AP Physics Constants Sheet: Every Value, All 4 Courses

The AP Physics constants box differs by course. Physics 1 prints four entries, C: Mechanics three, C: Electricity and Magnetism thirteen, and Physics 2 twenty. All four print gravity at Earth's surface twice, as 9.8 meters per second squared and as 9.8 newtons per kilogram.

Read off the CONSTANTS AND CONVERSION FACTORS box in the Table of Information appendix of all four Course and Exam Descriptions, effective Fall 2024: AP Physics 1 (printed appendix page 211), AP Physics 2 (page 218), AP Physics C: Mechanics (page 204) and AP Physics C: Electricity and Magnetism (page 178). Values are reproduced at the precision the booklet prints them.

Every constant printed in any AP Physics booklet

There is no single AP Physics constants sheet. There are four, one per Course and Exam Description, and they are not the same. The table below is the union of all four, written at the precision the booklet actually prints, with the last column saying which courses get it.

QuantityAs the booklet prints itPrinted in
Avogadro's numberN0=6.02×1023 mol1N_0 = 6.02 \times 10^{23}\ \text{mol}^{-1}Physics 2 only
Universal gas constantR=8.31 J/(molK)R = 8.31\ \text{J}/(\text{mol} \cdot \text{K})Physics 2 only
Boltzmann's constantkB=1.38×1023 J/Kk_B = 1.38 \times 10^{-23}\ \text{J}/\text{K}Physics 2 only
1 atmosphere of pressure1 atm=1.0×105 N/m2=1.0×105 Pa1\ \text{atm} = 1.0 \times 10^{5}\ \text{N}/\text{m}^2 = 1.0 \times 10^{5}\ \text{Pa}Physics 1 and Physics 2
Coulomb constantk=14πε0=9.0×109 Nm2/C2k = \dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N} \cdot \text{m}^2/\text{C}^2Physics 2 and C: E&M
Proton massmp=1.67×1027 kgm_p = 1.67 \times 10^{-27}\ \text{kg}Physics 2 and C: E&M
Neutron massmn=1.67×1027 kgm_n = 1.67 \times 10^{-27}\ \text{kg}Physics 2 and C: E&M
Electron massme=9.11×1031 kgm_e = 9.11 \times 10^{-31}\ \text{kg}Physics 2 and C: E&M
Elementary chargee=1.60×1019 Ce = 1.60 \times 10^{-19}\ \text{C}Physics 2 and C: E&M
Vacuum permittivityε0=8.85×1012 C2/(Nm2)\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N} \cdot \text{m}^2)Physics 2 and C: E&M
Vacuum permeabilityμ0=4π×107 (Tm)/A\mu_0 = 4\pi \times 10^{-7}\ (\text{T} \cdot \text{m})/\text{A}Physics 2 and C: E&M
1 electron volt1 eV=1.60×1019 J1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}Physics 2 and C: E&M
Planck's constanth=6.63×1034 Js=4.14×1015 eVsh = 6.63 \times 10^{-34}\ \text{J} \cdot \text{s} = 4.14 \times 10^{-15}\ \text{eV} \cdot \text{s}Physics 2 only
Planck's constant, second printed linehc=1.99×1025 Jm=1240 eVnmhc = 1.99 \times 10^{-25}\ \text{J} \cdot \text{m} = 1240\ \text{eV} \cdot \text{nm}Physics 2 only
Speed of lightc=3.00×108 m/sc = 3.00 \times 10^{8}\ \text{m}/\text{s}Physics 2 and C: E&M
Wien's constantb=2.90×103 mKb = 2.90 \times 10^{-3}\ \text{m} \cdot \text{K}Physics 2 only
Stefan-Boltzmann constantσ=5.67×108 W/(m2K4)\sigma = 5.67 \times 10^{-8}\ \text{W}/(\text{m}^2 \cdot \text{K}^4)Physics 2 only
1 unified atomic mass unit1 u=1.66×1027 kg=931 MeV/c21\ \text{u} = 1.66 \times 10^{-27}\ \text{kg} = 931\ \text{MeV}/c^2Physics 2 and C: E&M
Universal gravitational constantG=6.67×1011 m3/(kgs2)=6.67×1011 Nm2/kg2G = 6.67 \times 10^{-11}\ \text{m}^3/(\text{kg} \cdot \text{s}^2) = 6.67 \times 10^{-11}\ \text{N} \cdot \text{m}^2/\text{kg}^2all four
Acceleration due to gravity at Earth's surfaceg=9.8 m/s2g = 9.8\ \text{m}/\text{s}^2all four
Gravitational field strength at Earth's surfaceg=9.8 N/kgg = 9.8\ \text{N}/\text{kg}all four

Twenty one rows, and twenty labels, because Planck's constant is one labeled entry that runs onto a second printed line to give you hchc. Nothing else in any of the four boxes appears that is not in this table.

