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Physics fundamentals

Formulas for first-year physics in one place: mechanics, waves, fluids, heat, electromagnetism, optics and pointers into modern physics, with SI units throughout. Constant values are in constants, units and conversions, symbols in notation, and the algebra in math fundamentals.

Units and dimensions

Every equation must balance in dimensions as well as numbers. Write the dimension of a quantity as [ ⋅ ][\,\cdot\,] built from M\mathsf{M} (mass), L\mathsf{L} (length), T\mathsf{T} (time), I\mathsf{I} (current) and Θ\Theta (temperature). The SI base units and derived-unit definitions are on the constants and units sheet.

QuantityDimensionSI unit
velocityL T−1\mathsf{L\,T^{-1}}m/s
accelerationL T−2\mathsf{L\,T^{-2}}m/s²
forceM L T−2\mathsf{M\,L\,T^{-2}}N = kg·m/s²
energy, workM L2 T−2\mathsf{M\,L^2\,T^{-2}}J = N·m
powerM L2 T−3\mathsf{M\,L^2\,T^{-3}}W = J/s
pressureM L−1 T−2\mathsf{M\,L^{-1}\,T^{-2}}Pa = N/m²
momentumM L T−1\mathsf{M\,L\,T^{-1}}kg·m/s = N·s
chargeI T\mathsf{I\,T}C = A·s
voltageM L2 T−3 I−1\mathsf{M\,L^2\,T^{-3}\,I^{-1}}V = J/C
resistanceM L2 T−3 I−2\mathsf{M\,L^2\,T^{-3}\,I^{-2}}Ω = V/A
magnetic fieldM T−2 I−1\mathsf{M\,T^{-2}\,I^{-1}}T = N/(A·m)
angle, straindimensionlessrad, none
CheckRule
adding, subtracting, ==every term has the same dimension
inside sin⁡\sin, exe^x, ln⁡\lnargument is dimensionless (ωt\omega t, t/τt/\tau, x/x0x/x_0)
guessing a lawT=C LagbT = C\,L^a g^b with [T]=T[T] = \mathsf{T} forces a=12a = \tfrac12, b=−12b = -\tfrac12: T∝L/gT \propto \sqrt{L/g}
units in calculationsconvert to SI first (km/h to m/s: divide by 3.63.6)

Vectors quickly

OperationFormula
components from polarvx=vcos⁡θv_x = v\cos\theta, vy=vsin⁡θv_y = v\sin\theta
magnitude, directionv=vx2+vy2v = \sqrt{v_x^2 + v_y^2}, θ=atan2⁡(vy,vx)\theta = \operatorname{atan2}(v_y, v_x)
add, subtractcomponent-wise
dot product (scalar)a⃗⋅b⃗=abcos⁡θ\vec a \cdot \vec b = ab\cos\theta: work, flux, power
cross product (vector)∥a⃗×b⃗∥=absin⁡θ\lVert \vec a \times \vec b \rVert = ab\sin\theta, right-hand rule: torque, Lorentz force

Which equation?

You have or wantReach forSection
constant acceleration, time not givenv2=u2+2asv^2 = u^2 + 2asKinematics
forces on a body, want acceleration∑F⃗=ma⃗\sum \vec F = m\vec a on a free-body diagramNewton's laws
speeds at two heights, no time, no frictionK1+U1=K2+U2K_1 + U_1 = K_2 + U_2Energy
collision or explosion∑p⃗\sum \vec p conservedMomentum
spinning body under a torqueτ=Iα\tau = I\alphaRotation
spinning body changes shapeI1ω1=I2ω2I_1\omega_1 = I_2\omega_2Rotation
orbit period against radiusT2=4π2r3/GMT^2 = 4\pi^2 r^3 / GMGravitation
mass on a spring or pendulum periodT=2πm/kT = 2\pi\sqrt{m/k}, T=2πL/gT = 2\pi\sqrt{L/g}Osc. and waves
wave speed, frequency, wavelengthv=fλv = f\lambdaOsc. and waves
moving source or observer changes pitchDopplerOsc. and waves
pressure at a depthp=p0+ρghp = p_0 + \rho g hFluids
floating or sinkingFb=ρfluidVsubgF_b = \rho_\text{fluid} V_\text{sub} gFluids
flow speeds up in a narrowing pipeA1v1=A2v2A_1v_1 = A_2v_2 then BernoulliFluids
heating or melting a substanceQ=mcΔTQ = mc\Delta T, Q=mLQ = mLThermodynamics
gas pressure, volume and temperaturepV=nRTpV = nRTThermodynamics
best possible engine efficiencyηC=1−Tc/Th\eta_C = 1 - T_c/T_hThermodynamics
current in a circuitV=IRV = IR plus Kirchhoff's rulesE&M
capacitor voltage over timeVC=V0(1−e−t/RC)V_C = V_0(1 - e^{-t/RC})E&M
charged particle in a magnetic fieldr=mv/qBr = mv/qBE&M
changing flux, want voltageE=−N dΦB/dt\mathcal{E} = -N\,d\Phi_B/dtE&M
light crossing a boundaryn1sin⁡θ1=n2sin⁡θ2n_1\sin\theta_1 = n_2\sin\theta_2Optics
where a lens puts the image1/f=1/do+1/di1/f = 1/d_o + 1/d_iOptics
photon energy, particle wavelengthE=hfE = hf, λ=h/p\lambda = h/pModern physics

