8.1 Electric Charge and Electric Force

1. ABoth contain many charged particles, whose charges add to zero. Correct
   Charge is a property of the particles in all matter. The ruler and the rod each contain an enormous number of protons (+e) and electrons (−e); with no net charge, the positive and negative charge in each are equal and add to zero.
2. BNeither of them contains any charge until charge is given to it.
   A student who takes 'no net charge' to mean 'no charge' picks this. Both objects are made of atoms whose protons and electrons carry charge; zero net charge means that the positive and negative charges are equal.
3. COnly the copper rod contains charged particles; the ruler does not.
   A student who thinks insulators contain no charges picks this. Plastic, like copper, is made of atoms containing protons and electrons; the materials differ in how easily charge carriers move, not in whether charged particles are present.
4. DBoth contain charged electrons, but their protons carry no charge.
   A student who thinks only electrons carry charge picks this. Each proton carries charge +e; the electrons' negative charge is balanced by the protons' positive charge.

CED 8.1.A.1 · Read this in Fix

1. A24F
   A student who takes the force to be inversely proportional to r, not r², picks this: 6 × 4 = 24. Reducing the separation to one-quarter multiplies the force by 4² = 16.
2. B32F
   A student who thinks the force on sphere 2 depends only on the charge of sphere 1, which exerts it, picks this: 2 × 16 = 32. The force depends on both charges, so tripling sphere 2's charge also triples the force.
3. C96F Correct
   Coulomb's law gives |F⃗_E| ∝ |q₁q₂|/r². The product of the charges grows by 2 × 3 = 6, and reducing r to r/4 multiplies 1/r² by 16: 6 × 16 = 96, so the force is 96F.
4. D48F
   A student who thinks the force on sphere 2 depends only on sphere 2's own charge picks this: 3 × 16 = 48. The force depends on the product of both charges, so doubling sphere 1's charge doubles it too.

Working |F⃗_E| ∝ |q₁q₂|/r². Charges: ×2 × 3 = ×6. Separation ×1/4, so 1/r² ×16. New force = 6 × 16 F = 96F.

CED 8.1.A.2 · Read this in Fix

1. A2.1 × 10⁻⁶ N Correct
   A attracts B toward A with k(2.0 nC)(3.0 nC)/(0.30 m)² = 6.0 × 10⁻⁷ N; C attracts B toward C with k(4.0 nC)(3.0 nC)/(0.20 m)² = 2.7 × 10⁻⁶ N. The forces point in opposite directions, so the net force is 2.7 × 10⁻⁶ N − 0.60 × 10⁻⁶ N = 2.1 × 10⁻⁶ N, toward C.
2. B3.3 × 10⁻⁶ N
   A student who adds the magnitudes of the two forces picks this. A and C both attract B, but they are on opposite sides of it, so the two forces point in opposite directions and subtract.
3. C2.7 × 10⁻⁶ N
   A student who considers only the nearer sphere, C, picks this. Sphere A also attracts B, in the opposite direction, reducing the net force by 6.0 × 10⁻⁷ N.
4. D3.6 × 10⁻⁷ N
   A student who divides by r instead of r² picks this: 5.4 × 10⁻⁷ − 1.8 × 10⁻⁷. Each force falls off as 1/r², using 0.20 m and 0.30 m squared.

Working A on B (attractive, toward A): k(2.0 × 10⁻⁹)(3.0 × 10⁻⁹)/(0.30)² = 6.0 × 10⁻⁷ N. C on B (attractive, toward C): k(4.0 × 10⁻⁹)(3.0 × 10⁻⁹)/(0.20)² = 2.7 × 10⁻⁶ N. Opposite directions: net = 2.7 × 10⁻⁶ − 0.60 × 10⁻⁶ = 2.1 × 10⁻⁶ N, toward C.

CED 8.1.A.3 · Read this in Fix

1. ACharged particles in the surfaces of the book and table repel each other electrically. Correct
   The normal force is a contact force: the net effect of an enormous number of electric forces between the charged particles (electrons and nuclei) of the atoms at the two surfaces. Treating it as a single normal force is far simpler than adding all of those interactions.
2. BA separate, non-electric force acts between surfaces only once they are touching.
   A student who thinks contact forces are a separate, fundamental kind of force picks this. The normal force is not fundamental; it summarizes the electric forces between the particles at the two surfaces.
3. CThe table's atoms simply block the book's path without exerting any force on it.
   A student who thinks a rigid table is passive picks this. For the book to stay at rest something must exert an upward force balancing its weight; the table does, through electric forces between the surfaces.
4. DMagnetic forces between the atoms in the two surfaces push the book upward.
   A student who thinks the forces between atoms are magnetic picks this. The forces between the atoms in contact are electric forces between their charged particles.

