AP Physics C: E&M · Topic 10.2

Topic 10.2: Redistribution of Charge between Conductors

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

When two conductors touch or are joined by a wire, charge moves until both surfaces sit at the same electric potential. Their charges do not become equal unless the conductors are identical. Ground is an idealised body at zero potential that can absorb or supply any amount of charge.

AP Physics: Unit 10 (topics 10.2 Redistribution of Charge between Conductors). Topic 10.2 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, 10.2.A, using the task verb describe, with three essential-knowledge statements: 10.2.A.1, that conductors in electrical contact redistribute charge so the surfaces of each conductor are at the same electric potential; 10.2.A.2, that ground is an idealized reference point at zero electric potential which can absorb or provide an infinite amount of charge without changing its potential; and 10.2.A.3, that charge can be induced on a conductor by grounding it in the presence of an external electric field. Topic 10.2 prints no boundary statement and no equations, yet lists skill 2.A, derive a symbolic expression, so the two-sphere charge split is expected to be derived rather than recalled. Suggested skills are 1.A, 2.A, 2.C and 3.B. There is no AP Physics 2 topic of this name; the algebra-based course carries the equal-potential statement inside its Topic 10.5 as statement 10.5.A.4, worded with electrons rather than charges, and describes grounding physically in statement 10.2.A.3 as connection to a much larger approximately neutral system rather than as an idealized zero-potential reservoir.

What Topic 10.2 requires

Topic 10.2 has one learning objective and three essential-knowledge statements. Like Topic 10.1, it prints no equations at all.

10.2.A, describe the movement of charge and the resulting interactions when conductors physically contact each other.

10.2.A.1 states that when conductors are in electrical contact, charges will be redistributed such that the surfaces of each conductor are at the same electric potential.
10.2.A.2 states that ground is an idealized reference point that has zero electric potential and can absorb or provide an infinite amount of charge without changing its electric potential.
10.2.A.3 states that charge can be induced on a conductor by grounding the conductor in the presence of an external electric field.

Topic 10.2 prints no boundary statement. 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.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Skill 2.A is the one that gives the game away. A topic with no printed equations still lists derive a symbolic expression, which means the exam expects you to build the relations here yourself from Topic 10.1's equipotential statement and the potential of a charged sphere. That is what most of this page is about.

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

What equalises is the potential, and only the potential

Everything in this topic hangs on one clause of statement 10.2.A.1: charges redistribute such that the surfaces of each conductor are at the same electric potential.

Not the same charge. Not the same charge density. Not the same field. The potential.

The reason is Topic 10.1 doing its work. Join two conductors with a wire and you have made one conductor. Statement 10.1.A.3.ii says a conductor at electrostatic equilibrium is an equipotential surface, so once the charge stops moving, every point of the combined object is at one potential. Statement 10.1.A.3.i says that happens in a time so short as to be negligible, so you never have to model the transient.

The mechanism is worth stating too, because it is the answer to "which way does the charge flow?". While the two conductors are still at different potentials, there is a potential difference along the connecting wire, so there is a field in the wire, so the free charges in it feel a force and move. Conventional current runs from high potential to low. Electrons, being negative, physically move the other way. Charge keeps flowing precisely until the potential difference is gone, at which point the driving field is gone too.

That gives you a two-line recipe for every problem in this topic:

Charge is conserved. The total on the connected objects afterwards equals the total before, a principle Unit 8 establishes, and it is the equation the exam wants to see written down.
The final potentials are equal. Write each conductor's potential in terms of its final charge, set them equal, and solve simultaneously with the first line.

Two conclusions fall straight out, and both are examined.

Identical conductors split the charge equally. If the two objects have the same shape and size, equal potentials force equal charges. This is the only case where "the charge divides in half" is correct, and it is correct because the geometries match, not because contact does that in general.

Different conductors do not. A large object and a small one reach the same potential with very different charges on them, and the smaller one ends up with the higher surface charge density. That is Topic 10.1's statement 10.1.A.3.iii, that density is greater where the surface curves sharply, made quantitative.

Two spheres on a wire, done as algebra

The standard model problem is two conducting spheres, far apart, joined by a long thin wire. "Far apart" and "thin wire" are doing real work: they let you treat each sphere's potential as though the other were not there, and let you ignore any charge that ends up sitting on the wire itself. Say so when you set the problem up, because skill 2.A rewards naming your assumptions.

