9.1 Electric Potential Energy

1. AIt equals the work the electric force does on the spheres as they are brought from infinitely far apart to distance r.
   A student who equates the potential energy with the work done by the electric force picks this. The electric force does work W_E = −ΔU_E: for like charges pushed together, the electric force does negative work while the potential energy increases.
2. BIt is stored in whichever sphere was moved last, not in the system made up of both of the spheres.
   A student who thinks potential energy belongs to one object picks this. Electric potential energy belongs to the interacting pair; it depends on both charges and their separation, whichever sphere was moved.
3. CIt equals the work an external force must do to bring the spheres from infinitely far apart to distance r. Correct
   With zero potential energy at infinite separation, the potential energy of the pair is the work an external force must do to assemble it: to bring the spheres from infinitely far apart to their positions without giving them kinetic energy. It belongs to the system of both spheres.
4. DIt equals the electric potential that one of the spheres produces at the location of the other sphere.
   A student who confuses electric potential with electric potential energy picks this. The potential produced by one sphere at the other, kq₁/r, is in volts; the potential energy of the pair, kq₁q₂/r, is in joules and depends on both charges.

CED 9.1.A.1 · Read this in Fix

1. AThe particles repel each other, so they tend to move farther apart if they are released.
   A student who thinks a negative potential energy means repulsion picks this. Opposite charges attract; it is a repelling pair of like charges that has positive potential energy.
2. BThe potential energy points from the positive particle toward the negative particle.
   A student who treats potential energy as a vector picks this. Energy is a scalar: the negative sign compares the pair’s energy with its energy at infinite separation and gives no direction.
3. CAn external force must do positive work to pull the particles infinitely far apart. Correct
   With U = 0 at infinite separation, U_E is the work an external force does to assemble the pair. For opposite charges that work is negative, since their attraction pulls them together; separating them again takes +kq²/r of positive work.
4. DThe potential energy is stored in the negative particle, not in the positive particle.
   A student who thinks potential energy belongs to one of the particles picks this. Electric potential energy belongs to the pair as a system; its sign comes from the product q₁q₂, not from either particle alone.

CED 9.1.A.2 · Read this in Fix

1. A4.00 kq²/s
   A student who counts only neighboring particles picks this, leaving out the two diagonal pairs. Every pair of charges interacts, so the diagonal pairs add 2kq²/(√2 s) = √2 kq²/s.
2. B10.8 kq²/s
   A student who adds, for each particle, its potential energy with each of the other three picks this, which counts every pair twice. Each pair has one potential energy, kq²/r, so there are six terms, not twelve.
3. C2.71 kq²/s
   A student who takes the system’s potential energy to be that of one particle with the other three picks this: kq²(2/s + 1/(√2 s)). That leaves out the three pairs that do not include the chosen particle.
4. D5.41 kq²/s Correct
   Each pair of particles is counted once: four pairs along the sides, a distance s apart, and two pairs along the diagonals, √2 s apart. U = kq²(4/s + 2/(√2 s)) = (4 + √2)kq²/s ≈ 5.41 kq²/s.

Working Six pairs: four along the sides, a distance s apart, and two along the diagonals, √2 s apart. U = kq²(4/s + 2/(√2 s)) = (4 + √2)kq²/s ≈ 5.41 kq²/s. Distractors: sides only (diagonal pairs left out), 4.00 kq²/s; each pair counted twice (twelve terms), 2(4 + √2) ≈ 10.8 kq²/s; one particle with the other three only, 2 + 1/√2 ≈ 2.71 kq²/s.

