Must-know sheet

Physics C: E&M must-know sheet

The real AP Physics C: Electricity and Magnetism exam gives you an equation sheet with constants, the main equations and the exam's default assumptions, and you can use a calculator on both sections. This sheet covers what that sheet doesn't tell you: when each law applies, the field and circuit results you derive from it, sign rules, graph shapes and free-response habits.

Units, constants and conversions

Units 8, 9, 10, 11, 12, 13

Constants you'll use most: Elementary charge e = 1.60 × 10⁻¹⁹ C, $k = \frac{1}{4\pi\varepsilon_0}$ ≈ 9.0 × 10⁹ N·m²/C², ε₀ = 8.85 × 10⁻¹² C²/(N·m²) and μ₀ = 4π × 10⁻⁷ T·m/A (about 1.26 × 10⁻⁶). Electron mass is 9.11 × 10⁻³¹ kg and proton mass 1.67 × 10⁻²⁷ kg. The exam's sheet gives you the values; your job is to recognize which one a problem needs.
k and ε₀ are the same constant written two ways: Because $k = \frac{1}{4\pi\varepsilon_0}$ , you can switch freely: results from Gauss's law usually come out with ε₀, and Coulomb-style results with k. For example, $\frac{\lambda}{2\pi\varepsilon_0 r}$ and $\frac{2k\lambda}{r}$ are the same field.
Electric units: Field: N/C, which is the same as V/m. Potential: V = J/C. Capacitance: F = C/V. Current: A = C/s. Resistance: Ω = V/A. Resistivity: Ω·m. Charge densities: λ in C/m, σ in C/m², ρ in C/m³.
Magnetic units: Magnetic field: tesla, T = N/(A·m). Magnetic flux: weber, Wb = T·m². Inductance: henry, H = V·s/A = Wb/A. Earth's field is only about 5 × 10⁻⁵ T, and a strong lab magnet is around 1 T.
Time constants come out in seconds: For RC circuits, Ω·F = (V/A)(C/V) = C/A = s. For LR circuits, H/Ω = (V·s/A)/(V/A) = s. If your τ doesn't come out in seconds, recheck the setup.
Prefixes you'll see constantly: Milli (m) = 10⁻³, micro (μ) = 10⁻⁶, nano (n) = 10⁻⁹, pico (p) = 10⁻¹², kilo (k) = 10³, mega (M) = 10⁶. Charges are often in μC or nC, capacitors in μF to pF, inductors in mH, and resistors in Ω to MΩ.
Electron volt (eV): 1 eV = 1.60 × 10⁻¹⁹ J, the kinetic energy an electron or proton gains moving through a potential difference of 1 V. Convert to joules before using ½mv².
What the exam assumes unless it says otherwise: Batteries, wires and meters are ideal; V = 0 infinitely far from an isolated point charge; current points the way positive charge would drift; resistors and light bulbs are ohmic; capacitors are air-filled, so κ = 1; edge effects at the ends of charged plates are ignored; and solenoids are long, with a uniform field inside and almost none outside. If a question gives internal resistance, a dielectric or a bulb whose resistance changes, use what it gives.
Check units and limiting cases: Before you trust an expression, check its units, then test an extreme: far from any charge distribution the field should look like $\frac{kQ}{r^2}$ , at t = 0 and t → ∞ an RC or LR result should match what the circuit does right away and after a long time, and an answer should never blow up where the physics doesn't.

Signs, vectors and right-hand rules

Units 8, 9, 10, 11, 12, 13

Electric field direction: E points away from positive charges and toward negative charges. The force on a positive charge is along E; the force on a negative charge (like an electron) is opposite to E.
Fields add as vectors, potentials add as numbers: To find a net electric or magnetic field, break each field into components and add, using symmetry to cancel what you can. To find a net potential, just add the signed values of V; there are no components.
Cross product: magnitude and direction: $\lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta$ , largest when the vectors are perpendicular and zero when they're parallel. The result is perpendicular to both vectors, with its direction set by the right-hand rule.
Right-hand rule for the force on a moving charge: For $\vec{F} = q\vec{v} \times \vec{B}$ : point your fingers along v, curl them toward B, and your thumb gives F for a positive charge. For a negative charge, the force points the opposite way. The same rule with I in place of v gives the force on a wire.
Right-hand rule for a straight wire's field: Point your thumb along the conventional current; your fingers curl the way the magnetic field circles the wire. The field is tangent to circles centered on the wire, never pointing toward or away from it.
Right-hand rule for a loop or solenoid: Curl your fingers the way the current goes around; your thumb points along the magnetic field inside the loop, toward its north-pole end. Use the same rule to decide which way an induced current flows.
Into and out of the page: ⊙ (a dot) means a vector points out of the page, toward you; ⊗ (an X) means into the page. Exam diagrams use these for B fields and for currents.
Sign of flux: For a closed surface the area vector points outward, so field leaving gives positive flux and field entering gives negative flux. For an open loop you choose the area vector's direction, and the right-hand rule then sets which way around counts as positive current.
Kirchhoff loop-rule signs: Going across a resistor in the direction of the current, the potential drops by IR; against the current, it rises by IR. Going across a battery from − to +, it rises by ℰ. Going from a capacitor's + plate to its − plate, it drops by $\frac{Q}{C}$ , and going across an inductor in the direction of the current, the change is $-L\frac{dI}{dt}$ .
Energy and work signs: $\Delta U_E = q\Delta V$ , so a positive charge loses potential energy moving to lower V and a negative charge loses it moving to higher V. The work done by the electric field is $W = -\Delta U_E$ , and the work an outside agent does to move a charge slowly (no change in kinetic energy) is $+\Delta U_E$ .

