AP Physics C: E&M · Topic 10.1

Topic 10.1: Electrostatics with Conductors

Unit 10: Conductors and Capacitors10-15% of the multiple-choice section

A conductor in electrostatic equilibrium has no electric field inside it, all of its excess charge sitting on the surface, and every point on it at the same potential. The outside field meets the surface at right angles, and the charge crowds where the surface curves most sharply.

AP Physics: Unit 10 (topics 10.1 Electrostatics with Conductors). Topic 10.1 of the current AP Physics C: Electricity and Magnetism course and exam description, inside Unit 10, weighted 10 to 15% of the multiple-choice section at about 8 to 16 class periods. One learning objective, 10.1.A, using the task verb describe, with twelve essential-knowledge statements: seven at the top level, 10.1.A.1 through 10.1.A.7, plus five sub-statements, 10.1.A.2.i, 10.1.A.2.ii, 10.1.A.3.i, 10.1.A.3.ii and 10.1.A.3.iii. Topic 10.1 prints no boundary statement and no equations at all; Unit 10's only boundary statement sits at the end of Topic 10.3. Suggested skills are 1.A, 2.C, 3.B and 3.C, two of which are justification skills. The statements cover the definition of an ideal conductor, excess charge residing entirely on the surface at electrostatic equilibrium, the negligible time to reach equilibrium, the conductor becoming an equipotential surface, greater charge density at points and edges than at planar areas, zero net charge and zero electric field in the interior, the field being perpendicular to the outer surface, polarization as a consequence of remaining an equipotential surface, and electrostatic shielding by a closed conducting shell. There is no AP Physics 2 topic of this name; the algebra-based course carries the two headline facts inside its Topic 10.3 at learning objective 10.3.B, along with the isolated-sphere result and an insulator contrast that AP Physics C Topic 10.1 does not restate.

What Topic 10.1 requires

Topic 10.1 has one learning objective and twelve essential-knowledge statements under it, seven at the top level and five sub-statements. That is more statements than any other topic in Unit 10, and none of them carries an equation.

10.1.A, describe the charge distribution within a conductor.

10.1.A.1 states that an ideal conductor is a material in which electrons are able to move freely.
10.1.A.2 states that when a conductor is in electrostatic equilibrium, mutual repulsion of excess charge carriers results in those charge carriers residing entirely on the surface of the conductor.
10.1.A.2.i states that in a conductor with a negative net charge, excess electrons reside on the surface of the conductor.
10.1.A.2.ii states that in a conductor with a positive net charge, the surface becomes deficient in electrons, and can be modeled as if positive charge carriers reside on the surface of the conductor.
10.1.A.3 states that excess charges will move to the surface of a conductor to create a state of electrostatic equilibrium within the conductor.
10.1.A.3.i states that the time interval over which charges reach electrostatic equilibrium within a conductor is so short as to be negligible.
10.1.A.3.ii states that when a conductor reaches electrostatic equilibrium, all points on the surface of the conductor have the same electric potential, and the conductor becomes an equipotential surface.
10.1.A.3.iii states that the charge density on the surface of a conductor will be greater where there are points or edges compared to planar areas.
10.1.A.4 states that all excess charges reside on the surface of a conductor, which means there is no net charge in the interior of the conductor, and the electric field is zero within the conductor.
10.1.A.5 states that the electric field is perpendicular to the outer surface of a conductor.
10.1.A.6 states that a conductor can be polarized in the presence of an external electric field, and that this is a consequence of the conductor remaining an equipotential surface.
10.1.A.7 states that electrostatic shielding is the process of surrounding an area with a closed, conducting shell to create a region inside the conductor that is free from external electric fields.

Topic 10.1 prints no boundary statement, and no equations. Unit 10's only boundary statement sits at the end of Topic 10.3 and names three capacitor shapes.

Suggested skills: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Two of the four are justification skills. That is the shape of the topic: it is a list of claims about a conductor, and what you get marked on is the reasoning that produces them.

Unit 10 is weighted 10 to 15% of the multiple-choice section over a suggested 8 to 16 class periods.

Twelve statements, two facts

This topic is usually taught as a list to memorise, and taught that way it is both boring and fragile. It is not a list. Every one of the twelve statements follows from two things, and if you hold onto those two you can reconstruct the rest under exam pressure.

