Permittivity vs Permeability: What Is the Difference?

Permittivity, epsilon nought, measures how strongly a material polarizes in an electric field, and it appears wherever an electric field or force is calculated. Permeability, mu nought, measures how strongly a material magnetizes in a magnetic field. Their product fixes the speed of light.

AP Physics: Unit 10 (topics 10.1 Electric Charge and Electric Force, 12.1 Magnetic Fields, 13.2 Electromagnetic Induction). The two definitions are AP Physics 2 essential knowledge 10.1.C.1 (electric permittivity is a measurement of the degree to which a material or medium is polarized in the presence of an electric field) and 12.1.C.1 (magnetic permeability is a measurement of the amount of magnetization in a material in response to an external magnetic field), under learning objectives 10.1.C and 12.1.C. AP Physics C: Electricity and Magnetism repeats both word for word at 8.1.C.1 and 12.1.C.1. Free-space values come from 10.1.C.3 and 12.1.C.2 and are printed in the Constants and Conversion Factors box of both sheets. Behaviour in matter is 10.1.C.4 with 10.1.C.4.i and 10.1.C.4.ii on the electric side and 12.1.C.3 on the magnetic side, and only the magnetic statement says the quantity is not a constant for a material. The product of the two constants is AP Physics C: E&M essential knowledge 13.2.A.4, which labels c equals one over the root of epsilon nought mu nought a derived equation; that topic's boundary statement says the course does not expect students to mathematically derive the speed of light in free space from Maxwell's equations and that the relationship is included solely as an indication of further applications, implications, and connections that students may study in more advanced courses. Topic 10.1 carries its own boundary statement limiting electric force calculations to four or fewer interacting charged objects or systems, with more charges allowed in situations of high symmetry. AP Physics 2 Unit 10 is 15 to 18 percent of the multiple-choice section across about 14 to 21 class periods and Unit 12 is 12 to 15 percent across about 10 to 14, and AP Physics C: E&M Unit 13 is 10 to 20 percent.

The distinction, stated once

These two constants are easy to swap because they look alike, sound alike, and sit next to each other in the constants box. The College Board defines them in two sentences that are worth reading side by side, because the sentences are parallel and the two words in them that differ are the whole answer.

Permittivity. AP Physics 2 essential knowledge 10.1.C.1: electric permittivity is a measurement of the degree to which a material or medium is polarized in the presence of an electric field.

Permeability. AP Physics 2 essential knowledge 12.1.C.1: magnetic permeability is a measurement of the amount of magnetization in a material in response to an external magnetic field.

Both statements sit under a learning objective of the same shape, 10.1.C "Describe the electric permittivity of a material or medium" and 12.1.C "Describe the magnetic permeability of a material". AP Physics C: Electricity and Magnetism repeats both word for word, at 8.1.C.1 and 12.1.C.1.

So the distinction is not subtle and it is not about size or units. Permittivity is the electric one and permeability is the magnetic one, and each answers the same question in its own half of electromagnetism: how much does matter respond to a field of this kind?

The symbols follow the same split, and they are the reason the pair gets confused in the first place. Permittivity is $\varepsilon$ , a Greek epsilon; permeability is $\mu$ , a Greek mu. The vacuum values carry a nought: $\varepsilon_0$ and $\mu_0$ . Both are printed in the Constants and Conversion Factors box of the AP Physics 2 sheet and of the AP Physics C: E&M sheet, on adjacent lines, labelled "Vacuum permittivity" and "Vacuum permeability".

\varepsilon_0 = 8.85 \times 10^{-12} \ \frac{\mathrm{C}^2}{\mathrm{N}\cdot\mathrm{m}^2} \qquad \mu_0 = 4\pi \times 10^{-7} \ \frac{\mathrm{T}\cdot\mathrm{m}}{\mathrm{A}}

Read the units and the pair separates itself before you have learned anything. Coulombs and newtons are electric currency; teslas and amperes are magnetic currency. Neither constant borrows a symbol from the other's side.