Charge, Coulomb's law and electric fields

Unit 8

Charge is conserved and comes in multiples of e: Any object's charge is q = ne for a whole number n. Net charge changes only when charge moves onto or off an object, usually as electrons, and the total charge of an isolated system never changes.
Coulomb's law: $F = \frac{k\lvert q_1 q_2 \rvert}{r^2}$ along the line joining two point charges: like charges repel, opposite charges attract. It's an inverse-square law, so doubling the distance makes the force one-fourth as large, and by Newton's third law both charges feel the same size force even when their charges differ.
Net force on a charge: Draw each force on the charge from every other charge separately, find components, then add. Look for symmetry first: equal charges at equal distances often cancel one direction. The course keeps these problems to four or fewer charges unless the setup is highly symmetric.
Electric field: force per unit charge: $\vec{E} = \frac{\vec{F}}{q}$ , measured with a small positive test charge. Once you know E at a point, the force on any charge placed there is $\vec{F} = q\vec{E}$ . A point charge makes a field of size $\frac{k\lvert q \rvert}{r^2}$ .
Where can the net field be zero?: For two like charges, the zero point is between them, closer to the smaller charge. For two opposite charges, it's outside the pair on the side of the smaller-magnitude charge. It can never be closer to the bigger charge, and two equal and opposite charges have no zero point at all (except very far away).
Field-line rules: Lines start on positive charges and end on negative charges (or go to infinity), never cross, and are closer together where the field is stronger. The number of lines leaving a charge is proportional to its charge, and lines meet the surface of a conductor in equilibrium at right angles.
A charge in a uniform field moves like a projectile: The acceleration $a = \frac{qE}{m}$ is constant, so the constant-acceleration equations apply, and a charge fired across the field follows a parabola. For electrons and protons the electric force is so much larger than gravity that you can ignore mg; check by comparing qE with mg.
Charging by friction, contact and induction: Friction moves electrons from one material to another. Contact shares charge between conductors (identical conducting spheres end up with equal shares). Induction uses a nearby charged object to push charge around, then a ground connection lets charge flow in or out; break the ground before removing the object, and the conductor keeps a charge opposite to that object's.
Polarization: A nearby charge shifts the charges inside a neutral object, so the side closer to it gets the opposite sign. Because the closer side feels a stronger force, a charged object attracts a neutral one, whether the neutral object is a conductor or an insulator.
Conductors and insulators: In a conductor, like a metal, charges move freely; in an insulator, like glass or plastic, they mostly stay where they're put. That's why excess charge on a conductor spreads over its surface, while an insulator can hold charge anywhere in its volume.

Fields and potentials of spread-out charge (with calculus)

Units 8, 9

Charge densities: Linear density λ = Q/L (C/m), surface density σ = Q/A (C/m²), volume density ρ = Q/volume (C/m³). A small piece of charge is dq = λ dx, σ dA or ρ dV; if the density varies, integrate it to get the total, for example $Q = \int \lambda(x)\,dx$ .
Recipe for a field by integration: Pick a small piece dq, write its field $dE = \frac{k\,dq}{r^2}$ , draw it, and use symmetry to decide which components cancel. Write the surviving component with sin or cos, express everything in one variable, and integrate over the whole object.
Ring of charge, on its axis: $E = \frac{kQx}{(x^2 + R^2)^{3/2}}$ along the axis, where x is the distance from the center. It's zero at the center, peaks at $x = \frac{R}{\sqrt{2}}$ , and becomes $\frac{kQ}{x^2}$ far away. Its potential is $V = \frac{kQ}{\sqrt{x^2 + R^2}}$ .
Semicircular arc, at its center: $E = \frac{2k\lambda}{R} = \frac{2kQ}{\pi R^2}$ , pointing along the arc's line of symmetry (away from the arc for positive charge). The sideways components cancel. A full ring gives zero field at its center.
Finite rod, on its perpendicular bisector: At distance y from the middle of a rod of length L, $E = \frac{kQ}{y\sqrt{y^2 + L^2/4}}$ , perpendicular to the rod. For a very long rod this becomes $\frac{2k\lambda}{y} = \frac{\lambda}{2\pi\varepsilon_0 y}$ , the infinite-line result.
Finite rod, at a point on its own line: At distance a beyond one end of a rod of length L, $E = \frac{kQ}{a(a + L)}$ , along the rod's line, and $V = k\lambda \ln\frac{a + L}{a}$ . Far away (a ≫ L) the field becomes $\frac{kQ}{a^2}$ , like a point charge.
Potential by integration is easier: $V = \int \frac{k\,dq}{r}$ is a scalar, so there are no components to cancel. Every bit of charge on a ring or arc is the same distance R from the center, so V = kQ/R there no matter how the charge is spread around it.
Potential difference near a long line charge: Integrating the line's field $\frac{\lambda}{2\pi\varepsilon_0 r}$ from distance a out to distance b gives $V_a - V_b = \frac{\lambda}{2\pi\varepsilon_0}\ln\frac{b}{a}$ . Because the line is infinitely long, V = 0 at infinity doesn't work here, so you can only find differences in V.
Getting V from E and E from V: $\Delta V = -\int \vec{E} \cdot d\vec{r}$ along any path, and $E_x = -\frac{dV}{dx}$ (or $E_r = -\frac{dV}{dr}$ for spherical symmetry). Use the first when you know the field, and the second when you know V as a function of position.