Fact one: the charges can move. Statement 10.1.A.1 defines an ideal conductor as a material in which electrons move freely. Not "conducts well". Free to move, with nothing holding them in place.

Fact two: they have stopped. "Electrostatic equilibrium" means exactly that: no charge is moving any more. Statement 10.1.A.3.i adds that they get there in a time so short as to be negligible, so for any question that does not say otherwise, equilibrium is where you start.

Put those together and you get a rule with real teeth:

If a free charge is not moving, no net force acts on it, so the electric field at its location is zero.

That is the entire content of statement 10.1.A.4's second half, and it is a conclusion, not an assumption. The field inside a conductor is zero because if it were not, the free charges would feel a force, they would move, and by definition you would not be in electrostatic equilibrium yet. Wait a negligible interval and they will have rearranged themselves until the field they produce exactly cancels whatever field is trying to push them.

Everything else on this page is that argument applied to a different question. Where does the charge end up? Why is the surface an equipotential? Why does the outside field arrive at right angles? Why does a metal box shield its interior? Same argument each time.

There is a modelling nicety in statements 10.1.A.2.i and 10.1.A.2.ii worth catching. Physically, only electrons move: a negatively charged conductor has extra electrons on its surface, and a positively charged one has a surface deficient in electrons. The CED says a positive conductor "can be modeled as if positive charge carriers reside on the surface". That is a modelling choice the framework licenses explicitly, and it is why you may draw plus signs on a diagram without anyone objecting.

Why the excess charge ends up on the surface, from Gauss's law

Statement 10.1.A.2 gives a physical reason: mutual repulsion of the excess charge carriers pushes them apart until they are all on the surface. That is true and it is the intuition. It is not a proof, and it does not tell you the interior charge is exactly zero rather than merely small.

Gauss's law gives you both. Take any closed Gaussian surface drawn entirely inside the metal, however oddly shaped and however close to the outer boundary. From Unit 8, statement 8.6.A.1:

\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}

We have already established that $\vec{E} = 0$ at every point inside the metal. So the integrand is zero everywhere on that surface, so the flux is zero, so $q_{\text{enc}} = 0$ .

Now shrink the Gaussian surface to a point. Grow it to just inside the outer boundary. Wrap it around any interior region you like. Every one of those choices encloses zero net charge, and the only charge distribution for which that is true everywhere is zero charge density throughout the interior. So the excess charge is not merely mostly on the surface: there is exactly none of it anywhere else, which is what statement 10.1.A.4 asserts.

This is worth doing rather than reciting, because the same argument answers the harder version of the question.

A conductor with a cavity. Draw a Gaussian surface inside the metal that completely surrounds the cavity. The flux is zero, so the net charge it encloses is zero, so the total charge on the cavity wall plus any charge inside the cavity must be zero.

If the cavity is empty, the cavity wall carries zero net charge, and the field in the cavity is zero.
If a charge $+q$ sits in the cavity, the cavity wall must carry exactly $-q$ , wherever the charge sits and whatever shape the cavity is.

That is a result you can get in two lines and cannot get by memorising a list, and it is the standard multiple-choice question on this topic.

One useful extra follows from a Gaussian pillbox straddling the surface: with zero field on the inside face and a perpendicular field $E$ on the outside face, the flux is $EA$ and the enclosed charge is $\sigma A$ , so

E = \frac{\sigma}{\varepsilon_0}

just outside a conductor's surface. The CED does not print this as an equation in Topic 10.1, and it is not on the equation sheet. It is a derivation, which is what skill 2.A is for, and Topic 10.3 makes the same move for capacitor plates in statement 10.3.A.3.i.

Why the conductor is one equipotential, and why the field meets it square on

Statement 10.1.A.3.ii says all points on the surface of a conductor at equilibrium have the same electric potential, and the conductor becomes an equipotential surface. Statement 10.1.A.5 says the electric field is perpendicular to the outer surface. Both come out of the same zero-field fact, using the line integral from Topic 9.2.

The whole conductor is at one potential. Take any two points $a$ and $b$ in or on the conductor and walk between them along a path that stays inside the metal:

V_b - V_a = -\int_a^b \vec{E} \cdot d\vec{r} = 0

because $\vec{E} = 0$ at every point on that path. So $V_b = V_a$ , for any two points. Note what this does not say: the potential is not zero, it is merely the same everywhere. A charged sphere sits at a perfectly good nonzero potential, and that constant potential is what Topic 10.2 equalises when two conductors are connected.