Kinematics

v=dxdtv = \dfrac{dx}{dt}, a=dvdta = \dfrac{dv}{dt}; displacement is the area under v(t)v(t), change in velocity the area under a(t)a(t).

Constant acceleration (SUVAT)

ss displacement, uu initial velocity, vv final velocity, aa acceleration, tt time. Pick the equation that leaves out the variable you neither know nor want.

EquationMissing
v=u+atv = u + atss
s=ut+12at2s = ut + \tfrac12 at^2vv
s=vt−12at2s = vt - \tfrac12 at^2uu
v2=u2+2asv^2 = u^2 + 2astt
s=12(u+v) ts = \tfrac12(u + v)\,taa

Free fall: a=−ga = -g with g=9.81 m/s2g = 9.81\ \text{m/s}^2 (up positive).

Projectile (no air resistance)

Horizontal and vertical motion are independent; only yy accelerates.

x=v0cos⁡θ t,y=v0sin⁡θ t−12gt2,y=xtan⁡θ−gx22v02cos⁡2θx = v_0\cos\theta\,t, \qquad y = v_0\sin\theta\,t - \tfrac12 g t^2, \qquad y = x\tan\theta - \frac{g x^2}{2 v_0^2\cos^2\theta}
Level ground resultFormula
time of flightT=2v0sin⁡θgT = \dfrac{2v_0\sin\theta}{g}
maximum heightH=v02sin⁡2θ2gH = \dfrac{v_0^2\sin^2\theta}{2g}
rangeR=v02sin⁡2θgR = \dfrac{v_0^2\sin 2\theta}{g}, largest at 45∘45^\circ
speed at apexv0cos⁡θv_0\cos\theta (vertical part is zero)

Relative velocity: v⃗A/C=v⃗A/B+v⃗B/C\vec v_{A/C} = \vec v_{A/B} + \vec v_{B/C}.

Newton's laws and forces

LawStatement
firstno net force: velocity constant (inertial frames)
second∑F⃗=ma⃗=dp⃗dt\sum \vec F = m\vec a = \dfrac{d\vec p}{dt}
thirdF⃗AB=−F⃗BA\vec F_{AB} = -\vec F_{BA}: equal, opposite, on different bodies

Free-body diagrams

  1. Isolate one body; draw it as a dot or box.
  2. Draw every force on it: weight, normal, tension, friction, spring, drag, applied.
  3. Choose axes along the acceleration (along an incline: xx down the slope).
  4. Write ∑Fx=max\sum F_x = ma_x and ∑Fy=may\sum F_y = ma_y; solve.
  5. Several bodies: one diagram each, linked by shared tension or contact forces.

Common forces

ForceFormulaDirection
weightW=mgW = mgdown
normalfrom ∑F⊥=0\sum F_\perp = 0 (mgcos⁡θmg\cos\theta on a slope)perpendicular to the surface
static frictionfs≤μsNf_s \le \mu_s Nopposes impending slip
kinetic frictionfk=μkNf_k = \mu_k Nopposes sliding
spring (Hooke)F=−kxF = -kxtoward equilibrium
drag (fast)FD=12ρCDAv2F_D = \tfrac12 \rho C_D A v^2against velocity
centripetalFc=mv2r=mω2rF_c = \dfrac{mv^2}{r} = m\omega^2 rtoward the center

On an incline of angle θ\theta: along the slope mgsin⁡θmg\sin\theta, into it mgcos⁡θmg\cos\theta; sliding with friction gives a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta). Springs in series: 1/k=∑1/ki1/k = \sum 1/k_i; in parallel: k=∑kik = \sum k_i.

