire (V_C = 0 if uncharged) — all voltage drops across R. At t → ∞: the capacitor is fully charged and acts like an open circuit — no current, all voltage across C.

Common mistake: The exponential solution applies only to first-order RC circuits with a DC source and a single time constant. Do not try to apply these formulas to RLC circuits or AC sources — those require second-order differential equations or phasor analysis.

Unit 417–20% of exam

Magnetism and Sources of Magnetic Field

Magnetic force on moving charges and current-carrying conductors, the Biot–Savart law, Ampère's law, and magnetic flux.

Magnetism is the hardest unit conceptually because the forces are three-dimensional cross products. The magnetic force does no work — it only changes the direction of velocity, not speed. This is the single most important qualitative fact about magnetism.

Lorentz Force and Particle Motion

A particle with charge q moving at velocity v in magnetic field B experiences F = qv × B. The direction is found by the right-hand rule: point fingers along v, curl toward B — thumb points in the direction of F for positive charge (reverse for negative).

If v is perpendicular to B, the particle moves in a circle (cyclotron motion). The radius of circular motion: r = mv/(qB). This is directly tested in FRQs about mass spectrometers and velocity selectors.

Biot–Savart vs. Ampère's Law

Two laws let you find B from current:

Biot–Savart law: always valid; integrate dB = (μ₀/4π) I dl × r̂ / r² over the current path. Use for irregular geometries or a single wire segment.
Ampère's law: ∮B·dl = μ₀I_enc — analogous to Gauss's law. Only simplifies calculation when the field has symmetry (long straight wire, solenoid, toroid). Choose a path where B is constant.

\text{Lorentz Force: } \vec{F} = q\vec{v} \times \vec{B} \qquad |\vec{F}| = qvB\sin\theta \quad (\theta \text{ between } v \text{ and } B) \\[14pt] \text{Cyclotron radius: } r = \frac{mv}{qB} \qquad \text{Period: } T = \frac{2\pi m}{qB} \quad (\text{independent of } v) \\[14pt] \text{Force on wire: } \vec{F} = I\vec{L} \times \vec{B} \\[14pt] \text{Biot–Savart: } d\vec{B} = \frac{\mu_0 I}{4\pi}\frac{d\vec{\ell} \times \hat{r}}{r^2} \qquad \text{Long straight wire: } B = \frac{\mu_0 I}{2\pi r} \\[14pt] \text{Ampère's Law: } \oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}} \qquad \text{Solenoid: } B = \mu_0 n I \quad (n = N/L)

Magnets and Electromagnets

Interactive · PhET

Magnets and Electromagnets

Explore magnetic field lines around a bar magnet and a current-carrying coil. Toggle the field meter to observe how B varies with distance — confirms the 1/r dependence for a long wire and shows the nearly uniform field inside a solenoid.

Open simulation

Opens in a new tab · Powered by PhET · University of Colorado Boulder

Key Concepts

Magnetic forceF = qv × B; perpendicular to both v and B; does zero work on the charge.

Right-hand ruleFingers point along first vector, curl toward second — thumb gives direction of cross product.

Cyclotron radiusr = mv/(qB); larger mass/speed → larger circle; larger B or q → smaller circle.

Biot–Savart LawdB = (μ₀I/4π)(dl × r̂/r²); use to integrate B from a current distribution.

Ampère's Law∮B·dl = μ₀I_enc; valid always but only simplifies with high symmetry — solenoid, long wire.

Magnetic flux (Φ_B)Φ_B = ∫B·dA; the 'amount' of B through a surface — central to Faraday's law.

Exam tip: For Ampère's law problems: always state your Amperian loop choice, explain why B is constant on the loop (symmetry), and pull B outside the integral. For a solenoid of N turns and length L: ∮B·dl = B·L (only the inside segment contributes) = μ₀NI → B = μ₀nI where n = N/L. The outside path and end caps contribute zero because B ≈ 0 outside and B ⊥ dl at the ends. Show this reasoning explicitly.

Common mistake: The magnetic force on a moving charge is NOT in the direction of v or B — it is perpendicular to both. A common error is to write F = qvB (missing the cross product direction). Always use the right-hand rule to determine direction; write F = qvB sin θ for magnitude and specify direction separately. Also: stationary charges feel no magnetic force; the charge must be moving.

Unit 511–14% of exam

Electromagnetism — Faraday's Law & Inductance

Electromagnetic induction, Lenz's law, motional EMF, self-inductance, RL circuits, and the conceptual foundations of Maxwell's equations.

Faraday's law is the culmination of E&M — it reveals that a changing magnetic flux induces an EMF, linking electricity and magnetism into a unified electromagnetic theory. This unit also introduces inductance, the magnetic analog of capacitance.