The outside field is perpendicular to the surface. Suppose it were not. Then it would have a component lying along the surface, and that component would exert a force along the surface on the free charges there, and they would move, and you would not be in electrostatic equilibrium. So the tangential component is zero and only the normal component survives.

The same conclusion arrives from the isoline statements in Unit 9. Statement 9.2.B.3.ii says isolines are perpendicular to electric field vectors, and 9.2.B.3.iv says there is no component of an electric field along an isoline. A conductor's surface is an isoline, so the field crosses it at right angles. Two routes, one answer, and being able to give either is what skill 3.C asks for.

Two consequences that show up in questions:

Field lines terminate on a conductor, they do not skim along it. When you sketch a field map with a conductor in it, every line must arrive square on the metal and none may run parallel to the surface just outside it.
Moving a charge anywhere on or in a conductor takes no work. $\Delta V = 0$ , so $\Delta U_E = q\Delta V = 0$ , which is Topic 9.3.

Where the charge crowds: points, edges and curvature

Statement 10.1.A.3.iii is the one statement in this topic that most treatments skip: the charge density on the surface of a conductor will be greater where there are points or edges compared to planar areas.

Read carefully what it does and does not say.

It says charge density, not charge. A sharp tip may hold a tiny total charge and still have the largest charge per unit area on the object.

It compares points and edges to planar areas. The CED gives no formula, no proportionality to curvature, and no numbers. It is a comparative statement, and skill 2.C, comparing physical quantities at different locations in a single scenario, is exactly what it supports.

The reasoning behind it is the equipotential requirement. Every part of the surface has to sit at the same potential. A region that curves sharply is close to a lot of its own charge, so a given amount of charge there produces a larger potential than the same amount spread over a flat region. To bring the sharply curved region down to the common potential, less charge sits there in total, but concentrated into far less area, so the density is higher.

The clean quantitative version of this appears in Topic 10.2: two conducting spheres of different radii joined by a wire come to the same potential, so $kQ_1/R_1 = kQ_2/R_2$ , and the smaller sphere ends up with the greater surface charge density and the stronger field at its surface. That is the two-sphere model of a point on a conductor, and it is the way the exam tests 10.1.A.3.iii quantitatively.

Combine it with $E = \sigma/\varepsilon_0$ just outside the surface and you get the physical payoff: the external field is strongest just outside the sharpest parts of a charged conductor. That is why lightning rods are pointed and why high-voltage hardware is built with rounded corners.

Polarization and electrostatic shielding

The last two statements are about what a conductor does to a field that came from somewhere else.

Statement 10.1.A.6: a conductor can be polarized in the presence of an external electric field, and this is a consequence of the conductor remaining an equipotential surface. That second sentence is the part worth noticing, because it inverts the usual explanation. The usual story is that the external field pushes charges around and, as a side effect, the conductor stays at one potential. The CED says the equipotential condition is the reason: the charges rearrange precisely as far as they must to keep the interior field zero and the surface at a single potential, and no further.

So put a neutral conducting slab into a uniform external field and the free electrons drift against the field, leaving a negative face on one side and a positive face on the other. Those induced sheets produce their own field inside the metal, and they keep drifting only until that internal field exactly cancels the external one. The net charge on the slab is still zero; the distribution is not.

Statement 10.1.A.7: electrostatic shielding is the process of surrounding an area with a closed, conducting shell to create a region inside the conductor that is free from external electric fields. The mechanism is the previous statement applied to a hollow shape. Charge rearranges on the shell until the field it produces cancels the external field everywhere inside the metal, and, for a cavity with no charge in it, everywhere inside the cavity too.

Notice the word closed. The statement is about a closed conducting shell, and the shielding is of the interior from external fields. Two things it does not claim:

It does not say the shell hides a charge that is inside the cavity from the outside world. Put a charge in the cavity of an isolated shell and it still induces charge on the outer surface, which produces a field outside. That is the first worked example below.
It does not say a solid conductor blocks fields the way a wall blocks light. The exterior field is not absorbed; it is cancelled inside by the induced charge.