CED 8.1.A.4 · Read this in Fix

1. ABoth forces push the protons apart, because the two particles are exactly alike in every way.
   A student who thinks gravity, like the electric force, can repel similar objects picks this. Gravitational forces are always attractive; only the electric force pushes the protons apart.
2. BBoth forces pull the protons together, since electric forces between particles are attractive.
   A student who thinks electric forces between charged particles attract picks this. Like charges repel, so the electric force pushes the protons apart.
3. CThe electric force pushes the protons apart; the gravitational force pulls them together. Correct
   The protons have charges of the same sign, so the electric force between them is repulsive. Both have mass, and gravitational forces are always attractive, so gravity pulls them together; the electric force is far larger.
4. DOnly an electric force acts, as particles as small as protons exert no gravitational force.
   A student who thinks gravity acts only between large objects picks this. The protons have mass, so they attract each other gravitationally, though far more weakly than they repel electrically.

CED 8.1.B.1 · Read this in Fix

1. Am√(4πε₀G) Correct
   Setting the magnitudes equal: q²/(4πε₀r²) = Gm²/r². The r² cancels, so q² = 4πε₀Gm² and q = m√(4πε₀G), whatever the separation.
2. Bm√(ε₀G)
   A student who writes Coulomb's law as q²/(ε₀r²), leaving out the 4π, picks this. With k = 1/(4πε₀), the balance is q²/(4πε₀r²) = Gm²/r², which gives m√(4πε₀G).
3. Cm√(4πε₀G/r)
   A student who takes the electric force to fall off as 1/r picks this, so that r does not cancel. Both forces vary as 1/r², so the required charge does not depend on the separation.
4. Dm√(G/ε₀)
   A student who puts ε₀ where k goes, writing the electric force as ε₀q²/r², picks this. ε₀ appears in the denominator of Coulomb's law, q²/(4πε₀r²); the chosen expression does not even have units of charge.

Working q²/(4πε₀r²) = Gm²/r² ⇒ q² = 4πε₀Gm² ⇒ q = m√(4πε₀G); r cancels. Distractors (sympy-checked): 4π dropped, q²/(ε₀r²) = Gm²/r² → m√(ε₀G); electric force taken as 1/r, q²/(4πε₀r) = Gm²/r² → m√(4πε₀G/r); ε₀ used in place of k, ε₀q²/r² = Gm²/r² → m√(G/ε₀).

CED 8.1.B.2 · Read this in Fix

1. A1.6 × 10¹³ C
   A student who writes Coulomb's law as Q²/(ε₀r²), leaving out the 4π, picks this: Q = √(ε₀GMm). With k = 1/(4πε₀), Q = √(4πε₀GMm) = 5.7 × 10¹³ C.
2. B3.3 × 10²⁷ C
   A student who writes the electric force with one charge, kQ/r², picks this: Q = GMm/k. Both bodies' charges enter Coulomb's law, so Q appears squared and a square root is needed.
3. C4.0 × 10¹³ C
   A student who doubles the electric force because each body exerts one on the other picks this: 2kQ²/r² = GMm/r². Each body experiences one electric force, kQ²/r².
4. D5.7 × 10¹³ C Correct
   Setting Q²/(4πε₀r²) = GMm/r², the separation cancels: Q = √(4πε₀GMm) = √(4π(8.85 × 10⁻¹²)(6.67 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²)) C = 5.7 × 10¹³ C. The given distance is not needed.

Working Q²/(4πε₀r²) = GMm/r²; r cancels. Q = √(4πε₀GMm) = √(4π(8.85 × 10⁻¹²)(6.67 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²)) C = √(3.25 × 10²⁷) C = 5.7 × 10¹³ C. Distractors: two forces added → 4.0 × 10¹³ C; one charge (Q/(4πε₀) = GMm) → 3.3 × 10²⁷ C; 4π left out (Q²/(ε₀r²) = GMm/r²) → Q = √(ε₀GMm) = 1.6 × 10¹³ C.

CED 8.1.B.3 · Read this in Fix

1. AX conducts electric current more easily than Y, so X is the better conductor.
   A student who confuses permittivity with conductivity picks this. Permittivity describes polarization, the separation of charge within a material, not how easily charge flows through it.
2. BIn the same field, X is polarized more strongly, so its charges separate more. Correct
   Electric permittivity measures the degree to which a material is polarized in the presence of an electric field. A greater permittivity means X is polarized more strongly than Y in the same field, so X's positive and negative charges are separated more than Y's.
3. CAn electric field passes through X more easily, so the field inside X is stronger.
   A student who reads 'permittivity' as how easily a field passes through a material picks this. Permittivity measures how strongly the material is polarized by a field, not how much field it lets in.
4. DX can store more excess charge than Y can, so X keeps hold of its charge better.
   A student who thinks permittivity measures how much excess charge a material holds picks this. Polarization separates a material's own charges without changing its net charge; permittivity describes that response.

