Formula list

vavg = Δx / Δt

vavg

average velocitym/s

Δx

displacement, not distance travelledm

Δt

time intervals

A definition rather than a result. Displacement on top is the whole difference between velocity and speed.

aavg = Δv / Δt

aavg

average accelerationm/s²

Δv

change in velocitym/s

Δt

time intervals

Also a definition. Acceleration is a change in velocity, so a turn at constant speed is still an acceleration.

v = v₀ + at

v

velocity at time tm/s

v₀

velocity at t = 0m/s

a

constant accelerationm/s²

t

times

x = x₀ + v₀t + ½at²

x

position at time tm

x₀

position at t = 0m

v₀

velocity at t = 0m/s

a

constant accelerationm/s²

t

times

v² = v₀² + 2a(x − x₀)

v, v₀

final and initial velocitym/s

a

constant accelerationm/s²

x − x₀

displacementm

The one to reach for when the question never mentions time.

ax = 0 · ay = −g

ax

horizontal acceleration, with drag neglectedm/s²

ay

vertical acceleration, downwardsm/s²

t

the same time for both componentss

Not an equation the table needs to print: it is the method. Horizontal and vertical motion are solved separately and share only the time.

velocity = slope of x–t · acceleration = slope of v–t · Δx = area under v–t

slope

of a tangent where the graph is curved

area

counted negative below the time axis

AP asks for these as often as it asks for the equations, and a graph question rarely gives you numbers to substitute.

a = Fnet / m

RearrangedFnet = ma

Fnet

vector sum of the forces on the objectN

m

masskg

a

acceleration, along Fnetm/s²

Applied one axis at a time, off a free-body diagram.

Fg = mg

Fg

gravitational force on the objectN

m

masskg

g

gravitational field strengthN/kg or m/s²

Ff ≤ μs FN (static) · Ff = μk FN (kinetic)

Ff

friction force, opposing relative slidingN

μs, μk

coefficients of static and kinetic friction

FN

normal forceN

Static friction is an inequality: it takes whatever value up to the maximum keeps the surfaces from sliding.

Fs = −kx

Fs

force the spring exertsN

k

spring constantN/m

x

displacement from the natural lengthm

The minus sign says the force points back towards the natural length; in magnitude, Fs = k|x|.

Fg = G m₁m₂ / r²

Fg

attractive force on each massN

G

universal gravitational constantN m²/kg²

m₁, m₂

the two masseskg

r

distance between their centresm

g = Fg / m = G M / r²

g

field strength at that pointN/kg

M

mass producing the fieldkg

r

distance from its centrem

Why g at the surface has the value it does, and why it falls off with altitude.

ac = v² / r · Fnet = m v² / r

ac

centripetal acceleration, towards the centrem/s²

v

speed along the circlem/s

r

radiusm

Centripetal force is not an extra force to add to the diagram: it is the name for whatever real forces already point towards the centre.

v = √(G M / r) · T² = 4π² r³ / (G M)

v

orbital speedm/s

T

orbital periods

M

mass of the central bodykg

r

radius of the orbit, measured from the centrem

Gravity supplying the centripetal force: G M m / r² = m v² / r, and the orbiting mass cancels, so everything at that radius orbits at the same speed whatever its own mass. The period relation is Kepler's third law for a circular orbit. Neither is printed — rebuild them from the two equations above.

W = F d cos θ

W

work doneJ

F

forceN

d

displacementm

θ

angle between the force and the displacement°

A force perpendicular to the motion does no work, which is why the centripetal force never appears in an energy equation.

Wnet = ΔK

Wnet

work done by the net forceJ

ΔK

change in kinetic energyJ

The bridge between the force unit and this one.

K = ½mv²

K

kinetic energyJ

m

masskg

v

speedm/s

ΔUg = mgΔy

ΔUg

change in gravitational potential energyJ

Δy

change in heightm

For a field that can be treated as uniform. Only the change has meaning; you choose where zero sits.

Ug = − G m₁m₂ / r

Ug

potential energy of the pairJ

m₁, m₂

the two masseskg

r

separation of their centresm

Negative because the zero is taken at infinite separation. This is the form orbit and escape-speed questions need.

Us = ½kx²

Us

energy stored in the springJ

k

spring constantN/m

x

displacement from the natural lengthm

The area under a force–displacement graph.