Electric flux and Gauss's law

Units 8, 12

Electric flux: For a uniform field through a flat surface, $\Phi_E = EA\cos\theta$ , where θ is the angle between E and the area vector (the surface's normal), not the surface itself. In general $\Phi_E = \int \vec{E} \cdot d\vec{A}$ ; units are N·m²/C.
Gauss's law: $\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}$ for any closed surface. Charges outside the surface add zero net flux but still change E at each point, so the law gives you E only when symmetry makes E the same size everywhere you need it. It's the first of Maxwell's four equations.
Choosing a Gaussian surface: Match the symmetry: a concentric sphere for spherical charge, a coaxial cylinder for a long line or cylinder (its flat end caps have zero flux), and a short cylinder or box straddling a plane for planar charge. On each part of the surface E must be either constant and straight through it, or running along it.
Spherical symmetry, outside: Outside any spherically symmetric charge, $E = \frac{kQ}{r^2}$ , exactly as if all the charge sat at the center. This holds for shells, solid spheres and charged conductors.
Spherical symmetry, inside: Inside a uniformly charged insulating sphere, $E = \frac{kQr}{R^3}$ , growing linearly from zero at the center to its maximum at the surface. Inside a thin charged shell, or inside a conductor, E = 0. If the density varies, find $q_{\text{enc}} = \int_0^r \rho(r')\,4\pi r'^2\,dr'$ first, then $E = \frac{kq_{\text{enc}}}{r^2}$ .
Cylindrical symmetry: Outside a long line or cylinder, $E = \frac{\lambda}{2\pi\varepsilon_0 r}$ , falling off as 1/r. Inside a uniformly charged solid insulating cylinder, $E = \frac{\rho r}{2\varepsilon_0}$ . For a varying density, $q_{\text{enc}} = \int \rho\,2\pi r L\,dr$ .
Planar symmetry: A large sheet of charge makes $E = \frac{\sigma}{2\varepsilon_0}$ on each side, the same at every distance. Two large oppositely charged plates make $\frac{\sigma}{\varepsilon_0}$ between them and almost zero outside. Just outside a charged conductor's surface, $E = \frac{\sigma}{\varepsilon_0}$ .
Thick slab of charge: For a large slab of thickness d with uniform density ρ, $E = \frac{\rho y}{\varepsilon_0}$ at distance y from the middle plane inside, and $E = \frac{\rho d}{2\varepsilon_0}$ everywhere outside. The field points away from the middle for positive charge.
Gauss's law for magnetism: $\oint \vec{B} \cdot d\vec{A} = 0$ for every closed surface, because there are no magnetic monopoles: magnetic field lines always form closed loops. Whatever field enters a closed surface also leaves it.

Electric potential energy, potential and energy conservation

Unit 9

Potential energy of two point charges: $U_E = \frac{kq_1q_2}{r}$ with signs included, taking U = 0 when the charges are infinitely far apart. It's positive for like charges and negative for opposite charges, and it belongs to the pair, not to either charge.
Potential energy of a group of charges: Add the energy of every pair: 3 charges have 3 pairs, and 4 charges have 6. This total equals the work an outside agent must do to bring the charges together from very far apart, starting and ending at rest.
Electric potential: V is potential energy per unit charge, measured in volts. A point charge makes $V = \frac{kq}{r}$ with the sign of q included, and V from several charges is just the sum. V can be zero where E isn't (midway between equal and opposite charges), and E can be zero where V isn't (midway between equal like charges).
Uniform field: ΔV = −Ed: Moving a distance d along the field direction, the potential drops by Ed; moving perpendicular to the field, V doesn't change. This is how you get the field between parallel plates: E = ΔV/d.
The field points downhill in potential: Since $E_x = -\frac{dV}{dx}$ , E points toward lower potential, and it's strongest where V changes fastest with distance. On a graph of V against x, the field is the negative of the slope.
Equipotentials: Equipotential lines or surfaces always cross field lines at right angles, and moving a charge along one takes no work. Where equipotentials are drawn at equal steps of V, closer spacing means a stronger field.
Energy conservation for a moving charge: If only the electric force does work, $\Delta K = -q\Delta V$ . From rest, a charge crossing a potential difference ΔV reaches $\frac{1}{2}mv^2 = \lvert q\Delta V \rvert$ . Positive charges speed up heading toward lower V, and electrons speed up heading toward higher V.
Work is path-independent: The electric force is conservative, so the work it does between two points depends only on their potentials, not the path. A charge that returns to its starting point has zero net work done on it by the static field.
Choosing where V = 0: For finite charge distributions, V = 0 at infinity is standard. For an infinite line or plane that choice doesn't work, so pick a convenient reference point; only differences in V matter physically.

