8.2 Conservation of Electric Charge and the Process of Charging

1. AZero, since the rubbing produced the rod’s charge
   A student who thinks rubbing creates charge on the rubbed object picks this. Rubbing only moves electrons from one material to the other; the rod became positive because the silk took electrons from it, so the silk is negative.
2. BPositive, with the same magnitude as the rod’s
   A student who expects both rubbed objects to get the same kind of charge picks this. The electrons the rod lost went to the silk, so the silk is negative, not positive.
3. CNegative, but of a smaller magnitude than the rod’s
   A student who thinks some charge is used up or lost during rubbing picks this. No charge went anywhere else, and charge does not disappear: the silk has every electron the rod lost, so the magnitudes are equal.
4. DNegative, with a magnitude equal to the rod’s Correct
   Rubbing transfers electrons from the glass to the silk. The rod's positive charge is the charge of the electrons it lost, and the silk gained exactly those electrons, so its charge is negative and equal in magnitude. The pair's net charge is still zero.

CED 8.2.A.1.i · Read this in Fix

1. ATouching the rod to a metal pipe that is connected to the ground Correct
   The system's net charge changes only if charge crosses its boundary. A grounded pipe gives the rod's excess electrons a path out of the system to Earth, so the system's net charge changes.
2. BRubbing the charged rod and the fur together again for a longer time
   A student who thinks rubbing creates charge picks this. More rubbing moves more electrons from the fur to the rod, but both are inside the system, so its net charge stays zero.
3. CHolding a charged balloon near the rod without letting them touch
   A student who thinks a nearby charged object passes charge across without contact picks this. The balloon can shift charges within the rod, but no charge crosses the system's boundary.
4. DMoving the charged rod and the fur much farther apart from each other
   A student who thinks separated charge adds up to a net charge picks this. Moving the rod and fur apart changes where the charges are, not their sum, which stays zero.

CED 8.2.A.2 · Read this in Fix

1. AAbout 1.5 × 10¹⁰ electrons, from the sphere to ground
   A student who thinks the sphere and Earth share the charge equally picks this. Earth is a much larger system than the sphere, so the charge is not split equally: nearly all 3.0 × 10¹⁰ excess electrons leave the sphere.
2. BAbout 3.0 × 10¹⁰ electrons, from the sphere to the ground Correct
   Earth is a much larger, approximately neutral system, so when the sphere is grounded far from other charges, nearly all of its excess charge leaves it. That charge is carried by electrons flowing from the sphere to the ground: N = (4.8 × 10⁻⁹ C)/(1.60 × 10⁻¹⁹ C) = 3.0 × 10¹⁰.
3. CAbout 3.0 × 10¹⁰ protons, flowing up from the ground to the sphere
   A student who thinks positive charge flows up from the ground to neutralize the sphere picks this. Protons stay in the nuclei; the sphere is neutralized by losing its excess electrons to the ground.
4. DAbout 3.0 × 10¹⁰ electrons, up from the ground to the sphere
   A student who thinks a negative object is short of electrons picks this. A negative charge means excess electrons, so electrons flow out of the sphere into the ground.

Working Grounded far from other charges: the sphere ends essentially neutral, so its excess charge −4.8 nC flows to Earth as electrons. N = 4.8 × 10⁻⁹ C / 1.60 × 10⁻¹⁹ C = 3.0 × 10¹⁰ electrons, from sphere to ground. Earth's net charge changes by −4.8 nC, negligible for Earth.