ΔK + ΔU + ΔEthermal = 0

ΔK, ΔU

changes in kinetic and potential energyJ

ΔEthermal

energy that has gone to heating the surfacesJ

For an isolated system. Defining the system carefully is half of an AP energy question.

ΔEthermal = Ff d

Ff

kinetic friction forceN

d

distance slid along the surfacem

Not a printed statement: it is W = Fd cos θ with θ = 180°, counted as energy leaving the mechanical system. Distance slid, not displacement.

P = ΔE / Δt · P = F v cos θ

P

powerW

ΔE

energy transferredJ

v

speedm/s

p = mv

p

momentum, a vector along the velocitykg m/s

m

masskg

v

velocitym/s

J = FΔt = Δp

J

impulseN s

F

average force during the interactionN

Δt

how long it actss

Also the area under a force–time graph, which is how AP usually asks for it.

Σp before = Σp after

Σp

vector sum over the whole systemkg m/s

Holds when no external force acts on the system, and holds component by component.

elastic: ΣK before = ΣK after · perfectly inelastic: the objects move off together

K

kinetic energyJ

p

conserved in both caseskg m/s

Momentum is conserved in every collision here; kinetic energy only in the elastic one. The distinction is the point of the unit.

xcm = Σ(m x) / Σm

xcm

position of the centre of massm

m

mass of each partkg

x

position of each partm

The point that moves as if all the mass and all the external force acted there. AP Physics C: Mechanics turns this sum into an integral.

τ = r F sin θ

τ

torque about the chosen axisN m

r

distance from the axis to where the force actsm

F

forceN

θ

angle between r and F°

r sin θ is the lever arm — the perpendicular distance from the axis to the line of the force.

α = Στ / I

α

angular accelerationrad/s²

Στ

net torque about the axisN m

I

rotational inertia about that axiskg m²

I = Σ m r²

I

rotational inertiakg m²

m

each masskg

r

its distance from the axism

It depends on the axis, so it is meaningless to quote one without saying about what.

ω = ω₀ + αt · θ = θ₀ + ω₀t + ½αt²

ω, ω₀

final and initial angular velocityrad/s

θ, θ₀

final and initial angular positionrad

α

constant angular accelerationrad/s²

t

times

The straight-line equations with every symbol swapped for its angular twin.

ω² = ω₀² + 2α(θ − θ₀)

ω, ω₀

final and initial angular velocityrad/s

θ − θ₀

angle turned throughrad

The twin of v² = v₀² + 2aΔx. Worth deriving once rather than counting on it being printed.

s = rθ · v = rω · a = rα

s

arc lengthm

r

distance from the axism

θ

angle, in radiansrad

Only true with the angle in radians, which is why AP works in radians throughout.

ΣF = 0 · Στ = 0

Στ

about any axis you chooseN m

Both conditions are needed. Choosing the axis through an unknown force removes it from the torque equation.

K = ½Iω²

K

kinetic energy of rotationJ

I

rotational inertiakg m²

ω

angular velocityrad/s

L = Iω

L

angular momentum about the axiskg m²/s

I

rotational inertiakg m²

ω

angular velocityrad/s

ΔL = τ Δt

ΔL

change in angular momentumkg m²/s

τ

average external torqueN m

Δt

how long it actss

I₁ω₁ = I₂ω₂

I₁, I₂

rotational inertia before and afterkg m²

ω₁, ω₂

angular velocity before and afterrad/s

When no external torque acts. Pulling mass towards the axis lowers I, so ω rises — and K rises with it, because the pulling does work.

vcm = rω · K = ½m vcm² + ½ Icm ω²

vcm

speed of the centre of massm/s

Icm

rotational inertia about the centre of masskg m²

r

radius of the rolling objectm

The condition and the energy split follow from the definitions rather than being printed. It is why a hoop loses a race to a solid cylinder.

T = 1 / f · ω = 2πf

T

periods

f

frequencyHz

ω

angular frequencyrad/s

x = A cos(2πft)

x

displacement from equilibriumm

A

amplitudem

f

frequencyHz

t

times

Written for an oscillator released from maximum displacement at t = 0; a sine describes one started from equilibrium.

T = 2π √(m / k)

T

periods

m

oscillating masskg

k

spring constantN/m

No amplitude in it: the period of a simple harmonic oscillator does not depend on how far you pull it.