Circular motion

QuantityFormula
speed, periodv=ωr=2πrTv = \omega r = \dfrac{2\pi r}{T}
centripetal accel.ac=v2r=ω2ra_c = \dfrac{v^2}{r} = \omega^2 r
flat curve, frictionvmax=μsgrv_\text{max} = \sqrt{\mu_s g r}
banked curve, no frictiontan⁡θ=v2rg\tan\theta = \dfrac{v^2}{rg}
top of vertical loopvmin=grv_\text{min} = \sqrt{gr}

Centripetal force is not a new force: it is the net inward part of the forces already on the diagram.

Work, energy and power

QuantityFormulaUnit
work (constant force)W=F⃗⋅d⃗=Fdcos⁡θW = \vec F \cdot \vec d = Fd\cos\thetaJ
work (varying force)W=∫F⃗⋅dr⃗W = \int \vec F \cdot d\vec rJ
kinetic energyK=12mv2=p22mK = \tfrac12 mv^2 = \dfrac{p^2}{2m}J
gravitational PE (near surface)Ug=mghU_g = mghJ
elastic PEUs=12kx2U_s = \tfrac12 kx^2J
work-energy theoremWnet=ΔKW_\text{net} = \Delta KJ
force from potentialFx=−dUdxF_x = -\dfrac{dU}{dx}N
powerP=dWdt=F⃗⋅v⃗P = \dfrac{dW}{dt} = \vec F \cdot \vec vW
efficiencyη=Pout/Pin\eta = P_\text{out}/P_\text{in}none

Conservation of energy, with WncW_\text{nc} the work by non-conservative forces (friction, drag, a motor):

K1+U1+Wnc=K2+U2K_1 + U_1 + W_\text{nc} = K_2 + U_2

Friction on a flat path: Wnc=−μkNdW_\text{nc} = -\mu_k N d. A drop from rest through height hh: v=2ghv = \sqrt{2gh}.

Momentum and collisions

QuantityFormula
momentump⃗=mv⃗\vec p = m\vec v
impulseJ⃗=∫F⃗ dt=F⃗avgΔt=Δp⃗\vec J = \int \vec F\,dt = \vec F_\text{avg}\Delta t = \Delta \vec p
conservationno external net force: ∑p⃗before=∑p⃗after\sum \vec p_\text{before} = \sum \vec p_\text{after}
center of massr⃗cm=∑mir⃗i∑mi\vec r_\text{cm} = \dfrac{\sum m_i \vec r_i}{\sum m_i},  F⃗ext=Ma⃗cm\ \vec F_\text{ext} = M\vec a_\text{cm}
CollisionMomentumKinetic energy1D result (m2m_2 at rest)
elasticconservedconservedv1′=m1−m2m1+m2v1v_1' = \dfrac{m_1 - m_2}{m_1 + m_2}v_1,  v2′=2m1m1+m2v1\ v_2' = \dfrac{2m_1}{m_1 + m_2}v_1
inelasticconservedsome lostneed one more fact, such as the coefficient of restitution
perfectly inelasticconservedmost loststick together: v′=m1v1m1+m2v' = \dfrac{m_1 v_1}{m_1 + m_2}

Coefficient of restitution: e=v2′−v1′v1−v2e = \dfrac{v_2' - v_1'}{v_1 - v_2} (11 elastic, 00 stuck).

Rotation

LinearRotationalLink
xx (m)θ\theta (rad)arc s=rθs = r\theta
vvω\omega (rad/s)v=ωrv = \omega r
aaα\alpha (rad/s²)at=αra_t = \alpha r
mmII (kg·m²)I=∑miri2I = \sum m_i r_i^2
FFτ\tau (N·m)τ⃗=r⃗×F⃗\vec\tau = \vec r \times \vec F,  τ=rFsin⁡ϕ\ \tau = rF\sin\phi
F=maF = maτnet=Iα\tau_\text{net} = I\alpha
p=mvp = mvL=IωL = I\omega (kg·m²/s)L⃗=r⃗×p⃗\vec L = \vec r \times \vec p
K=12mv2K = \tfrac12 mv^2K=12Iω2K = \tfrac12 I\omega^2
P=FvP = FvP=τωP = \tau\omega

SUVAT carries over with θ,ω0,ω,α,t\theta, \omega_0, \omega, \alpha, t. Static equilibrium: ∑F⃗=0\sum \vec F = 0 and ∑τ⃗=0\sum \vec \tau = 0 about any point. Rolling without slipping: v=ωRv = \omega R, K=12mv2+12Iω2K = \tfrac12 mv^2 + \tfrac12 I\omega^2. With no external torque, LL is conserved.