An isolated conducting sphere of radius $R$ carrying charge $Q$ has, at its surface and throughout its interior,

V = \frac{kQ}{R}

which comes from Gauss's law outside the sphere plus the line integral in from infinity. Now apply the recipe. Charge conservation gives $Q_1 + Q_2 = Q_{\text{total}}$ , and equal potentials give

\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \quad \Rightarrow \quad \frac{Q_1}{Q_2} = \frac{R_1}{R_2}

The charge divides in the ratio of the radii. Solving with the conservation equation,

Q_1 = Q_{\text{total}}\frac{R_1}{R_1 + R_2}, \qquad Q_2 = Q_{\text{total}}\frac{R_2}{R_1 + R_2}

Three follow-up quantities get asked, and each answers a different question about the same configuration.

Surface charge density. $\sigma = \dfrac{Q}{4\pi R^2}$ , so $\dfrac{\sigma_1}{\sigma_2} = \dfrac{Q_1}{Q_2}\left(\dfrac{R_2}{R_1}\right)^2 = \dfrac{R_2}{R_1}$ . The smaller sphere carries the higher density, in inverse ratio to the radii.

Surface field. $E = \dfrac{kQ}{R^2} = \dfrac{\sigma}{\varepsilon_0}$ , so the field ratio is the same as the density ratio. The smaller sphere has the stronger field at its surface even though it holds less charge.

Common potential. $V = \dfrac{kQ_{\text{total}}}{R_1 + R_2}$ , which you can read off by substituting either result.

That third relation is a small piece of algebra worth doing once and remembering, and the two-sphere system is the AP-level model of a conductor with a bump on it: the bump is the small sphere, at the same potential as the rest and carrying a fiercer field.

One warning about the ratio $Q_1/Q_2 = R_1/R_2$ . It is a derived result for spheres, not a printed equation and not a general rule. Two conductors that are not spheres still reach equal potentials, but the way charge splits depends on their shapes, which the course does not ask you to compute.

Ground, as this course defines it

Statement 10.2.A.2 defines ground as an idealized reference point that (a) has zero electric potential and (b) can absorb or provide an infinite amount of charge without changing its electric potential.

Read that as two independent properties, because questions test them separately.

Property (a) fixes a number. Grounding a conductor does not merely connect it to something; it sets its potential to zero. Every potential in the problem is then measured against that. Note that this is a second zero in the course, sitting alongside the exam convention printed with the equation sheet that the electric potential is zero at an infinite distance from an isolated point charge. They are consistent, because Earth is treated as an infinite conductor at the reference potential, but if a problem gives you both, use the one it names.

Property (b) says you never run out. Charge can flow to ground or from ground in any quantity and ground's potential does not budge. So a grounded conductor can shed all of its charge, or gain as much as it needs, and no bookkeeping about "where the charge went" is required. Charge is still conserved globally; it just left the system you are tracking.

Contrast this with the definition the algebra-based course gives. AP Physics 2's statement 10.2.A.3 says grounding involves electrically connecting a charged system to a much larger and approximately neutral system, for example Earth. That is a physical description: a big neutral thing. AP Physics C's is an idealisation: a body at exactly zero potential with unlimited capacity. The Physics C version is more useful in a calculation, because it hands you a boundary condition, $V = 0$ , rather than a comparison.

The practical consequences:

Ground an isolated charged conductor and it loses all its excess charge. Its potential must go to zero, and for an isolated sphere $V = kQ/R$ , so $Q = 0$ .
Ground a conductor that is near other charges and it generally does not end up neutral. The condition is $V = 0$ , not $Q = 0$ , and those differ whenever something else is contributing to the potential. This is the case statement 10.2.A.3 is about.
Grounding a shell around a charge kills the exterior field. The outer surface charge drains away until the outside sees zero potential and zero field, which finishes the shielding story that Topic 10.1 left one-sided.

Charging by induction, which needs the ground

Statement 10.2.A.3 is short: charge can be induced on a conductor by grounding the conductor in the presence of an external electric field. Unpacked, it is a four-step procedure, and getting the order right is the whole question.

