10.4 Electric Potential Energy

1. AU_E of the two-sphere system equals the positive work done by the external force. Correct
   U_E is zero when the spheres are infinitely far apart. The spheres repel, so the external force pushes sphere 2 in the direction it moves and does positive work; because sphere 2 gains no kinetic energy, all of this work is stored as the electric potential energy of the two-sphere system.
2. BU_E of the two-sphere system equals the work done on sphere 2 by the electric force.
   A student who links U_E to the work of the electric force picks this. The electric force on sphere 2 points away from sphere 1, opposite to the motion, so its work is negative; U_E equals the external force's positive work, which is the same size with the opposite sign.
3. CU_E of the two-sphere system is negative, since the spheres were brought closer together.
   A student who expects potential energy to fall as objects come together, as it does for gravity, picks this. These charges repel: work had to be done to push them together, so U_E rose from zero and is positive.
4. DU_E is stored in sphere 2 alone, since sphere 2 is the only sphere that was moved.
   A student who assigns potential energy to one object picks this. U_E = k(q₁q₂/r) depends on both charges and on the distance between them: it belongs to the system of the two spheres, whichever sphere was moved.

CED 10.4.A.1 · Read this in Fix

1. A+0.30 J
   A student who uses the magnitudes of the charges picks this. U_E = k(q₁q₂/r) takes the charges with their signs: the product (+4.0 μC)(−5.0 μC) is negative, so U_E is −0.30 J.
2. B+0.50 J
   A student who uses the Coulomb's-law expression k|q₁q₂|/r² for the energy picks this: (9.0 × 10⁹)(4.0 × 10⁻⁶)(5.0 × 10⁻⁶)/(0.60)² = 0.50. That expression gives the magnitude of the force, in newtons. Electric potential energy is k(q₁q₂/r), with the signs of the charges and r to the first power, giving −0.30 J.
3. C−0.60 J
   A student who gives each particle its own share of the interaction energy and adds them picks this, counting the one pair twice. A pair of charges has a single interaction energy, k(q₁q₂/r) = −0.30 J.
4. D−0.30 J Correct
   U_E = k(q₁q₂/r) = (9.0 × 10⁹ N·m²/C²)(4.0 × 10⁻⁶ C)(−5.0 × 10⁻⁶ C)/(0.60 m) = −0.30 J. It is negative because the charges have opposite signs.

Working U_E = k q₁q₂/r = (9.0 × 10⁹ N·m²/C²)(+4.0 × 10⁻⁶ C)(−5.0 × 10⁻⁶ C)/(0.60 m) = −1.8 × 10⁻¹ N·m²/(0.60 m) = −0.30 J.

CED 10.4.A.2 · Read this in Fix

1. A−0.54 J
   A student who counts only neighbouring pairs picks this, adding pairs 1–2 and 2–3 only. Particles 1 and 3 also interact, whatever lies between them: their pair adds +0.09 J at a separation of 0.40 m.
2. B−0.90 J
   A student who adds each particle's interactions with the other two picks this, which counts every pair twice. Each of the three pairs has one interaction energy and is counted once, giving −0.45 J.
3. C−0.45 J Correct
   Add the energies of all three pairs. Pairs 1–2 and 2–3 are each (9.0 × 10⁹)(2.0 × 10⁻⁶)(−3.0 × 10⁻⁶)/(0.20) = −0.27 J; pair 1–3 is (9.0 × 10⁹)(2.0 × 10⁻⁶)(2.0 × 10⁻⁶)/(0.40) = +0.09 J. Total: −0.27 − 0.27 + 0.09 = −0.45 J.
4. D+0.63 J
   A student who uses the magnitudes of the charges picks this, making every pair positive: 0.27 + 0.27 + 0.09 = 0.63 J. Particle 2 has a charge opposite to the others, so pairs 1–2 and 2–3 have negative energies.