Side by side

Property	Permittivity	Permeability
Symbol	$\varepsilon$ , vacuum value $\varepsilon_0$	$\mu$ , vacuum value $\mu_0$
Which field it describes	Electric	Magnetic
CED definition	Degree to which a material is polarized in an electric field (10.1.C.1)	Amount of magnetization in a material in response to an external magnetic field (12.1.C.1)
Vacuum value on the sheet	$8.85 \times 10^{-12} \ \mathrm{C}^2/(\mathrm{N}\cdot\mathrm{m}^2)$	$4\pi \times 10^{-7} \ (\mathrm{T}\cdot\mathrm{m})/\mathrm{A}$
How the C: E&M symbol list names it	Electric permittivity	Magnetic permeability
Where it sits in a force law	In the denominator, under $4\pi$	In the numerator, above $2\pi$ or $4\pi$
Effect of increasing it	Weaker electric force and field for the same charge	Stronger magnetic field for the same current
Named laws that carry it	Coulomb's law, Gauss's law	Ampere's law, Biot-Savart law
For matter	Different from free space, set by how easily electrons rearrange (10.1.C.4.i)	Different from free space, and not a constant for a material (12.1.C.3)
Ratio to the vacuum value	The dielectric constant, $\kappa = \varepsilon/\varepsilon_0$ , printed on the C: E&M sheet	No corresponding printed ratio in either CED
AP Physics 2 unit	10, Electric Force, Field, and Potential	12, Magnetism and Electromagnetism

Two rows do most of the work in a question.

The where it sits row is the fastest sanity check available on an equation you half remember. Permittivity is always underneath: $1/(4\pi\varepsilon_0)$ in Coulomb's law, $q_{\text{enc}}/\varepsilon_0$ in Gauss's law. Permeability is always on top: $\mu_0 I/(2\pi r)$ for a wire, $\mu_0 I_{\text{enc}}$ in Ampere's law. If you have written $\mu_0$ in a denominator on its own, something has gone wrong.

The for matter row is where the symmetry breaks, and it is the part almost nobody notices. Both constants change value inside matter, but the CED says something about permeability that it does not say about permittivity: 12.1.C.3 states that the permeability of matter is not a constant for a material and varies based on many factors, including temperature, orientation, and strength of the external field. A dielectric constant can be quoted in a table. A magnetic permeability, in general, cannot.

Every printed equation that carries one of them

The clearest way to hold the pair apart is to see exactly where each one is printed, so here is the census, taken line by line off the two equation sheets rather than from memory.

AP Physics 2, the Electricity group. Six of its twenty printed lines carry $\varepsilon_0$ :

Equation	What it gives
$\lvert \vec{F}_E \rvert = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\lvert q_1 q_2 \rvert}{r^2} = k\dfrac{\lvert q_1 q_2 \rvert}{r^2}$	Coulomb's law
$\lvert \vec{E} \rvert = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\lvert q \rvert}{r^2} = k\dfrac{\lvert q \rvert}{r^2}$	Field of a point charge
$U_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r} = k\dfrac{q_1 q_2}{r}$	Potential energy of a pair
$V = \dfrac{1}{4\pi\varepsilon_0}\sum_i \dfrac{q_i}{r_i}$	Potential of a set of charges
$C = \kappa\varepsilon_0 \dfrac{A}{d}$	Parallel-plate capacitance
$E_c = \dfrac{Q}{\kappa\varepsilon_0 A}$	Field inside a capacitor

AP Physics 2, the Magnetism group. One of its seven printed lines carries $\mu_0$ , and that single line is $B = \dfrac{\mu_0 I}{2\pi r}$ , the field around a long straight wire. On this course, that is the only place the constant appears in an equation you are given.

AP Physics C: Electricity and Magnetism. The census grows and the split holds. Seven printed lines carry $\varepsilon_0$ : Coulomb's law, the integral field of a charge distribution, Gauss's law $\oint \vec{E} \cdot d\vec{A} = q_{\text{enc}}/\varepsilon_0$ , the pair potential energy, the integral potential $V$ , the parallel-plate capacitance, and the definition $\kappa = \varepsilon/\varepsilon_0$ . Three carry $\mu_0$ : the Biot-Savart law $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I(d\vec{\ell} \times \hat{r})}{r^2}$ , Ampere's law $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$ , and the solenoid field $B_{\text{sol}} = \mu_0 n I$ . One more carries a material permeability rather than the vacuum one: the solenoid's inductance, $L_{\text{sol}} = \dfrac{\mu_{\text{core}} N^2 A}{\ell}$ , where the subscript is the sheet telling you the core is not free space.