Every value above is copied at the precision the College Board prints. Do not upgrade any of them. The sheet says 9.89.8, not 9.819.81. It says 9.0×1099.0 \times 10^{9}, not 8.99×1098.99 \times 10^{9}. It says 3.00×1083.00 \times 10^{8}, not 2.998×1082.998 \times 10^{8}. Those are the numbers a grader expects your arithmetic to be consistent with.

What each course actually gets

AP Physics 1: four entries. The whole box is the universal gravitational constant, one atmosphere of pressure, and gravity at Earth's surface written out twice. That is it. If a Physics 1 problem needs a mass, a density or a spring constant, the problem gives it to you, because the sheet will not.

AP Physics C: Mechanics: three entries. The same as Physics 1 minus the atmosphere. C: Mechanics has no fluids unit, so the pressure conversion has nothing to do. This is the smallest constants box of the four.

AP Physics C: Electricity and Magnetism: thirteen entries. The Coulomb constant, the two vacuum constants, the three particle masses, the elementary charge, the electron volt conversion, the speed of light, the atomic mass unit, and then the same three gravitational lines every course gets.

AP Physics 2: twenty entries. Everything C: E&M has, plus one atmosphere, plus the six that belong to nobody else: Avogadro's number, the universal gas constant, Boltzmann's constant, Planck's constant, Wien's constant and the Stefan-Boltzmann constant. Those six are exactly the thermal physics and modern physics content that only Physics 2 carries.

Work through that and a clean relationship falls out: the AP Physics 2 constants box contains every constant printed in any of the four booklets. The other three boxes are subsets of it. Check it against the table above if you want to be sure, quantity by quantity, because it is the sort of tidy claim that is usually false.

One consequence catches people out. AP Physics C: Electricity and Magnetism does not print Planck's constant. Neither hh nor hchc is in that box, because the C: E&M course does not cover photons. If you are working a photon problem, you are working in AP Physics 2.

Why gravity is printed twice

Every one of the four booklets spends two of its scarce lines on the same number:

g=9.8 m/s2andg=9.8 N/kgg = 9.8\ \text{m}/\text{s}^2 \qquad \text{and} \qquad g = 9.8\ \text{N}/\text{kg}

Those are the same quantity in algebraically identical units. A newton is a kgm/s2\text{kg} \cdot \text{m}/\text{s}^2, so a N/kg\text{N}/\text{kg} reduces to m/s2\text{m}/\text{s}^2 exactly. Nothing numerical changes.

What changes is what the number is being used for, and the redesigned courses care about that distinction. Written as m/s2\text{m}/\text{s}^2, gg is the acceleration of a freely falling object: an effect. Written as N/kg\text{N}/\text{kg}, gg is the gravitational field strength: force per unit mass, a property of the region of space, which exists whether or not anything is falling through it. The label on each line says so. The first reads "Magnitude of the acceleration due to gravity at Earth's surface" and the second reads "Magnitude of the gravitational field strength at Earth's surface".

The printing is a hint about how the field is meant to be discussed, not a second number to keep track of.

The precision trap, and the g = 9.8 versus g = 10 question

The Table of Information prints g=9.8g = 9.8. It does not print 9.819.81, and it does not print 1010.

That is not the whole story, and pretending it is has misled a lot of students. The AP Physics 1 Course and Exam Description states elsewhere, in the Unit 1 material, that the exam will use g10 m/s2g \approx 10\ \text{m}/\text{s}^2 where a numerical value for gg is required, and in the same breath says students will not be penalized for correctly using the more precise commonly accepted values of 9.81 m/s29.81\ \text{m}/\text{s}^2 or 9.8 m/s29.8\ \text{m}/\text{s}^2. So both live in the same document: the exam questions may round to 1010 to keep the arithmetic mental, and 9.89.8 is explicitly sanctioned. Pick one, say which you picked, and use it to the end of the problem.