The CED suggests a demonstration for this topic that is worth picturing: find a metal wire mesh container that will fit a neon gas discharge tube, touch the tube to a Tesla coil to show it lighting up, then put the tube inside the mesh and ask students to predict what happens. It does not light, no matter how much you try, and the students have to justify their prediction with evidence. That activity is skills 3.B and 3.C in one demonstration.

Where AP Physics 2 covers this, and how far it goes

There is no AP Physics 2 topic called Electrostatics with Conductors. That does not mean the algebra-based course omits the ideas, and it is worth being precise about where they live, both because it is honest and because a Physics 2 student arriving here should know what is new.

AP Physics 2 puts conductors inside Topic 10.3, Electric Fields, under learning objective 10.3.B, "describe the electric field generated by charged conductors or insulators". It has three statements relevant here:

10.3.B.1, that while in electrostatic equilibrium, the excess charge of a solid conductor is distributed on the surface of the conductor, and the electric field within the conductor is zero.
10.3.B.1.i, that at the surface of a charged conductor, the electric field is perpendicular to the surface.
10.3.B.1.ii, that the electric field outside an isolated sphere with spherically symmetric charge distribution is the same as the electric field due to a point charge with the same net charge as the sphere located at the center of the sphere.

So the two headline facts, surface charge and zero interior field, are in both courses. What AP Physics C's Topic 10.1 adds:

the definition of an ideal conductor (10.1.A.1),
the electron-deficiency modelling note for positive conductors (10.1.A.2.i and ii),
that equilibrium is reached in a negligible time (10.1.A.3.i),
that the conductor is an equipotential surface (10.1.A.3.ii), which is the statement that makes Topic 10.2 possible at all,
the charge density at points and edges (10.1.A.3.iii),
polarization as a consequence of the equipotential condition (10.1.A.6),
electrostatic shielding by a closed conducting shell (10.1.A.7).

Two things travel the other way and are not in Physics C's Topic 10.1. The isolated-sphere result, 10.3.B.1.ii, which Physics C obtains from Gauss's law in Unit 8 instead of stating separately. And the insulator contrast, Physics 2's 10.3.B.2, that an insulator's excess charge is distributed throughout its interior as well as at its surface and the field within it may be nonzero. Physics 2 also caps that contrast in its boundary statement: students will only be expected to perform qualitative analysis of electric fields within insulators.

The practical difference in exam terms is the shift from stating to deriving. Physics 2's version can be quoted. Physics C's version has to be argued from Gauss's law and the line integral, because Topic 10.1 prints no equations at all and its two justification skills are what the free-response question rewards. If you want the algebra-based treatment, the AP Physics 2 electric fields topic is the page for it. See also conductor vs insulator.

Traps, and how Topic 10.1 is assessed

Saying the potential inside a conductor is zero. It is constant, not zero. A charged isolated sphere has a nonzero potential everywhere inside and on it, equal to $kQ/R$ measured from the printed zero at infinity.

Saying the field is zero everywhere near a conductor. Zero within the conductor, and zero in an empty cavity inside it. Just outside the surface it is generally not zero, and it is largest where the surface curves most.

Forgetting the induced charge on a cavity wall. A charge $+q$ in a cavity forces exactly $-q$ onto the cavity wall. If the shell is neutral overall, $+q$ then appears on the outer surface.

Thinking a neutral conductor stays uncharged everywhere. Net charge zero and charge density zero are different claims. Statement 10.1.A.6 is about a neutral conductor with a very unequal surface distribution.

Applying these results while charge is still moving. Every statement in this topic is conditioned on electrostatic equilibrium. The moment a circuit is carrying a steady current, there is a field inside the conducting wire, which is what the printed relation $\vec{E} = \rho\vec{J}$ describes in Unit 11. The two are not in conflict; they are different conditions.

Assuming the outer surface charge is uniform on any shape. For a sphere it is, once nothing is inside the cavity. For an irregular conductor, statement 10.1.A.3.iii says it is not.

Treating shielding as symmetric. A closed conducting shell keeps external fields out of its interior. It does not automatically keep an interior charge's field from reaching outside, unless the shell is grounded, which is Topic 10.2.

On assessment: the exam is 3 hours long, 42 multiple-choice questions worth 50% in 85 minutes, then 4 free-response questions worth 50% in 95 minutes. The Unit 10 opener names skills 2.A, 3.B and 3.C as the ones the unit builds, and points at the fourth free-response question, the Qualitative/Quantitative Translation, worth 8 points with a suggested time of 15 to 20 minutes. That question asks for a claim with justification, then a derivation, then a connection between the two. A topic made entirely of claims with no printed equations is built for it. The Unit 10 Progress Check is about 18 multiple-choice questions and 4 free-response questions.