Faraday's Law

The induced EMF in a loop equals the negative rate of change of magnetic flux through the loop: ε = −dΦ_B/dt. The negative sign encodes Lenz's law: the induced current opposes the change that caused it.

Three ways to induce an EMF:

Change the magnetic field B through a fixed loop (electromagnet switching on/off)
Change the area of the loop in a fixed B field (expanding loop)
Change the angle between B and the loop normal (rotating coil — the basis of AC generators)

Motional EMF

A conducting rod of length L moving at velocity v perpendicular to a field B generates EMF = BLv. Charge carriers in the rod experience F = qv × B, separating charge to create a potential difference.

Self-Inductance and RL Circuits

An inductor (coil) stores energy in a magnetic field. Self-inductance L is defined by ε = −L(dI/dt) — a larger L means a larger back-EMF opposing changes in current. An RL circuit has time constant τ = L/R; current grows or decays exponentially, analogous to the RC circuit.

\text{Faraday's Law: } \varepsilon = -\frac{d\Phi_B}{dt} \qquad \Phi_B = \int \vec{B} \cdot d\vec{A} = BA\cos\theta \\[14pt] \text{Motional EMF: } \varepsilon = BLv \qquad \text{(rod moving in uniform field, rod } \perp B \perp v\text{)} \\[14pt] \text{Self-inductance: } \varepsilon_L = -L\frac{dI}{dt} \qquad \text{Solenoid: } L = \mu_0 n^2 A \ell = \frac{\mu_0 N^2 A}{\ell} \\[14pt] \text{Energy stored in inductor: } U = \frac{1}{2}LI^2 \\[10pt] \text{RL current (rising): } I(t) = \frac{\varepsilon}{R}\!\left(1 - e^{-Rt/L}\right) \qquad \tau = \frac{L}{R}

Faraday's Law — Loop Moving Through a Magnetic Field

Interactive · Custom

Drag the conducting loop left and right through the B-field region (blue). Watch the magnetic flux Φ(t) and induced EMF ε(t) = −dΦ/dt update in real time. When the loop is fully inside the field, flux is constant and EMF is zero. When entering or exiting, flux changes and EMF is nonzero. The ↺/↻ symbol shows Lenz's law: the induced current always opposes the flux change.

Lenz's Law: Loop is entering B-field — flux ↑, induced current opposes (creates B into page inside loop).

Φ = 0.040 (norm.)ε = −dΦ/dt = -3.60 (norm.)I: counter-clockwise

Magnetic flux Φ

Induced EMF ε = −dΦ/dt

EMF is zero when flux is constant

Faraday's Law and Lenz's Law — AP Physics C

AP Physics C · YouTube

Key Concepts

Magnetic flux (Φ_B)Φ_B = ∫B·dA = BAcosθ — the 'amount' of B threading through a surface.

Faraday's Lawε = −dΦ_B/dt — EMF induced by any change in flux through a loop.

Lenz's LawThe induced current opposes the change in flux that caused it (the negative sign in Faraday's law).

Motional EMFε = BLv — EMF from a conductor moving perpendicular to a magnetic field.

Self-inductance Lε = −L(dI/dt); L for solenoid = μ₀n²Aℓ. Energy stored: U = ½LI².

RL time constantτ = L/R — current rises or falls exponentially; larger L means slower change.

Exam tip: Faraday's law FRQs often show a rectangular loop with one side moving at velocity v in a uniform field B — motional EMF = BLv where L is the length of the moving side. To find the induced current direction, use Lenz's law: ask 'is flux through the loop increasing or decreasing?' If increasing (loop entering field), the induced current creates B that opposes the increase (field out of page → induced B into page → current clockwise). Always state Lenz's law explicitly in your answer.

Common mistake: EMF is induced only by a CHANGE in flux — not by a large flux. If the loop is stationary in a uniform field, or fully inside a uniform field, the flux is constant and ε = 0. Students frequently assume that being inside a strong magnetic field automatically induces an EMF. It does not. Only dΦ/dt ≠ 0 matters.

Exam prediction: This topic frequently appears on the AP Physics C: E&M exam. See our full AP Physics C: E&M predictions →

AP Physics C: E&M Study Guide

Practice Quiz

AP Physics C: E&M Practice Quiz

Electrostatics

Electric Force and Field

Gauss's Law

Electric Potential

Conductors, Capacitors & Dielectrics

Capacitance

Energy in Capacitors

Dielectrics

Electric Circuits

Kirchhoff's Rules

RC Circuits

Magnetism and Sources of Magnetic Field

Lorentz Force and Particle Motion

Biot–Savart vs. Ampère's Law

Electromagnetism — Faraday's Law & Inductance

Faraday's Law

Motional EMF

Self-Inductance and RL Circuits

Know exactly what to study for AP Physics C: E&M