CED 9.1.A.3 · Read this in Fix

1. A5.4 × 10⁻⁴ J
   A student who takes the work to be the potential energy at the final position, kq₁q₂/(0.10 m), picks this. The pair already had potential energy kq₁q₂/(0.30 m) at the start; the work is the change, U_final − U_initial.
2. B3.6 × 10⁻⁴ J Correct
   With no change in kinetic energy, the external force’s work equals the change in the pair’s potential energy: W = kq₁q₂(1/r_f − 1/r_i) = (9.0 × 10⁹)(3.0 × 10⁻⁶ C)(2.0 × 10⁻⁹ C)(1/0.10 m − 1/0.30 m) = 3.6 × 10⁻⁴ J. It is positive because the like charges repel, so the external force pushes along the displacement.
3. C4.8 × 10⁻³ J
   A student who uses Coulomb’s-law form, kq₁q₂/r², for the potential energy picks this: kq₁q₂(1/0.10² − 1/0.30²). That expression gives a force in newtons; the potential energy is kq₁q₂/r, in joules.
4. D7.2 × 10⁻⁴ J
   A student who gives each sphere its own potential energy, kq₁q₂/r, and so doubles the change picks this. The potential energy belongs to the pair as a whole, and there is one term, kq₁q₂/r, for the pair.

Working With no change in kinetic energy, W_ext = ΔU_E = kq₁q₂(1/r_f − 1/r_i) = (9.0 × 10⁹)(3.0 × 10⁻⁶ C)(2.0 × 10⁻⁹ C)(1/0.10 m − 1/0.30 m) = (5.4 × 10⁻⁵ J·m)(6.67 m⁻¹) = 3.6 × 10⁻⁴ J. Distractors: potential energy at the final position only, kq₁q₂/r_f = 5.4 × 10⁻⁴ J; Coulomb’s-law form, kq₁q₂(1/r_f² − 1/r_i²) = 4.8 × 10⁻³; each sphere given its own change, 2ΔU = 7.2 × 10⁻⁴ J.

CED 9.1.A.1 · Read this in Fix

1. A−3kq²/(2d) Correct
   Add the potential energy of each pair once, with the signs of the charges. The two neighboring pairs are each (+q)(−q) at distance d, −kq²/d each; the outer pair is (+q)(+q) at 2d, +kq²/(2d). U = −2kq²/d + kq²/(2d) = −3kq²/(2d).
2. B5kq²/(2d)
   A student who treats every pair’s potential energy as positive, using the magnitudes of the charges, picks this: kq²/d + kq²/d + kq²/(2d). A pair of opposite charges has negative potential energy, because positive work is needed to pull them apart.
3. C−2kq²/d
   A student who thinks the middle particle blocks the interaction between the two outer particles picks this, leaving out the pair 2d apart. The outer particles still interact, and their pair adds +kq²/(2d).
4. D−7kq²/(4d²)
   A student who uses Coulomb’s-law form, kq₁q₂/r², for each pair picks this: −kq²/d² − kq²/d² + kq²/(4d²). That expression gives forces; potential energy is kq₁q₂/r, in joules.

Working Three pairs: (+q, −q) at d, (−q, +q) at d and (+q, +q) at 2d. U = k[(q)(−q)/d + (−q)(q)/d + (q)(q)/(2d)] = −2kq²/d + kq²/(2d) = −3kq²/(2d). Distractors: signs ignored, kq²(1/d + 1/d + 1/(2d)) = 5kq²/(2d); outer pair left out, −2kq²/d; Coulomb’s-law form kq₁q₂/r², −2kq²/d² + kq²/(4d²) = −7kq²/(4d²).

CED 9.1.A.3 · Read this in Fix

1. A−0.91 kq²/a
   A student who takes the work to be the total potential energy of the final three-particle system picks this, including the fixed pair’s +kq²/(2a). That pair’s potential energy was there before the third particle moved, so it is not part of this work.
2. B−1.41 kq²/a Correct
   The external force’s work equals the change in the system’s potential energy. The fixed pair’s own potential energy does not change, so only the two new pairs count: each is (+q)(−q) at distance √2 a, so W = 2k(q)(−q)/(√2 a) = −√2 kq²/a ≈ −1.41 kq²/a. It is negative because the attraction pulls the −q particle in and the external force holds it back.
3. C−2.00 kq²/a
   A student who uses a, the distance of each fixed particle from O, instead of its distance from P picks this. In kq₁q₂/r, r is the separation of the two particles in the pair, here √2 a.
4. D−2.83 kq²/a
   A student who counts each new pair twice, once for each particle in it, picks this. Each pair has a single potential energy, k(q)(−q)/(√2 a), so the two new pairs give −√2 kq²/a.