Bring a charged object near the neutral conductor, without touching. The conductor polarizes, by statement 10.1.A.6. Its near face acquires charge opposite to the object, its far face the same sign. Net charge is still zero.
Ground the conductor while the charged object stays where it is. The far-face charge, which is repelled by the nearby object, now has somewhere to go, and it leaves. The conductor is left with a net charge opposite to the nearby object, and it sits at $V = 0$ .
Remove the ground first, while the object is still nearby. This is the step that gets reversed. Breaking the ground connection traps the charge that is now on the conductor.
Then remove the charged object. The trapped charge redistributes over the surface according to statement 10.1.A.2, and the conductor is left permanently charged, with the opposite sign to the object you never touched it with.

Reverse steps 3 and 4 and you get nothing: remove the object first and the trapped charge is no longer held in place, so it flows back to ground and the conductor ends neutral.

Two features of this are worth naming.

No charge was transferred from the charged object. It has exactly as much charge at the end as at the start. The induced charge came from ground. Compare charging by contact, which is Topic 10.2's other half: there, the two objects share their combined charge and the source object ends with less. See charging by induction for the standalone definition.

The sign is opposite, and that is the diagnostic. Induction with a negative rod leaves a positive conductor. Contact with a negative rod leaves a negative conductor. A question that tells you the final sign is telling you which process happened.

The AP Physics 2 course covers the polarization half of this in its Topic 10.2, at statements 10.2.A.1.ii and 10.2.A.1.iii, and the grounding half in 10.2.A.3, but it never combines them into the induced-charging procedure the way statement 10.2.A.3 does here.

Where AP Physics 2 puts this content

There is no AP Physics 2 topic called Redistribution of Charge between Conductors. The core statement does exist in the algebra-based course, and it is worth knowing exactly where, because the wording differs in a way that matters.

AP Physics 2 statement 10.5.A.4, buried in its Topic 10.5, Electric Potential: "When conductors are in electrical contact, electrons will be redistributed such that the surfaces of the conductors are at the same electric potential."

AP Physics C statement 10.2.A.1: "When conductors are in electrical contact, charges will be redistributed such that the surfaces of each conductor are at the same electric potential."

The change from electrons to charges matches the modelling licence in statement 10.1.A.2.ii, which lets you treat a positively charged conductor as if positive carriers reside on its surface. Physics 2 keeps the physical carrier in the sentence; Physics C generalises it so you can do the bookkeeping in whichever sign is convenient.

Beyond that one sentence, AP Physics C makes three additions that AP Physics 2 does not have anywhere:

A topic of its own, which means the exam can build a whole question on it rather than mentioning it inside a potential question.
The idealised definition of ground (10.2.A.2), against Physics 2's physical description of grounding as connection to a much larger, approximately neutral system.
Induced charging by grounding in an external field (10.2.A.3), which Physics 2 does not state as a procedure.

And it adds skill 2.A, derive a symbolic expression, which Physics 2's Topic 10.5 also lists but for a different purpose. Here it is aimed at the two-sphere algebra above.

If you want the algebra-based treatment of contact and charging, read AP Physics 2's Topic 10.2 on conservation of charge and the process of charging and Topic 10.5. This page is for the calculus-based course, where you are expected to derive the split rather than be told it.

Traps

Splitting the charge equally between unequal conductors. Equal potentials, not equal charges. Equal charges only when the conductors are identical.

Assuming the bigger sphere has the stronger surface field. It has the most charge and the weakest surface field. Density and field both go inversely with radius once the potentials are equal.

Deciding the flow direction from the charge. Charge flows from high potential to low potential, and the object with more charge is not necessarily at higher potential. A small sphere with $8$ nC can be at a higher potential than a large one with $16$ nC, and then charge flows from the small one to the large one.

Thinking grounding always neutralises. Grounding sets $V = 0$ . That gives $Q = 0$ only for an isolated conductor with nothing else nearby. With an external charge present, a grounded conductor ends up charged, which is the entire content of statement 10.2.A.3.

Reversing the order of the induction steps. Ground off, then object away. Object away first and the charge escapes back to ground.

Forgetting that the wire is part of the conductor. Two spheres joined by a wire are one conductor at one potential, including the wire, which is why "the wire's own capacitance is negligible" is an assumption worth stating rather than an obvious truth.