Conductors in equilibrium

Units 8, 10

Four facts about a conductor in electrostatic equilibrium: The field inside the metal is zero; any excess charge sits on its surface (the outer surface unless a charge sits inside a cavity); the field just outside is perpendicular to the surface with size σ/ε₀; and the whole conductor, inside and surface, is at one potential.
Charge crowds at sharp points: On an irregular conductor the surface charge density, and so the field just outside, is largest where the surface curves most sharply. That's why lightning rods are pointed and why sparks jump from sharp edges.
A charged conducting sphere: Outside, $E = \frac{kQ}{r^2}$ and $V = \frac{kQ}{r}$ . Inside, E = 0 and V stays at $\frac{kQ}{R}$ all the way to the center. The field jumps to zero at the surface, but the potential never jumps.
Charge in a cavity: If a charge q sits in a hollow inside a conductor, a charge of −q collects on the cavity's wall, and the outer surface carries the conductor's own net charge plus q. The field outside depends only on that outer-surface charge.
Electrostatic shielding: An empty cavity inside a conductor has zero field no matter what charges or fields are outside. This is how a metal box (a Faraday cage) protects what's inside it.
Connected conductors reach the same potential: Charge flows through a connecting wire until V is equal everywhere, and the total charge is conserved. For two far-apart spheres, $\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$ : the bigger sphere ends up with more charge, but the smaller one has the higher surface charge density and the stronger field at its surface.
Identical conductors that touch share equally: Two identical conducting spheres that touch each end up with half the total charge, (Q₁ + Q₂)/2, signs included.
Grounding: The ground acts as a huge conductor at V = 0 that can give or take any amount of charge. A grounded conductor takes whatever charge keeps it at zero potential, which is how charging by induction works: with a charged rod nearby, ground, disconnect the ground, then remove the rod.

Capacitors and dielectrics

Units 10, 11

Capacitance: $C = \frac{Q}{\Delta V}$ , where Q is the size of the charge on either plate (the plates hold +Q and −Q, so the net charge is zero). C depends only on the capacitor's shape, size and the material between the plates, not on Q or ΔV.
Parallel plates: $C = \frac{\kappa\varepsilon_0 A}{d}$ . The field between the plates is nearly uniform with $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$ in vacuum, and $\Delta V = Ed$ . Bigger plates or a smaller gap give more capacitance.
Spherical and cylindrical capacitors: Concentric spheres (radii a < b): $C = \frac{4\pi\varepsilon_0 ab}{b - a}$ , and an isolated sphere of radius R (b → ∞) has $C = 4\pi\varepsilon_0 R$ . Coaxial cylinders of length L: $C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}$ . To derive either, use Gauss's law for E between the conductors, integrate to get ΔV, then divide Q by ΔV.
Energy stored: $U_C = \frac{1}{2}Q\Delta V = \frac{1}{2}C(\Delta V)^2 = \frac{Q^2}{2C}$ . It equals the work done to separate the charge, and the energy is stored in the electric field between the plates.
Isolated or connected? Decide first: An isolated capacitor (disconnected) keeps its charge Q fixed. A capacitor connected to a battery keeps its ΔV fixed. Every 'what happens if…' question starts by deciding which one stays constant.
Changing the plate separation: Isolated (Q fixed): E stays the same, while ΔV and U both grow in proportion to d. Connected (ΔV fixed): C, Q, E and U all shrink as $\frac{1}{d}$ .
Inserting a dielectric: C becomes κC (κ ≥ 1; vacuum is 1 and air is about 1). Isolated: Q stays the same, while ΔV, E and U all drop by a factor of κ. Connected: ΔV and E stay the same, while Q and U both grow by a factor of κ.
Why a dielectric works: The insulator's molecules polarize in the field, building up a thin layer of opposite charge against each plate. That opposing field weakens the net field, so the same charge sits at a lower ΔV, and the capacitance rises.
Capacitors in parallel and in series: Parallel: same ΔV on each, charges add, and $C_{\text{eq}} = C_1 + C_2 + \cdots$ . Series: same Q on each, voltages add, and $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$ , so the total is less than the smallest one. These are the opposite of the resistor rules.
Partly filled capacitors: A dielectric slab that fills part of the gap (layered across the whole plate area) acts like two capacitors in series. A slab that fills the whole gap over only part of the plate area acts like two capacitors in parallel. A metal slab of thickness t that doesn't touch the plates leaves only the gap d − t with any field, so $C = \frac{\varepsilon_0 A}{d - t}$ .