T = 2π √(L / g)

L

length of the pendulumm

g

gravitational field strengthm/s²

For small angles only, and independent of the mass hung on the end.

E = ½kA²

E

total energy of the oscillationJ

A

amplitudem

k

spring constantN/m

Us = ½kx² read at maximum displacement, where the kinetic energy is zero. Energy goes as amplitude squared.

ρ = m / V

ρ

densitykg/m³

m

masskg

V

volumem³

P = F / A

P

pressurePa

F

force perpendicular to the surfaceN

A

aream²

P = P₀ + ρgh

P

absolute pressure at that depthPa

P₀

pressure at the surfacePa

ρ

density of the fluidkg/m³

h

depth below the surfacem

Depth, not the shape of the container. ρgh on its own is the gauge pressure.

Fb = ρfluid Vdisplaced g

Fb

upward force on the submerged objectN

ρfluid

density of the fluid, not the objectkg/m³

Vdisplaced

volume of fluid pushed asidem³

Archimedes' principle. For a floating object the displaced weight equals the object's whole weight.

A₁v₁ = A₂v₂

A

cross-sectional aream²

v

flow speed therem/s

Conservation of volume for an incompressible fluid: a narrower pipe means a faster flow.

P₁ + ρgy₁ + ½ρv₁² = P₂ + ρgy₂ + ½ρv₂²

P

pressure at that pointPa

y

height of that pointm

v

flow speed therem/s

Conservation of energy per unit volume, for steady non-viscous incompressible flow along one streamline.

PV = nRT = N kB T

P

pressurePa

V

volumem³

n

amount of gasmol

N

number of molecules

T

absolute temperatureK

The same law counted two ways, by moles or by molecules. T must be in kelvin.

Kavg = (3/2) kB T

Kavg

mean translational kinetic energy of one moleculeJ

kB

Boltzmann's constantJ/K

T

absolute temperatureK

What temperature actually measures. It does not depend on which gas.

vrms = √(3 kB T / m) = √(3RT / M)

vrms

root-mean-square molecular speedm/s

m

mass of one moleculekg

M

molar masskg/mol

Heavier molecules move more slowly at the same temperature.

ΔU = Q + W

ΔU

change in the internal energy of the gasJ

Q

energy added by heatingJ

W

work done on the gasJ

AP takes W as the work done on the gas, so compressing it is positive. Signs are where marks are lost here.

W = − P ΔV

W

work done on the gasJ

P

pressure, held constantPa

ΔV

change in volumem³

At constant pressure. In general it is minus the area under the process on a PV diagram.

Q = m c ΔT

Q

energy transferred by heatingJ

m

masskg

c

specific heat capacityJ/(kg K)

ΔT

temperature changeK

FE = (1 / 4πε₀) q₁q₂ / r²

FE

force on each charge, along the line joining themN

q₁, q₂

the two chargesC

r

separationm

ε₀

permittivity of free spaceC²/(N m²)

1 / 4πε₀ is the constant written k = 9.0 × 10⁹ N m²/C². Like charges repel, so signs tell you direction, not magnitude.

E = FE / q

E

electric field, a vectorN/C or V/m

FE

force on the test chargeN

q

test chargeC

The definition. Field points along the force on a positive charge.

E = (1 / 4πε₀) q / r²

E

field strength at distance rN/C

q

charge producing the fieldC

r

distance from itm

Away from a positive charge, towards a negative one. Fields from several charges add as vectors.

E = ΔV / d

E

field strength, uniform between the platesV/m

ΔV

potential difference across the gapV

d

separation of the platesm

The field points from high potential to low.

V = (1 / 4πε₀) q / r

V

potential at that pointV

q

charge producing it, sign includedC

r

distance from itm

A scalar: potentials from several charges add by ordinary arithmetic, with their signs.

UE = (1 / 4πε₀) q₁q₂ / r

UE

potential energy of the pairJ

q₁, q₂

the two charges, signs includedC

Zero at infinite separation. Negative for opposite charges, which is why they are bound.

ΔUE = q ΔV

ΔUE

change in electric potential energyJ

q

charge movedC

ΔV

potential difference it moves throughV

The definition of the electronvolt, and the way accelerated-charge questions are set.

E = 0 inside · excess charge sits on the surface · the whole conductor is one equipotential

E

field within the conducting materialN/C

V

the same everywhere on and in the conductorV

Not formulae, but the facts every conductor question turns on. Any field inside would move the free charges, so it cannot persist.