CED 8.1.C.1 · Read this in Fix

1. AElectrons in the ball's molecules shift slightly toward the rod, so the side nearer the rod becomes slightly negative. Correct
   The rod is positive. Its field causes an induced rearrangement of electrons within each molecule of the ball, toward the rod. The ball is polarized: the near side becomes slightly negative and the far side slightly positive, while its net charge stays zero.
2. BNothing changes, because the electrons in an insulator are held fixed and do not respond to the rod.
   A student who thinks electrons in an insulator cannot move at all picks this. They cannot move through the material, but they shift slightly within each molecule, which polarizes the ball.
3. CProtons in the ball move away from the rod, so the side of the ball farther from the rod becomes positive.
   A student who thinks positive charge moves by protons moving picks this. The far side does become slightly positive, but because electrons shift toward the rod; the protons in the nuclei stay in place.
4. DPositive charge crosses the gap from the rod, giving the ball a small net positive charge of its own.
   A student who thinks a nearby charged object passes charge across a gap picks this. With no contact, the ball's net charge stays zero; the rod only rearranges the ball's own charges.

CED 8.1.C.2 · Read this in Fix

1. A1.1 × 10⁻¹⁰ C²/(N·m²)
   A student who writes Coulomb's law as q²/(ε₀r²), leaving out the 4π, picks this: ε₀ = q²/(Fr²). The constant is k = 1/(4πε₀), so dividing by 4π is needed.
2. B2.7 × 10⁻¹² C²/(N·m²)
   A student who takes the force to fall off as 1/r picks this: ε₀ = q²/(4πFr). The force varies as 1/r², so r² = 0.090 m² belongs in the denominator.
3. C8.8 × 10⁻¹² C²/(N·m²) Correct
   Coulomb's law with ε₀ is |F⃗_E| = q²/(4πε₀r²), so ε₀ = q²/(4πFr²) = (2.0 × 10⁻⁹ C)²/(4π × 4.0 × 10⁻⁷ N × (0.30 m)²) = 8.8 × 10⁻¹² C²/(N·m²).
4. D1.8 × 10⁻¹¹ C²/(N·m²)
   A student who takes the force between the spheres to be twice q²/(4πε₀r²), because each sphere pushes on the other, picks this: ε₀ = 2q²/(4πFr²). The measured force is the single force one sphere exerts on the other, q²/(4πε₀r²).

Working |F⃗_E| = q²/(4πε₀r²) ⇒ ε₀ = q²/(4πFr²) = (2.0 × 10⁻⁹)²/(4π × 4.0 × 10⁻⁷ × 0.090) C²/(N·m²) = 4.0 × 10⁻¹⁸/4.52 × 10⁻⁷ = 8.8 × 10⁻¹² C²/(N·m²).

CED 8.1.C.3 · Read this in Fix

1. A|Q_Z| > |Q_X| > |Q_Y|
   A student who adds the sizes of the charges and ignores their signs picks this: Z gives 10 nC, X gives 9 nC and Y gives 6 nC. Charge is a signed scalar, so positive and negative charges partly cancel: the net charges are −2 nC, +1 nC and −4 nC.
2. B|Q_Y| > |Q_Z| > |Q_X| Correct
   Add each system's charges with their signs: X has +5 − 2 − 2 = +1 nC, Y has +1 − 5 = −4 nC and Z has +4 − 6 = −2 nC. The magnitudes are 4 nC, 2 nC and 1 nC, so |Q_Y| > |Q_Z| > |Q_X|.
3. C|Q_X| > |Q_Z| > |Q_Y|
   A student who ranks the signed net charges, +1 nC > −2 nC > −4 nC, as if a negative charge had a smaller magnitude picks this. The magnitude ignores the sign: |−4 nC| = 4 nC is the largest.
4. D|Q_X| = |Q_Y| = |Q_Z|
   A student who thinks any system containing both positive and negative charges is neutral picks this, making all three net charges zero. A system is neutral only if its positive and negative charges are equal; here the sums are +1 nC, −4 nC and −2 nC.

Working Net charge is the algebraic sum. X: +5 − 2 − 2 = +1 nC. Y: +1 − 5 = −4 nC. Z: +4 − 6 = −2 nC. Magnitudes: |Q_Y| = 4 nC > |Q_Z| = 2 nC > |Q_X| = 1 nC.

CED 8.1.A.1.i · Read this in Fix

1. AThe smallest possible charge is 3.2 × 10⁻¹⁹ C, since that is the smallest value measured here.
   A student who takes the smallest measured value as the basic unit picks this. 4.8 × 10⁻¹⁹ C is 1.5 times 3.2 × 10⁻¹⁹ C, which is impossible if 3.2 × 10⁻¹⁹ C were the smallest charge; all five values are whole-number multiples of 1.6 × 10⁻¹⁹ C.
2. BCharge can take any value at all; these five drops simply happened to carry these five amounts.
   A student who thinks charge is continuous picks this. If any value were possible, five random charges would be unlikely all to be whole-number multiples of 1.6 × 10⁻¹⁹ C; the data point to charge coming in units of e.
3. CAll five are whole-number multiples of 1.6 × 10⁻¹⁹ C, consistent with charge in units of e. Correct
   Dividing by 1.6 × 10⁻¹⁹ C gives 2, 3, 4, 6 and 7: every value is a whole number of elementary charges. The data are consistent with net charge coming in whole-number multiples of e, the smallest indivisible amount of charge.
4. DThe drops carry different charges because a larger oil drop carries a larger charge than a small one.
   A student who links an object's charge to its size picks this. The data give no drop sizes, so they cannot support this; a drop's net charge depends on its number of excess or missing electrons, not on its size.