Working Each fixed particle is √(a² + a²) = √2 a from P. W_ext = U_final − U_initial; the fixed pair’s kq²/(2a) is the same before and after, so only the two new pairs count: W_ext = 2k(q)(−q)/(√2 a) = −√2 kq²/a ≈ −1.41 kq²/a. Distractors: total final U including the fixed pair, −√2 kq²/a + kq²/(2a) ≈ −0.91 kq²/a; distance a (from O) used, −2kq²/a = −2.00 kq²/a; each new pair counted twice, −2√2 kq²/a ≈ −2.83 kq²/a.

CED 9.1.A.1 · Read this in Fix

1. A+0.12 J Correct
   The initial potential energy is the sum over the three pairs, all 0.30 m apart: one (+,+) pair and two (+,−) pairs, so U = −kq²/s = −(9.0 × 10⁹)(2.0 × 10⁻⁶ C)²/(0.30 m) = −0.12 J. Far apart, U = 0. The external force’s work equals the change in potential energy: 0 − (−0.12 J) = +0.12 J.
2. B−0.12 J
   A student who takes the work needed to separate the particles to be their potential energy picks this. The potential energy, −0.12 J, is the work needed to assemble them from far apart; separating them reverses the process, so the work is +0.12 J.
3. C−0.36 J
   A student who adds the three pairs’ potential energies as positive quantities picks this: +3kq²/s = +0.36 J, which separating the particles would appear to release. Two of the pairs have opposite charges and negative potential energy; the total is −0.12 J.
4. D+0.40 J
   A student who uses Coulomb’s-law form, kq₁q₂/r², for each pair picks this: the pairs then total −kq²/s², of magnitude 0.40, and the work appears to be +0.40 J. That expression gives forces in newtons; potential energy is kq₁q₂/r.

Working U_initial = k[(q)(q) + (q)(−q) + (q)(−q)]/s = −kq²/s = −(9.0 × 10⁹)(2.0 × 10⁻⁶ C)²/(0.30 m) = −0.12 J; U_final = 0. W_ext = U_final − U_initial = +0.12 J. Distractors: work set equal to U itself, −0.12 J; signs ignored (U = +3kq²/s = +0.36 J), W = −0.36 J; Coulomb’s-law form (U = −kq²/s², magnitude 0.40), W = +0.40 J.

CED 9.1.A.3 · Read this in Fix

1. AU₃ > U₁ = U₂
   A student who ignores the signs of the charges picks this: Pairs 1 and 2 both get kq²/d, and Pair 3 gets 3kq²/(2d) = 1.5kq²/d, the largest. Pairs 1 and 3 have opposite charges, so their potential energies are negative: −kq²/d and −1.5kq²/d.
2. BU₂ > U₃ > U₁
   A student who uses Coulomb’s-law form, kq₁q₂/r², picks this: Pair 3 would get −3kq²/(4d²), less negative than Pair 1’s −kq²/d². Potential energy is proportional to 1/r, so doubling the distance only halves it: Pair 3 has −1.5kq²/d, below Pair 1’s −kq²/d.
3. CU₂ > U₁ > U₃ Correct
   U = kq₁q₂/r, with the signs of the charges. Pair 2, like charges: +kq²/d, the highest. Pair 1, opposite charges: −kq²/d. Pair 3: k(−3q)(q)/(2d) = −1.5kq²/d, the lowest. So U₂ > U₁ > U₃.
4. DU₃ > U₁ > U₂
   A student who thinks attracting pairs have positive potential energy and repelling pairs negative picks this, reversing every sign: Pair 3 +1.5kq²/d, Pair 1 +kq²/d, Pair 2 −kq²/d. Positive work is needed to push like charges together, so a repelling pair has positive potential energy and an attracting pair negative.

Working U = kq₁q₂/r with signs. Pair 1: −kq²/d. Pair 2: +kq²/d. Pair 3: k(−3q)(q)/(2d) = −1.5kq²/d. So U₂ > U₁ > U₃. Distractors: magnitudes (1, 1, 1.5) → U₃ > U₁ = U₂; Coulomb’s-law form with signs (−1, +1, −0.75) → U₂ > U₃ > U₁; signs reversed (+1, −1, +1.5) → U₃ > U₁ > U₂.