I = ΔQ / Δt

I

currentA

ΔQ

charge passing a pointC

Δt

time takens

Conventional current is the direction positive charge would move.

R = ρ L / A

R

resistanceΩ

ρ

resistivity of the materialΩ m

L

lengthm

A

cross-sectional aream²

I = ΔV / R

RearrangedΔV = IR · R = ΔV / I

ΔV

potential difference across the componentV

I

current through itA

R

resistanceΩ

A property of ohmic components only; a filament lamp or a diode does not obey it.

P = I ΔV = I²R = (ΔV)² / R

P

power deliveredW

I

currentA

ΔV

potential differenceV

The last two forms hold only where ΔV = IR does.

Rs = R₁ + R₂ + …

Rs

equivalent resistanceΩ

Same current through each; the potential differences add.

1 / Rp = 1 / R₁ + 1 / R₂ + …

Rp

equivalent resistanceΩ

Same potential difference across each; the currents add. The result is always smaller than the smallest resistor.

Σ I into a junction = Σ I out of it · Σ ΔV round any loop = 0

I

current at a junctionA

ΔV

potential change across each element in the loopV

Conservation of charge and conservation of energy. Stated as rules rather than printed as formulae, and needed for any circuit that will not reduce to series and parallel.

C = Q / ΔV

C

capacitanceF

Q

charge on either plateC

ΔV

potential difference between the platesV

C = κ ε₀ A / d

κ

dielectric constant of the material between the plates

A

plate aream²

d

plate separationm

κ = 1 for a vacuum and is greater than 1 for any dielectric, so a dielectric always raises the capacitance.

UC = ½ Q ΔV = ½ C (ΔV)²

UC

energy storedJ

Q

charge storedC

C

capacitanceF

The half is there because the potential difference grows as the charge arrives.

1 / Cs = 1 / C₁ + 1 / C₂ + … · Cp = C₁ + C₂ + …

C

equivalent capacitanceF

The opposite way round from resistors, which is the mistake this pair exists to catch.

FM = q v B sin θ

FM

magnetic forceN

q

chargeC

v

speedm/s

B

magnetic flux densityT

θ

angle between v and B°

The force is perpendicular to both v and B, found with the right hand, so it does no work and only changes direction.

FM = B I L sin θ

I

currentA

L

length of wire in the fieldm

θ

angle between the wire and the field°

B = μ₀ I / (2π r)

B

flux density at distance rT

μ₀

permeability of free spaceT m/A

r

perpendicular distance from the wirem

The field circles the wire; grip it with the right hand, thumb along the current.

ΦB = B A cos θ

ΦB

magnetic flux through the loopWb

A

area of the loopm²

θ

angle between the field and the normal to the loop°

ε = − ΔΦB / Δt

ε

induced emfV

ΔΦB

change in flux through the loopWb

Δt

time it takess

The minus sign is Lenz's law: the induced current opposes the change that caused it. For N turns, multiply by N.

ε = B L v

L

length of the moving conductorm

v

speed across the fieldm/s

Faraday's law for a bar sliding on rails, where the area of the circuit changes at Lv. Worth being able to rebuild rather than recall.

n = c / v

n

index of refraction of the medium

c

speed of light in a vacuumm/s

v

speed of light in the mediumm/s

n is never less than 1. The frequency does not change when light crosses a boundary; the wavelength does.

n₁ sin θ₁ = n₂ sin θ₂

n₁, n₂

indices either side of the boundary

θ₁, θ₂

angles to the normal°

Angles are measured from the normal, not the surface.

sin θc = n₂ / n₁

θc

critical angle°

n₁

index of the denser medium the light is in

n₂

index of the medium beyond

Snell's law with the refracted ray at 90°. Only possible when n₁ is greater than n₂; beyond θc the light is totally internally reflected.

1 / so + 1 / si = 1 / f

so

object distancem

si

image distance, negative for a virtual imagem

f

focal length, negative for a diverging lens or convex mirrorm

The sign convention carries the physics: get it wrong and a virtual image comes out real.