Working Dividing each value by 1.6 × 10⁻¹⁹ C gives 2, 3, 4, 6 and 7, all whole numbers. 4.8/3.2 = 1.5 is not a whole number, so 3.2 × 10⁻¹⁹ C cannot be the basic unit.

CED 8.1.A.1.ii · Read this in Fix

1. A7.2 × 10⁻⁸ N
   A student who counts all 7 nucleons as charged picks this: k(7e)(e)/r². Neutrons carry no charge, so the nucleus's charge is 3e.
2. B6.1 × 10⁻⁸ N
   A student who adds the force the nucleus exerts on the electron to the force the electron exerts on the nucleus picks this, doubling the result. Those two forces act on different objects; the electron experiences one force, k(3e)(e)/r².
3. C1.9 × 10¹¹ N
   A student who uses only the charge of the nucleus, the object exerting the force, picks this: k(3e)/r². Coulomb's law contains both charges, so the electron's charge e must multiply this.
4. D3.1 × 10⁻⁸ N Correct
   Only the 3 protons are charged, so the nucleus has charge +3e. |F⃗_E| = k(3e)(e)/r² = (9.0 × 10⁹)(3)(1.60 × 10⁻¹⁹ C)²/(1.5 × 10⁻¹⁰ m)² = 3.1 × 10⁻⁸ N.

Working Nuclear charge = 3e (neutrons are uncharged). |F⃗_E| = k(3e)(e)/r² = (9.0 × 10⁹)(3)(1.60 × 10⁻¹⁹)²/(1.5 × 10⁻¹⁰)² = 6.912 × 10⁻²⁸/2.25 × 10⁻²⁰ N = 3.1 × 10⁻⁸ N.

CED 8.1.A.1.iii · Read this in Fix

1. ABeads 2 mm across, 3 mm apart: any object just a few millimeters across can be treated as a point
   A student who thinks 'small' means small in everyday terms picks this. What matters is size compared with separation: beads 2 mm across that are 3 mm apart are not small compared with their separation.
2. BBeads 2 mm across, 0.50 m apart: each bead is tiny compared with the distance between them Correct
   The point-charge model applies when an object's size is negligible compared with the distances in the situation. A 2 mm bead is 1/250 of the 0.50 m separation, so every part of each bead is at nearly the same distance from the other.
3. CSpheres 20 cm across, 25 cm apart: each of the spheres carries a charge of just a few nanocoulombs
   A student who links point charges with small charges picks this. The model depends on the objects' size compared with their separation, not on how much charge they carry; spheres 20 cm across that are 25 cm apart are far from point-like.
4. DA 1 m rod and a bead 0.10 m from its center: all of the rod's charge acts at its center
   A student who replaces any charged object by a point at its center picks this. The ends of the 1 m rod are several times farther from the bead than its center is, so the rod cannot be treated as a point charge here.

Working The model needs each object's size to be negligible compared with the separation. Beads: 2 mm/500 mm = 0.004 of the separation. Beads 3 mm apart: size is 2/3 of the separation. Spheres: 20 cm/25 cm = 0.8 of the separation. Rod: 1 m long, 10 times the 0.10 m distance to the bead.

CED 8.1.A.1.iv · Read this in Fix

1. A1.6 × 10⁻⁶ N Correct
   Each sphere's charge is q = Ne = (2.5 × 10¹⁰)(1.60 × 10⁻¹⁹ C) = 4.0 × 10⁻⁹ C. Then |F⃗_E| = q²/(4πε₀r²) = (4.0 × 10⁻⁹ C)²/(4π × 8.85 × 10⁻¹² C²/(N·m²) × (0.30 m)²) = 1.6 × 10⁻⁶ N.
2. B4.8 × 10⁻⁷ N
   A student who divides by r instead of r² picks this: q²/(4πε₀ × 0.30 m). The force falls off as 1/r², so (0.30 m)² = 0.090 m² belongs in the denominator.
3. C3.2 × 10⁻⁶ N
   A student who adds the two forces of the interaction picks this, doubling q²/(4πε₀r²). The two forces act on different spheres; each sphere experiences one force of magnitude 1.6 × 10⁻⁶ N.
4. D2.0 × 10⁻⁵ N
   A student who writes Coulomb's law as q²/(ε₀r²), leaving out the 4π, picks this. With ε₀ the law is q²/(4πε₀r²); the 4π reduces the result by a factor of about 12.6.

