Skip to the content
Find a formula

AP Physics C E&M Formula Sheet, Explained

Physics formula sheetsAP Physics CCalculus basedUpdated 1 Nov 2026

∮E⃗⋅dA⃗=qencε0\oint \vec{E}\cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}
where
  • E electric field N/C or V/m
  • dA small outward area element of the closed surface m²
  • q_enc net charge inside the surface C
  • ε₀ permittivity of free space, 8.854 × 10⁻¹² C²/(N m²)

This AP Physics C E&M formula sheet is a study guide to the calculus based electricity and magnetism course, anchored by Gauss's law, ∮E⃗⋅dA⃗=qencε0\oint \vec{E}\cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}. Students also search for it as the AP Physics C E and M formula sheet; it is the same course. Every big idea here is a statement about fields: how charges make them, how currents make them, and how a magnetic field that varies in time creates an electric one.

This is not the exam board's sheet. The board provides its own equation tables and constants with the exam. What follows groups the relations by topic and adds the conditions and symmetry choices that the tables leave out. For the mechanics half of the C program, see the C mechanics sheet; for the algebra version of these topics, see AP Physics 2; and for every course, the formula sheets hub.

Quick facts

  • What this is

    A study guide, not the official tables
  • Gauss's law

    ∮E⃗⋅dA⃗=qenc/ε0\oint \vec{E}\cdot d\vec{A} = q_{enc}/\varepsilon_0
  • Ampère's law

    ∮B⃗⋅dℓ⃗=μ0Ienc\oint \vec{B}\cdot d\vec{\ell} = \mu_0 I_{enc}
  • Faraday's law

    ε=−dΦBdt\varepsilon = -\frac{d\Phi_B}{dt}
  • Constants

    ε₀ = 8.854 × 10⁻¹² C²/(N m²), μ₀ = 4π × 10⁻⁷ T m/A
  • Math used

    Integrals over lines, surfaces and volumes; first order differential equations

Electrostatics: fields, Gauss's law and potential

Charges and the fields they create
Law or relationPurposeUnitPage
E⃗=14πε0∫dqr2r^\vec{E} = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r^2}\hat{r}field of a continuous chargeN/CElectric field
ΦE=∫E⃗⋅dA⃗\Phi_E = \int \vec{E}\cdot d\vec{A}electric fluxN m²/CElectric flux
∮E⃗⋅dA⃗=qencε0\oint \vec{E}\cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}Gauss's law for symmetric chargeN m²/CElectric flux
ΔV=−∫abE⃗⋅dr⃗\Delta V = -\int_a^b \vec{E}\cdot d\vec{r}potential difference along a pathVVoltage
Ex=−dVdxE_x = -\frac{dV}{dx}field from a potential functionV/mElectric field
λ=dqdℓ, σ=dqdA, ρ=dqdV\lambda = \frac{dq}{d\ell},\ \sigma = \frac{dq}{dA},\ \rho = \frac{dq}{dV}line, surface and volume charge densityC/m, C/m², C/m³Charge density

Example 1: field of a long line of charge

A long straight wire carries a uniform charge of 3.0 μC per metre. Find the field 20 cm from it.

Wrap a Gaussian cylinder (radius r, length ℓ) round the wire. Only the curved side has flux: E(2πrℓ)=λℓε0E(2\pi r\ell) = \frac{\lambda\ell}{\varepsilon_0}, so E=λ2πε0rE = \frac{\lambda}{2\pi\varepsilon_0 r}.

E=3.0×10−62π(8.854×10−12)(0.20)=3.0×10−61.113×10−11=2.7×105 N/CE = \frac{3.0 \times 10^{-6}}{2\pi(8.854 \times 10^{-12})(0.20)} = \frac{3.0 \times 10^{-6}}{1.113 \times 10^{-11}} = 2.7 \times 10^{5}\ \text{N/C}.

Answer: about 2.7 × 10⁵ N/C, pointing away from the wire.

Conductors, capacitors and dielectrics

Storing charge and energy
Law or relationPurposeUnitPage
C=QΔVC = \frac{Q}{\Delta V}definition of capacitanceFCapacitance
C=κε0AdC = \frac{\kappa\varepsilon_0 A}{d}parallel plates with a dielectricFCapacitance
UC=12C(ΔV)2U_C = \frac{1}{2}C(\Delta V)^2stored energyJEnergy density
uE=12ε0E2u_E = \frac{1}{2}\varepsilon_0 E^2energy per unit volume of an electric fieldJ/m³Energy density

Inside a conductor in equilibrium the field is zero, all excess charge sits on the surface, and the whole conductor is at one potential. A dielectric slab lowers the field between plates by κ, which raises the capacitance by κ.