CED 8.2.A.3 · Read this in Fix

1. ABoth redistribute: the sphere’s electrons shift away from the rod, and the rod’s shift toward the sphere Correct
   The rod's excess electrons repel the free electrons in the sphere, which move to its far side, leaving the near side positive. That positive side in turn attracts the free electrons in the metal rod, which gather at the end nearer the sphere. Each object's charge distribution changes in response to the other; neither object's net charge changes, because no charge crosses the gap.
2. BThe sphere’s charges shift away from the rod, while the rod’s excess charge stays where it was on the rod
   A student who thinks only the neutral object responds picks this. The forces act both ways: the sphere's positive near side attracts the rod's free electrons, so the charge distribution of the metal rod changes too.
3. CSome of the rod’s excess electrons cross over to the sphere, so both objects now carry negative charge
   A student who thinks a nearby charged object passes charge across without contact picks this. With no contact there is no path for the electrons, so the sphere's net charge stays zero; only its charge distribution changes.
4. DNeither changes: the sphere is neutral, so it has no charges in it that the rod could move
   A student who reads 'neutral' as 'having no charge' picks this. The neutral sphere contains equal, very large amounts of positive and negative charge; its free electrons are pushed away by the rod, so its distribution changes.

CED 8.2.A.1 · Read this in Fix

1. A1.0 × 10¹⁰
   A student who adds the charges as amounts, 6.4 nC + 3.2 nC = 9.6 nC, and gives each sphere +4.8 nC picks this: A's charge would fall by 1.6 nC. With signs the total is +3.2 nC, so each sphere ends with +1.6 nC and A's charge falls by 4.8 nC.
2. B3.0 × 10¹⁰ Correct
   The total charge is +6.4 nC + (−3.2 nC) = +3.2 nC, and it is shared equally, so each sphere ends with +1.6 nC. A's charge changes from +6.4 nC to +1.6 nC, a change of −4.8 nC, which is carried by the electrons it gains: N = (4.8 × 10⁻⁹ C)/(1.60 × 10⁻¹⁹ C) = 3.0 × 10¹⁰.
3. C2.0 × 10¹⁰
   A student who thinks charge flows only until B is neutral picks this: B's 3.2 nC of excess electrons move to A, which is left with +3.2 nC. The flow continues until the two spheres have equal charges, +1.6 nC each, so A gains electrons carrying 4.8 nC.
4. D6.0 × 10¹⁰
   A student who takes the charge that moves to be the difference between the charges, 6.4 nC − (−3.2 nC) = 9.6 nC, picks this. Each electron that moves changes both spheres' charges, so only 4.8 nC needs to move to bring both to +1.6 nC.

Working Total charge (conserved): +6.4 nC + (−3.2 nC) = +3.2 nC. Equal shares: +1.6 nC each. Change in A: +1.6 nC − 6.4 nC = −4.8 nC, carried by electrons gained. N = 4.8 × 10⁻⁹ C / 1.60 × 10⁻¹⁹ C = 3.0 × 10¹⁰ electrons.

CED 8.2.A.1.i · Read this in Fix

1. ACorrect: electrons pass from the rod through the air to the knob and then to the leaves
   A student who thinks charge crosses from a nearby charged object without contact picks this. The rod does not touch the knob, so no electrons cross; the leaves spread because electrons already in the electroscope were pushed down to them.
2. BCorrect: the leaves carry negative charge, so the electroscope as a whole is negative
   A student who judges the whole system by one part picks this. The leaves are negative, but the knob is positive by the same amount; adding the charges of all the parts gives zero.
3. CIncorrect: electrons go from knob to leaves, but the electroscope’s net charge is zero Correct
   The rod's excess electrons repel free electrons in the knob, which move down the stem to the leaves. Both leaves become negative and repel each other, while the knob becomes positive by the same amount. No charge crossed the gap, so the electroscope's net charge is still zero.
4. DIncorrect: the rod attracts the neutral leaves directly, so no charge in them needs to move
   A student who thinks a charged object attracts neutral matter without any change in it picks this. The leaves spread because they carry charge of the same sign and repel each other; that charge was pushed down from the knob.