Working q = Ne = (2.5 × 10¹⁰)(1.60 × 10⁻¹⁹ C) = 4.0 × 10⁻⁹ C on each. |F⃗_E| = q²/(4πε₀r²) = (4.0 × 10⁻⁹)²/(4π × 8.85 × 10⁻¹² × 0.090) N = 1.6 × 10⁻¹⁷/1.00 × 10⁻¹¹ N = 1.6 × 10⁻⁶ N (repulsive).

CED 8.1.A.2 · Read this in Fix

1. A3.3 × 10⁻⁹ C
   A student who doubles the force because each sphere pushes on the other picks this, taking the slope to be 2q²/(4πε₀). Each sphere experiences one force, q²/(4πε₀r²), so the slope is q²/(4πε₀).
2. B1.3 × 10⁻⁹ C
   A student who writes Coulomb's law as q²/(ε₀r²), leaving out the 4π, picks this: q = √(ε₀ × slope). The law is q²/(4πε₀r²), so q = √(4πε₀ × slope).
3. C1.9 × 10⁻⁶ C
   A student who reads 400 on the horizontal axis as r² = 400 m² picks this: q² = 4πε₀Fr² with F = 8.0 × 10⁻⁵ N. The axis shows 1/r², so at that point r² = 1/400 m², and the slope of the line gives q = 4.7 × 10⁻⁹ C.
4. D4.7 × 10⁻⁹ C Correct
   Coulomb's law gives F = (q²/(4πε₀))(1/r²), so the slope of the line is q²/(4πε₀): 8.0 × 10⁻⁵ N/400 m⁻² = 2.0 × 10⁻⁷ N·m². Then q = √(4πε₀ × 2.0 × 10⁻⁷ N·m²) = 4.7 × 10⁻⁹ C.

Working F = (1/(4πε₀))q²(1/r²), so the slope is q²/(4πε₀). From the line: slope = 8.0 × 10⁻⁵ N/400 m⁻² = 2.0 × 10⁻⁷ N·m². q = √(4πε₀ × slope) = √(4π × 8.85 × 10⁻¹² × 2.0 × 10⁻⁷) C = √(2.22 × 10⁻¹⁷) C = 4.7 × 10⁻⁹ C.

CED 8.1.A.2 · Read this in Fix

1. A(L sin θ)√(mg tan θ/k)
   A student who uses the distance from one sphere to the vertical line, L sin θ, as r picks this. In Coulomb's law r is the distance between the two spheres, which is 2L sin θ.
2. B(2L cos θ)√(mg/(k tan θ))
   A student who takes horizontal components as L cos θ and T cos θ picks this: the separation becomes 2L cos θ and the electric force mg/tan θ. θ is measured from the vertical, so the horizontal components are L sin θ and T sin θ, giving separation 2L sin θ and F_E = mg tan θ.
3. C(2L sin θ)√(mg tan θ/k) Correct
   For each sphere, the vertical part of the tension balances the weight, T cos θ = mg, and the horizontal part balances the electric force, T sin θ = F_E, so F_E = mg tan θ. The spheres are 2L sin θ apart, so kq²/(2L sin θ)² = mg tan θ, giving q = 2L sin θ √(mg tan θ/k).
4. D(L sin θ)√(2 mg tan θ/k)
   A student who doubles the electric force because each sphere exerts one on the other picks this: 2kq²/(2L sin θ)² = mg tan θ, so q = (L sin θ)√(2mg tan θ/k). Each sphere feels one electric force, kq²/(2L sin θ)², so q = 2L sin θ √(mg tan θ/k).

Working Each sphere: T cos θ = mg (vertical), T sin θ = F_E (horizontal), so F_E = mg tan θ. Separation d = 2L sin θ. kq²/(2L sin θ)² = mg tan θ ⇒ q = 2L sin θ √(mg tan θ/k). Distractors (sympy-checked): distance to the vertical line, L sin θ, used as r → L sin θ √(mg tan θ/k); sin/cos swapped for both horizontal components → d = 2L cos θ, F_E = mg/tan θ → 2L cos θ √(mg/(k tan θ)); force doubled (m11), 2kq²/d² = mg tan θ → (L sin θ)√(2mg tan θ/k).

CED 8.1.A.2 · Read this in Fix

1. A√(ke/r)
   A student who adapts the gravitational-orbit result v = √(GM/r), replacing GM with ke, picks this. In a gravitational orbit the orbiting mass cancels; here the force depends on charge while mv²/r depends on mass, so m stays and e appears squared.
2. Be·√(2k/(mr))
   A student who adds the force the proton exerts on the electron to the force the electron exerts on the proton picks this, using 2ke²/r². Only the force exerted by the proton acts on the electron: ke²/r² = mv²/r.
3. C√(ke/(mr))
   A student who writes the force on the electron with only the proton's charge, ke/r², picks this. Coulomb's law contains both charges, so the force is ke²/r² and v = e√(k/(mr)); the chosen expression does not even have units of speed.
4. De·√(k/(mr)) Correct
   The electric force on the electron, ke·e/r², is the centripetal force mv²/r. So ke²/r² = mv²/r, v² = ke²/(mr), and v = e√(k/(mr)).