M = hi / ho = − si / so

M

magnification

hi, ho

image and object heightm

A negative M means the image is inverted.

f = R / 2

f

focal lengthm

R

radius of curvaturem

v = f λ

Rearrangedf = v / λ · λ = v / f

v

wave speed, set by the mediumm/s

f

frequency, set by the sourceHz

λ

wavelengthm

λn = 2L / n, n = 1, 2, 3 …

L

length of the string or pipem

n

harmonic number

A string fixed at both ends, or a pipe open at both ends. Read off the boundary conditions rather than recalled — draw the pattern and count.

λn = 4L / n, n = 1, 3, 5 …

L

length of the pipem

n

odd harmonics only

A node at the closed end and an antinode at the open one, which is why the even harmonics are missing.

d sin θ = m λ

d

slit separation, or grating spacingm

θ

angle of the bright fringe from the centre°

m

order of the fringe, 0, 1, 2 …

Constructive interference, where the path difference is a whole number of wavelengths.

Δy = λ L / d

Δy

distance between neighbouring bright fringesm

L

distance from the slits to the screenm

d

slit separationm

For small angles, with L much greater than d.

a sin θ = m λ, m = 1, 2, 3 …

a

width of the single slitm

θ

angle of the dark fringe°

The same-looking equation as the two-slit one but locating dark fringes, not bright. Note which it is before substituting.

E = h f = h c / λ

E

energy of one photonJ or eV

h

Planck's constantJ s

f

frequency of the lightHz

Kmax = h f − φ

Kmax

maximum kinetic energy of an emitted electronJ or eV

φ

work function of the surfaceJ or eV

Below the threshold frequency nothing is emitted however bright the light — the observation that made the photon necessary.

λ = h / p

λ

wavelength associated with the particlem

p

momentumkg m/s

h f = |ΔE|

ΔE

difference between the two levelsJ or eV

f

frequency of the photonHz

Levels are discrete, so the emitted frequencies are too. That is what a line spectrum is.

E = m c²

E

energy equivalent of the massJ

m

masskg

c

speed of light in a vacuumm/s

Used as ΔE = Δm c² for the mass defect of a nucleus and the energy released in a decay.

Σ A unchanged · Σ Z unchanged

A

mass number, the count of nucleons

Z

atomic number, the count of protons

Charge and nucleon number balance across the arrow; energy and momentum are conserved as well. Not a formula, but it is what the equations are checked against.

v = dx / dt

v

instantaneous velocity, a vectorm/s

x

positionm

t

times

The slope of the position graph, said exactly. In two dimensions each component is differentiated separately.

a = dv / dt = d²x / dt²

a

instantaneous accelerationm/s²

v

velocitym/s

v = v₀ + ∫ a dt · x = x₀ + ∫ v dt

v₀, x₀

the initial values, which the integration constants fixm/s, m

∫ … dt

over the interval in question

The same two definitions read backwards, and the reason this course can set motion where the acceleration is not constant. The algebra-based equations of motion are what these give when a is constant.

Fnet = dp / dt

Fnet

net forceN

p

momentumkg m/s

Newton's second law in the form he wrote it. Fnet = ma is the special case of constant mass.

m dv / dt = m g − b v

b

drag coefficient for a force proportional to speedkg/s

v

speedm/s

m g

the driving force, here the weightN

Written for a falling object with drag proportional to v; a drag proportional to v² is set the same way. This separable differential equation is the piece of calculus this unit exists to use.

vterm = m g / b

vterm

speed at which the acceleration reaches zerom/s

b

drag coefficient, for drag proportional to speedkg/s

Set dv/dt = 0 in the equation above. The speed is approached but never quite reached.

W = ∫ F · dr

W

work doneJ

F · dr

the component of the force along the path, times the step along itJ

The area under a force–displacement graph, and the general statement W = Fd cos θ is the constant-force case of.

P = dW / dt = F · v

P

power at that instantW

F · v

force times the component of velocity along itW

F = − dU / dx

U

potential energy as a function of positionJ

F

conservative force, along xN

The force points down the slope of the U–x graph. Equilibrium is where the slope is zero, stable where the curve is a minimum.

ΔU = − ∫ F · dr

ΔU

change in potential energyJ

F

the conservative forceN

Where ½kx² and −Gm₁m₂/r come from, and only defined for a force whose work does not depend on the path.

J = ∫ F dt = Δp

J

impulseN s

F

force, varying through the interactionN

The area under a force–time graph, without needing the force to be constant.

rcm = (1 / M) ∫ r dm

rcm

position of the centre of massm

M

total masskg

dm

mass of an element at position rkg

The discrete sum turned into an integral. Writing dm in terms of a density and a length, area or volume element is the whole technique.

