Skip to content
✍️

AP Physics C: Electricity and Magnetism FRQ Practice Guide

How to attack free-response questions and earn easy partial credit.

⏱️ ~3-4 hours of focused work⚡ AP Physics C: Electricity & Magnetism

The AP Physics C: E&M exam has 3 free-response questions in 45 minutes. Nearly every FRQ falls into one of three archetypes. Master these setups, and you'll know exactly what the rubric demands.

Archetype 1: Gauss's Law

Setup: "A uniformly charged [sphere/cylinder/sheet] has [charge/charge density]. Using Gauss's law, find the electric field E(r)E(r)."

Rubric demands:

  1. State Gauss's law: ∮E⃗⋅dA⃗=Qencϵ0\oint \vec E \cdot d\vec A = \frac{Q_{enc}}{\epsilon_0}.
  2. Choose a Gaussian surface (sphere, cylinder, or pillbox) and justify symmetry.
  3. Explain why EE is perpendicular (or parallel) to dA⃗d\vec A at your surface.
  4. Evaluate the dot product; simplify to E⋅A=Qencϵ0E \cdot A = \frac{Q_{enc}}{\epsilon_0}.
  5. Solve for E(r)E(r). Give separate answers inside and outside the charge distribution.
  6. Often: relate to potential V(r)=−∫E(r) drV(r) = -\int E(r) \, dr or calculate field energy.

Symmetry table (memorize)

Charge distributionGaussian surfaceQencQ_{enc}Dot productResult
Point charge qqSphere, radius rrqqE⋅4πr2E \cdot 4\pi r^2E=kqr2E = \frac{kq}{r^2} or q4πϵ0r2\frac{q}{4\pi\epsilon_0 r^2}
Uniformly charged sphere, radius RR, total charge QQSphere r<Rr < RQr3R3Q \frac{r^3}{R^3} (volume ratio)E⋅4πr2E \cdot 4\pi r^2E=kQrR3E = \frac{kQr}{R^3} (linear in rr)
Uniformly charged sphere, radius RR, total charge QQSphere r>Rr > RQQE⋅4πr2E \cdot 4\pi r^2E=kQr2E = \frac{kQ}{r^2} (like point charge)
Infinite line charge, λ\lambda C/mCylinder, radius rr, length LLλL\lambda LE⋅2πrLE \cdot 2\pi r LE=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r} or kλr\frac{kλ}{r} (using k=1/(4πϵ0)k = 1/(4\pi\epsilon_0), adjust)
Infinite sheet, charge density σ\sigmaPillbox (cylinder, both caps at area AA)σA\sigma AE⋅2AE \cdot 2A (two caps)E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0} (uniform, independent of distance)
Parallel-plate capacitor, surface charge +σ+\sigma and −σ-\sigmaPillbox straddling one plateσAcap\sigma A_{cap} (one side)E⋅AcapE \cdot A_{cap} (one cap inside, one outside)E=σϵ0E = \frac{\sigma}{\epsilon_0} (between plates)

Why symmetry matters: Gauss's law is always true, but only when symmetry lets you pull EE out of the dot product integral do you avoid integration. If you don't cite symmetry, the rubric won't give you full credit for "directly" applying the law.

Archetype 2: Ampère's Law & Magnetic Circuits

Setup: "A long wire (or solenoid) carries current II. Using Ampère's law, find the magnetic field B(r)B(r)."

Rubric demands:

  1. State Ampère's law: ∮B⃗⋅dℓ⃗=μ0Ienc\oint \vec B \cdot d\vec\ell = \mu_0 I_{enc}.
  2. Choose an Amperian loop (circle for wire, rectangle for solenoid) and justify symmetry.
  3. Argue that B⃗\vec B is parallel (or perpendicular) to dℓ⃗d\vec\ell over your loop.
  4. Evaluate the line integral; simplify to B⋅(perimeter)=μ0IencB \cdot (\text{perimeter}) = \mu_0 I_{enc}.
  5. Solve for B(r)B(r). Clearly state inside vs outside (or note if B=0B = 0 outside).
  6. Often: calculate the force on a charge moving through the field, or the energy per unit length in an inductor.