Working The electric force provides the centripetal force: ke²/r² = mv²/r ⇒ v² = ke²/(mr) ⇒ v = e√(k/(mr)). Distractors (sympy-checked): orbit result √(GM/r) with GM → ke: √(ke/r); forces of the pair added, 2ke²/r² = mv²/r: e√(2k/(mr)); one charge, ke/r² = mv²/r: √(ke/(mr)).

CED 8.1.A.2 · Read this in Fix

1. A2.50kq²/s²
   A student who adds the three magnitudes, 1 + 1 + 1/2 in units of kq²/s², picks this. The two neighbors' forces are perpendicular, so they combine to √2 kq²/s², not 2kq²/s².
2. B1.91kq²/s² Correct
   The two neighbors, each a distance s away, push P with forces kq²/s² at right angles; together they give √2 kq²/s² along the diagonal, away from the center. The opposite corner is √2 s away and pushes along the same diagonal with kq²/(2s²). Net: (√2 + 1/2)kq²/s² = 1.91kq²/s².
3. C1.41kq²/s²
   A student who considers only the two nearest particles picks this: √2 kq²/s². The particle at the opposite corner also repels P, adding kq²/(2s²) along the same diagonal.
4. D2.41kq²/s²
   A student who uses the side s as the distance to the opposite corner picks this: √2 + 1. That particle is a diagonal, √2 s, away, so its force is kq²/(2s²).

Working Adjacent corners: two forces kq²/s², perpendicular, sum √2 kq²/s² along the diagonal away from the center. Opposite corner: distance √2 s, force kq²/(2s²) along the same diagonal. Net = (√2 + 1/2)kq²/s² = 1.91kq²/s². Distractors: magnitudes added 1 + 1 + 1/2 = 2.50; opposite corner ignored √2 = 1.41; opposite corner at distance s, √2 + 1 = 2.41.

CED 8.1.A.2 · Read this in Fix

1. A1.30kqQ/R² Correct
   Each ring particle is √(R² + 3R²) = 2R from Q and exerts kqQ/(4R²) along the line joining them. By symmetry the components across the axis cancel in pairs, and each axial component is √3R/2R = √3/2 of the force. Net: 6 × (kqQ/4R²)(√3/2) = 1.30kqQ/R².
2. B1.50kqQ/R²
   A student who adds the six magnitudes, 6kqQ/(4R²), picks this. The forces point in different directions: their components across the axis cancel, and only the axial components, √3/2 of each force, add.
3. C0.75kqQ/R²
   A student who takes the component along the axis as the force times R/2R picks this. The side adjacent to the axis direction is √3R, so the axial component is √3R/2R = √3/2 of each force.
4. D2.00kqQ/R²
   A student who treats the six particles as one point charge 6q at the center picks this: 6kqQ/(√3R)². The particles are 2R from Q, not √3R, and their forces are not along the axis, so the ring does not act as a point charge at its center.

Working Each ring particle is √(R² + 3R²) = 2R from Q, so each force is kqQ/(4R²), directed along the line from that particle to Q. By symmetry the components perpendicular to the axis cancel in pairs; the axial component of each is (√3R/2R) = √3/2 of it. Net = 6 × kqQ/(4R²) × √3/2 = (3√3/4)kqQ/R² = 1.30kqQ/R². Distractors: magnitudes added 6/4 = 1.50; perpendicular (R/2R) component used 6/4 × 1/2 = 0.75; ring as a point charge 6q at its center, 6/(3R²) → 2.00.

CED 8.1.A.2 · Read this in Fix

1. A3.6 × 10⁻³ m/s²
   A student who divides by r instead of r² picks this: q_AqB/(4πε₀ × 0.15 m), divided by the mass. The force falls off as 1/r².
2. B2.4 × 10⁻² m/s² Correct
   Coulomb's law: |F⃗_E| = q_AqB/(4πε₀r²) = (20 × 10⁻⁹ C)(12 × 10⁻⁹ C)/(4π × 8.85 × 10⁻¹² C²/(N·m²) × (0.15 m)²) = 9.6 × 10⁻⁵ N. With no friction this is the net force, so a = F/m = 9.6 × 10⁻⁵ N/(4.0 × 10⁻³ kg) = 2.4 × 10⁻² m/s².
3. C4.8 × 10⁻² m/s²
   A student who adds the force A exerts on B to the force B exerts on A picks this, doubling the force. Only the force exerted by A acts on B, so a = 9.6 × 10⁻⁵ N/(4.0 × 10⁻³ kg).
4. D3.0 × 10⁻¹ m/s²
   A student who writes Coulomb's law as q_AqB/(ε₀r²), leaving out the 4π, picks this. With ε₀ the law is q_AqB/(4πε₀r²), which is smaller by a factor of about 12.6.