The same discipline applies to the rest of the box:

  • The Coulomb constant is given as 9.0×1099.0 \times 10^{9}, two significant figures. Using 8.99×1098.99 \times 10^{9} is not wrong physics, but it can shift a two-figure answer, and it is not what the sheet handed you.
  • The speed of light is 3.00×1083.00 \times 10^{8}, three figures.
  • GG is 6.67×10116.67 \times 10^{-11}, three figures.
  • The particle masses are 1.67×10271.67 \times 10^{-27} kg for the proton and the neutron, printed identically, and 9.11×10319.11 \times 10^{-31} kg for the electron. The proton and neutron differ in reality, and the sheet does not distinguish them.

If your answer changes at the second significant figure depending on which version of a constant you used, the grader is not going to mind. If you mixed a two-figure constant into a computation and then reported six figures, that is a different problem.

The three entries that save the most time

Three entries in the AP Physics 2 box are pre-converted for you, and each exists to remove an arithmetic step people routinely get wrong.

hc=1240 eVnmhc = 1240\ \text{eV} \cdot \text{nm}. Wavelengths of visible light are quoted in nanometers and photon energies in electron volts, so this form lets you divide directly and read an answer in eV with no unit conversion at all. The box also gives hc=1.99×1025 Jmhc = 1.99 \times 10^{-25}\ \text{J} \cdot \text{m} for when you want joules and meters.

h=4.14×1015 eVsh = 4.14 \times 10^{-15}\ \text{eV} \cdot \text{s}. The same idea applied to E=hfE = hf. If a frequency is in hertz and you want energy in electron volts, use this rather than converting through joules.

1 u=931 MeV/c21\ \text{u} = 931\ \text{MeV}/c^2. Mass defect in a nuclear problem is naturally expressed in atomic mass units, and the answer is wanted in MeV. This line is the conversion, already carrying the c2c^2.

Notice also what the Coulomb constant line does. It prints kk and 14πε0\dfrac{1}{4\pi\varepsilon_0} as the same object, on one line, because the AP Physics 2 equation sheet writes Coulomb's law and the point charge field both ways and students otherwise treat them as two different constants.

Notation the booklet uses that textbooks do not

A few of the printed symbols do not match what a general physics textbook would show, and copying the textbook version into an AP answer is a small avoidable friction.

  • Avogadro's number is N0N_0, not NAN_A. The subscript is a zero. On the same sheet, NN on its own is the number of atoms in the thermal physics list and the number of particles in the modern physics list.
  • Boltzmann's constant is kBk_B with a subscript B, which matters because plain kk is already doing duty as the Coulomb constant, the spring constant and thermal conductivity elsewhere in the same booklet.
  • Wien's constant is bb. It is stated as a plain constant with units mK\text{m} \cdot \text{K}, and the displacement law itself, λmax=b/T\lambda_{max} = b/T, appears in the modern physics equations rather than in the constants box.
  • The universal gravitational constant is written two ways on one line, as m3/(kgs2)\text{m}^3/(\text{kg} \cdot \text{s}^2) and as Nm2/kg2\text{N} \cdot \text{m}^2/\text{kg}^2. Same constant. The second form is the one that cancels cleanly in Newton's law of gravitation.

The constants box also holds three unit conversions rather than constants of nature: one atmosphere, one electron volt, and one unified atomic mass unit. The header says "CONSTANTS AND CONVERSION FACTORS" for that reason.

What is not in the box

Students look for these and they are not there, in any of the four booklets:

  • Densities of anything, including water. A fluids problem states the density it wants.
  • Specific heats, latent heats and thermal conductivities. Physics 2 defines cc as specific heat and kk as thermal conductivity in its symbol list and gives numerical values in the problem.
  • Coefficients of friction for named surfaces.
  • Resistivities, dielectric constants and indices of refraction. The index of refraction of water and of glass are not printed.
  • Moments of inertia of standard shapes. AP Physics 1 and C: Mechanics print the definition I=miri2I = \sum m_i r_i^2 and the parallel axis theorem, and expect a specific problem to supply or derive the rest.
  • The mass or radius of the Earth, the Sun or the Moon. A gravitation problem gives them.
  • The Rydberg constant, and any table of atomic energy levels.

The pattern is consistent: the sheet gives you universal constants and unit conversions, and every material property comes from the question. If you are stuck because you think a number should be on the sheet, reread the problem instead of hunting for it.

