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AP Physics C: E&M — Core flashcards

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Calculus-based electricity and magnetism for AP Physics C: E&M: fields of charge distributions, Gauss's law, potential, conductors and capacitors, RC transients, Biot–Savart and Ampère's law, induction, and RL and LC circuits.

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Learn deck · part of AP Physics C: Electricity and Magnetism · written against: College Board AP Physics C: Electricity and Magnetism Course and Exam Description (2024 revision). Updated 2026-10-10.

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  1. How is the Coulomb constant kk related to the permittivity of free space ε₀, and what is its value?
    k = 1/(4πε₀) ≈ 8.99 × 10⁹ N·m²/C², so Coulomb's law F=kq1q2r2F = \frac{kq_1q_2}{r^2} can equally be written with 4πε₀ in the denominator.
    Electric Charges Fields and Gauss's Law › Coulomb's Law and Fields
  2. Charges +q+q and −q-q are a distance dd apart. What is the electric field at the midpoint?
    Both fields point toward −q-q and add: E=2⋅kq(d/2)2=8kqd2E = 2\cdot\frac{kq}{\left(d/2\right)^2} = \frac{8kq}{d^2}⁠, directed toward the negative charge.
    Electric Charges Fields and Gauss's Law › Coulomb's Law and Fields
  3. How do you find a potential difference from the electric field?
    ΔV=−∫ABE⃗⋅dl⃗\Delta V = -\int_A^B \vec{E}\cdot d\vec{l}⁠. Moving along the field lowers the potential.
    Electric Potential
  4. The potential in a region is V(x)=3x2−2xV(x) = 3x^2 - 2x (volts, meters). What is the electric field at x=1x = 1 m?
    Ex=−dVdx=−6x+2E_x = -\frac{dV}{dx} = -6x + 2⁠, so Ex=−4E_x = -4 V/m at x=1x = 1 m — 4 V/m pointing in the −x-x direction.
    Electric Potential
  5. Why is the field zero inside an empty cavity in a conductor, whatever charges sit outside?
    A Gaussian surface inside the metal around the cavity encloses no net charge, and because the whole conductor is one equipotential, no field line can run from one point of the cavity wall to another. This is how a Faraday cage shields its contents.
    Conductors and Capacitors › Conductors
  6. Why is the field just outside a conductor σ/ε₀ while a nonconducting sheet gives σ/(2ε₀)?
    Inside the conductor the field is zero, so all the flux through the Gaussian pillbox leaves through the outside face; the sheet sends half its flux out each side.
    Conductors and Capacitors › Conductors
  7. The charge passing a point in a wire is Q(t)=4t2Q(t) = 4t^2 (coulombs, seconds). What is the current at t=3t = 3 s?
    I=dQdt=8t=24I = \frac{dQ}{dt} = 8t = 24 A at t=3t = 3 s.
    Electric Circuits › Current Resistance and Power
  8. How is the current in a wire related to the drift velocity of its charge carriers?
    I=nqvdAI = nqv_dA⁠, with nn the carrier density, qq the carrier charge and AA the cross-sectional area. Drift speeds are tiny, typically fractions of a millimeter per second.
    Electric Circuits › Current Resistance and Power
  9. What is the magnetic field a perpendicular distance dd from a finite straight wire segment carrying current II⁠?
    Integrating Biot–Savart along the segment gives B=μ0I4πd(sin⁡θ1+sin⁡θ2)B = \frac{\mu_0 I}{4\pi d}\left(\sin\theta_1 + \sin\theta_2\right)⁠, where the angles are measured at the field point from the perpendicular to each end, and an end's angle counts as negative when the foot of the perpendicular lies beyond that end. Letting both angles reach 90° gives the infinite-wire result.
    Magnetic Fields and Electromagnetism › Biot–Savart and Ampère's Law
  10. The flux through each turn of a 10-turn coil is 0.5t20.5t^2 (webers, seconds). What emf is induced at t=2t = 2 s?
    Faraday's law, emf = −N dΦ/dt: the flux changes at ddt(0.5t2)=t=2\frac{d}{dt}\left(0.5t^2\right) = t = 2 Wb/s, so the emf has magnitude 10×2=10\times 2 = 20 V.
    Electromagnetic Induction › Faraday's and Lenz's Laws

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  • Topic outline follows the College Board AP Physics C: Electricity and Magnetism Course and Exam Description (2024 revised framework, Units 8–13); the AP name is used nominatively and no course or exam text is reproduced.
  • OpenStax University Physics Volume 2, consulted as a reference; no text copied.

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