Working |F⃗_E| = q_AqB/(4πε₀r²) = (20 × 10⁻⁹)(12 × 10⁻⁹)/(4π × 8.85 × 10⁻¹² × 0.0225) N = 2.4 × 10⁻¹⁶/2.50 × 10⁻¹² N = 9.6 × 10⁻⁵ N. a = F/m = 9.6 × 10⁻⁵ N/4.0 × 10⁻³ kg = 2.4 × 10⁻² m/s².

CED 8.1.A.2 · Read this in Fix

1. AF₂ > F₃ > F₁
   A student who takes the force to fall off as 1/r picks this: 1, 4/2 = 2 and 4/3. The force falls off as 1/r², which gives 1, 1 and 4/9.
2. BF₂ = F₃ > F₁
   A student who compares only the products of the charges, 1, 4 and 4, picks this. The separations differ: pair 2's product is divided by 2² and pair 3's by 3².
3. CF₁ > F₃ > F₂
   A student who treats pair 2's attraction as a negative, and so smaller, force picks this: +1, +4/9 and −1. Magnitudes ignore sign: pair 2's force has magnitude k(2q)(2q)/(2d)² = kq²/d², equal to pair 1's.
4. DF₁ = F₂ > F₃ Correct
   Using |F⃗_E| = k|q₁q₂|/r² in units of kq²/d²: pair 1 gives (1)(1)/1² = 1; pair 2 gives (2)(2)/2² = 1; pair 3 gives (4)(1)/3² = 4/9. So F₁ = F₂ > F₃; the sign of pair 2's charges affects only the direction.

Working |F⃗_E| = k|q₁q₂|/r² in units of kq²/d²: pair 1: 1·1/1 = 1; pair 2: 2·2/4 = 1; pair 3: 4·1/9 ≈ 0.44. F₁ = F₂ > F₃. Distractors: 1/r: 1, 2, 1.33 → F₂ > F₃ > F₁; distance ignored: 1, 4, 4 → F₂ = F₃ > F₁; signed products: +1, −1, +0.44 → F₁ > F₃ > F₂.

CED 8.1.A.2 · Read this in Fix

1. A8.00r
   A student who takes the force to fall off as 1/r picks this, making r grow by the same factor, 8, as the charges. The force falls off as 1/r², so r² must grow by 8 and r by √8.
2. B1.41r
   A student who thinks the force on sphere 2 depends only on sphere 2's own charge picks this: that charge doubles, so r² doubles. The force depends on both charges, whose product grows by 8.
3. C2.83r Correct
   The charges multiply the product |q₁q₂| by 4 × 2 = 8. For the force to stay F, r² must also grow by a factor of 8: r' = √8 r = 2.83r.
4. D2.00r
   A student who thinks the force on sphere 2 depends only on the charge of sphere 1, which exerts it, picks this: that charge grows by 4, so r doubles. Sphere 2's charge also doubles, so the product grows by 8.

Working |F⃗_E| ∝ |q₁q₂|/r². The charges multiply the force by 4 × 2 = 8, so r² must grow by 8: r' = √8 r = 2.83r. Distractors: 1/r → 8r; only q₂ (×2) → √2 r = 1.41r; only q₁ (×4) → 2r.

CED 8.1.A.2 · Read this in Fix

1. A2.8 × 10⁻⁶ N
   A student who adds the two magnitudes picks this. The forces are at right angles, so they combine by the Pythagorean theorem, giving 2.1 × 10⁻⁶ N.
2. B8.0 × 10⁻⁷ N
   A student who thinks the forces from a positive and a negative sphere oppose each other picks this, subtracting them: 1.8 × 10⁻⁶ − 1.0 × 10⁻⁶. The two forces are perpendicular, not opposite.
3. C2.1 × 10⁻⁶ N Correct
   A repels P horizontally, away from A: k(5.0 nC)(2.0 nC)/(0.30 m)² = 1.0 × 10⁻⁶ N. B attracts P vertically, toward B: k(4.0 nC)(2.0 nC)/(0.20 m)² = 1.8 × 10⁻⁶ N. The forces are perpendicular: √((1.0 × 10⁻⁶)² + (1.8 × 10⁻⁶)²) N = 2.1 × 10⁻⁶ N.
4. D1.8 × 10⁻⁶ N
   A student who considers only the nearer sphere, B, picks this. A also exerts a force on P, at right angles to B's, so the net force is larger than either one.

Working A on P: repulsive, along A→P (horizontal): k(5.0 × 10⁻⁹)(2.0 × 10⁻⁹)/(0.30)² = 1.0 × 10⁻⁶ N. B on P: attractive, toward B (vertical): k(4.0 × 10⁻⁹)(2.0 × 10⁻⁹)/(0.20)² = 1.8 × 10⁻⁶ N. Perpendicular: net = √((1.0 × 10⁻⁶)² + (1.8 × 10⁻⁶)²) = 2.06 × 10⁻⁶ N = 2.1 × 10⁻⁶ N.