CED 9.1.A.2 · Read this in Fix

1. A+1.50
   A student who ignores the sign of the new charge picks this: 3 × 1/2 = 1.50. A pair of opposite charges has negative potential energy, so U₂ is negative while U₁ is positive.
2. B+0.75
   A student who uses the force expression k|q₁q₂|/r² in place of the potential energy picks this: 3 × 1/4 = 0.75. Potential energy is proportional to 1/r, not 1/r², and it carries the sign of q₁q₂.
3. C−6.00
   A student who thinks potential energy grows in proportion to the particles’ separation picks this: (−3)(2) = −6.00. For point charges U = kq₁q₂/r, so doubling r halves U.
4. D−1.50 Correct
   U = kq₁q₂/r. Replacing q₁ by −3q₁ multiplies U by −3, and doubling the separation multiplies it by 1/2, so U₂/U₁ = (−3)(1/2) = −1.50. The new pair attracts, so its potential energy is negative.

Working U ∝ q₁q₂/r, so U₂/U₁ = (−3)(1/2) = −1.50. Distractors: sign of the new charge ignored, +1.50; force expression k|q₁q₂|/r² used, 3 × 1/4 = +0.75; U taken to grow in proportion to the separation, (−3)(2) = −6.00.

CED 9.1.A.2 · Read this in Fix

1. ANegative: the attraction pulls it in, so the external force acts opposite to its displacement. Correct
   The fixed −Q attracts the +q particle, so to keep it from speeding up the external force must pull back on it, opposite to its displacement: its work is negative. It equals the change in potential energy, U = −kQq/r − 0 < 0.
2. BPositive: an external force does positive work whenever it moves an object from one place to another.
   A student who thinks the work done to move an object is always positive picks this. Work is F⃗ · d⃗; here the external force points away from the fixed particle while the displacement is toward it, so the work is negative.
3. CPositive: the potential energy of any pair of charges increases as the charges get closer together.
   A student who thinks potential energy always increases as charges approach picks this. That is true for like charges; for opposite charges U = −kQq/r becomes more negative as r decreases, so the external work, ΔU, is negative.
4. DZero: the particle starts and ends at rest, so the external force does no work on it.
   A student who confuses the external force’s work with the net work picks this. The net work is zero because the kinetic energy does not change, but the electric force does positive work and the external force does an equal amount of negative work.

CED 9.1.A.1 · Read this in Fix

1. A−0.33 q
   A student who writes each pair’s potential energy as kq₁q₂/r², like the force, picks this: 3kq²/s² + 9kqQ/s² = 0. Potential energy falls off as 1/r, so each corner–center pair contributes kqQ√3/s, not 3kqQ/s².
2. B−1.73 q
   A student who takes a pair’s potential energy to grow in proportion to its separation, as mgh grows with height, picks this: 3kq²s + √3kqQs = 0. The potential energy of a pair is kq₁q₂/r, which is inversely proportional to the separation.
3. C−0.58 q Correct
   There are six pairs, each counted once. The three corner–corner pairs, a distance s apart, contribute 3kq²/s; the three corner–center pairs, s/√3 apart, contribute 3√3kqQ/s. Setting the sum to zero gives Q = −q/√3 ≈ −0.58 q.
4. D−1.00 q
   A student who uses each particle’s distance from the center, s/√3, as r for every pair, the corner–corner pairs included, picks this. In kq₁q₂/r, r is the separation of the two particles in the pair, which is s for two corners.

Working Six pairs, each counted once. Three corner–corner pairs, each a distance s apart: 3kq²/s. Three corner–center pairs, each s/√3 apart: 3kqQ/(s/√3) = 3√3kqQ/s. U = 3kq²/s + 3√3kqQ/s = 0 gives Q = −q/√3 ≈ −0.58 q. Distractors (sympy-checked): kq₁q₂/r² used, 3kq²/s² + 9kqQ/s² = 0, Q = −q/3 ≈ −0.33 q; pair energy taken to grow in proportion to separation, 3kq²s + √3kqQs = 0, Q = −√3q ≈ −1.73 q; every pair’s r taken as the distance s/√3 from the center, 3√3kq²/s + 3√3kqQ/s = 0, Q = −1.00 q.

CED 9.1.A.3 · Read this in Fix