Moments of inertia

Body (mass MM)AxisII
point mass at distance rranyMr2Mr^2
thin ring or thin-walled cylindercentral axisMR2MR^2
solid disc or cylindercentral axis12MR2\tfrac12 MR^2
thick-walled cylindercentral axis12M(R12+R22)\tfrac12 M(R_1^2 + R_2^2)
solid spherediameter25MR2\tfrac25 MR^2
thin spherical shelldiameter23MR2\tfrac23 MR^2
thin rod, length LLcenter, perpendicular112ML2\tfrac1{12} ML^2
thin rod, length LLend, perpendicular13ML2\tfrac13 ML^2
rectangular plate a×ba \times bcenter, perpendicular to plate112M(a2+b2)\tfrac1{12} M(a^2 + b^2)

Parallel-axis theorem: I=Icm+Md2I = I_\text{cm} + Md^2. Perpendicular-axis (flat plates): Iz=Ix+IyI_z = I_x + I_y.

Gravitation

QuantityFormula
Newton's lawF=GMmr2F = \dfrac{GMm}{r^2}, attractive
field strengthg=GMr2g = \dfrac{GM}{r^2} (9.81 m/s29.81\ \text{m/s}^2 at Earth's surface)
potential energyU=−GMmrU = -\dfrac{GMm}{r} (zero at infinity)
circular orbit speedv=GM/rv = \sqrt{GM/r}
escape speedvesc=2GM/rv_\text{esc} = \sqrt{2GM/r} (11.2 km/s11.2\ \text{km/s} from Earth)
Kepler's third lawT2=4π2GMr3T^2 = \dfrac{4\pi^2}{GM}r^3 (rr = semi-major axis for ellipses)
orbital energyE=−GMm2rE = -\dfrac{GMm}{2r}

Kepler: orbits are ellipses with the Sun at a focus; the radius sweeps equal areas in equal times (angular momentum is conserved); T2∝a3T^2 \propto a^3.

Oscillations and waves

Simple harmonic motion

Simple harmonic motion (SHM): restoring force proportional to displacement, x¨=−ω2x\ddot x = -\omega^2 x.

x(t)=Acos⁡(ωt+ϕ),v(t)=−Aωsin⁡(ωt+ϕ),a(t)=−ω2xx(t) = A\cos(\omega t + \phi), \quad v(t) = -A\omega\sin(\omega t + \phi), \quad a(t) = -\omega^2 x
QuantityFormula
angular frequencyω=2πf=2π/T\omega = 2\pi f = 2\pi/T
mass on springω=k/m\omega = \sqrt{k/m},  T=2πm/k\ T = 2\pi\sqrt{m/k}
simple pendulum (small angle)ω=g/L\omega = \sqrt{g/L},  T=2πL/g\ T = 2\pi\sqrt{L/g}
physical pendulumT=2πI/(mgd)T = 2\pi\sqrt{I/(mgd)}, dd = pivot to center of mass
maximavmax=Aωv_\text{max} = A\omega,  amax=Aω2\ a_\text{max} = A\omega^2
energyE=12kA2=12kx2+12mv2E = \tfrac12 kA^2 = \tfrac12 kx^2 + \tfrac12 mv^2
speed at xxv=ωA2−x2v = \omega\sqrt{A^2 - x^2}
damped (light)x=Ae−bt/2mcos⁡(ω′t)x = Ae^{-bt/2m}\cos(\omega' t),  ω′=ω2−(b/2m)2\ \omega' = \sqrt{\omega^2 - (b/2m)^2}

With A=ω=1A = \omega = 1 and ϕ=0\phi = 0: velocity leads displacement by a quarter cycle and acceleration is always opposite to displacement.

π2π-11
SHM: x, v and a (A = ω = 1) x = cos t v = −sin t a = −cos t

Driven oscillator: resonance when the driving frequency is near ω0\omega_0; lighter damping gives a taller, narrower peak.