Two things fall out of the census that are worth carrying into an exam.

First, $k$ and $\varepsilon_0$ are one constant in two costumes. The constants box prints the identity itself: $k = \dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9 \ \mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2$ . Every AP Physics 2 line above gives both forms, so you may substitute whichever is less work. There is no equivalent shorthand for $\mu_0$ on either sheet.

Second, the $4\pi$ that sits with $\varepsilon_0$ and the $4\pi$ inside $\mu_0$ point in opposite directions. Permittivity is divided by $4\pi$ ; permeability already contains a factor of $4\pi$ , so dividing it by $2\pi$ or $4\pi$ produces a clean power of ten. That is a calculation shortcut, and the next section spends it.

The case that separates them, and the product that joins them

Take one current-carrying wire and one point charge, both at the same distance, and work each with its own constant. Nothing crosses over: the charge problem never sees $\mu_0$ and the wire problem never sees $\varepsilon_0$ . Worked example two below does exactly that.

Then multiply the two constants together, and something appears that is in neither problem.

AP Physics C: Electricity and Magnetism essential knowledge 13.2.A.4 states it: Maxwell's equations can be used to show that electric and magnetic fields obey wave equations and that electromagnetic waves travel at a constant speed in free space. The CED labels what follows a derived equation rather than a relevant equation, and that label is defined in the course framework's Required Equations note: not all equations in the framework appear on the equation sheet, and many are provided for reference and guidance or to demonstrate the final results of derivations expected of students, with those denoted as derived equations. So the label is the CED saying in advance that this line is not on your sheet:

c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}

Substitute the two numbers off the constants box and you get $2.9986 \times 10^8 \ \mathrm{m/s}$ , which is the printed speed of light $c = 3.00 \times 10^8 \ \mathrm{m/s}$ to three significant figures. Worked example one does the arithmetic and the unit algebra.

This is the historical point and it is why the pair is worth a page rather than two glossary lines. Both constants were measured in laboratories that had nothing to do with light: $\varepsilon_0$ from forces between charges, $\mu_0$ from forces between currents. Combining them produced a speed that matched the measured speed of light, and that agreement is what identified light as an electromagnetic wave.

Read the boundary statement before you plan any revision around this. Topic 13.2 of AP Physics C: E&M reads: AP Physics C: Electricity and Magnetism does not expect students to mathematically derive the speed of light in free space from Maxwell's equations. This relationship is included above solely as an indication of the further applications, implications, and connections to physical phenomena that students may study in more advanced physics courses.

That is unusually explicit. The relation is in the framework so that you know it exists and know why the two constants belong together. It is not a derivation you will be asked to produce, and it is not printed on any of the four equation sheets. Knowing the sentence is the assessable part; deriving it is not.

AP Physics 2 does not carry the relation at all. Its Unit 14 gives $c = 3.00 \times 10^8 \ \mathrm{m/s}$ as a measured universal constant and leaves it there, so on that course the two constants never meet in a single expression.

What each one does inside matter

Free space is the easy case, and the CED handles both constants identically there. Essential knowledge 10.1.C.3 says free space has a constant value of electric permittivity, $\varepsilon_0$ , that appears in physical relationships, and 12.1.C.2 says free space has a constant value of magnetic permeability, known as the vacuum permeability $\mu_0$ , that appears in equations representing physical relationships. Same sentence, two fields.

Matter is where they stop being mirror images.

Permittivity in matter has a mechanism the CED spells out. 10.1.C.4 says the permittivity of matter has a value different from that of free space that arises from the matter's composition and arrangement, and 10.1.C.4.i names the mechanism: in a given material, electric permittivity is determined by the ease with which electrons can change configurations within the material. 10.1.C.2 supplies the picture behind the definition, that electric polarization can be modeled as the induced rearrangement of electrons by an external electric field, resulting in a separation of positive and negative charges within a material or medium. And 10.1.C.4.ii closes the loop to a familiar distinction: conductors are made from electrically conducting materials in which charge carriers move easily, while insulators are made from electrically nonconducting materials in which charge carriers cannot move easily.