Reading gravity two ways from the same printed line

A 4.0 kg object hangs at rest from a spring scale near Earth's surface. Use the constants box to find the gravitational field strength at the object, and the reading on the scale.

  1. The box prints the field strength directly: g=9.8 N/kgg = 9.8\ \text{N}/\text{kg}. That is the answer to the first part with no calculation. The field does not depend on the 4.0 kg object hanging in it.

  2. For the scale reading, the object is in equilibrium, so the spring force equals the gravitational force in magnitude.

  3. Fg=mg=(4.0 kg)(9.8 N/kg)=39.2 NF_g = mg = (4.0\ \text{kg})(9.8\ \text{N}/\text{kg}) = 39.2\ \text{N}

  4. The kilograms cancel against the kilograms in the field-strength unit and leave newtons, which is the point of printing gg in N/kg\text{N}/\text{kg}.

  5. To two significant figures, the scale reads 39 N.

  6. The other printed line, g=9.8 m/s2g = 9.8\ \text{m}/\text{s}^2, would give the same 39.2 N through F=maF = ma, because 1 N/kg1\ \text{N}/\text{kg} and 1 m/s21\ \text{m}/\text{s}^2 are the same unit.

The field strength is 9.8 N/kg9.8\ \text{N}/\text{kg} and the scale reads 39 N.

How much the Coulomb constant's precision actually matters

Two point charges of +3.0 μC+3.0\ \mu\text{C} and 2.0 μC-2.0\ \mu\text{C} sit 0.15 m apart. Find the magnitude of the force between them using the AP sheet value of kk, then check what the more precise value would have given.

  1. Strip the prefixes first. 3.0 μC=3.0×106 C3.0\ \mu\text{C} = 3.0 \times 10^{-6}\ \text{C} and 2.0 μC=2.0×106 C2.0\ \mu\text{C} = 2.0 \times 10^{-6}\ \text{C}.

  2. The AP Physics 2 and C: E&M boxes both print k=9.0×109 Nm2/C2k = 9.0 \times 10^{9}\ \text{N} \cdot \text{m}^2/\text{C}^2.

  3. F=kq1q2r2=(9.0×109)(3.0×106)(2.0×106)(0.15)2F = k\dfrac{q_1 q_2}{r^2} = \left(9.0 \times 10^{9}\right)\dfrac{\left(3.0 \times 10^{-6}\right)\left(2.0 \times 10^{-6}\right)}{(0.15)^2}

  4. Numerator: (9.0×109)(6.0×1012)=5.4×102\left(9.0 \times 10^{9}\right)\left(6.0 \times 10^{-12}\right) = 5.4 \times 10^{-2}

  5. Denominator: (0.15)2=2.25×102(0.15)^2 = 2.25 \times 10^{-2}

  6. F=5.4×1022.25×102=2.4 NF = \dfrac{5.4 \times 10^{-2}}{2.25 \times 10^{-2}} = 2.4\ \text{N}

  7. With 8.99×1098.99 \times 10^{9} instead: (8.99×109)(6.0×1012)=5.394×102\left(8.99 \times 10^{9}\right)\left(6.0 \times 10^{-12}\right) = 5.394 \times 10^{-2}, and 5.394×102/2.25×102=2.397 N5.394 \times 10^{-2}/2.25 \times 10^{-2} = 2.397\ \text{N}, which is 2.4 N to two figures.

  8. Same answer at the precision the givens support. Use the printed value and stop worrying about it.

F=2.4F = 2.4 N, and the more precise constant gives 2.4 N as well.

Getting an energy in electron volts without leaving electron volts

A metal has a work function of 2.3 eV. Light of wavelength 400 nm strikes it. Find the maximum kinetic energy of the ejected electrons in eV, using only values printed in the AP Physics 2 constants box.

  1. The box prints hc=1240 eVnmhc = 1240\ \text{eV} \cdot \text{nm}, which is designed for exactly this.

  2. Photon energy: E=hcλ=1240 eVnm400 nm=3.10 eVE = \dfrac{hc}{\lambda} = \dfrac{1240\ \text{eV} \cdot \text{nm}}{400\ \text{nm}} = 3.10\ \text{eV}

  3. The nanometers cancel and the answer arrives in eV with no conversion.

  4. The AP Physics 2 equation sheet gives Kmax=hfϕK_{max} = hf - \phi, and hfhf is the photon energy just computed.

  5. Kmax=3.10 eV2.3 eV=0.80 eVK_{max} = 3.10\ \text{eV} - 2.3\ \text{eV} = 0.80\ \text{eV}

  6. In joules, using the printed 1 eV=1.60×1019 J1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}: Kmax=0.80×1.60×1019=1.28×1019 JK_{max} = 0.80 \times 1.60 \times 10^{-19} = 1.28 \times 10^{-19}\ \text{J}, or 1.3×10191.3 \times 10^{-19} J to two figures.