Applying this while current is still flowing. Every statement here describes the final equilibrium state, reached in a negligible time by statement 10.1.A.3.i. A capacitor discharging through a resistor in Unit 11 is the case where the transient is the point, and there the time constant is not negligible at all.

How Topic 10.2 is assessed

The AP Physics C: Electricity and Magnetism exam is 3 hours long, with 42 multiple-choice questions worth 50% in 85 minutes and 4 free-response questions worth 50% in 95 minutes. A four-function, scientific or graphing calculator is allowed throughout.

The Unit 10 opener names skills 2.A, 3.B and 3.C as the ones this 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. The CED describes that question as asking students to make and justify a claim about a scenario, then derive an equation related to it, and finally connect the two, for instance by justifying why the claim and the derivation agree or by predicting how the representations would change if properties of the scenario were altered.

That is a precise description of a Topic 10.2 question. Claim: the smaller sphere ends with the stronger surface field. Derivation: set the potentials equal, get $Q_1/Q_2 = R_1/R_2$ , then form $E = kQ/R^2$ . Connection: explain why the derived ratio supports the claim, and predict what happens if one radius is doubled. That last part is functional-dependence reasoning, and it is why the unit lists skill 2.D on its other topics.

The framework's Unit 10 Progress Check is about 18 multiple-choice questions and 4 free-response questions.

On multiple choice, this topic tends to appear as a ranking question, which is skill 2.C: rank the final charges, the potentials, the surface densities or the fields for several connected conductors. The potentials are always all equal, which is the free mark, and the other three orderings follow from the radii.

For the conductor properties this topic depends on, see Topic 10.1; for the potential of a charged object, Topic 9.2; and for the Gauss's law step in the sphere derivation, the Gauss's law guide.

Two unequal spheres joined by a wire: which way does the charge go?

Sphere 1 is a conducting sphere of radius $R_1 = 0.12$ m carrying $+16$ nC. Sphere 2 is a conducting sphere of radius $R_2 = 0.040$ m carrying $+8.0$ nC. They are far apart, and are then joined by a long thin conducting wire whose own capacitance is negligible. Find (a) the potential of each before connection and the direction of charge flow, (b) the final charge on each, (c) the common final potential, and (d) the ratio of their surface charge densities and surface field magnitudes.

(a) Potentials before. Each is isolated, so $V = kQ/R$ . Sphere 1: $V_1 = \dfrac{(9.0 \times 10^9)(16 \times 10^{-9})}{0.12} = \dfrac{144}{0.12} = 1200$ V. Sphere 2: $V_2 = \dfrac{72}{0.040} = 1800$ V.
Read the direction from the potentials, not the charges. Sphere 2 has half the charge and is at the higher potential, so conventional current flows from sphere 2 to sphere 1 along the wire, and electrons move from sphere 1 to sphere 2.
(b) Write the two conditions. Charge conservation: $Q_1 + Q_2 = 24$ nC. Equal final potentials: $\dfrac{kQ_1}{R_1} = \dfrac{kQ_2}{R_2}$ , so $\dfrac{Q_1}{Q_2} = \dfrac{R_1}{R_2} = \dfrac{0.12}{0.040} = 3$ .
Solve. $Q_1 = 3Q_2$ and $Q_1 + Q_2 = 24$ nC give $4Q_2 = 24$ nC, so $Q_2 = 6.0$ nC and $Q_1 = 18$ nC.
Check against the predicted direction. Sphere 1 went from $16$ to $18$ nC, gaining $2.0$ nC; sphere 2 went from $8.0$ to $6.0$ nC, losing $2.0$ nC. Positive charge moved from 2 to 1, exactly as the potentials said it would.
(c) Common potential. $V = \dfrac{kQ_1}{R_1} = \dfrac{(9.0 \times 10^9)(18 \times 10^{-9})}{0.12} = \dfrac{162}{0.12} = 1350$ V. Check with the other sphere: $\dfrac{54}{0.040} = 1350$ V. Equal, as required. It also equals $\dfrac{kQ_{\text{total}}}{R_1+R_2} = \dfrac{216}{0.16} = 1350$ V.
(d) Densities. $\sigma_1 = \dfrac{18 \times 10^{-9}}{4\pi(0.12)^2} = 9.95 \times 10^{-8}\ \mathrm{C/m^2}$ and $\sigma_2 = \dfrac{6.0 \times 10^{-9}}{4\pi(0.040)^2} = 2.98 \times 10^{-7}\ \mathrm{C/m^2}$ , so $\sigma_2/\sigma_1 = 3$ , the inverse ratio of the radii.
Surface fields. $E_1 = \dfrac{kQ_1}{R_1^2} = \dfrac{162}{0.0144} = 1.13 \times 10^4$ V/m and $E_2 = \dfrac{54}{0.0016} = 3.38 \times 10^4$ V/m, also a factor of 3. The small sphere has three times the field with one third the charge.