That is why the electric side gets a tidy number. Both sheets print capacitance as $C = \kappa\varepsilon_0 A/d$ , and the C: E&M sheet prints the definition of the multiplier outright, $\kappa = \varepsilon/\varepsilon_0$ . The dielectric constant is the material's permittivity measured in units of the vacuum's, so a slab with $\kappa = 3.0$ has $\varepsilon = 2.655 \times 10^{-11}$ in the sheet's units, and worked example three uses it.

Permeability in matter refuses to behave. 12.1.C.3 says the permeability of matter has values different from that of free space and arises from the matter's composition and arrangement, and then adds the clause that has no counterpart on the electric side: it is not a constant for a material and varies based on many factors, including temperature, orientation, and strength of the external field.

Take that sentence at its word. There is no magnetic analogue of $\kappa$ on any AP sheet, no table of magnetic permeabilities anywhere in the four CEDs, and no printed equation in AP Physics 2 that uses a material permeability. The one place a material permeability is printed at all is the C: E&M inductance line, $L_{\text{sol}} = \mu_{\text{core}} N^2 A/\ell$ , where the value is handed to you in the problem because it could not be looked up.

Both constants are measurable, and the CED shows how for one of them. A sample free-response question in the AP Physics C: E&M exam information section gives students force and current data for two parallel wires at fixed separation and asks them to graph the quantities so as to produce a straight line, then to calculate the magnetic permeability of free space from the line and the measured $L = 1.4 \ \mathrm{m}$ and $d = 0.005 \ \mathrm{m}$ . That is $\mu_0$ arriving as an experimental result rather than a printed value, which is the honest way to think about both constants.

When it costs a mark

Writing $\mu_0$ where $\varepsilon_0$ belongs, or the reverse. The check that catches it in one second is the units. If the expression is producing newtons or volts from charges, the constant in it has coulombs in its units, so it is $\varepsilon_0$ . If it is producing teslas from amperes, it is $\mu_0$ . Nothing on any AP sheet mixes them, except the relation for $c$ that the C: E&M boundary statement removes from assessment.
Converting $\mu_0$ to a decimal before substituting. $4\pi \times 10^{-7}$ is exact as printed and the $\pi$ almost always cancels. In $B = \mu_0 I/(2\pi r)$ the whole constant collapses to $2 \times 10^{-7}$ , and in the Biot-Savart law it arrives as $\mu_0/4\pi = 1 \times 10^{-7}$ already. Turning it into $1.26 \times 10^{-6}$ first adds a rounding step and hides the cancellation.
Using $8.99 \times 10^9$ for the Coulomb constant. The AP sheets print $k = 9.0 \times 10^9 \ \mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2$ . Textbook values differ in the third figure, and the sheet is what the exam supplies.
Confusing $\mu$ with the coefficient of friction. The same Greek letter carries both, plus the prefix micro. Context separates them, but a symbol list on a mixed-topic free response will not.
Treating a material permeability as a lookup value. 12.1.C.3 says it is not a constant for a material. A question that needs one will give it to you, as the C: E&M sheet does with $\mu_{\text{core}}$ .
Claiming a bigger permittivity means a bigger force. It is in the denominator, so raising it lowers the force and lowers the field. Sliding a dielectric between capacitor plates raises the permittivity, weakens the field at fixed charge, and therefore raises the capacitance. Permeability runs the other way: raising it raises the field a given current produces.
Trying to derive $c$ from Maxwell's equations in a free response. The Topic 13.2 boundary statement rules that out explicitly and says the relation is there as a pointer to later courses. Quote the relation, do not attempt the derivation.
Expecting the relation for $c$ on AP Physics 2. It is not in that CED and not on that sheet. Worked example one is a C: E&M idea being checked numerically, not an AP Physics 2 procedure.