Working Pairs: U₁₂ = k q₁q₂/r₁₂ = (9.0 × 10⁹)(2.0 × 10⁻⁶)(−3.0 × 10⁻⁶)/(0.20 m) = −0.27 J; U₂₃ = −0.27 J (same charges and distance); U₁₃ = (9.0 × 10⁹)(2.0 × 10⁻⁶)(2.0 × 10⁻⁶)/(0.40 m) = +0.090 J. U_total = −0.27 − 0.27 + 0.090 = −0.45 J.

CED 10.4.A.3 · Read this in Fix

1. A+2.00U₀
   A student who ignores the signs of the charges picks this, keeping only the effect of the smaller distance. Replacing a charge with one of the opposite sign turns a repelling pair into an attracting one, so U_E becomes negative.
2. B−2.00U₀ Correct
   U_E = k(q₁q₂/r). Reversing the sign of one charge reverses the sign of q₁q₂, a factor of −1; halving r doubles 1/r, a factor of 2. The new energy is (−1)(2)U₀ = −2.00U₀.
3. C+4.00U₀
   A student who uses the Coulomb's-law expression k|q₁q₂|/r² for the energy picks this: the absolute values hide the change of sign, and halving r quadruples 1/r². U_E = k(q₁q₂/r) keeps the signs, so the energy becomes negative, and it depends on 1/r, so halving r only doubles its magnitude: −2.00U₀.
4. D−0.50U₀
   A student who takes U_E to be proportional to the separation, as U_g = mgh is to height, picks this. U_E = k(q₁q₂/r) has r in the denominator, so a smaller separation gives a larger magnitude, not a smaller one.

Working U_E ∝ q₁q₂/r. New/old = [(q₁)(−q₂)/(r/2)] / [q₁q₂/r] = (−1)(2) = −2, so U_new = −2.00U₀.

CED 10.4.A.2 · Read this in Fix

1. A+2.00kq²/a
   A student who thinks moving a particle into place always takes positive work picks this. The fixed particles attract the −q particle, so the external force must point away from them, opposite to the motion, and its work is negative.
2. B+1.73kq²/a
   A student who adds the two pair energies as vectors, each of size kq²/a along a side of the triangle, picks this: two equal vectors 60° apart have a resultant of magnitude 2 cos 30° kq²/a = √3 kq²/a ≈ 1.73kq²/a, and a magnitude is positive. Energies are scalars with signs: the two contributions, −kq²/a each, simply add to −2kq²/a.
3. C−1.00kq²/a
   A student who equates the work for the last step with the total U_E of the final system picks this: kq²/a − 2kq²/a = −kq²/a. The energy of the pair already in place, +kq²/a, was supplied when those particles were assembled; this step adds only the two new pairs.
4. D−2.00kq²/a Correct
   The work done by the external force equals the increase in the system's U_E, which is the energy of the two new pairs: k(+q)(−q)/a + k(+q)(−q)/a = −2kq²/a. The work is negative because the particles attract, so the external force holds the third particle back as it comes in.

Working W_ext = ΔU_E = (new pairs only) = k(+q)(−q)/a + k(+q)(−q)/a = −2kq²/a = −2.00kq²/a.

CED 10.4.A.1 · Read this in Fix

1. AU₁ = U₂ = 0
   A student who thinks a system with zero net charge has zero U_E picks this. Both arrangements are neutral, but U_E depends on which charges are near which: summing the six pairs gives about −2.59kq²/s and −1.41kq²/s.
2. BU₁ < U₂ < 0 Correct
   Arrangement 1: the four sides join opposite charges (4 × −kq²/s) and the two diagonals join like charges (2 × +kq²/(√2 s)), so U₁ ≈ −2.59kq²/s. Arrangement 2: the top and bottom sides join like charges (+2kq²/s), the vertical sides join opposite charges (−2kq²/s), and both diagonals join opposite charges (2 × −kq²/(√2 s)), so U₂ ≈ −1.41kq²/s. Both are negative and U₁ is lower.
3. CU₁ < U₂ = 0
   A student who counts only the four sides of each square picks this, giving −4kq²/s and 0. The diagonal pairs interact too: in arrangement 2 both diagonals join opposite charges, so U₂ is negative, about −1.41kq²/s.
4. DU₁ = U₂ > 0
   A student who drops the signs of the charges picks this: both arrangements then have four side pairs and two diagonal pairs, all positive, and the same total. With signs kept, the pairs of opposite charges give negative terms and the totals differ.