A point charge inside a spherical conducting shell: three regions, two induced surfaces

A point charge $q = +5.0$ nC sits at the centre of an initially neutral spherical conducting shell with inner radius $a = 0.10$ m and outer radius $b = 0.14$ m. Find (a) the charge on each surface of the shell, (b) the field magnitude at $r = 0.050$ m, $r = 0.12$ m and $r = 0.20$ m, and (c) the surface charge densities. Then (d) repeat part (a) and the field at $r = 0.20$ m if the shell instead carries a net charge of $-8.0$ nC.

(a) Use a Gaussian surface inside the metal. Draw a sphere of radius $0.12$ m, which lies entirely within the conductor. The field is zero there, so the flux is zero, so the enclosed charge is zero: $q + Q_{\text{inner}} = 0$ , giving $Q_{\text{inner}} = -5.0$ nC on the inner surface.
Conserve the shell's own charge. The shell started neutral, so whatever left the inner surface went to the outer one: $Q_{\text{outer}} = 0 - (-5.0) = +5.0$ nC.
(b) At $r = 0.050$ m, inside the cavity. A Gaussian sphere there encloses only $q$ , so $E = \dfrac{kq}{r^2} = \dfrac{(9.0 \times 10^9)(5.0 \times 10^{-9})}{(0.050)^2} = \dfrac{45}{2.5 \times 10^{-3}} = 1.8 \times 10^4$ N/C, radially outward. The shell contributes nothing here.
At $r = 0.12$ m, inside the metal. $E = 0$ , by statement 10.1.A.4. No calculation needed, and saying why is the mark.
At $r = 0.20$ m, outside. The enclosed charge is $q + Q_{\text{inner}} + Q_{\text{outer}} = +5.0$ nC, so $E = \dfrac{45}{(0.20)^2} = \dfrac{45}{0.040} = 1.1 \times 10^3$ N/C, outward. From outside, the neutral shell is invisible.
(c) Densities. Inner surface area $4\pi a^2 = 0.1257\ \mathrm{m^2}$ , so $\sigma_{\text{inner}} = \dfrac{-5.0 \times 10^{-9}}{0.1257} = -3.98 \times 10^{-8}\ \mathrm{C/m^2}$ . Outer area $4\pi b^2 = 0.2463\ \mathrm{m^2}$ , so $\sigma_{\text{outer}} = \dfrac{+5.0 \times 10^{-9}}{0.2463} = +2.03 \times 10^{-8}\ \mathrm{C/m^2}$ . Check the outer one against $E = \sigma/\varepsilon_0$ at $r = b$ : $\dfrac{2.03 \times 10^{-8}}{8.85 \times 10^{-12}} = 2.29 \times 10^3$ N/C, and $kq/b^2 = 45/0.0196 = 2.30 \times 10^3$ N/C. They agree.
(d) Shell carrying $-8.0$ nC. The inner-surface argument is unchanged, because it depends only on the enclosed charge being zero: $Q_{\text{inner}} = -5.0$ nC still. The rest of the shell's charge goes outside: $Q_{\text{outer}} = -8.0 - (-5.0) = -3.0$ nC.
Field at $r = 0.20$ m now. Enclosed charge $= +5.0 - 5.0 - 3.0 = -3.0$ nC, so $E = \dfrac{(9.0 \times 10^9)(3.0 \times 10^{-9})}{0.040} = 675$ N/C, directed radially inward.

(a) Inner surface $-5.0$ nC, outer surface $+5.0$ nC. (b) $1.8 \times 10^4$ N/C outward in the cavity, exactly zero in the metal, $1.1 \times 10^3$ N/C outward at $0.20$ m. (c) $\sigma_{\text{inner}} = -3.98 \times 10^{-8}\ \mathrm{C/m^2}$ , $\sigma_{\text{outer}} = +2.03 \times 10^{-8}\ \mathrm{C/m^2}$ . (d) With a $-8.0$ nC shell, the inner surface is still $-5.0$ nC and the outer becomes $-3.0$ nC, giving $675$ N/C inward at $0.20$ m.