Where the symmetry is real, and where it lulls you

The symmetry between the two constants is genuine, and that is exactly the problem: it is close enough to make the pair feel interchangeable and incomplete enough to punish anyone who treats them that way.

What really is symmetric. Both are defined by parallel CED statements about how matter responds to a field. Both have a fixed free-space value that the constants box prints. Both appear as the single constant of proportionality in their side's force and field laws. Both change inside matter for the same stated reason, the matter's composition and arrangement. Both are measurable in a school laboratory. And their product carries the speed of light, which is as symmetric a role as two constants can share.

Where the symmetry stops. Four places, and each has produced a wrong answer.

Position in the algebra. Permittivity divides, permeability multiplies. A formula recalled with the constant on the wrong level of the fraction inverts the dependence you were asked about.
The shorthand. $\varepsilon_0$ has an alias, $k$ , that the sheet prints on the same line. $\mu_0$ has none, so every magnetic substitution carries the $4\pi$ through.
Behaviour in matter. Permittivity gets a dielectric constant and a table. Permeability gets 12.1.C.3's warning that it is not a constant for a material.
Course coverage. AP Physics 2 puts $\varepsilon_0$ in six printed equations and $\mu_0$ in one. Weight your practice accordingly.

The last lull is the units. Both look like the same kind of object because both are "a constant of free space", so it is tempting to compare their sizes. Do not. $8.85 \times 10^{-12} \ \mathrm{C}^2/(\mathrm{N}\cdot\mathrm{m}^2)$ and $4\pi \times 10^{-7} \ (\mathrm{T}\cdot\mathrm{m})/\mathrm{A}$ measure different physical things and cannot be ranked against each other. The only meaningful thing to do with the two numbers together is multiply them, and what comes out is not a constant of either field. It is a speed.

The speed of light out of the two constants, digits and units

Using only the values printed in the Constants and Conversion Factors box, $\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C}^2/(\mathrm{N}\cdot\mathrm{m}^2)$ and $\mu_0 = 4\pi \times 10^{-7} \ (\mathrm{T}\cdot\mathrm{m})/\mathrm{A}$ : (a) evaluate $c = 1/\sqrt{\varepsilon_0\mu_0}$ ; (b) compare it with the printed $c = 3.00 \times 10^8 \ \mathrm{m/s}$ ; (c) show that the units of $1/\sqrt{\varepsilon_0\mu_0}$ really are metres per second.

(a) Write $\mu_0$ as a decimal for this one calculation, since nothing is going to cancel it: $\mu_0 = 4\pi \times 10^{-7} = 1.2566 \times 10^{-6} \ (\mathrm{T}\cdot\mathrm{m})/\mathrm{A}$ .
Form the product. $\varepsilon_0\mu_0 = (8.85 \times 10^{-12})(1.2566 \times 10^{-6}) = 1.1121 \times 10^{-17}$ .
Take the square root: $\sqrt{1.1121 \times 10^{-17}} = 3.3348 \times 10^{-9}$ .
Invert it: $c = 1/(3.3348 \times 10^{-9}) = 2.9986 \times 10^8$ .
(b) To three significant figures that is $3.00 \times 10^8 \ \mathrm{m/s}$ , the value printed two lines below the two constants in the same box. The two agree. Carrying more digits, the sheet's rounded $\varepsilon_0$ gives $2.99863 \times 10^8$ against the accepted $2.99792 \times 10^8 \ \mathrm{m/s}$ , a difference of $0.024$ percent. That is exactly half of the $0.047$ percent by which $8.85$ falls short of the unrounded permittivity, as it has to be, since $c$ goes as one over the square root of $\varepsilon_0$ .
(c) The unit algebra. Start from $\varepsilon_0\mu_0$ , which has units $\dfrac{\mathrm{C}^2}{\mathrm{N}\cdot\mathrm{m}^2} \cdot \dfrac{\mathrm{T}\cdot\mathrm{m}}{\mathrm{A}}$ .
Replace the tesla using the magnetic force law $F_B = qvB\sin\theta$ from the sheet, which makes $1 \ \mathrm{T} = 1 \ \mathrm{N}/(\mathrm{A}\cdot\mathrm{m})$ .
Substituting, $\dfrac{\mathrm{C}^2}{\mathrm{N}\cdot\mathrm{m}^2} \cdot \dfrac{\mathrm{N}\cdot\mathrm{m}}{\mathrm{A}\cdot\mathrm{m}\cdot\mathrm{A}} = \dfrac{\mathrm{C}^2}{\mathrm{A}^2 \cdot \mathrm{m}^2}$ . The newtons cancel and one metre cancels.
Now use $1 \ \mathrm{C} = 1 \ \mathrm{A}\cdot\mathrm{s}$ , which is the sheet's $I = \Delta q/\Delta t$ read backwards. Then $\dfrac{\mathrm{A}^2\mathrm{s}^2}{\mathrm{A}^2\mathrm{m}^2} = \dfrac{\mathrm{s}^2}{\mathrm{m}^2}$ .
So $\sqrt{\varepsilon_0\mu_0}$ has units of $\mathrm{s}/\mathrm{m}$ , and its reciprocal has units of $\mathrm{m}/\mathrm{s}$ . The expression is a speed before any number is put into it.
State the limit alongside the answer. The AP Physics C: E&M Topic 13.2 boundary statement says students are not expected to mathematically derive the speed of light in free space from Maxwell's equations, and this check is arithmetic on a relation the CED supplies at 13.2.A.4, not a derivation of it.

