10.2 Conservation of Electric Charge and the Process of Charging

1. AThe whole electroscope has now become negatively charged, like the rod.
   A student who thinks an object near a charged rod takes on the rod's sign picks this. The rod does not touch the electroscope, so no charge passes to it and its net charge stays zero. The rod only moves electrons within it: the leaves become negative and the knob becomes positive by the same amount.
2. BThe rod's presence has made new negative charge appear on the leaves.
   A student who thinks charging makes new charge picks this. Charge is never created. The negative charge on the leaves consists of electrons that were already in the electroscope and have been pushed down from the knob, which is left positive by the same amount.
3. CElectrons have been pushed from the knob down onto the pair of leaves. Correct
   The rod's negative charge repels the free electrons in the metal. Some move from the knob down to the leaves, so the leaves carry excess negative charge and repel each other, while the knob is left with an equal positive charge. No charge enters or leaves the electroscope, so its net charge is still zero.
4. DPositive charges have moved up from the leaves into the knob.
   A student who thinks positive charges move in a metal picks this. The protons are fixed in the atomic nuclei; only electrons move. The knob becomes positive because electrons have left it for the leaves, not because positive charge has arrived.

CED 10.2.A.1.ii · Read this in Fix

1. AProtons moved from the rod across to the wool.
   A student who thinks positive charges move during charging picks this. Protons are held in the atomic nuclei and do not move from one solid to another when they are rubbed; the rod became negative by gaining electrons.
2. BRubbing made new electrons on the rod's surface.
   A student who thinks rubbing makes charge picks this. Charge is never created: the extra electrons on the rod came from the wool, which is left positively charged by the same amount.
3. CElectrons moved off the rod and over to the wool.
   A student who links 'negative' with losing something picks this. Electrons are negative, so an object that loses them becomes positive. The rod became negative, so it must have gained electrons, from the wool.
4. DElectrons moved from the wool onto the rod. Correct
   Charging a solid moves electrons. The rod ends up with more electrons than protons, so it is negative; the wool has lost the same number of electrons, so it is positive by the same amount.

CED 10.2.A.2.i · Read this in Fix

1. APositive charge flows off the sphere and down the wire into the Earth.
   A student who thinks positive charges move in metals picks this. The protons stay in the nuclei of the sphere's atoms; the sphere loses its positive charge because electrons flow onto it from Earth.
2. BCharge flows until the sphere and the Earth hold equal amounts of charge.
   A student who thinks objects in contact always end with equal charges picks this. Equal sharing holds only for identical conductors. Earth is enormous compared with the sphere, so the sphere ends nearly neutral and Earth's charge barely changes.
3. CNo charge flows, as the neutral Earth exerts no electric force.
   A student who thinks a neutral object cannot interact with a charged one picks this. Earth contains vast numbers of electrons, and the positive sphere attracts them: once the wire connects the two, electrons flow onto the sphere.
4. DElectrons flow from Earth onto the sphere until it is almost neutral. Correct
   The sphere has fewer electrons than protons. When it is connected to Earth, a very large, approximately neutral object, electrons flow up the wire onto the sphere until almost none of its excess charge is left. Earth's own charge hardly changes, because Earth is so large.

CED 10.2.A.3 · Read this in Fix

1. A+5.0 nC
   A student who adds the charges' sizes and ignores their signs picks this: (8.0 nC + 2.0 nC)/2 = 5.0 nC. Y's charge is negative, so it reduces the total: (+8.0 nC − 2.0 nC)/2 = +3.0 nC.
2. B+3.0 nC Correct
   The total charge is (+8.0 nC) + (−2.0 nC) = +6.0 nC. Charge is conserved, and two identical conducting spheres share it equally, so each ends with +6.0 nC/2 = +3.0 nC. Electrons moved from Y to X until the charges were equal.
3. C+6.0 nC
   A student who thinks each sphere ends up with the whole charge picks this. The +6.0 nC total is shared, not copied: if each sphere had +6.0 nC, the pair would have +12.0 nC, more than it started with. Each has +3.0 nC.
4. D−2.0 nC
   A student who thinks charge moves only when objects are rubbed picks this. Charge moves between conductors that simply touch: here electrons move from Y to X until the two identical spheres carry equal charges.

Working Total charge before contact: (+8.0 nC) + (−2.0 nC) = +6.0 nC. Charge is conserved, and identical conducting spheres share the total equally: each ends with (+6.0 nC)/2 = +3.0 nC.