Waves and sound

QuantityFormula
wave speedv=fλ=λ/Tv = f\lambda = \lambda/T
traveling wavey=Asin⁡(kx−ωt)y = A\sin(kx - \omega t),  k=2π/λ\ k = 2\pi/\lambda,  v=ω/k\ v = \omega/k
stringv=FT/μv = \sqrt{F_T/\mu}, μ\mu = mass per length
sound in airv≈331+0.6 T∘Cv \approx 331 + 0.6\,T_{^\circ\text{C}}, about 343 m/s343\ \text{m/s} at 20 °C
intensityI=P/(4πr2)I = P/(4\pi r^2) from a point source
sound levelβ=10log⁡10(I/I0)\beta = 10\log_{10}(I/I_0) dB,  I0=10−12 W/m2\ I_0 = 10^{-12}\ \text{W/m}^2
beatsfbeat=∣f1−f2∣f_\text{beat} = \lvert f_1 - f_2 \rvert

Doppler effect (sound)

f′=f v±vov∓vsf' = f\,\frac{v \pm v_o}{v \mp v_s}

Top signs when observer and source move toward each other: the observed frequency rises. vv is the speed of sound in the medium. Light, for v≪cv \ll c: Δf/f≈vr/c\Delta f / f \approx v_r / c.

Superposition and standing waves

SituationCondition or modes
constructive interferencepath difference Δ=mλ\Delta = m\lambda
destructive interferenceΔ=(m+12)λ\Delta = (m + \tfrac12)\lambda
double slit, bright fringesdsin⁡θ=mλd\sin\theta = m\lambda, spacing Δy=λL/d\Delta y = \lambda L/d
single slit, first dark fringeasin⁡θ=λa\sin\theta = \lambda
string fixed both ends, open pipeλn=2L/n\lambda_n = 2L/n,  fn=nv/2L\ f_n = nv/2L, n=1,2,3,…n = 1, 2, 3, \ldots
pipe closed one endλn=4L/n\lambda_n = 4L/n,  fn=nv/4L\ f_n = nv/4L, n=1,3,5,…n = 1, 3, 5, \ldots

Fluids

QuantityFormula
density, pressureρ=m/V\rho = m/V,  p=F/A\ p = F/A (1 atm =101.325= 101.325 kPa)
hydrostatic pressurep=p0+ρghp = p_0 + \rho g h; gauge pressure is p−patmp - p_\text{atm}
Pascal (hydraulic press)F1/A1=F2/A2F_1/A_1 = F_2/A_2
buoyancy (Archimedes)Fb=ρfluidVsubgF_b = \rho_\text{fluid} V_\text{sub} g
floating fractionVsub/V=ρobject/ρfluidV_\text{sub}/V = \rho_\text{object}/\rho_\text{fluid}
continuityA1v1=A2v2A_1 v_1 = A_2 v_2 (volume flow Q=AvQ = Av)
Bernoulli (along a streamline)p+12ρv2+ρgh=constp + \tfrac12\rho v^2 + \rho g h = \text{const}
Torricelliv=2ghv = \sqrt{2gh} out of a hole at depth hh
Poiseuille (laminar pipe)Q=πr4Δp8ηLQ = \dfrac{\pi r^4 \Delta p}{8\eta L}
Reynolds numberRe=ρvD/ηRe = \rho v D/\eta; pipes are laminar below about 2000

Bernoulli assumes steady, incompressible, non-viscous flow. The atmosphere is the compressible, rotating case: see numerical weather modeling.

Thermodynamics

QuantityFormula
temperatureTK=T∘C+273.15T_\text{K} = T_{^\circ\text{C}} + 273.15
linear expansionΔL=αL ΔT\Delta L = \alpha L\,\Delta T
heat to change temperatureQ=mc ΔTQ = mc\,\Delta T (cwater≈4186c_\text{water} \approx 4186 J/(kg·K))
latent heat (phase change)Q=mLQ = mL
conductionP=kA ΔT/dP = kA\,\Delta T / d
radiationP=εσAT4P = \varepsilon\sigma A T^4
ideal gaspV=nRT=NkBTpV = nRT = Nk_BT
mean kinetic energy⟨K⟩=32kBT\langle K \rangle = \tfrac32 k_B T,  vrms=3kBT/m\ v_\text{rms} = \sqrt{3k_BT/m}
internal energy, monatomicU=32nRTU = \tfrac32 nRT

Laws

LawStatement
zerothtwo bodies each in equilibrium with a third are in equilibrium with each other
firstΔU=Q−W\Delta U = Q - W (QQ into the system, WW done by it)
secondtotal entropy of an isolated system never decreases; heat flows hot to cold
thirdentropy approaches a minimum as T→0T \to 0; absolute zero is unreachable