(a) $V_1 = 1200$ V and $V_2 = 1800$ V, so charge flows from sphere 2 to sphere 1 even though sphere 1 already holds more of it. (b) $Q_1 = 18$ nC and $Q_2 = 6.0$ nC, in the ratio of the radii. (c) Both at $1350$ V. (d) The smaller sphere ends with three times the surface charge density and three times the surface field, which is statement 10.1.A.3.iii made quantitative.

Grounding a shell around a charge: what leaves and what stays

A point charge $q = +6.0$ nC sits at the centre of a conducting spherical shell with inner radius $0.10$ m and outer radius $0.15$ m. The shell is initially neutral and isolated. It is then connected to ground. Find the charge on each surface and the field outside, both before and after grounding, and state how much charge flowed to or from ground.

Before grounding, inner surface. A Gaussian sphere drawn inside the metal encloses zero net charge because the field there is zero, so the inner surface carries $-6.0$ nC.
Before grounding, outer surface. The shell is neutral, so the outer surface carries $+6.0$ nC.
Before grounding, field at $r = 0.30$ m. Enclosed charge $= +6.0 - 6.0 + 6.0 = +6.0$ nC, so $E = \dfrac{(9.0 \times 10^9)(6.0 \times 10^{-9})}{(0.30)^2} = \dfrac{54}{0.090} = 600$ N/C, radially outward. Note this: an uncharged shell does not hide the charge inside it.
Now ground it, and apply statement 10.2.A.2. Grounding fixes the shell's potential at zero. Since the shell is a conductor, that is the potential of every point on and in it, including points on its outer surface.
Work out the outer surface charge from that condition. Just outside the outer surface, the potential due to everything enclosed is $\dfrac{k\,q_{\text{enc,total}}}{0.15}$ measured from zero at infinity. Setting it to zero requires $q_{\text{enc,total}} = 0$ , so $q + Q_{\text{inner}} + Q_{\text{outer}} = 0$ . With $q = +6.0$ nC and $Q_{\text{inner}} = -6.0$ nC still fixed by the Gaussian argument, $Q_{\text{outer}} = 0$ .
Field outside after grounding. Any Gaussian sphere outside now encloses zero net charge, so $E = 0$ everywhere outside the shell. The shielding is now two-way.
Charge flowed to ground. The outer surface went from $+6.0$ nC to $0$ , so $+6.0$ nC left the shell for ground, or equivalently $6.0$ nC of electrons came up from ground. The shell now carries a net charge of $-6.0$ nC and is at zero potential, which are compatible statements: grounding fixes the potential, not the charge.
Confirm nothing changed inside the cavity. The inner surface still holds $-6.0$ nC and the field in the cavity is still $\dfrac{kq}{r^2}$ . Grounding the outside has no effect there, because the metal shields the cavity from whatever happens beyond it.

Before grounding: inner surface $-6.0$ nC, outer surface $+6.0$ nC, and $600$ N/C outward at $r = 0.30$ m. After grounding: inner surface still $-6.0$ nC, outer surface $0$ , and zero field everywhere outside. Exactly $6.0$ nC of positive charge flowed to ground, leaving the shell with a net charge of $-6.0$ nC while sitting at zero potential.

Three identical spheres, touched in sequence

Sphere A is an isolated conducting sphere carrying $+32$ nC. Spheres B and C are identical to A and both neutral, and all three are far apart. A is touched to B and separated; then A is touched to C and separated; then B is touched to C and separated. Find the charge on each sphere at the end, and state the general rule you used at each step.