CED 10.2.A.1.i · Read this in Fix

1. A(8/9)kQq/r², toward the rod Correct
   The rod attracts −q with a force of magnitude kQq/r² toward the rod and repels +q with a force of magnitude kQq/(3r)² = kQq/(9r²) away from it. The forces are opposite, so the net force is kQq/r² − kQq/(9r²) = (8/9)kQq/r², toward the rod: the neutral sphere is attracted because its opposite charge is nearer.
2. B(10/9)kQq/r², toward the rod
   A student who adds the two forces' magnitudes picks this: kQq/r² + kQq/(9r²) = (10/9)kQq/r². The force on −q points toward the rod and the force on +q points away from it, so they subtract.
3. C2kQq/(3r²), toward the rod
   A student who thinks the force falls off as 1/r rather than 1/r² picks this: at three times the distance they take the force on +q to be one-third of kQq/r², so the net force is kQq/r² − (1/3)kQq/r² = 2kQq/(3r²). Coulomb's law is an inverse-square law, so the force on +q is kQq/(3r)² = kQq/(9r²), one-ninth of the near force; the net force is (8/9)kQq/r².
4. DZero, as the two forces cancel
   A student who thinks a neutral object feels no net electric force picks this. The induced charges are equal and opposite, but −q is nearer to the rod than +q, so the attraction on −q is larger than the repulsion on +q and the sphere is pulled toward the rod.

Working Force on −q (attraction, toward the rod): kQq/r². Force on +q (repulsion, away from the rod): kQq/(3r)² = kQq/(9r²). Opposite directions, so |F_net| = kQq/r² − kQq/(9r²) = (8/9)kQq/r², directed toward the rod.

CED 10.2.A.1.iii · Read this in Fix

1. A+2.0 nC
   A student who thinks contact, like rubbing, leaves the two objects with equal and opposite charges picks this, giving the sphere the opposite of the rod's −2.0 nC. The total would then be zero, not the −6.0 nC that existed. The charge came from the negative rod, so the sphere's charge has the rod's sign: q_sphere = −6.0 nC − (−2.0 nC) = −4.0 nC.
2. B−2.0 nC
   A student who thinks objects that touch always end with equal charges picks this. Equal sharing holds only for identical conductors; here the charges must add up to the original −6.0 nC, so the sphere has −6.0 nC − (−2.0 nC) = −4.0 nC.
3. C−4.0 nC Correct
   The rod and sphere together form an isolated system, so their total charge stays −6.0 nC. After contact, −2.0 nC + q_sphere = −6.0 nC, so q_sphere = −4.0 nC: the rod gave 4.0 nC of negative charge (electrons) to the sphere.
4. D+4.0 nC
   A student who takes the change in the rod's charge, −2.0 nC − (−6.0 nC) = +4.0 nC, as the sphere's charge picks this. Conservation of charge makes the sphere's change the opposite of the rod's: the rod's charge rose by 4.0 nC, so the sphere's fell by 4.0 nC, from 0 to −4.0 nC.

Working The rod and sphere form an isolated system: total charge before = (−6.0 nC) + 0 = −6.0 nC. After: (−2.0 nC) + q_sphere = −6.0 nC, so q_sphere = −6.0 nC − (−2.0 nC) = −4.0 nC.

CED 10.2.A.2.ii · Read this in Fix

1. ANegative, the same sign as the rod's charge
   A student who thinks a nearby rod gives an object its own sign picks this. No charge passes between the rod and the sphere. The rod pushes electrons out of the sphere into Earth, so the sphere is left positive, opposite to the rod.
2. BPositive, spread over the whole surface Correct
   In step 1 the rod repels electrons from the sphere through the wire into Earth, so the sphere has fewer electrons than protons. In step 2 the wire is removed, so those electrons cannot return. In step 3 the rod is taken away, and the sphere's positive charge, no longer held near the rod, spreads out over its surface.
3. CNeutral, as grounding removed its excess charge
   A student who thinks grounding always leaves an object neutral picks this. While the rod was near, electrons were pushed off the sphere into Earth, and the wire was removed before the rod was taken away, so they could not flow back. The sphere keeps a positive charge.
4. DPositive, on the side that had faced the rod
   A student who thinks charge stays where it gathered picks this. The positive charge collected on the near side only while the rod held it there. Once the rod is removed, nothing holds it, and the like charges push one another apart until they are spread over the surface.