CED 8.2.A.1.ii · Read this in Fix

1. AElectrons flow through the wall, away from the balloon, so the wall surface near it becomes positive
   A student who pictures the electrons in any material as free to travel picks this. In an insulator the electrons cannot flow through the wall; each molecule is polarized in place, which is enough to leave the nearby surface slightly positive.
2. BIts molecules’ electron clouds shift slightly away, so the nearby wall surface is slightly positive Correct
   In an insulator the electrons stay with their molecules, but the balloon's negative charge pushes each molecule's electron cloud slightly away. The wall's surface nearest the balloon is left slightly positive and the negative ends are a little farther away, so the attraction is stronger than the repulsion. The wall's net charge stays zero.
3. CThe wall becomes positively charged overall, and this net positive charge holds the balloon
   A student who counts the positive surface as the charge of the whole wall picks this. The shifted electron clouds leave the far ends of the molecules negative by the same amount, so the wall's net charge is still zero.
4. DA charged balloon attracts any neutral object directly, so nothing in the wall has to change
   A student who thinks charged objects attract neutral matter directly picks this. A neutral object is attracted only because its charges are rearranged, with the opposite charge nearer the balloon.

CED 8.2.A.1.ii · Read this in Fix

1. AA is positive and B is negative, with A’s charge larger in magnitude
   A student who thinks the part nearer the rod receives more charge picks this. Every electron that left A arrived at B, so the two charges are equal in magnitude, and the pair's net charge is still zero.
2. BA and B are both negative, sharing electrons given off by the rod
   A student who thinks the rod passes charge across the gap picks this. The rod never touches A, so no electrons leave it; the spheres' own electrons are only moved from A to B.
3. CA and B are both neutral, as induced charge vanishes once the rod is gone
   A student who thinks induced charge always disappears when the rod is removed picks this. That happens when the spheres are still touching. Here they were separated first, so B's extra electrons have no path back to A.
4. DA is positive and B is negative, with charges of equal magnitude Correct
   The rod's excess electrons repel free electrons from A into B. When B is moved away, those electrons are trapped on B. The pair started neutral and no charge crossed to or from it, so A's positive charge and B's negative charge are equal in magnitude.

CED 8.2.A.1.iii · Read this in Fix

1. AIncreases by 2qm_e/e
   A student who takes the final charge to be the charge transferred, ignoring the sphere's initial +q, picks this. The change in charge is −2q − (+q) = −3q, so 3q/e electrons are gained.
2. BDecreases by 3qm_e/e
   A student who thinks becoming more negative means losing electrons picks this. Electrons are negative, so the sphere's charge became more negative because it gained 3q/e electrons, and its mass increased.
3. CIncreases by 3qm_e/e Correct
   The sphere's charge changes by −2q − (+q) = −3q. In a solid the charge is carried by electrons, each with charge −e, so the sphere gains 3q/e electrons. Its mass increases by (3q/e)m_e = 3qm_e/e.
4. DDecreases by 3qm_p/e
   A student who thinks the sphere became negative by losing protons picks this. In a metal the protons are held in the nuclei; the sphere gained 3q/e electrons, each of mass m_e.

Working Δq = q_final − q_initial = −2q − (+q) = −3q. Charge is carried by electrons (charge −e each): number gained N = 3q/e. Δm = N m_e = 3qm_e/e, an increase. Distractors: final charge taken as the transfer → 2q/e electrons → 2qm_e/e; 'more negative = lost electrons' → decrease 3qm_e/e; protons lost → decrease 3qm_p/e.

CED 8.2.A.2.i · Read this in Fix

1. A+(3/4)Q
   A student who adds B's and C's charges as amounts, Q/2 + Q = 3Q/2, and shares that picks this. With signs the total is +Q/2 − Q = −Q/2, so each sphere ends with −Q/4.
2. B−(1/2)Q
   A student who thinks charge flows only until B is neutral picks this: B's +Q/2 cancels half of C's −Q. The flow continues until B and C have equal charges, −Q/4 each.
3. C−(3/4)Q
   A student who thinks B, the sphere touched to C, passes half of its own charge, +Q/4, to C whatever C carries picks this: −Q + Q/4 = −3Q/4. B and C share their total, −Q/2, so each ends with −Q/4.
4. D−(1/4)Q Correct
   First contact: A and B share +Q, so each has +Q/2. Second contact: B and C have a total of +Q/2 + (−Q) = −Q/2, shared equally, so each ends with −Q/4. C's final charge is −(1/4)Q.