M acm = Σ Fexternal

acm

acceleration of the centre of massm/s²

Σ Fexternal

sum of the external forces onlyN

Internal forces cancel in pairs, so an exploding shell's centre of mass carries on along the same parabola.

τ = r × F

τ

torque, a vector along the axisN m

r

vector from the axis to the point of applicationm

F

forceN

Magnitude rF sin θ, direction from the right hand. The vector form is what lets torques about different axes be added properly.

I = ∫ r² dm

I

rotational inertia about the chosen axiskg m²

r

perpendicular distance of the element from the axism

dm

mass elementkg

Deriving I for a rod, a disc and a sphere from this is standard work in this course, not a lookup.

I = Icm + M d²

Icm

rotational inertia about a parallel axis through the centre of masskg m²

M

total masskg

d

distance between the two axesm

Only works between parallel axes, one of which passes through the centre of mass.

L = r × p

L

angular momentum about the chosen pointkg m²/s

r

vector from that point to the particlem

p

momentumkg m/s

Magnitude m v r sin θ. A particle travelling in a straight line still has angular momentum about a point off its line of motion, and that angular momentum is constant.

τ = dL / dt

τ

net external torque about the pointN m

L

angular momentum about the same pointkg m²/s

The rotational twin of Fnet = dp/dt, and the cleanest statement of why L is conserved when no external torque acts.

d²x / dt² = − ω² x

x

displacement from equilibriumm

ω

angular frequencyrad/s

The definition of simple harmonic motion in this course: show a restoring force proportional to displacement and this is what you have. ω² is read straight off it.

x = A cos(ω t + φ)

A

amplitudem

φ

phase constant, fixed by the initial conditionsrad

ω

angular frequencyrad/s

Differentiate it for v = −Aω sin(ωt + φ) and a = −Aω² cos(ωt + φ), which is where the maximum speed Aω and maximum acceleration Aω² come from.

T = 2π √(I / (m g d))

I

rotational inertia about the pivotkg m²

d

distance from the pivot to the centre of massm

m

mass of the bodykg

From Στ = Iα for small angles. The simple pendulum is the case I = mL² and d = L.

E = (1 / 4πε₀) ∫ dq / r²

dq

charge on an element of the distributionC

r

distance from that element to the field pointm

E

field, summed as a vectorN/C

Add the components, not the magnitudes: symmetry usually kills one component and is what makes the integral doable.

λ = dq / dl · σ = dq / dA · ρ = dq / dV

λ

charge per unit lengthC/m

σ

charge per unit areaC/m²

ρ

charge per unit volumeC/m³

Definitions rather than results, but choosing the right one is what turns dq into something you can integrate.

∮ E · dA = Qenc / ε₀

∮ E · dA

flux of the field out through a closed surfaceN m²/C

Qenc

net charge enclosed by that surfaceC

ε₀

permittivity of free spaceC²/(N m²)

Always true, but only useful where symmetry makes E constant and parallel to dA over the surface — spherical, cylindrical or planar.

sphere, outside: E = (1 / 4πε₀) Q / r² · infinite line: E = λ / (2πε₀ r) · infinite plane: E = σ / (2ε₀)

Q

total charge on the sphereC

λ, σ

line and surface charge densityC/m, C/m²

r

distance from the centre or the axism

Results of the derivation, not statements to quote in place of it. A uniformly charged sphere pulls from outside as if all its charge sat at the centre; inside a uniform shell the field is zero.

ΔV = − ∫ E · dr

ΔV

potential difference between the two pointsV

E · dr

component of the field along the path, times the stepV

The path does not matter, only the endpoints. Moving against the field raises the potential.

Ex = − ∂V / ∂x

Ex

component of the field along xV/m

∂V / ∂x

rate of change of potential in that directionV/m

The field points the way the potential falls fastest, and is perpendicular to the equipotentials.

V = (1 / 4πε₀) ∫ dq / r

V

potential at the point, taken as zero at infinityV

dq

charge elementC

r

its distance from the pointm

A scalar integral with no components to resolve, which is why it is usually easier to find V first and differentiate for E.

UE = (1 / 4πε₀) Σ qi qj / rij

qi, qj

each pair of charges, signs includedC

rij

separation of that pairm

Summed over each pair once — the work needed to assemble the configuration from infinity.