(a) $c = 2.9986 \times 10^8 \ \mathrm{m/s}$ . (b) That is $3.00 \times 10^8 \ \mathrm{m/s}$ to three significant figures, matching the printed value; the remaining $0.024$ percent gap is the rounding in $\varepsilon_0 = 8.85 \times 10^{-12}$ . (c) The units reduce to $\mathrm{s}^2/\mathrm{m}^2$ under the root, so $1/\sqrt{\varepsilon_0\mu_0}$ is in $\mathrm{m/s}$ .

One charge problem and one current problem, at the same distance

At a distance $r = 0.030 \ \mathrm{m}$ : (a) find the magnitude of the electrostatic force between two point charges of $+2.0 \ \mu\mathrm{C}$ each; (b) find the magnitude of the magnetic field produced by a long straight wire carrying $I = 5.0 \ \mathrm{A}$ ; (c) say which constant each part used and why the other could not have appeared.

(a) The printed relation is $\lvert \vec{F}_E \rvert = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\lvert q_1 q_2 \rvert}{r^2} = k\dfrac{\lvert q_1 q_2 \rvert}{r^2}$ . The sheet supplies $k = 9.0 \times 10^9 \ \mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2$ directly, so use that form.
$q_1 q_2 = (2.0 \times 10^{-6})(2.0 \times 10^{-6}) = 4.0 \times 10^{-12} \ \mathrm{C}^2$ , and $r^2 = (0.030)^2 = 9.0 \times 10^{-4} \ \mathrm{m}^2$ .
$\lvert \vec{F}_E \rvert = (9.0 \times 10^9)\dfrac{4.0 \times 10^{-12}}{9.0 \times 10^{-4}} = (9.0 \times 10^9)(4.444 \times 10^{-9}) = 40 \ \mathrm{N}$ , repulsive because both charges are positive (10.1.A.3.i).
(b) The printed relation is $B = \dfrac{\mu_0 I}{2\pi r}$ , the only AP Physics 2 equation containing $\mu_0$ .
Do not decimalise the constant. $\dfrac{\mu_0}{2\pi} = \dfrac{4\pi \times 10^{-7}}{2\pi} = 2 \times 10^{-7}$ exactly, because the $\pi$ cancels.
$B = (2 \times 10^{-7})\dfrac{5.0}{0.030} = (2 \times 10^{-7})(166.7) = 3.3 \times 10^{-5} \ \mathrm{T}$ .
(c) Part (a) turned charges into a force, so its constant has coulombs and newtons in its units and it is $\varepsilon_0$ , wearing its $k$ costume. Part (b) turned a current into a magnetic field, so its constant has teslas and amperes in its units and it is $\mu_0$ .
Sanity check the position of each constant in its formula. In (a) the constant sits under $4\pi$ , so a larger permittivity would give a smaller force. In (b) the constant sits on top, so a larger permeability would give a larger field. Both dependences match the definitions in 10.1.C.1 and 12.1.C.1.