Working Contact 1 (A, B): total +Q + 0 = +Q → +Q/2 each. Contact 2 (B, C): total +Q/2 + (−Q) = −Q/2 → −Q/4 each. C = −Q/4. Distractors: magnitudes (Q/2 + Q)/2 = +3Q/4; flow stops when B is neutral → C = −Q + Q/2 = −Q/2; B gives half its charge → C = −Q + Q/4 = −3Q/4.

CED 8.2.A.2.ii · Read this in Fix

1. A+3.0 nC Correct
   X and Y form an isolated system, so q_X + q_Y stays constant: at t = 0 it is +6.0 nC + (−2.0 nC) = +4.0 nC. At t = 4.0 s the graph gives q_X = +1.0 nC, so q_Y = +4.0 nC − 1.0 nC = +3.0 nC. The 5.0 nC that X lost, Y gained.
2. B−1.0 nC
   A student who thinks two objects that exchange charge end with equal and opposite charges picks this. That holds only if their total is zero; here the total is +4.0 nC, so Y has +3.0 nC.
3. C−2.0 nC
   A student who thinks the charge X lost simply disappeared picks this. X and Y exchange charge only with each other, so the 5.0 nC that X lost went to Y.
4. D+5.0 nC
   A student who takes Y's final charge to be the charge it received, ignoring its initial −2.0 nC, picks this. Y's charge is −2.0 nC + 5.0 nC = +3.0 nC.

Working Isolated pair: q_X + q_Y = constant = +6.0 nC + (−2.0 nC) = +4.0 nC (graph: q_X(0) = +6.0 nC). At t = 4.0 s, graph gives q_X = +1.0 nC, so q_Y = +4.0 − 1.0 = +3.0 nC.

CED 8.2.A.2.ii · Read this in Fix

1. APositive, because the rod pulled positive charge up from the ground through the wire
   A student who thinks positive charges move in metals and in the ground picks this. The protons are held in nuclei; the sphere becomes positive because electrons leave it through the wire.
2. BNeutral, because connecting the sphere to the ground removed any charge it may have had
   A student who thinks grounding always leaves an object neutral picks this. With the negative rod nearby, electrons are pushed off the sphere to the ground, and removing the wire first keeps them from coming back.
3. CPositive, because the rod pushed electrons off the sphere through the wire into the ground Correct
   The rod is negative, as its minus signs show. It repels free electrons in the sphere, and the wire lets some of them escape to the ground. Breaking the connection before removing the rod traps the sphere with fewer electrons than protons, so it is left positive.
4. DNegative, because electrons passed from the rod to the sphere while the rod was held near
   A student who thinks a nearby charged object passes charge across without contact picks this. The rod never touches the sphere; the sphere loses its own electrons to the ground and ends up positive.

CED 8.2.A.3 · Read this in Fix

1. AThe balloon exchanged charge with its surroundings, with a net −5.0 nC leaving it. Correct
   The balloon's net charge can change only by transfer across its boundary. It became 5.0 nC less negative, so a net −5.0 nC passed to its surroundings, for example to moisture in the air; that charge still exists outside the balloon.
2. BA charge of −5.0 nC wore off the balloon overnight and no longer exists anywhere.
   A student who thinks charge wears off and disappears picks this. Charge is not destroyed; the −5.0 nC left the balloon for its surroundings, such as the damp air.
3. CThe balloon’s own positive charges cancelled 5.0 nC of its excess electrons.
   A student who thinks an object's own opposite charges can cancel its excess picks this. The balloon's protons and electrons are all counted in its net charge; rearranging them cannot change it. Only a transfer can.
4. DThe rubbing made the charge, so most of that charge could later be unmade again.
   A student who thinks rubbing creates charge picks this. Rubbing only moved electrons from the sweater to the balloon, and the later decrease is another transfer, from the balloon to its surroundings.

Working Conservation: the balloon's net charge changes only by transfer. Δq = −1.0 nC − (−6.0 nC) = +5.0 nC, so a net −5.0 nC left the balloon for its surroundings (for example, electrons to moist air); the charge still exists there.

CED 8.2.A.2 · Read this in Fix