Amperian loop table

GeometryLoopSymmetryB⋅ℓB \cdot \ellResult
Infinite wire, current II at centerCircle, radius rr, parallel to pageB⃗\vec B tangent; perpendicular to radiusB⋅2πrB \cdot 2\pi rB=μ0I2πrB = \frac{\mu_0 I}{2\pi r}
Long solenoid, nn turns/meter, current IIRectangle inside (width ww, height inside)B⃗\vec B parallel to axis inside; B=0B = 0 outsideB⋅wB \cdot w (one side)B=μ0nIB = \mu_0 n I (uniform inside)
Toroid, NN total turns, current II, inner radius r1r_1, outer r2r_2Circle, radius rr, r1<r<r2r_1 < r < r_2B⃗\vec B tangent, varies with rrB(r)⋅2πrB(r) \cdot 2\pi rB(r)=μ0IN2πrB(r) = \frac{\mu_0 I N}{2\pi r}
Thick conducting cylinder, radius RR, uniform current density, II totalCircle inside (r<Rr < R)B⃗\vec B tangent, proportional to enclosed currentB⋅2πrB \cdot 2\pi rB(r)=μ0Ir2πR2B(r) = \frac{\mu_0 I r}{2\pi R^2} (linear in rr)
Thick conducting cylinder, uniform II, r>Rr > R outsideCircle outside (r>Rr > R)B⃗\vec B tangent; all current is enclosedB⋅2πrB \cdot 2\pi rB(r)=μ0I2πrB(r) = \frac{\mu_0 I}{2\pi r} (like thin wire)

💡 Key insight: Ampère's law replaces integration when your loop has symmetry. Choose your loop so that B⃗\vec B is either parallel or perpendicular to dℓ⃗d\vec\ell everywhere on the loop.

Archetype 3: Faraday's Law & Electromagnetic Induction

Setup: "A magnetic field through a loop changes. Find the induced EMF and current, and state the current direction by Lenz's law."

Rubric demands:

  1. State Faraday's law: ε=−dΦBdt\varepsilon = -\frac{d\Phi_B}{dt} (the negative sign is Lenz's law).
  2. Define the magnetic flux: ΦB=∫B⃗⋅dA⃗\Phi_B = \int \vec B \cdot d\vec A (or BAcos⁡θBA\cos\theta if B⃗\vec B is uniform).
  3. Calculate dΦBdt\frac{d\Phi_B}{dt} (is BB changing? Is the area changing? Is the angle changing?).
  4. Write the induced EMF: ε=−dΦBdt\varepsilon = -\frac{d\Phi_B}{dt} (with the sign).
  5. State Lenz's law: "The induced current flows [direction] to oppose the [increasing/decreasing] flux through the loop."
  6. Calculate induced current if resistance is given: I=∣ε∣RI = \frac{|\varepsilon|}{R}.

Flux scenarios & derivatives

ScenarioΦB\Phi_BdΦBdt\frac{d\Phi_B}{dt}ε\varepsilon
Uniform B⃗\vec B perpendicular to area, AA is constant, BB decreases linearly over time Δt\Delta tBABAAdBdt=AΔBΔtA \frac{dB}{dt} = A \frac{\Delta B}{\Delta t}ε=−AΔBΔt\varepsilon = -A \frac{\Delta B}{\Delta t}
Conducting rod of length LL slides perpendicular to B⃗\vec B at velocity vvB⋅x(t)B \cdot x(t) (where x=vtx = vt)BvBv (motional EMF)ε=BLv\varepsilon = BLv
Loop area changes at rate dAdt\frac{dA}{dt}; B⃗\vec B is constantBA(t)B A(t)BdAdtB \frac{dA}{dt}ε=−BdAdt\varepsilon = -B \frac{dA}{dt}
Loop rotates through B⃗\vec B at angular velocity ω\omega, B⃗⊥\vec B \perp loop normal at t=0t=0BAcos⁡(ωt)BA\cos(\omega t)−BAωsin⁡(ωt)-BA\omega \sin(\omega t)ε=BAωsin⁡(ωt)\varepsilon = BA\omega \sin(\omega t) (AC generator)

⚠️ Critical: The negative sign in ε=−dΦBdt\varepsilon = -\frac{d\Phi_B}{dt} is not just a math sign; it embodies Lenz's law. Always cite Lenz when answering the direction question.

Worked Example 1: Gauss's Law for a Uniformly Charged Sphere

Problem: A uniformly charged insulating sphere has radius R=0.10 mR = 0.10 \, m and total charge Q=1.0×10−6 CQ = 1.0 \times 10^{-6} \, C. Using Gauss's law, find the electric field (a) inside at r=0.05 mr = 0.05 \, m, and (b) outside at r=0.20 mr = 0.20 \, m. (c) Calculate the electric potential at r=0.20 mr = 0.20 \, m using V=−∫∞rE drV = -\int_\infty^r E \, dr (for r>Rr > R).

Solution:

(a) Inside the sphere (r<Rr < R):

By symmetry, the charge is distributed uniformly in the volume. Use a Gaussian sphere of radius r=0.05 mr = 0.05 \, m.