CED 10.2.A.3 · Read this in Fix

1. ABoth are attracted, the metal sphere more strongly than the plastic one. Correct
   Both spheres are polarized, so both are attracted. In the metal, electrons move across the whole sphere toward the rod, giving a large separation of charge; in the plastic, the charges of each molecule shift only slightly. The larger induced charges on the metal sphere give it the larger attraction.
2. BNeither is attracted, since neither has a charge opposite to that of the rod.
   A student who thinks only an oppositely charged object can be attracted picks this. A charged rod attracts neutral objects too: it polarizes them, and the induced charge nearer the rod, which is opposite to the rod's, is attracted more strongly than the farther charge is repelled.
3. CThe metal sphere is attracted, and the plastic sphere feels no force.
   A student who thinks an insulator cannot be polarized picks this. Electrons cannot move through the plastic, but within each molecule they shift slightly toward the rod, so the plastic sphere is polarized and attracted too, though less strongly.
4. DBoth are attracted equally, as electrons flow through each alike.
   A student who thinks electrons flow through an insulator as through a metal picks this. In the plastic, electrons stay with their own molecules and shift only slightly, so its induced charges, and the force on it, are smaller than for the metal sphere.

CED 10.2.A.1.ii · Read this in Fix

1. AIt is negative, since the rubbing produced negative charge on the balloon.
   A student who thinks rubbing makes new charge picks this. The balloon's extra electrons came from the sweater, which is left positive by the same amount; no charge was produced, so the total is still zero.
2. BIt is positive, since some electrons were used up as they were transferred.
   A student who thinks charge is used up when it moves picks this. Charge is conserved: every electron that left the sweater arrived on the balloon, so the balloon's negative charge equals the sweater's positive charge in size.
3. CIt is twice the balloon's charge, as both parts are now charged.
   A student who adds the charges' sizes without their signs picks this. The balloon's charge is negative and the sweater's is positive and equal in size, so they add to zero.
4. DIt is zero, since electrons only moved from one part of the system to another. Correct
   The system's net charge can change only if charge crosses its boundary. The rubbing moved electrons from the sweater to the balloon, both inside the system, so the balloon's negative charge is exactly matched by the sweater's positive charge and the system's net charge is still zero.

CED 10.2.A.2 · Read this in Fix

1. A+3.00Q
   A student who thinks each sphere in a contact takes the pair's whole charge picks this: Y gets −Q, and then Y and Z each get −Q + 4Q = +3Q. The spheres would then carry −Q + 3Q + 3Q = +5Q in all, more than the +3Q they started with. The charge is shared, and Y ends with +1.75Q.
2. B+1.75Q Correct
   At the first contact the total, −Q, is shared equally, so Y gets −Q/2. At the second contact Y and Z have a total of −Q/2 + 4Q = +3.5Q, which they share equally, so Y ends with +1.75Q. Z also has +1.75Q and X keeps −Q/2, so the three charges still add to +3Q.
3. C+2.25Q
   A student who adds the sizes of the charges at the second contact, ignoring that Y's charge is negative, picks this: (0.50Q + 4Q)/2 = 2.25Q. Y's −Q/2 cancels part of Z's +4Q, so the pair's total is +3.5Q and each gets +1.75Q.
4. D+1.00Q
   A student who shares the total charge of all three spheres, +3Q, equally among them picks this. The three spheres never touch at the same time: X is moved away after the first contact and keeps −Q/2, so the second contact shares only the charge of Y and Z, and Y ends with +1.75Q.

Working Identical conducting spheres in contact share their total charge equally, and charge is conserved. X (−Q) with Y (0): total −Q, each −Q/2. Y (−Q/2) with Z (+4Q): total +3.5Q, each +1.75Q = +7Q/4. Check: X −0.5Q, Y +1.75Q, Z +1.75Q, total +3Q as before. Errors: charge copied → Y takes −Q, then Y and Z each take −Q + 4Q = +3.00Q; signs ignored at the second contact → (0.5Q + 4Q)/2 = +2.25Q; all three shared at once → (−Q + 0 + 4Q)/3 = +1.00Q.