Current, resistance and DC circuits

Unit 11

Current: $I = \frac{dq}{dt}$ , so the charge that passes is $Q = \int I\,dt$ , the area under an I–t graph. Conventional current points the way positive charge would move; in metal wires, electrons actually drift the opposite way.
Drift velocity and current density: $I = nqv_dA$ , where n is the number of charge carriers per cubic meter. Drift speeds are tiny (typically a fraction of a millimeter per second), even though a circuit responds almost instantly. Current density is $J = \frac{I}{A}$ ; if it varies across a wire, $I = \int J\,dA$ , for example $\int_0^R J(r)\,2\pi r\,dr$ .
Resistance of a wire: $R = \frac{\rho\ell}{A}$ , with resistivity ρ in Ω·m set by the material (and rising with temperature for metals). Doubling the length doubles R, and doubling the diameter makes R one-fourth as large. If the resistivity changes along the length of a uniform wire, add up thin slices: $R = \frac{1}{A}\int_0^{\ell} \rho(x)\,dx$ .
Ohm's law and ohmic materials: ΔV = IR. For an ohmic resistor, a graph of I against ΔV is a straight line through the origin with slope 1/R. On the exam, resistors and light bulbs count as ohmic unless a question says otherwise. A real bulb filament isn't: its resistance rises as it heats, so its graph curves.
Power: $P = I\Delta V$ for any element; for a resistor, $P = I^2R = \frac{(\Delta V)^2}{R}$ . A battery delivers Iℰ. Energy = Pt, and 1 kWh = 3.6 × 10⁶ J.
Ranking bulb brightness: Brightness follows power. In series (same I), the bigger resistance is brighter because P = I²R. In parallel (same ΔV), the smaller resistance is brighter because $P = \frac{(\Delta V)^2}{R}$ .
Series and parallel resistors: Series: same current, voltages add, $R_{\text{eq}} = R_1 + R_2 + \cdots$ . Parallel: same ΔV, currents add, $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$ , so the total is less than the smallest branch. With an ideal battery, adding a parallel branch raises the total current but doesn't change the current in the other branches.
Open and short circuits: An open switch or a break means no current in that path. A wire placed across part of a circuit (a short) has no potential difference across it, so current takes the wire and skips the elements it goes around, while the rest of the circuit still works.
Real batteries: internal resistance: A real battery is an ideal emf ℰ in series with a small internal resistance r, so its terminal voltage is $\Delta V = \mathcal{E} - Ir$ . The terminal voltage equals ℰ only when no current flows, and it drops as you draw more current.
Meters: Ammeters go in series with the element and ideally have zero resistance; voltmeters go in parallel across it and ideally have infinite resistance. A real ammeter adds a little resistance, and a real voltmeter draws a little current, so each slightly changes what it measures.
Kirchhoff's rules: Loop rule: the potential changes around any closed loop add to zero (conservation of energy). Junction rule: current into a junction equals current out (conservation of charge). Label a current in each branch with a guessed direction, write enough junction and loop equations to match the unknowns, and solve; a negative answer just means the current flows the other way. The exam won't ask about batteries of different emf connected in parallel.
Potential around a loop: If you graph V while walking around a loop, it rises across a battery (− to +), drops across each resistor in the direction of the current, and stays flat along ideal wires. Every point on one ideal wire is at the same potential.

RC circuits

Unit 11

Just after the switch closes: A capacitor's voltage can't change instantly, so an uncharged capacitor acts like a plain wire at first (no voltage across it), and the starting current in a series RC circuit is $\frac{\mathcal{E}}{R}$ . A charged capacitor at that instant acts like a battery with $\Delta V = \frac{Q}{C}$ .
After a long time: A fully charged capacitor acts like a break in the circuit: no current flows in its branch. Its voltage equals the voltage across whatever it is in parallel with, which you find by analyzing the rest of the circuit as if the capacitor's branch weren't there.
Charging: $Q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right)$ , $I(t) = \frac{\mathcal{E}}{R}e^{-t/RC}$ . The capacitor's voltage rises as $\mathcal{E}\left(1 - e^{-t/RC}\right)$ while the resistor's voltage falls as $\mathcal{E}e^{-t/RC}$ ; the two always add to ℰ.
Discharging: $Q(t) = Q_0e^{-t/RC}$ and $I(t) = \frac{Q_0}{RC}e^{-t/RC} = \frac{\Delta V_0}{R}e^{-t/RC}$ . Charge, voltage and current all decay with the same time constant, and the current flows the opposite way to the charging current.
Time constant τ = RC: After one τ, a charging capacitor has about 63% of its final charge, and a discharging one keeps about 37% of its starting charge. After about 5τ the process is essentially finished, and the half-life is RC ln 2 ≈ 0.69RC.
Which R goes in τ?: Use the equivalent resistance (and, with several capacitors, the equivalent capacitance) in the path the capacitor charges or discharges through. If it charges through one set of resistors and discharges through another, the two time constants are different.
Deriving the RC equation: Apply the loop rule with $I = \frac{dq}{dt}$ . Charging: $\mathcal{E} - R\frac{dq}{dt} - \frac{q}{C} = 0$ . Discharging: $R\frac{dq}{dt} + \frac{q}{C} = 0$ . Separate the variables, integrate with the starting and ending values as limits, and check that your answer matches the t = 0 and long-time behavior.
Energy when charging: Charging an uncharged capacitor fully from a battery of emf ℰ: the battery supplies $C\mathcal{E}^2$ , the capacitor stores $\frac{1}{2}C\mathcal{E}^2$ , and the resistor turns the other $\frac{1}{2}C\mathcal{E}^2$ into thermal energy, whatever the value of R.