Ideal gas processes

ProcessConstantWork by gas WWNote
isobaricppp ΔVp\,\Delta VQ=nCpΔTQ = nC_p\Delta T
isochoricVV00Q=ΔU=nCVΔTQ = \Delta U = nC_V\Delta T
isothermalTTnRTln⁡(V2/V1)nRT\ln(V_2/V_1)ΔU=0\Delta U = 0, Q=WQ = W
adiabaticpVγpV^\gammap1V1−p2V2γ−1\dfrac{p_1V_1 - p_2V_2}{\gamma - 1}Q=0Q = 0, γ=Cp/CV\gamma = C_p/C_V

Cp−CV=RC_p - C_V = R; monatomic γ=5/3\gamma = 5/3, diatomic (air) γ≈7/5\gamma \approx 7/5.

Entropy and heat engines

QuantityFormula
entropy change (reversible)ΔS=∫dQ/T\Delta S = \int dQ/T; at constant TT: ΔS=Q/T\Delta S = Q/T
engine efficiencyη=W/Qh=1−Qc/Qh\eta = W/Q_h = 1 - Q_c/Q_h
Carnot (maximum) efficiencyηC=1−Tc/Th\eta_C = 1 - T_c/T_h (kelvin)
refrigerator COPK=Qc/WK = Q_c/W,  KC=Tc/(Th−Tc)\ K_C = T_c/(T_h - T_c)
heat pump COPQh/WQ_h/W,  Th/(Th−Tc)\ T_h/(T_h - T_c)

Electricity and magnetism

Electrostatics

QuantityFormulaUnit
Coulomb's lawF=kq1q2r2F = k\dfrac{q_1q_2}{r^2},  k=14πε0\ k = \dfrac{1}{4\pi\varepsilon_0}N
fieldE⃗=F⃗/q\vec E = \vec F/q; point charge E=kq/r2E = kq/r^2N/C = V/m
potentialV=kq/rV = kq/r;  ΔU=qΔV\ \Delta U = q\Delta VV
field from potentialEx=−dVdxE_x = -\dfrac{dV}{dx}; parallel plates E=V/dE = V/dV/m
Gauss's law∮E⃗⋅dA⃗=Qenc/ε0\oint \vec E \cdot d\vec A = Q_\text{enc}/\varepsilon_0
capacitanceC=Q/VC = Q/V; parallel plates C=ε0κA/dC = \varepsilon_0 \kappa A/dF
energy in a capacitorU=12CV2=Q22CU = \tfrac12 CV^2 = \dfrac{Q^2}{2C}J

Energy unit: 1 eV=1.602×10−19 J1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}, the energy an electron gains across 1 V.

Circuits

QuantityFormula
currentI=dQ/dtI = dQ/dt (A)
Ohm's lawV=IRV = IR
resistance of a wireR=ρL/AR = \rho L/A
powerP=IV=I2R=V2/RP = IV = I^2R = V^2/R
resistors in seriesR=R1+R2+⋯R = R_1 + R_2 + \cdots (same current)
resistors in parallel1R=1R1+1R2+⋯\dfrac1R = \dfrac1{R_1} + \dfrac1{R_2} + \cdots (same voltage)
capacitors in series1C=1C1+1C2+⋯\dfrac1C = \dfrac1{C_1} + \dfrac1{C_2} + \cdots
capacitors in parallelC=C1+C2+⋯C = C_1 + C_2 + \cdots
real batteryV=E−IrV = \mathcal{E} - Ir
AC RMSVrms=V0/2V_\text{rms} = V_0/\sqrt2,  Irms=I0/2\ I_\text{rms} = I_0/\sqrt2

Kirchhoff: currents into a junction sum to zero (charge); voltage changes around any closed loop sum to zero (energy).

RC circuit, time constant τ=RC\tau = RC (63% of the way after one τ\tau, over 99% after 5τ5\tau):

charging: VC=V0(1−e−t/RC),discharging: VC=V0 e−t/RC,I=V0Re−t/RC\text{charging: } V_C = V_0\left(1 - e^{-t/RC}\right), \qquad \text{discharging: } V_C = V_0\,e^{-t/RC}, \qquad I = \frac{V_0}{R}e^{-t/RC}