Circuits with resistors and capacitors

Current and RC behaviour
Law or relationPurposeUnitPage
I=dqdtI = \frac{dq}{dt}current as a rateACurrent
J⃗=IA, R=ρℓA\vec{J} = \frac{I}{A},\ R = \frac{\rho\ell}{A}current density and resistanceA/m², ΩCurrent density
P=IΔV=I2RP = I\Delta V = I^2Rpower dissipatedWResistance
q(t)=Cε(1−e−t/RC)q(t) = C\varepsilon\left(1 - e^{-t/RC}\right)charge while charging a capacitorCCapacitance
q(t)=Q0e−t/RCq(t) = Q_0 e^{-t/RC}charge while dischargingCCapacitance
τ=RC\tau = RCtime constantsTime

Example 2: charging a capacitor

A 50 μF capacitor charges through a 2.0 kΩ resistor from a 12 V battery. Find the time constant and the capacitor voltage after one time constant.

τ=RC=(2.0×103)(50×10−6)=0.10 s\tau = RC = (2.0 \times 10^{3})(50 \times 10^{-6}) = 0.10\ \text{s}.

VC=12(1−e−1)=12(0.632)=7.6 VV_C = 12\left(1 - e^{-1}\right) = 12(0.632) = 7.6\ \text{V}.

Answer: 0.10 s and about 7.6 V. After about five time constants the capacitor is over 99% charged.

Magnetic fields and their sources

Forces from and sources of magnetic field
Law or relationPurposeUnitPage
F⃗=qv⃗×B⃗\vec{F} = q\vec{v}\times\vec{B}push on a charge moving through BNLorentz force
dF⃗=I dℓ⃗×B⃗d\vec{F} = I\,d\vec{\ell}\times\vec{B}force on a piece of current carrying wireNMagnetic force
dB⃗=μ04πI dℓ⃗×r^r2d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{\ell}\times\hat{r}}{r^2}Biot Savart lawTMagnetic field
∮B⃗⋅dℓ⃗=μ0Ienc\oint \vec{B}\cdot d\vec{\ell} = \mu_0 I_{enc}Ampère's law for symmetric currentsT mMagnetic field
B=μ0nIB = \mu_0 nIinside a long solenoid, n turns per metreTMagnetic flux density

Use Biot Savart for loops, arcs and short segments. Use Ampère's law only when symmetry makes B constant along a loop: long wires, solenoids, toroids and thick cylinders.

Electromagnetic induction and inductors

Changing flux and self inductance
Law or relationPurposeUnitPage
ΦB=∫B⃗⋅dA⃗\Phi_B = \int \vec{B}\cdot d\vec{A}magnetic fluxWbMagnetic flux
ε=−dΦBdt\varepsilon = -\frac{d\Phi_B}{dt}Faraday's law; Lenz's law gives the signVElectromotive force
εL=−LdIdt\varepsilon_L = -L\frac{dI}{dt}back emf of an inductorVSelf inductance
UL=12LI2U_L = \frac{1}{2}LI^2energy stored in an inductorJSelf inductance
I(t)=εR(1−e−Rt/L)I(t) = \frac{\varepsilon}{R}\left(1 - e^{-Rt/L}\right)current growth in an RL circuit, τ = L/RACurrent
Lenz's law: the induced current always pushes back against whatever change in flux produced it. That one sentence fixes every direction question.

Using the AP Physics C E&M formula sheet in exams

  • Name the symmetry first. Spherical, cylindrical or planar symmetry decides the Gaussian surface.
  • Integrate potentials, not fields, when you can. Potential is a scalar, so there are no components to track.
  • Check limiting behaviour. RC and RL answers should match the circuit at t = 0 and at very long times.
  • Keep ε₀ and μ₀ straight. ε₀ appears in electric results, μ₀ in magnetic ones.

The single quantity pages linked in the tables, such as electric flux and self inductance, work through more examples.

Frequently asked questions

Is this the official AP Physics C E&M equation table?

No. It is a study guide that explains the relations in plain words. The exam board provides the official tables with the exam.

What topics are in AP Physics C E&M?

In general terms: electrostatics and Gauss's law, electric potential, conductors and capacitors, circuits including RC circuits, magnetic fields and their sources, and electromagnetic induction. Your course outline has the precise list.

When should I use Gauss's law instead of integrating Coulomb's law?

Use Gauss's law when the charge has spherical, cylindrical or planar symmetry, so E is constant over your chosen surface. Otherwise, integrate the contributions of small pieces of charge.

Is E and M harder than AP Physics C mechanics?

Many students find it more abstract because fields cannot be seen and the integrals are over surfaces and loops. Strong mechanics and calculus skills make it much easier.