Magnetic fields and magnetic forces

Unit 12

Where magnetic fields come from: Moving charges and currents make magnetic fields, and so do magnets, which always have both a north and a south pole. Outside a magnet, field lines run from north to south; inside, they continue from south to north, closing the loop. Earth's geographic north is near a magnetic south pole, which is why compass needles point north.
Magnetic materials: Ferromagnetic materials (iron, nickel, cobalt) are strongly attracted and can become permanent magnets. Paramagnetic materials are weakly attracted, and every material has a weak diamagnetic response that pushes it away. Permeability μ measures how strongly a material magnetizes; μ₀ is the value for vacuum.
Magnets and current loops are dipoles: A bar magnet and a current loop both act as magnetic dipoles with a north and a south side; cutting a magnet just makes two smaller dipoles, since no lone poles exist. In a uniform field a dipole feels no net force but a torque that turns it to line up with the field, like a compass needle. In a nonuniform field it also feels a net force.
Force on a moving charge: $\vec{F}_B = q\vec{v} \times \vec{B}$ , size $\lvert q \rvert vB\sin\theta$ . A charge at rest, or one moving parallel to the field, feels no magnetic force.
Magnetic forces do no work: The force is always perpendicular to the velocity, so it changes a charge's direction but never its speed or kinetic energy. To speed up a charge you need an electric field.
Circular motion in a uniform field: A charge moving perpendicular to B travels in a circle with $r = \frac{mv}{\lvert q \rvert B}$ . Its period $T = \frac{2\pi m}{\lvert q \rvert B}$ doesn't depend on speed. If the velocity also has a component along B, the path is a helix.
Velocity selector: When perpendicular E and B fields push a charge in opposite directions, it passes straight through only if qE = qvB, so $v = \frac{E}{B}$ , whatever its charge or mass.
Force on a current-carrying wire: $\vec{F} = I\vec{\ell} \times \vec{B}$ , size $I\ell B\sin\theta$ , where $\vec{\ell}$ points along the current. For a curved wire, integrate $I\,d\vec{\ell} \times \vec{B}$ ; in a uniform field, a closed loop feels zero net force, though it can still feel a torque.
Force between parallel wires: Per unit length, $\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2\pi d}$ . Currents in the same direction attract, and currents in opposite directions repel; the forces on the two wires are equal and opposite.
Hall effect: In a current-carrying strip in a magnetic field, the magnetic force pushes the moving charges to one side until the electric field from the built-up charge balances it: $qE = qv_dB$ . The Hall voltage across the strip's width w is $\Delta V_H = v_dBw$ , and which side ends up positive shows the sign of the charge carriers.

Magnetic fields from currents: Biot–Savart and Ampère's law

Unit 12

Biot–Savart law: $d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{\ell} \times \hat{r}}{r^2}$ : each small piece of current adds a bit of field perpendicular to both the current and the line to the point. Pieces lined up with the point add nothing, since the cross product is zero. The course uses it for the center of an arc, the axis of a loop and the bisector of a straight wire.
Field of a moving point charge: $\vec{B} = \frac{\mu_0}{4\pi}\frac{q\vec{v} \times \hat{r}}{r^2}$ , the point-charge version of Biot–Savart. It's zero straight ahead of and behind the charge, and largest to the side.
Long straight wire: $B = \frac{\mu_0 I}{2\pi r}$ , circling the wire, from either Ampère's law or Biot–Savart with an infinitely long wire. Double the distance and the field halves.
Finite and half-infinite straight wires: On the perpendicular bisector of a wire of length L, at distance r: $B = \frac{\mu_0 I L}{2\pi r\sqrt{L^2 + 4r^2}}$ , which becomes $\frac{\mu_0 I}{2\pi r}$ for a long wire. At distance r beside the end of a half-infinite wire, measured perpendicular to the wire, B is half the long-wire value, $\frac{\mu_0 I}{4\pi r}$ .
Circular loops and arcs: At the center of a loop, $B = \frac{\mu_0 I}{2R}$ (times N for N turns). At the center of an arc spanning angle θ in radians, $B = \frac{\mu_0 I\theta}{4\pi R}$ . On the axis of a loop, $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$ , pointing along the axis.
Ampère's law: $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$ around any closed loop. Like Gauss's law, it gives B only when symmetry lets you choose a loop where B is constant and along the path (or perpendicular to it) on each piece. Count enclosed currents as positive or negative using the right-hand rule.
Inside and outside a thick wire: For a wire of radius R carrying I spread evenly: inside, $B = \frac{\mu_0 I r}{2\pi R^2}$ , growing linearly from zero at the center; outside, $B = \frac{\mu_0 I}{2\pi r}$ . If the current density varies, find $I_{\text{enc}} = \int_0^r J\,2\pi r'\,dr'$ first. Outside a coaxial cable carrying equal and opposite currents, B = 0.
Long solenoid: $B = \mu_0 nI$ inside, where n = N/ℓ is the number of turns per meter. The field inside is uniform and doesn't depend on the radius, and the field outside is nearly zero. An iron core multiplies the field greatly (μ in place of μ₀).
Current slab or sheet: For a wide slab of thickness d with uniform current density J, the field outside is $B = \frac{\mu_0 Jd}{2}$ , the same at every distance but pointing opposite ways on the two sides. Inside, at distance y from the middle plane, $B = \mu_0 Jy$ .
Changing electric fields also make magnetic fields: Maxwell added a term to Ampère's law: a changing electric field, like the one between the plates of a charging capacitor, makes a magnetic field just as a current does. The course treats this idea qualitatively only.