Magnetism and induction

QuantityFormula
Lorentz forceF⃗=q(E⃗+v⃗×B⃗)\vec F = q(\vec E + \vec v \times \vec B)
force on a wireF⃗=IL⃗×B⃗\vec F = I\vec L \times \vec B
radius in a uniform fieldr=mvqBr = \dfrac{mv}{qB},  f=qB2πm\ f = \dfrac{qB}{2\pi m} (cyclotron)
long straight wireB=μ0I2πrB = \dfrac{\mu_0 I}{2\pi r} (circles, right-hand grip)
solenoidB=μ0nIB = \mu_0 n I, nn = turns per meter
fluxΦB=B⃗⋅A⃗=BAcos⁡θ\Phi_B = \vec B \cdot \vec A = BA\cos\theta (Wb)
Faraday + LenzE=−NdΦBdt\mathcal{E} = -N\dfrac{d\Phi_B}{dt}; induced current opposes the change
motional emfE=BLv\mathcal{E} = BLv
generatorE=NBAωsin⁡ωt\mathcal{E} = NBA\omega\sin\omega t
transformer (ideal)Vs/Vp=Ns/Np=Ip/IsV_s/V_p = N_s/N_p = I_p/I_s
inductorE=−L dI/dt\mathcal{E} = -L\,dI/dt, energy 12LI2\tfrac12 LI^2, RL τ=L/R\tau = L/R

Maxwell's equations tie all of this together and predict light at speed c=1/μ0ε0c = 1/\sqrt{\mu_0\varepsilon_0}.

Optics

QuantityFormula
reflectionθi=θr\theta_i = \theta_r (from the normal)
refractive indexn=c/vn = c/v; in a medium λn=λ/n\lambda_n = \lambda/n, ff unchanged
Snell's lawn1sin⁡θ1=n2sin⁡θ2n_1\sin\theta_1 = n_2\sin\theta_2
critical anglesin⁡θc=n2/n1\sin\theta_c = n_2/n_1 (needs n1>n2n_1 > n_2; beyond it: total internal reflection)
thin lens and mirror1f=1do+1di\dfrac1f = \dfrac1{d_o} + \dfrac1{d_i}
magnificationm=−dido=hihom = -\dfrac{d_i}{d_o} = \dfrac{h_i}{h_o}
spherical mirrorf=R/2f = R/2
lensmaker1f=(n−1)(1R1−1R2)\dfrac1f = (n - 1)\left(\dfrac1{R_1} - \dfrac1{R_2}\right)
lens powerP=1/fP = 1/f in dioptres (ff in meters)
diffraction gratingdsin⁡θ=mλd\sin\theta = m\lambda
Sign convention (real is positive)PositiveNegative
ffconverging lens, concave mirrordiverging lens, convex mirror
did_ireal imagevirtual image
mmuprightinverted

Modern physics

Pointers only; each line opens a whole subject.

IdeaFormula
mass-energyE=mc2E = mc^2;  E2=(pc)2+(mc2)2\ E^2 = (pc)^2 + (mc^2)^2
Lorentz factorγ=1/1−v2/c2\gamma = 1/\sqrt{1 - v^2/c^2}
time dilation, length contractionΔt=γΔt0\Delta t = \gamma\Delta t_0,  L=L0/γ\ L = L_0/\gamma
photon energyE=hf=hc/λE = hf = hc/\lambda; hc≈1240 eV⋅nmhc \approx 1240\ \text{eV·nm}
photon momentump=E/c=h/λp = E/c = h/\lambda
photoelectric effectKmax=hf−ϕK_\text{max} = hf - \phi
de Broglie wavelengthλ=h/p=h/mv\lambda = h/p = h/mv
uncertaintyΔx Δp≥ℏ/2\Delta x\,\Delta p \ge \hbar/2
hydrogen levelsEn=−13.6 eV/n2E_n = -13.6\ \text{eV}/n^2
Wien's lawλmaxT=2.898×10−3 m⋅K\lambda_\text{max}T = 2.898 \times 10^{-3}\ \text{m·K}
radioactive decayN=N0e−λtN = N_0e^{-\lambda t},  t1/2=ln⁡2/λ\ t_{1/2} = \ln 2/\lambda

Worked examples

Projectile range

A ball leaves level ground at v0=20v_0 = 20 m/s, θ=30∘\theta = 30^\circ. Find flight time, range and peak height.

vx=20cos⁡30∘=17.32 m/s,vy=20sin⁡30∘=10.0 m/sT=2vyg=2(10.0)9.81=2.04 sR=vxT=17.32×2.039=35.3 m(=v02sin⁡60∘g)H=vy22g=10019.62=5.10 m\begin{aligned} v_x &= 20\cos 30^\circ = 17.32\ \text{m/s}, \qquad v_y = 20\sin 30^\circ = 10.0\ \text{m/s} \\ T &= \frac{2v_y}{g} = \frac{2(10.0)}{9.81} = 2.04\ \text{s} \\ R &= v_xT = 17.32 \times 2.039 = 35.3\ \text{m} \quad \left(= \tfrac{v_0^2\sin 60^\circ}{g}\right) \\ H &= \frac{v_y^2}{2g} = \frac{100}{19.62} = 5.10\ \text{m} \end{aligned}

Block sliding down a rough incline

m=5.0m = 5.0 kg on a 30∘30^\circ slope, μk=0.20\mu_k = 0.20. Axes: xx down the slope.