  7. Doing it the long way, in SI throughout, needs hh, cc, a metre conversion on the wavelength and a joule conversion on the work function. Four more places to slip.

Kmax=0.80K_{max} = 0.80 eV, which is 1.3×10191.3 \times 10^{-19} J.

Frequently asked questions

What constants are given on the AP Physics 1 exam?

Four entries, and that is the entire box. The universal gravitational constant, 6.67 times ten to the minus eleventh in either of two unit forms. One atmosphere of pressure, 1.0 times ten to the fifth pascals. The acceleration due to gravity at Earth's surface, 9.8 meters per second squared. And the gravitational field strength at Earth's surface, 9.8 newtons per kilogram, which is the same number written for a different purpose. Any mass, density, spring constant or coefficient of friction a Physics 1 problem needs is stated in the problem.

Is the AP Physics constants sheet the same for all four courses?

No, and the differences are large. AP Physics 1 prints four entries, AP Physics C: Mechanics three, AP Physics C: Electricity and Magnetism thirteen, and AP Physics 2 twenty. The AP Physics 2 box contains every constant printed in any of the four, so it is a superset of the other three. Only three entries are common to all four courses: the universal gravitational constant and the two printings of g.

Is g 9.8 or 9.81 on the AP Physics exam?

The Table of Information prints 9.8 meters per second squared, so 9.8 is the sheet value. The AP Physics 1 Course and Exam Description separately states that the exam will use g of about 10 meters per second squared where a numerical value is required, and that students are not penalized for correctly using 9.81 or 9.8 instead. All three numbers appear in the official materials. Choose one, state it, and be consistent through the whole problem.

Is Planck's constant on the AP Physics C: E&M equation sheet?

No. Planck's constant appears in the AP Physics 2 constants box only, in three forms: h in joule seconds, h in electron volt seconds, and hc in both joule meters and electron volt nanometers. The AP Physics C: Electricity and Magnetism box has no h, because that course does not cover photons or quantum energy levels. It does print the speed of light, the electron volt conversion and the electron, proton and neutron masses.

Why does the AP sheet print g twice, in m/s^2 and in N/kg?

Because they are the same value labeled for two different jobs. A newton is a kilogram meter per second squared, so newtons per kilogram reduces exactly to meters per second squared and the number does not change. Written as an acceleration, g describes how fast a falling object speeds up. Written as newtons per kilogram, g is the gravitational field strength, a force per unit mass that exists at that point in space whether or not anything is there to fall. The redesigned courses want the field treated as a real thing, so both labels are printed.

Should I use 9.0 times ten to the ninth or 8.99 times ten to the ninth for k?

Use 9.0 times ten to the ninth newton meters squared per coulomb squared, which is what both the AP Physics 2 and the AP Physics C: Electricity and Magnetism boxes print. The difference from the more precise value is about one tenth of one percent, which almost never survives rounding to the two or three significant figures an AP answer carries. Using the printed value also keeps your arithmetic matching the number the question was designed around.

Are densities and specific heats on the AP Physics constants sheet?

No. No material property is printed in any of the four constants boxes: no densities including water, no specific heats, no latent heats, no thermal conductivities, no resistivities, no dielectric constants, no indices of refraction and no coefficients of friction. Every one of those is a measured property of a particular substance, and an AP question that needs one states it. The boxes hold universal constants and unit conversions only.

What is hc equals 1240 eV nm used for?

It converts a wavelength in nanometers straight into a photon energy in electron volts, with no unit conversion in between. Divide 1240 by the wavelength in nm and the answer is in eV. It is printed on the AP Physics 2 sheet because visible light is naturally quoted in nanometers and photon energies in electron volts, so the pre-multiplied form removes two conversions from every photoelectric effect and energy level problem. The same box also prints hc as 1.99 times ten to the minus twenty fifth joule meters for work in SI units.