State the rule for identical spheres. By statement 10.2.A.1, contact makes the potentials equal. Identical spheres have the same radius, so $V = kQ/R$ equal means $Q$ equal. Only for identical conductors does contact split the charge evenly.
Step 1, A touches B. Combined charge $32 + 0 = 32$ nC, shared equally: A has $16$ nC, B has $16$ nC, C has $0$ .
Step 2, A touches C. Combined charge $16 + 0 = 16$ nC, shared equally: A has $8.0$ nC, C has $8.0$ nC. B is untouched at $16$ nC.
Step 3, B touches C. Combined charge $16 + 8.0 = 24$ nC, shared equally: B has $12$ nC, C has $12$ nC. A is untouched at $8.0$ nC.
Check charge conservation. $8.0 + 12 + 12 = 32$ nC, the amount you started with. Nothing was grounded, so nothing left the system.
Notice that order matters. Sphere A, the one that started charged, ends with the least. Each contact halves whatever it is carrying at that moment, and it took part in two contacts: $32 \to 16 \to 8$ .
Contrast with grounding. If instead of touching C, sphere A had been grounded once while isolated from everything else, it would have gone to $V = 0$ , and since $V = kQ/R$ for an isolated sphere that means $Q = 0$ : all of it gone, not half.

A ends with $8.0$ nC, B with $12$ nC and C with $12$ nC, totalling the original $32$ nC. Equal splitting applies only because the spheres are identical; had they differed in radius, each contact would have divided the charge in the ratio of the two radii instead.

Frequently asked questions

What happens when two charged conductors touch?

Charge moves between them until both surfaces sit at the same electric potential, which is CED statement 10.2.A.1. Their charges do not become equal unless the conductors are identical in size and shape. While the potentials differ there is a field in the connecting path, which drives the charge; once the potentials match, the driving field is gone and the flow stops. The whole process takes a time so short as to be negligible, per statement 10.1.A.3.i.

How does charge divide between two conducting spheres connected by a wire?

In the ratio of their radii. Each isolated sphere has potential V = kQ/R, so equal potentials give Q1 over Q2 equal to R1 over R2, and charge conservation fixes the totals. The larger sphere ends with more charge but the lower surface charge density and the weaker surface field, both in inverse ratio to the radii. This result assumes the spheres are far apart and the wire's own capacitance is negligible; it is a derivation rather than a printed equation.

Which way does charge flow when two conductors are connected?

From higher potential to lower potential, taking conventional current as the flow of positive charge; electrons physically move the other way. It is the potentials that decide, not the charges, and those can disagree. A small sphere carrying 8 nC can sit at a higher potential than a large one carrying 16 nC, in which case charge flows from the small sphere to the large one even though the large one already holds more.

What does ground mean in AP Physics C?

Statement 10.2.A.2 defines ground as an idealized reference point that has zero electric potential and can absorb or provide an infinite amount of charge without changing its electric potential. Two properties, and they are used separately: it fixes a conductor's potential at zero, and it never runs out of charge. AP Physics 2 gives a physical description instead, grounding as connecting a charged system to a much larger and approximately neutral system such as Earth. The Physics C version supplies a boundary condition you can put into an equation.

Does grounding a conductor always make it neutral?

No. Grounding sets the potential to zero, not the charge. For an isolated conductor with nothing else nearby, zero potential does mean zero charge, since V = kQ/R for a sphere. But with another charge present, a grounded conductor generally ends up carrying net charge while still sitting at zero potential. That is exactly what statement 10.2.A.3 describes: charge can be induced on a conductor by grounding it in the presence of an external electric field.

How does charging by induction work?

Bring a charged object near a neutral conductor without touching it, so the conductor polarizes. Ground the conductor while the object is still there, and the charge repelled by the object leaves for ground. Disconnect the ground first, then remove the object. The conductor is left with a permanent charge of the opposite sign to the object, and the object itself lost nothing. The order matters: remove the object before the ground and the induced charge simply flows back, leaving the conductor neutral.

Is redistribution of charge between conductors on the AP Physics 2 exam?

The core statement is, but not as a topic of its own. AP Physics 2 puts it in statement 10.5.A.4, inside its Topic 10.5 on electric potential, worded as electrons being redistributed until the conductor surfaces are at the same potential. AP Physics C promotes it to Topic 10.2, changes electrons to charges, and adds two statements the algebra-based course does not have: the idealised definition of ground, and induced charging by grounding a conductor in an external field.