Magnetic flux, Faraday's law and Lenz's law

Unit 13

Magnetic flux: For a uniform field through a flat loop, $\Phi_B = BA\cos\theta$ , where θ is between B and the loop's area vector (its normal). Flux is largest when the field passes straight through and zero when the field runs along the loop. If B varies, integrate $\int \vec{B} \cdot d\vec{A}$ ; for a rectangle of length L beside a long wire, from distance a to b, $\Phi_B = \frac{\mu_0 IL}{2\pi}\ln\frac{b}{a}$ .
Faraday's law: $\mathcal{E} = -N\frac{d\Phi_B}{dt}$ for a coil of N turns. Flux can change three ways: the field changes, the area changes, or the angle changes. The induced current is $I = \frac{\lvert \mathcal{E} \rvert}{R}$ .
Only change matters: A huge but steady flux induces nothing. A loop sliding through a uniform field while staying completely inside it has no induced current, because its flux isn't changing.
Lenz's law: finding the direction: The induced current makes its own field that opposes the change in flux. Step 1: which way does the outside field point through the loop? Step 2: is the flux increasing or decreasing? Step 3: if increasing, the induced field points the opposite way; if decreasing, the same way. Step 4: use the right-hand rule to turn that field direction into a current direction.
Motional emf: A rod of length ℓ moving at speed v perpendicular to B has $\mathcal{E} = B\ell v$ between its ends. In a circuit of resistance R, $I = \frac{B\ell v}{R}$ , and the magnetic force on the rod, $\frac{B^2\ell^2v}{R}$ , opposes its motion.
Sliding bar on rails: With a constant applied force F, the bar speeds up until the magnetic drag balances F, at terminal speed $v = \frac{FR}{B^2\ell^2}$ ; on vertical rails, $v = \frac{mgR}{B^2\ell^2}$ . With no applied force, Newton's second law gives $v(t) = v_0e^{-B^2\ell^2t/(mR)}$ . At terminal speed, the power the applied force puts in, Fv, equals the I²R dissipated in the resistor.
Loop moving into and out of a field: An emf of Bℓv appears only while one edge is inside the field and the other is outside, so the flux is changing. Entering and leaving give currents in opposite directions, but the magnetic force opposes the motion both times, and with the loop fully inside there is no current at all.
Rotating coil (generator): A coil of N turns and area A spinning at angular speed ω in a uniform field has $\Phi_B = BA\cos\omega t$ per turn, so $\mathcal{E} = NBA\omega\sin\omega t$ , with a maximum of NBAω. The emf is largest when the coil's plane is parallel to the field, where the flux is zero but changing fastest.
Induced electric fields: A changing magnetic field makes a circulating electric field, $\oint \vec{E} \cdot d\vec{\ell} = -\frac{d\Phi_B}{dt}$ , even with no wire present. This field isn't conservative, so the idea of potential doesn't apply to it. Faraday's law is one of Maxwell's four equations.
Eddy currents and magnetic braking: Changing flux through a solid piece of metal drives swirling eddy currents, and their fields oppose the motion. A magnet dropped through a copper pipe falls slowly, and induction brakes slow trains without touching the track.