A neutral slab in a uniform field: how much charge the polarization needs

A large neutral conducting slab with flat parallel faces is placed in a uniform external electric field of magnitude $E_0 = 4.0 \times 10^4$ N/C, with the field perpendicular to the faces. Find the induced surface charge density on each face, and state the total charge on the slab. Take $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$ .

Name the requirement. Statement 10.1.A.6 says the conductor polarizes as a consequence of remaining an equipotential surface, and statement 10.1.A.4 says the field inside is zero. So the induced charge must produce a field of magnitude $E_0$ inside the metal, pointing opposite to the external field.
Model the induced charge. Electrons drift against the external field, so the face the field enters becomes negative with density $-\sigma$ and the face it leaves becomes positive with density $+\sigma$ . Two large oppositely charged sheets.
Get the field of the pair. Use a Gaussian pillbox with one end face inside the metal, where $E = 0$ , and one end face just outside, in the external region where the field is $E_0$ and perpendicular to the face. Only the outer face contributes flux, so $E_0 A = \dfrac{\sigma A}{\varepsilon_0}$ , giving $\sigma = \varepsilon_0 E_0$ .
Substitute. $\sigma = (8.85 \times 10^{-12})(4.0 \times 10^4) = 3.54 \times 10^{-7}\ \mathrm{C/m^2}$ .
Check it cancels. Two sheets of surface density $\pm\sigma$ produce, in the region between them, a field of magnitude $\sigma/\varepsilon_0 = \dfrac{3.54 \times 10^{-7}}{8.85 \times 10^{-12}} = 4.0 \times 10^4$ N/C, opposite to $E_0$ . Sum zero, as required.
State the total charge. Zero. The slab is neutral, and stays neutral: nothing was added or removed, only rearranged. $+\sigma$ on one face and $-\sigma$ on the other of equal area sum to zero.
Note what did not enter. The slab's thickness never appeared. The induced density is set by the external field alone, which is why a thin foil shields as well as a thick block.

$\sigma = \varepsilon_0 E_0 = 3.54 \times 10^{-7}\ \mathrm{C/m^2}$ , negative on the face the external field enters and positive on the face it leaves. The slab's total charge remains exactly zero: polarization redistributes charge without adding any. The result is independent of the slab's thickness.

A hollow charged conductor with an off-centre cavity

A solid conducting sphere of radius $R = 0.10$ m carries a net charge of $+12$ nC. An empty cavity of arbitrary shape is hollowed out somewhere inside it, not at the centre. Find (a) the field inside the cavity, (b) the charge on the cavity wall, (c) the field magnitude at $r = 0.25$ m from the sphere's centre, and (d) the surface charge density on the outer surface. Justify why the answer to (c) does not depend on where the cavity is.

(a) Show the cavity wall carries no charge, then that the cavity field is zero. Draw a Gaussian surface in the metal that completely surrounds the cavity. The field is zero on it, so the enclosed charge is zero, so the total charge on the cavity wall is zero.
Rule out a plus-and-minus arrangement on the wall. Zero total does not by itself forbid patches of both signs. Suppose there were: then a field line would have to run from a positive patch to a negative patch across the cavity. Close that path by returning through the metal, where $\vec{E} = 0$ . Then $\oint \vec{E} \cdot d\vec{r}$ around the loop would be nonzero, so the potential would not return to its starting value, which is impossible for an electrostatic field. So there are no patches, and the field in the cavity is zero everywhere.
(b) Charge on the cavity wall: zero, from the Gaussian argument in the first step. This is statement 10.1.A.7 in its cleanest form: the region inside the shell of metal is free from fields.
(c) Field at $r = 0.25$ m. All $+12$ nC sits on the outer surface. A Gaussian sphere of radius $0.25$ m encloses all of it, so $E = \dfrac{kQ}{r^2} = \dfrac{(9.0 \times 10^9)(12 \times 10^{-9})}{(0.25)^2} = \dfrac{108}{0.0625} = 1.73 \times 10^3$ N/C, radially outward.
Why the cavity's position is irrelevant. The cavity wall carries no charge at all, so it contributes nothing to any exterior field, and the outer surface is a sphere carrying $+12$ nC. Nothing about the interior geometry reaches the outside.
(d) Outer surface density. With the outer surface spherical and the interior contributing nothing asymmetric, the $+12$ nC spreads uniformly: $\sigma = \dfrac{12 \times 10^{-9}}{4\pi(0.10)^2} = \dfrac{12 \times 10^{-9}}{0.1257} = 9.55 \times 10^{-8}\ \mathrm{C/m^2}$ .
Cross-check. $E$ just outside the surface should be $\sigma/\varepsilon_0 = \dfrac{9.55 \times 10^{-8}}{8.85 \times 10^{-12}} = 1.08 \times 10^4$ N/C, and $kQ/R^2 = 108/0.010 = 1.08 \times 10^4$ N/C. They agree.
Contrast with the first example. There, a charge sat in the cavity and the wall had to carry $-q$ . Here the cavity is empty, so the wall carries nothing. The Gaussian argument is the same; only the enclosed charge changed.