Qenc=Qr3R3=1.0×10−6(0.05)3(0.10)3=1.0×10−6×18=1.25×10−7 CQ_{enc} = Q \frac{r^3}{R^3} = 1.0 \times 10^{-6} \frac{(0.05)^3}{(0.10)^3} = 1.0 \times 10^{-6} \times \frac{1}{8} = 1.25 \times 10^{-7} \, C

Apply Gauss's law: ∮E⃗⋅dA⃗=Qencϵ0\oint \vec E \cdot d\vec A = \frac{Q_{enc}}{\epsilon_0}

By spherical symmetry, E⃗\vec E is radial and constant on the sphere surface: E⋅4πr2=Qencϵ0E \cdot 4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}

E=Qenc4πϵ0r2=1.25×10−74π×8.85×10−12×(0.05)2E = \frac{Q_{enc}}{4\pi\epsilon_0 r^2} = \frac{1.25 \times 10^{-7}}{4\pi \times 8.85 \times 10^{-12} \times (0.05)^2}

E=1.25×10−72.78×10−14≈4500 N/CE = \frac{1.25 \times 10^{-7}}{2.78 \times 10^{-14}} \approx 4500 \, N/C

Or in the form E=kQrR3E = \frac{kQr}{R^3} with k=8.99×109k = 8.99 \times 10^9:

E=8.99×109×1.0×10−6×0.05(0.10)3≈4500 N/CE = \frac{8.99 \times 10^9 \times 1.0 \times 10^{-6} \times 0.05}{(0.10)^3} \approx 4500 \, N/C

(b) Outside the sphere (r>Rr > R):

Use a Gaussian sphere of radius r=0.20 mr = 0.20 \, m. All charge is enclosed: Qenc=Q=1.0×10−6 CQ_{enc} = Q = 1.0 \times 10^{-6} \, C

E=Q4πϵ0r2=1.0×10−64π×8.85×10−12×(0.20)2≈225,000 N/CE = \frac{Q}{4\pi\epsilon_0 r^2} = \frac{1.0 \times 10^{-6}}{4\pi \times 8.85 \times 10^{-12} \times (0.20)^2} \approx 225,000 \, N/C

Or: E=kQr2=8.99×109×1.0×10−6(0.20)2≈225,000 N/CE = \frac{kQ}{r^2} = \frac{8.99 \times 10^9 \times 1.0 \times 10^{-6}}{(0.20)^2} \approx 225,000 \, N/C

(c) Electric potential at r=0.20 mr = 0.20 \, m:

For r>Rr > R, the field is that of a point charge: V(r)=−∫∞rE dr=−∫∞rkQr′2 dr′=kQrV(r) = -\int_\infty^r E \, dr = -\int_\infty^r \frac{kQ}{r'^2} \, dr' = \frac{kQ}{r}

V=8.99×109×1.0×10−60.20=4.5×104 VV = \frac{8.99 \times 10^9 \times 1.0 \times 10^{-6}}{0.20} = 4.5 \times 10^4 \, V

Rubric scoring:

  • ✓ Named and applied Gauss's law
  • ✓ Chose Gaussian surface (sphere) and justified by symmetry
  • ✓ Calculated QencQ_{enc} correctly (inside used volume ratio)
  • ✓ Evaluated dot product (EE perpendicular to dA⃗d\vec A; pulled out of integral)
  • ✓ Solved for EE inside and outside separately
  • ✓ Linked to potential via integration
  • Score: 9/9 points

Worked Example 2: RC Circuit with Capacitor Charging

Problem: A capacitor C=100 μFC = 100 \, \mu F is charged through a resistor R=5000 ΩR = 5000 \, \Omega connected to a battery of EMF E=12 V\mathcal{E} = 12 \, V. (a) Find the time constant τ\tau. (b) Write the charge on the capacitor as a function of time Q(t)Q(t) during charging. (c) Find the current I(t)I(t) and evaluate at t=0t = 0 and t=τt = \tau. (d) Calculate the energy stored in the capacitor after 5 time constants.

Solution:

(a) Time constant:

τ=RC=5000×100×10−6=0.5 s\tau = RC = 5000 \times 100 \times 10^{-6} = 0.5 \, s

(b) Charging equation:

The capacitor charges from Q=0Q = 0 toward Q0=CEQ_0 = C\mathcal{E}:

Q0=CV=100×10−6×12=1.2×10−3 CQ_0 = CV = 100 \times 10^{-6} \times 12 = 1.2 \times 10^{-3} \, C

Q(t)=Q0(1−e−t/τ)=1.2×10−3(1−e−t/0.5)Q(t) = Q_0(1 - e^{-t/\tau}) = 1.2 \times 10^{-3} (1 - e^{-t/0.5})

(c) Current:

I(t)=dQdt=Q0τe−t/τ=1.2×10−30.5e−t/0.5=2.4×10−3e−t/0.5 AI(t) = \frac{dQ}{dt} = \frac{Q_0}{\tau} e^{-t/\tau} = \frac{1.2 \times 10^{-3}}{0.5} e^{-t/0.5} = 2.4 \times 10^{-3} e^{-t/0.5} \, A

At t=0t = 0: I(0)=2.4×10−3 A=2.4 mAI(0) = 2.4 \times 10^{-3} \, A = 2.4 \, mA

At t=τ=0.5 st = \tau = 0.5 \, s: I(τ)=2.4×10−3e−1≈0.88×10−3 A≈0.88 mAI(\tau) = 2.4 \times 10^{-3} e^{-1} \approx 0.88 \times 10^{-3} \, A \approx 0.88 \, mA

(d) Energy after 5 time constants:

After t=5τt = 5\tau, the charge is: Q(5τ)=Q0(1−e−5)≈Q0(1−0.0067)≈0.993Q0≈Q0Q(5\tau) = Q_0(1 - e^{-5}) \approx Q_0(1 - 0.0067) \approx 0.993 Q_0 \approx Q_0

The capacitor is nearly fully charged. The energy is:

U=12CV2=12×100×10−6×(12)2=7.2×10−4 J=0.72 mJU = \frac{1}{2}CV^2 = \frac{1}{2} \times 100 \times 10^{-6} \times (12)^2 = 7.2 \times 10^{-4} \, J = 0.72 \, mJ

Or equivalently: U=Q022C=(1.2×10−3)22×100×10−6=7.2×10−4 JU = \frac{Q_0^2}{2C} = \frac{(1.2 \times 10^{-3})^2}{2 \times 100 \times 10^{-6}} = 7.2 \times 10^{-4} \, J

Rubric scoring:

  • ✓ Calculated time constant correctly
  • ✓ Wrote charging equation with Q0=CVQ_0 = CV and exponential decay
  • ✓ Found current as derivative; evaluated at specific times
  • ✓ Recognized steady state (nearly full charge) and energy calculation
  • Score: 8/8 points

FRQ pacing & strategy

For each FRQ (15 min per question):

  1. Read carefully (2 min): Underline what you're asked to find. Is it asking for setup or numerical answer or direction (Lenz)?
  2. State the law (1 min): Write Gauss, Ampère, or Faraday. This is free points if correct.
  3. Choose your geometry (2 min): Gaussian surface, Amperian loop, or coordinate system.
  4. Set up the integral or derivative (3 min): Don't solve yet; make sure the rubric sees your expression.
  5. Solve (5 min): Algebra. If you get stuck, leave the expression and move to the next part.
  6. Check units (1 min): Coulombs, Tesla, Volts, Amperes — do your final answers have the right units?

💡 Pro tip: If you finish a part quickly and move to the next, write a sentence explaining your reasoning (e.g., "By Lenz's law, the induced current is clockwise"). Partial credit rubrics reward explicit reasoning even if the numerical answer is wrong.

Common FRQ errors & recovery

ErrorHow it costs pointsRecovery
"Used Gauss's law but didn't justify symmetry"−2 points (rubric expects "By spherical symmetry..." or "The Gaussian surface is a cylinder...")Always name the surface & say why symmetry holds
"Calculated Faraday's law but forgot the negative sign"−1 point (sign error); −1 more if no Lenz's law citedAlways write ε=−dΦBdt\varepsilon = -\frac{d\Phi_B}{dt}, then cite Lenz for direction
"RC circuit: solved for Q(t)Q(t) but used discharging formula by mistake"−2 points (wrong curve entirely)Check initial condition: charging starts at Q=0Q=0 with (1−e...)(1 - e^{...}); discharging is e−...e^{-...}
"Ampère's law: pulled BB out of integral but B(r)B(r) varies with rr"−1–2 points (invalid simplification)Use Ampère's law only for symmetric geometries where BB is constant on your chosen loop
"No units on final answer"−1 point each (exam asks for numerical + unit)Before moving to next part, write units on every answer

Before exam day

You've got this. 🎯

Start studying
AP Physics C: Electricity & Magnetism topics & practice →

Free interactive lessons, flashcards, and worked examples.

Find weak spots
Take the AP Physics C: Electricity & Magnetism diagnostic →

~10 minutes. Get a personalized study plan.

Other AP Physics C: Electricity & Magnetism study plans

🚨3-Day Cram PlanA focused 72-hour rescue plan when the exam is almost here.📆7-Day Cram PlanOne full week to lock in the highest-leverage topics and FRQ patterns.🗓️1-Month Study PlanA structured 4-week plan that builds mastery without burning out.🌙Last-Minute Review (Night Before)The night-before checklist: top formulas, common traps, and what NOT to do.