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This AP Physics C E&M formula sheet covers the core equations students need for electrostatics, electric potential, capacitance, circuits, magnetism, and electromagnetic induction. It is designed to help grades 11-12 students connect calculus-based formulas with the physical situations they describe. A clear reference is useful because E&M problems often require choosing between field, potential, energy, flux, and circuit models.

Key Facts

  • Coulomb’s law gives the electric force between point charges as F=14πϵ0q1q2r2r^\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\hat{r}.
  • The electric field is force per unit positive test charge, so E=Fq\vec{E} = \frac{\vec{F}}{q} and for a point charge E=14πϵ0qr2r^\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}.
  • Gauss’s law relates electric flux to enclosed charge by EdA=Qencϵ0\oint \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}.
  • Electric potential difference is related to electric field by ΔV=abEd\Delta V = -\int_a^b \vec{E}\cdot d\vec{\ell}, and potential energy is U=qVU = qV.
  • Capacitance is defined by C=QΔVC = \frac{Q}{\Delta V}, and a parallel-plate capacitor has C=κϵ0AdC = \kappa\epsilon_0\frac{A}{d}.
  • Circuit equations include Ohm’s law V=IRV = IR, power P=IV=I2R=V2RP = IV = I^2R = \frac{V^2}{R}, and capacitor charging q(t)=CE(1et/RC)q(t) = C\mathcal{E}(1 - e^{-t/RC}).
  • The magnetic force on a moving charge is FB=qv×B\vec{F}_B = q\vec{v}\times\vec{B}, and the force on a current-carrying wire is FB=IL×B\vec{F}_B = I\vec{L}\times\vec{B}.
  • Faraday’s law gives induced emf as E=dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}, where magnetic flux is ΦB=BdA\Phi_B = \int \vec{B}\cdot d\vec{A}.

Vocabulary

Electric field
The electric field E\vec{E} is the force per unit positive test charge at a point in space.
Electric flux
Electric flux ΦE=EdA\Phi_E = \int \vec{E}\cdot d\vec{A} measures how much electric field passes through a surface.
Electric potential
Electric potential VV is electric potential energy per unit charge, defined by V=UqV = \frac{U}{q}.
Capacitance
Capacitance CC is the ratio of stored charge to potential difference, given by C=QΔVC = \frac{Q}{\Delta V}.
Magnetic flux
Magnetic flux ΦB=BdA\Phi_B = \int \vec{B}\cdot d\vec{A} measures the amount of magnetic field passing through a surface.
Induced emf
Induced emf E\mathcal{E} is the voltage produced by a changing magnetic flux, described by E=dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}.

Common Mistakes to Avoid

  • Using E=kqr2E = \frac{kq}{r^2} inside any charge distribution, which is wrong because that formula applies directly to a point charge or a spherically symmetric distribution outside the charge.
  • Forgetting the negative sign in ΔV=Ed\Delta V = -\int \vec{E}\cdot d\vec{\ell}, which is wrong because electric potential decreases in the direction of the electric field.
  • Treating C=QΔVC = \frac{Q}{\Delta V} as meaning capacitance changes whenever QQ changes, which is wrong because capacitance depends on geometry and dielectric material for a fixed capacitor.
  • Using F=qvBF = qvB without checking direction or angle, which is wrong because the full magnitude is F=qvBsinθF = |q|vB\sin\theta and the direction comes from v×B\vec{v}\times\vec{B}.
  • Ignoring Lenz’s law in induction problems, which is wrong because the negative sign in E=dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt} determines the direction of the induced current.

Practice Questions

  1. 1 A charge of +3.0 μC+3.0\ \mu\text{C} is placed 0.20 m0.20\ \text{m} from a charge of 2.0 μC-2.0\ \mu\text{C}. Find the magnitude of the electric force using k=8.99×109 Nm2/C2k = 8.99\times10^9\ \text{N}\cdot\text{m}^2/\text{C}^2.
  2. 2 A parallel-plate capacitor has area A=0.015 m2A = 0.015\ \text{m}^2, plate separation d=1.0×103 md = 1.0\times10^{-3}\ \text{m}, and dielectric constant κ=2.5\kappa = 2.5. Find CC using C=κϵ0AdC = \kappa\epsilon_0\frac{A}{d}.
  3. 3 A circular loop with area 0.040 m20.040\ \text{m}^2 is perpendicular to a uniform magnetic field that changes from 0.20 T0.20\ \text{T} to 0.80 T0.80\ \text{T} in 0.30 s0.30\ \text{s}. Find the magnitude of the average induced emf.
  4. 4 Explain why a point on a conductor in electrostatic equilibrium can have E=0\vec{E} = 0 inside the material while excess charge remains on the surface.