E = σ / ε₀

σ

local surface charge densityC/m²

E

field just outside, perpendicular to the surfaceN/C

From Gauss's law on a small pillbox straddling the surface. Twice the field of an isolated charged sheet, because all the flux leaves on one side.

u = ½ ε₀ E²

u

energy stored per unit volume of the fieldJ/m³

E

field strength thereV/m

The energy in a capacitor located in the field between the plates rather than on them.

Q → E by Gauss's law → ΔV = − ∫ E · dr → C = Q / ΔV

Q

charge assumed on the conductorsC

C

capacitance, which the Q cancels out ofF

A method, not a printed formula, and the way this course gets the capacitance of a spherical or cylindrical capacitor rather than only the parallel-plate one.

I = dQ / dt

I

instantaneous currentA

Q

charge that has passedC

Needed because in an RC or LR circuit the current is not constant.

J = I / A · E = ρ J

J

current densityA/m²

ρ

resistivityΩ m

E

field driving the current inside the wireV/m

The local form of Ohm's law. R = ρL/A follows from it.

q(t) = C ε (1 − e^(−t / RC)) · I(t) = (ε / R) e^(−t / RC)

q

charge on the capacitor at time tC

ε

emf of the sourceV

RC

time constant of the circuits

The solution of the loop equation ε = IR + q/C. This course expects the differential equation to be set up and solved, so learn the shape of the curves as well as the algebra.

q(t) = Q₀ e^(−t / RC) · I(t) = (Q₀ / RC) e^(−t / RC)

Q₀

charge on the capacitor at t = 0C

R

resistance in the loopΩ

C

capacitanceF

Same exponential, no source. After one time constant RC the charge has fallen to about 37% of its starting value.

at t = 0 the capacitor behaves as a wire · after a long time it behaves as a break

t = 0

the instant the switch closes, when the capacitor is uncharged

t → ∞

when the capacitor is fully charged and no current flows into it

Not an equation, but it answers most of what an AP C circuit question actually asks, without solving anything.

F = q v × B

F

magnetic force on the moving chargeN

v

velocity of the chargem/s

B

magnetic fieldT

The vector form of the magnitude qvB sin θ the algebra-based course uses. It does no work: speed cannot change, only direction.

dF = I dl × B

dl

element of the wire, in the direction of the currentm

I

currentA

Integrated along the wire, which is how a curved or looped conductor is handled.

dB = (μ₀ / 4π) I (dl × r) / r³

dB

field contributed by one current elementT

r

vector from the element to the field pointm

μ₀

permeability of free spaceT m/A

The magnetic counterpart of adding up dE from dq. The r³ in the denominator is there because r on top is not a unit vector.

∮ B · dl = μ₀ Ienc

∮ B · dl

field summed round a closed loopT m

Ienc

current passing through the loopA

Gauss's law's magnetic twin, and useful under the same condition: enough symmetry that B is constant along the chosen loop.

B = μ₀ n I

n

turns per unit lengthm⁻¹

I

current in the windingA

Uniform along the axis and independent of the radius, which is what makes a solenoid useful.

r = m v / (q B)

r

radius of the circlem

m

mass of the particlekg

q

chargeC

From qvB = mv²/r, the magnetic force supplying the centripetal force. The period does not depend on the speed.

ΦB = ∫ B · dA

ΦB

flux through the surfaceWb

B · dA

component of the field normal to each area element, times that areaWb

Needed where the field varies across the loop, as it does near a straight wire.

ε = − dΦB / dt

ε

induced emfV

dΦB / dt

rate of change of flux, from B, from the area, or from the angleWb/s

The instantaneous form of the ΔΦ/Δt statement the algebra-based course uses. The minus sign is Lenz's law, and this unit is built on it.

ε = − L dI / dt

L

inductanceH

dI / dt

rate of change of currentA/s

An inductor opposes changes in current, not current itself: with a steady current it is just a wire.

UL = ½ L I²

UL

energy stored in the magnetic fieldJ

I

current through the inductorA

I(t) = (ε / R) (1 − e^(−R t / L))

ε

emf of the sourceV

L / R

time constant of the circuits

The solution of ε = IR + L dI/dt. As with RC, the differential equation is the examinable part; at t = 0 the inductor blocks the current entirely and after a long time it does nothing.