CED 10.2.A.2.ii · Read this in Fix

1. A3.20F
   A student who thinks each sphere takes the pair's whole charge, −4Q, picks this: k(4Q)(4Q)/r² = 16kQ²/r² = 3.20F. The two spheres would then hold −8Q together, twice what they started with. Each has −2Q, which gives 0.80F.
2. B1.80F
   A student who adds the sizes of the charges, 6Q, and shares that picks this: each gets 3Q, and k(3Q)(3Q)/r² = 9kQ²/r² = 1.80F. The charges have opposite signs, so the total is −4Q, each sphere gets −2Q, and the force is 0.80F.
3. C1.00F
   A student who thinks conductors that only touch keep their own charges picks this: nothing changes, so the force is still F. Charge moves between conductors in contact: electrons move from the −5Q sphere to the +Q sphere until each has −2Q, and the force falls to 0.80F.
4. D0.80F Correct
   Initially F = k(Q)(5Q)/r² = 5kQ²/r². When the identical spheres touch, their total charge, +Q − 5Q = −4Q, is shared equally, so each has −2Q. The new force has magnitude k(2Q)(2Q)/r² = 4kQ²/r² = (4/5)F = 0.80F, and it is now repulsive.

Working Before: F = k(Q)(5Q)/r² = 5kQ²/r². Contact: total +Q − 5Q = −4Q, shared equally, −2Q each. After: k(2Q)(2Q)/r² = 4kQ²/r² = (4/5)F = 0.80F (now repulsive). Errors: charge copied, −4Q each → 16kQ²/r² = 3.20F; magnitudes added, 6Q shared, 3Q each → 9kQ²/r² = 1.80F; no charge moves on touching → 1.00F.

CED 10.2.A.1.i · Read this in Fix

1. A7.5 × 10¹⁰
   A student who thinks each sphere ends with the pair's whole net charge, −10 nC, picks this: X would then change by 12 nC, and 12 nC/e = 7.5 × 10¹⁰. The two spheres would hold −20 nC together, twice the −10 nC they started with. They share it, −5.0 nC each, so X changes by 7.0 nC and 4.4 × 10¹⁰ electrons move.
2. B3.1 × 10¹⁰
   A student who adds the sizes of the charges, 2.0 nC + 12 nC = 14 nC, ignoring the minus sign, picks this: each sphere would get 7.0 nC, so X would change by 5.0 nC, and 5.0 nC/e = 3.1 × 10¹⁰. Y's negative charge cancels part of X's positive charge: the total is −10 nC, each sphere ends with −5.0 nC, and X changes by 7.0 nC.
3. C4.4 × 10¹⁰ Correct
   The total charge, (+2.0 nC) + (−12 nC) = −10 nC, is shared equally, so each sphere ends with −5.0 nC. X goes from +2.0 nC to −5.0 nC, a change of −7.0 nC (and Y goes from −12 nC to −5.0 nC, a change of +7.0 nC). X gains that charge as electrons that come from Y: (7.0 × 10⁻⁹ C)/(1.60 × 10⁻¹⁹ C) = 4.4 × 10¹⁰ electrons.
4. D1.9 × 10¹⁰
   A student who finds X's change by comparing sizes, 5.0 nC − 2.0 nC = 3.0 nC, ignoring the signs, picks this: 3.0 nC/e = 1.9 × 10¹⁰. X goes from +2.0 nC to −5.0 nC, passing through zero, so its change is (−5.0 nC) − (+2.0 nC) = −7.0 nC, and 4.4 × 10¹⁰ electrons move.

Working Charge is conserved: total = (+2.0 nC) + (−12 nC) = −10 nC. Identical conducting spheres share it equally: −5.0 nC each. X changes from +2.0 nC to −5.0 nC, Δq_X = −7.0 nC (Y changes by +7.0 nC, from −12 nC to −5.0 nC). The charge is carried by electrons moving from Y to X: N = (7.0 × 10⁻⁹ C)/(1.60 × 10⁻¹⁹ C) = 4.4 × 10¹⁰. Errors: charge copied, each −10 nC (m03): X changes by 12 nC, 7.5 × 10¹⁰; magnitudes added, (2.0 + 12)/2 = 7.0 nC each (m02): X changes by 5.0 nC, 3.1 × 10¹⁰; change taken as difference of magnitudes, 5.0 − 2.0 = 3.0 nC (new): 1.9 × 10¹⁰.

CED 10.2.A.2.i · Read this in Fix