Inductors, LR circuits and LC circuits

Unit 13

Self-induced emf: $\mathcal{E}_L = -L\frac{dI}{dt}$ : an inductor fights changes in current, not the current itself. When the current is increasing, the inductor's potential drops in the direction of the current by $L\frac{dI}{dt}$ ; with a steady current, it has no voltage across it at all.
Inductance of a solenoid: $L = \frac{N\Phi_B}{I}$ , which for a long solenoid gives $L = \frac{\mu_0 N^2 A}{\ell}$ (with a core, use the core material's permeability μ in place of μ₀). Doubling the number of turns, keeping the length the same, makes L four times as large.
Energy stored in an inductor: $U_L = \frac{1}{2}LI^2$ , stored in the magnetic field the current makes. This energy must go somewhere when the current stops, which is why opening an inductor's circuit suddenly can cause a spark.
Just after a switch, and after a long time: An inductor's current can't change instantly. Right after a switch closes, an inductor with no current acts like a break in the circuit, and the starting rate of change is $\frac{dI}{dt} = \frac{\mathcal{E}}{L}$ in a simple series circuit. After a long time the current is steady, and the inductor acts like a plain wire.
LR circuit: current growing: From the loop rule $\mathcal{E} = IR + L\frac{dI}{dt}$ : $I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right)$ with $\tau = \frac{L}{R}$ . The inductor's voltage starts at ℰ and decays as $\mathcal{E}e^{-t/\tau}$ .
LR circuit: current decaying: When the battery is removed and the inductor drives current through a resistor, $I(t) = I_0e^{-t/\tau}$ with $\tau = \frac{L}{R}$ . Note the difference from RC: more resistance makes an LR circuit faster but an RC circuit slower.
LC circuit oscillations: The loop rule gives $\frac{d^2q}{dt^2} = -\frac{1}{LC}q$ , so $q(t) = Q_{\max}\cos(\omega t + \phi)$ with $\omega = \frac{1}{\sqrt{LC}}$ and period $T = 2\pi\sqrt{LC}$ . The current is the derivative of q, so it is largest when the capacitor is empty.
LC energy: $\frac{q^2}{2C} + \frac{1}{2}LI^2$ stays constant. All the energy is in the capacitor when I = 0 and all in the inductor when q = 0, a quarter period later, so $\frac{Q_{\max}^2}{2C} = \frac{1}{2}LI_{\max}^2$ and $I_{\max} = \omega Q_{\max} = \frac{Q_{\max}}{\sqrt{LC}}$ .
The spring analogy: An LC circuit matches a mass on a spring: charge q is like position x, current I is like velocity v, inductance L is like mass m, and $\frac{1}{C}$ is like the spring constant k. That's why $\omega = \frac{1}{\sqrt{LC}}$ matches $\omega = \sqrt{\frac{k}{m}}$ .

Graphs, labs and free-response habits

Units 8, 9, 10, 11, 12, 13

The four free-response question types: Section II has one each, in this order: Mathematical Routines (10 points), Translation Between Representations (12 points), Experimental Design and Analysis (10 points) and Qualitative/Quantitative Translation (8 points). Calculators and the equation sheet are allowed throughout.
E and V graphs for spheres: Conducting sphere: E is zero inside, jumps to kQ/R² at the surface, then falls as 1/r²; V is flat at kQ/R inside, then falls as 1/r. Uniform insulating sphere: E rises linearly inside to a peak at the surface, then falls as 1/r². V is always continuous, even where E jumps.
Slopes and areas to know: Slope of V against x is $-E_x$ , and the area under E against x is −ΔV. Area under I against t is the charge, and the slope of q against t is the current. The slope of magnetic flux against t gives the emf (with a minus sign, times N), and the slope of I against ΔV for a resistor is 1/R.
Exponential graph shapes: Charging capacitor charge and growing LR current rise steeply, then level off toward a final value. RC current, discharging charge and decaying LR current start at their maximum and fall toward zero. Each curve's starting slope, extended in a straight line, reaches the final value after one time constant.
Linearizing data: Rearrange into the form y = (slope)x + b. Examples: $\ln(\Delta V_C)$ against t during discharge has slope $-\frac{1}{RC}$ ; B against 1/r near a long wire has slope $\frac{\mu_0 I}{2\pi}$ ; B against I in a solenoid has slope μ₀n; C against 1/d for plates has slope κε₀A; and R against length has slope ρ/A.
Derivations: Start from a basic law written in general form (Gauss's law, Ampère's law, Kirchhoff's rules, Faraday's law, Newton's second law or energy conservation), apply it to this situation, and show each substitution. Keep letters until the end, use only given quantities and constants, and finish with the requested quantity alone on one side.
Differential equations: Write the equation from the loop rule or Newton's second law, separate the variables, and integrate with the starting and ending values as limits. Then check that your answer gives the right value at t = 0 and after a long time.
'What if it doubles?' questions: Write the relationship, cancel what stays the same, then scale. In $B = \frac{\mu_0 I}{2\pi r}$ , doubling r halves B; in $E = \frac{kQ}{r^2}$ , doubling r makes E one-fourth as large. For capacitors and circuits, first decide whether Q or ΔV stays fixed.
Justifying a claim: Name the law or principle, apply it to the specific objects in the problem, and state the conclusion with a direction or trend. Make sure your words agree with your equations, diagrams and graphs.
Designing an experiment: Name what you'll change, what you'll measure and what you'll keep the same, and say which equipment measures each quantity (ammeter, voltmeter, stopwatch, ruler, magnetic field probe, multimeter for resistance). Test a wide range of values, repeat trials, and plan a graph that will be a straight line.
Best-fit lines: Label each axis with the quantity and unit, and use most of the grid. Draw one straight line with points scattered evenly on both sides, and find its slope from two points on the line that are far apart, not from data points; give the slope its units.