(a) Zero, everywhere in the cavity. (b) Zero net charge on the cavity wall, and no local patches of either sign. (c) $1.73 \times 10^3$ N/C radially outward, independent of where the cavity is. (d) $\sigma = 9.55 \times 10^{-8}\ \mathrm{C/m^2}$ , uniform over the spherical outer surface.

Frequently asked questions

Why is the electric field zero inside a conductor?

Because the charges in a conductor are free to move and, in electrostatic equilibrium, they have stopped moving. If any field remained inside, it would exert a force on those free charges and they would accelerate, so the state would not be equilibrium. The charges rearrange until the field they produce exactly cancels whatever field was pushing them, and CED statement 10.1.A.3.i says that takes a time so short as to be negligible. Zero interior field is a conclusion from equilibrium, not a separate assumption.

Why does all the excess charge sit on the surface of a conductor?

Draw any closed Gaussian surface entirely inside the metal. The field is zero at every point on it, so the flux through it is zero, so by Gauss's law the enclosed charge is zero. That holds for every such surface, however small or however close to the boundary, so the charge density in the interior is exactly zero and all the excess charge is on the surface. CED statement 10.1.A.2 gives the physical version, that mutual repulsion of the excess charge carriers pushes them outward, and 10.1.A.4 states the conclusion.

Is the electric potential inside a conductor zero?

No, it is constant, which is a different statement. Because the field is zero everywhere inside, the line integral of E along any path within the conductor is zero, so the potential difference between any two points in or on it is zero. The common value can be anything: an isolated sphere of radius R carrying charge Q sits at kQ over R relative to the printed zero at infinity. CED statement 10.1.A.3.ii puts it as the conductor becoming an equipotential surface.

Why is the electric field perpendicular to a conductor's surface?

Because a component along the surface would push the free surface charges sideways, and they would move, contradicting electrostatic equilibrium. So the tangential component must be zero and only the normal component survives, which is CED statement 10.1.A.5. The same result follows from Unit 9: the surface is an equipotential, statement 9.2.B.3.iv says there is no component of an electric field along an isoline, and 9.2.B.3.ii says isolines are perpendicular to field vectors.

Where is the charge density greatest on a charged conductor?

At points and edges, compared with planar areas, which is CED statement 10.1.A.3.iii. The reason is the equipotential requirement: a sharply curved region is close to more of its own charge, so a given amount of charge there raises the local potential more, and less charge is needed to bring that region to the common potential, concentrated into a much smaller area. Since the field just outside a conductor is proportional to the local surface charge density, the external field is strongest just outside the sharpest features.

What is electrostatic shielding?

CED statement 10.1.A.7 defines it as surrounding an area with a closed conducting shell to create a region inside the conductor that is free from external electric fields. Charge rearranges on the shell until its own field cancels the external field throughout the metal and, when the cavity contains no charge, throughout the cavity too. The shielding is one-way as stated: a charge placed inside the cavity of an isolated shell still induces charge on the outer surface and so still produces a field outside.

Does AP Physics 2 cover electrostatics with conductors?

Partly, and not as a topic of its own. AP Physics 2 puts it inside Topic 10.3, Electric Fields, at learning objective 10.3.B, which states that a conductor in electrostatic equilibrium has its excess charge on the surface with zero field inside, that the field at the surface is perpendicular to it, and that the field outside an isolated sphere matches that of a point charge at its centre. What AP Physics C Topic 10.1 adds is the equipotential-surface statement, the negligible equilibrium time, the greater charge density at points and edges, polarization as a consequence of the equipotential condition, and electrostatic shielding.