N=mgcos⁡θ=5.0(9.81)(0.866)=42.5 Nfk=μkN=0.20(42.5)=8.50 Nma=mgsin⁡θ−fk=24.5−8.50=16.0 Na=16.0/5.0=3.21 m/s2\begin{aligned} N &= mg\cos\theta = 5.0(9.81)(0.866) = 42.5\ \text{N} \\ f_k &= \mu_k N = 0.20(42.5) = 8.50\ \text{N} \\ ma &= mg\sin\theta - f_k = 24.5 - 8.50 = 16.0\ \text{N} \\ a &= 16.0/5.0 = 3.21\ \text{m/s}^2 \end{aligned}

Car power from kinetic energy

A 1200 kg car reaches 100 km/h from rest in 8.0 s. Average power, ignoring losses:

v=1003.6=27.8 m/s,K=12(1200)(27.8)2=4.63×105 J,P=Kt=57.9 kWv = \frac{100}{3.6} = 27.8\ \text{m/s}, \quad K = \tfrac12(1200)(27.8)^2 = 4.63 \times 10^5\ \text{J}, \quad P = \frac{K}{t} = 57.9\ \text{kW}

Elastic and sticky collisions

m1=2.0m_1 = 2.0 kg at 3.03.0 m/s hits m2=1.0m_2 = 1.0 kg at rest.

CaseAfterKK before → after
perfectly inelasticv′=2.0(3.0)3.0=2.0v' = \dfrac{2.0(3.0)}{3.0} = 2.0 m/s9.0 J → 6.0 J
elasticv1′=13(3.0)=1.0v_1' = \tfrac{1}{3}(3.0) = 1.0 m/s,  v2′=43(3.0)=4.0\ v_2' = \tfrac{4}{3}(3.0) = 4.0 m/s9.0 J → 9.0 J

Check momentum: 2.0(1.0)+1.0(4.0)=6.0 kg⋅m/s2.0(1.0) + 1.0(4.0) = 6.0\ \text{kg·m/s} both times.

Geostationary orbit radius

Period one sidereal day, T=86 164T = 86\,164 s; GM⊕=3.986×1014 m3/s2GM_\oplus = 3.986 \times 10^{14}\ \text{m}^3/\text{s}^2.

r=(GMT24π2)1/3=(3.986×1014 (86 164)239.48)1/3=4.22×107 mr = \left(\frac{GM T^2}{4\pi^2}\right)^{1/3} = \left(\frac{3.986 \times 10^{14}\,(86\,164)^2}{39.48}\right)^{1/3} = 4.22 \times 10^7\ \text{m}

Altitude above the equator: 42 164−6378≈35 78642\,164 - 6378 \approx 35\,786 km. Orbital speed 2πr/T=3.072\pi r/T = 3.07 km/s.

RC charging

R=10 kΩR = 10\ \text{k}\Omega, C=100 μFC = 100\ \mu\text{F}, supply 5.05.0 V.

τ=RC=(104)(10−4)=1.0 s,VC(2 s)=5.0(1−e−2)=4.32 V\tau = RC = (10^4)(10^{-4}) = 1.0\ \text{s}, \qquad V_C(2\ \text{s}) = 5.0\left(1 - e^{-2}\right) = 4.32\ \text{V}

Time to reach 99%: 1−e−t/τ=0.99⇒t=τln⁡100=4.61 - e^{-t/\tau} = 0.99 \Rightarrow t = \tau\ln 100 = 4.6 s.

Thin lens image

Converging lens f=10f = 10 cm, object at do=15d_o = 15 cm.

1di=110−115=130 ⇒ di=30 cm,m=−3015=−2\frac{1}{d_i} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30} \ \Rightarrow\ d_i = 30\ \text{cm}, \qquad m = -\frac{30}{15} = -2

Real, inverted, twice the size, on the far side of the lens.

References