(a) $\lvert \vec{F}_E \rvert = 40 \ \mathrm{N}$ , repulsive, using $k = 1/(4\pi\varepsilon_0)$ . (b) $B = 3.3 \times 10^{-5} \ \mathrm{T}$ , using $\mu_0/(2\pi) = 2 \times 10^{-7}$ . (c) The electric problem can only use permittivity and the magnetic problem can only use permeability; no printed AP Physics 2 equation contains both.

Permittivity of matter: a capacitor with and without a dielectric

A parallel-plate capacitor has plates of area $A = 0.020 \ \mathrm{m}^2$ separated by $d = 1.0 \ \mathrm{mm}$ . (a) Find its capacitance with air between the plates. (b) Find it with a dielectric of $\kappa = 3.0$ filling the gap. (c) State the permittivity of that dielectric, and say what the AP CED does and does not let you assume about the corresponding magnetic quantity.

(a) The printed relation is $C = \kappa\varepsilon_0 \dfrac{A}{d}$ . The AP Physics 2 exam conventions state that capacitors are air-filled with $\kappa = 1.0$ unless told otherwise, so with air the factor drops out.
$C = (8.85 \times 10^{-12})\dfrac{0.020}{1.0 \times 10^{-3}} = (8.85 \times 10^{-12})(20) = 1.77 \times 10^{-10} \ \mathrm{F} = 177 \ \mathrm{pF}$ .
(b) With $\kappa = 3.0$ the same expression scales: $C = 3.0 \times 1.77 \times 10^{-10} = 5.31 \times 10^{-10} \ \mathrm{F} = 531 \ \mathrm{pF}$ .
(c) The AP Physics C: E&M sheet prints $\kappa = \varepsilon/\varepsilon_0$ , so the dielectric's own permittivity is $\varepsilon = \kappa\varepsilon_0 = (3.0)(8.85 \times 10^{-12}) = 2.655 \times 10^{-11} \ \mathrm{C}^2/(\mathrm{N}\cdot\mathrm{m}^2)$ .
Read that against 10.1.C.4, which says the permittivity of matter has a value different from that of free space that arises from the matter's composition and arrangement, and 10.1.C.4.i, which attributes it to the ease with which electrons can change configurations within the material. The slab tripled the capacitance because its electrons rearrange more readily than a vacuum's, which is 10.1.C.2's polarization.
Now the asymmetry. Essential knowledge 12.1.C.3 says the permeability of matter has values different from that of free space and is not a constant for a material, varying with temperature, orientation, and strength of the external field. So there is no magnetic $\kappa$ to multiply by, and no analogous calculation to do here.
Check the direction of the field change while you are here. The sheet also prints $E_c = Q/(\kappa\varepsilon_0 A)$ , so at fixed charge the dielectric divides the field by $3.0$ . Weaker field for the same charge, higher capacitance: one effect, two descriptions.

(a) $C = 177 \ \mathrm{pF}$ . (b) $C = 531 \ \mathrm{pF}$ . (c) $\varepsilon = 2.655 \times 10^{-11} \ \mathrm{C}^2/(\mathrm{N}\cdot\mathrm{m}^2)$ , three times the vacuum value. No such fixed value exists on the magnetic side, because 12.1.C.3 states that the permeability of matter is not a constant for a material.

Frequently asked questions

What is the difference between permittivity and permeability?

Permittivity is the electric constant and permeability is the magnetic one. AP Physics 2 essential knowledge 10.1.C.1 defines electric permittivity as a measurement of the degree to which a material or medium is polarized in the presence of an electric field, and 12.1.C.1 defines magnetic permeability as a measurement of the amount of magnetization in a material in response to an external magnetic field. Permittivity is written as epsilon, with the free-space value 8.85 times 10 to the minus 12 in coulombs squared per newton square metre; permeability is written as mu, with the free-space value 4 pi times 10 to the minus 7 tesla metres per ampere. Both are printed in the constants box of the AP Physics 2 and AP Physics C: E&M equation sheets.