Working Pairs in a square: 4 sides (distance s), 2 diagonals (distance √2 s). Arrangement 1 (+ − + − around the square): sides all opposite: −4kq²/s; diagonals like: +2kq²/(√2 s) = +1.41kq²/s; U₁ = −2.59kq²/s. Arrangement 2 (+ + on top, − − below): top and bottom sides like: +2kq²/s; vertical sides opposite: −2kq²/s; diagonals opposite: −1.41kq²/s; U₂ = −1.41kq²/s. So U₁ < U₂ < 0.

CED 10.4.A.3 · Read this in Fix

1. ACorrect: the forces on the middle particle cancel, so its two pair energies cancel.
   A student who adds the pair energies as if they were vectors picks this. The forces on the middle particle do cancel, but its two pair energies are both positive scalars, +kq²/d each, and they add.
2. BIncorrect: each particle has its own share of each pair, so U_E is 5kq²/d.
   A student who adds every particle's interactions with the other two picks this, counting each pair twice. Each pair has one interaction energy, so the total is 5kq²/(2d), not twice that.
3. CIncorrect: U_E is a scalar, so all three pair energies add, giving 5kq²/(2d). Correct
   Each pair has a positive energy because all the charges are positive: kq²/d for each of the two neighbouring pairs and kq²/(2d) for the outer pair. Energies have no direction, so nothing cancels: U_E = kq²/d + kq²/d + kq²/(2d) = 5kq²/(2d).
4. DIncorrect: each pair's U_E depends on 1/r², so U_E is 9kq²/(4d²).
   A student who uses the inverse-square dependence of the force picks this: kq²/d² + kq²/d² + kq²/(4d²). The pair energy is k(q₁q₂/r), inversely proportional to r, and 9kq²/(4d²) does not even have units of energy.

Working Pairs: (−d, 0): kq²/d; (0, +d): kq²/d; (−d, +d): kq²/(2d). U_total = 2kq²/d + kq²/(2d) = 5kq²/(2d). The student's value is the outer pair alone.

CED 10.4.A.3 · Read this in Fix

1. AOnly for system Y, the proton and electron Correct
   U_E = k(q₁q₂/r). For system Y the charges are opposite, so U_E is negative and moves up toward zero as r increases: it increases. For system X, U_E is positive and falls toward zero as r increases.
2. BOnly for system X, the pair of protons
   A student who equates U_E with the work done by the electric force picks this: as the protons separate, their repulsion does positive work. U_E changes by the negative of that work, so for the protons it decreases; it increases for the attracting proton–electron pair.
3. CFor both systems, as the particles separate
   A student who expects potential energy to rise whenever objects are pulled apart, as it does for gravity, picks this. That holds for attracting pairs such as Y. The two protons repel, so their U_E, which is positive, decreases toward zero as they separate.
4. DFor neither system, since U_E falls as r grows
   A student who ignores the signs of the charges picks this, treating every U_E as a positive k|q₁q₂|/r that shrinks as r grows. The proton–electron pair has a negative U_E, which increases toward zero as they separate.

Working U_E = k q₁q₂/r. X: q₁q₂ = +e², U_E = +ke²/r, decreases as r increases. Y: q₁q₂ = −e², U_E = −ke²/r, increases (toward 0) as r increases.

CED 10.4.A.2 · Read this in Fix

1. A+0.67kq²/d Correct
   The particles start and end at rest, so the external force's work equals the increase in the system's U_E. U_E goes from k(+q)(−q)/d = −kq²/d to k(+q)(−q)/(3d) = −kq²/(3d), an increase of 2kq²/(3d) ≈ 0.67kq²/d. The work is positive because the external force pulls the particle in the direction it moves, against the attraction.
2. B−0.67kq²/d
   A student who drops the signs of the charges picks this: with U_E = +kq²/r the pair behaves as if it repelled, U_E falls from kq²/d to kq²/(3d), and the work comes out as −2kq²/(3d). The charges are opposite, so U_E = −kq²/r; moving the particles apart raises it, and the external force does positive work.
3. C−0.33kq²/d
   A student who equates the work with the U_E of the final arrangement picks this: k(+q)(−q)/(3d) = −kq²/(3d). That would be the work if the particles had started infinitely far apart. Here they started a distance d apart with U_E = −kq²/d, so the work is the change, +2kq²/(3d).
4. D+1.00kq²/d
   A student who thinks pulling the pair apart takes the full 'binding' energy |U_E,initial| = kq²/d picks this. That is the work needed to separate them completely. At a separation of 3d the system still has U_E = −kq²/(3d), so only 2kq²/(3d) of work is needed.

Working The particles start and end at rest, so W_ext = ΔU_E = U_E,f − U_E,i. U_E,i = k(+q)(−q)/d = −kq²/d. U_E,f = k(+q)(−q)/(3d) = −kq²/(3d). W_ext = −kq²/(3d) − (−kq²/d) = +2kq²/(3d) ≈ +0.67kq²/d.

CED 10.4.A.1 · Read this in Fix

1. A−0.14 J Correct
   The particle starts and ends at rest, so the external work equals the change in the system's electric potential energy. Only the two pairs that include the moving particle change: its distance from each fixed particle grows from 0.30 m to 0.50 m. ΔU_E = 2k(3.0 μC)(2.0 μC)(1/0.50 m − 1/0.30 m) = −0.14 J. The repulsion pushes the particle outward, so the external force, holding it back, does negative work.
2. B−0.29 J
   A student who counts each pair's energy once for each of its two charges picks this, double the change. Each pair of charges has one electric potential energy, k(q_iqj/r_ij), so the change is −0.14 J.
3. C−0.77 J
   A student who uses 1/r², as in the force law, picks this: 2(0.054)(1/0.50² − 1/0.30²) = −0.77. Electric potential energy varies as 1/r, U_E = k(q₁q₂/r), which gives −0.14 J.
4. D+0.35 J
   A student who takes the work to be the total electric potential energy of the final arrangement picks this: 0.216 J for the two changed pairs plus 0.135 J for the fixed pair. The arrangement already had electric potential energy before the move, so the work is the change, −0.14 J; the fixed pair's energy does not change at all.

Working The particle starts and ends at rest, so W_ext = ΔU_E of the three-particle system. Only the two pairs that include the third particle change: its distance from each fixed particle goes from 0.30 m to √(0.30² + 0.40²) = 0.50 m. kq₁q₃ = (9.0 × 10⁹)(3.0 × 10⁻⁶)(2.0 × 10⁻⁶) = 0.054 J·m. ΔU_E = 2(0.054 J·m)(1/0.50 m − 1/0.30 m) = 2(0.054)(2.0 − 3.33) J = −0.14 J. (The fixed pair's energy, k(3.0 μC)²/0.60 m = +0.135 J, does not change.) Errors: each pair counted twice (m09): −0.29 J; 1/r² (m07): 2(0.054)(1/0.25 − 1/0.09) = −0.77 J; work taken as the final arrangement's total U_E (m14): 2(0.054)/0.50 + 0.135 = +0.35 J.

CED 10.4.A.3 · Read this in Fix