Which equations use epsilon nought and which use mu nought?

Epsilon nought appears wherever charges produce forces, fields, potential energy or potential, and in capacitance. On the AP Physics 2 sheet that is six printed lines: Coulomb's law, the field of a point charge, the potential energy of a pair of charges, the potential of a set of charges, the parallel-plate capacitance, and the field inside a capacitor. On the AP Physics C: E&M sheet it also appears in Gauss's law as the enclosed charge divided by epsilon nought, and in the definition of the dielectric constant. Mu nought appears wherever currents produce magnetic fields. The AP Physics 2 sheet uses it in exactly one equation, the field around a long straight wire; the C: E&M sheet uses it in the Biot-Savart law, in Ampere's law, and in the solenoid field.

Why does the product of permittivity and permeability give the speed of light?

Because Maxwell's equations, which tie changing electric fields to magnetic fields and back, predict a wave whose speed is set by exactly those two constants. AP Physics C: Electricity and Magnetism essential knowledge 13.2.A.4 states that Maxwell's equations can be used to show that electric and magnetic fields obey wave equations and that electromagnetic waves travel at a constant speed in free space, and gives the derived equation c equals one over the square root of epsilon nought times mu nought. Substituting the printed constants gives 2.9986 times 10 to the 8 metres per second, which is the printed speed of light to three significant figures. The historical weight of this is that both constants had been measured in experiments on charges and currents, with no light involved, so the agreement is what identified light as an electromagnetic wave. The Topic 13.2 boundary statement adds that students are not expected to derive the relationship, which is included solely as an indication of further study.

Is permittivity or permeability on the AP Physics 2 equation sheet?

Both, in the Constants and Conversion Factors box, on adjacent lines labelled Vacuum permittivity and Vacuum permeability. The printed values are epsilon nought equals 8.85 times 10 to the minus 12 coulombs squared per newton square metre and mu nought equals 4 pi times 10 to the minus 7 tesla metres per ampere. The same box also prints the Coulomb constant as one over 4 pi epsilon nought equals 9.0 times 10 to the 9 newton square metres per coulomb squared, which makes epsilon nought and k the same constant in two forms. The relation between the two constants and the speed of light is not printed on any of the four AP equation sheets.

Does a higher permittivity mean a stronger or weaker electric field?

Weaker, for the same charge, because permittivity sits in the denominator. Coulomb's law is printed as one over 4 pi epsilon nought times the product of the charges over the separation squared, so raising epsilon nought lowers the force and lowers the field. This is what a dielectric does: sliding it between capacitor plates raises the permittivity of the gap, weakens the field at fixed charge, and therefore raises the capacitance, since the AP sheets print capacitance as kappa times epsilon nought times area over separation. Permeability runs the opposite way, because it sits in the numerator: raising it raises the magnetic field a given current produces.

Can you look up the magnetic permeability of a material like you look up a dielectric constant?

Not in AP Physics. AP Physics 2 essential knowledge 12.1.C.3 and the identical AP Physics C: E&M statement say the permeability of matter has values different from that of free space and arises from the matter's composition and arrangement, then add that it is not a constant for a material and varies based on many factors, including temperature, orientation, and strength of the external field. That clause has no counterpart on the electric side, where 10.1.C.4 attributes the permittivity of matter to composition and arrangement and stops there, and where the dielectric constant kappa equals epsilon over epsilon nought is a printed definition. So a question that needs a material permeability will supply it, exactly as the C: E&M sheet does in the solenoid inductance equation with its mu-core subscript.

Why is mu nought written as 4 pi times 10 to the minus 7 instead of a decimal?

Because the pi almost always cancels, and the form as printed is exact. In the field of a long straight wire, mu nought over 2 pi equals 2 times 10 to the minus 7 exactly. In the Biot-Savart law on the AP Physics C: E&M sheet, the constant arrives as mu nought over 4 pi, which equals 1 times 10 to the minus 7 exactly. Converting to the decimal 1.26 times 10 to the minus 6 before substituting throws away that cancellation and adds a rounding step for nothing. The only calculation that genuinely wants the decimal is the product with epsilon nought that gives the speed of light.