CED 8.1.A.3 · Read this in Fix

1. AArrow 1
   A student who considers only the nearer particle, +Q, picks this: it pushes the positive particle straight up. −Q also pulls it, toward the lower right, which tilts the net force to the right of vertical.
2. BArrow 2 Correct
   +Q repels the particle straight up with a force F. −Q, √2 d away, attracts it toward the lower right with F/2 at 45° below horizontal. Adding components: right 0.35F, up F − 0.35F = 0.65F. The net force points up and to the right, about 29° from vertical: arrow 2.
3. CArrow 3
   A student who takes each force to fall off as 1/r picks this: −Q's pull is then F/√2, whose components cancel exactly half of the upward push and give equal right and up parts, a 45° direction. With 1/r², −Q's pull is F/2 and the net force is closer to vertical.
4. DArrow 4
   A student who expects the net force from equal and opposite charges to be parallel to the line joining them picks this. That holds only on the perpendicular bisector of the two charges and on the line through them; at P the particle is closer to +Q, whose upward push is the larger force, so the net force points mostly upward.

Working Force from +Q: repulsive, straight up, magnitude F = kQq/d². Force from −Q: attractive, toward −Q (down and to the right at 45°), distance √2 d, magnitude F/2. Components: x = (F/2)(1/√2) = 0.354F, y = F − 0.354F = 0.646F. Direction: 28.7° to the right of vertical (arrow 2). Distractors: +Q only → straight up (arrow 1); 1/r magnitudes (F/√2) → x = y = 0.5F, 45° (arrow 3); net parallel to the line joining the charges → horizontal (arrow 4).

CED 8.1.A.3 · Read this in Fix

1. A9kqQ/a²
   A student who adds the three magnitudes, 3 × 3kqQ/a², picks this. The forces point in three different directions, so they must be added as vectors.
2. B2kqQ/a²
   A student who uses the side a as the distance from each corner to the center picks this. The center is a/√3 from each corner, so each force is 3kqQ/a², not kqQ/a².
3. C3kqQ/a²
   A student who thinks the −q particle's force cancels one of the +q particles' forces, leaving just one force, picks this. −q attracts +Q toward the lower right, while the lower-left +q pushes it away to the upper right; the forces are at 60° to each other and do not cancel.
4. D6kqQ/a² Correct
   Each corner is a/√3 from the center, so each force is kqQ/(a²/3) = 3kqQ/a². The top +q pushes +Q straight down, the lower-left +q pushes it up and to the right, and −q pulls it down and to the right, toward −q. Adding components gives 2 × 3kqQ/a² = 6kqQ/a², directed toward −q.

Working Center-to-corner distance: a/√3, so each force has magnitude kqQ/(a²/3) = 3kqQ/a². Top +q: repels, straight down. Lower-left +q: repels, up and to the right at 30° above horizontal. Lower-right −q: attracts, down and to the right at 30° below horizontal. x: 3(cos 30° + cos 30°) = 3√3; y: 3(−1 + 1/2 − 1/2) = −3. Magnitude 3√(3 + 1) = 6kqQ/a², directed toward the −q corner. Distractors: magnitudes added 9; distance a → each kqQ/a², net 2; −q taken to cancel one +q → 3.

CED 8.1.A.3 · Read this in Fix

1. A0.33d Correct
   At a distance x from +q, the two repulsions balance when kqQ/x² = k(4q)Q/(d − x)², so (d − x)² = 4x² and d − x = 2x: x = d/3 ≈ 0.33d. The point is nearer the smaller charge, and the result does not depend on the third particle's charge.
2. B0.20d
   A student who takes each force to fall off as 1/r picks this: q/x = 4q/(d − x) gives x = d/5. With 1/r², the balance is (d − x)² = 4x², so x = d/3.
3. C0.50d
   A student who expects the balance point to be midway picks this. At the midpoint +4q exerts four times the force of +q, so the particle is pushed toward +q; the balance point is nearer +q.
4. D0.67d
   A student who expects the balance point to be nearer the larger charge picks this, pairing each charge with the other's distance. At 2d/3 from +q the particle is nearer +4q, which then exerts sixteen times the force of +q.

Working Third charge Q at distance x from +q: kqQ/x² = k(4q)Q/(d − x)² ⇒ (d − x)² = 4x² ⇒ d − x = 2x ⇒ x = d/3 = 0.33d (independent of Q). Distractors (sympy-checked): 1/r, q/x = 4q/(d − x) → d/5; midpoint → d/2; distances swapped, q/(d − x)² = 4q/x² → 2d/3.

CED 8.1.A.3.i · Read this in Fix

1. A8.00kqQ/d²
   A student who thinks the forces from a positive and a negative charge oppose each other picks this: 12kqQ/d² − 4kqQ/d². At the midpoint +q pushes +Q toward −3q while −3q pulls it the same way, so the forces add.
2. B16.0kqQ/d² Correct
   Each particle is d/2 from +Q, so 1/r² = 4/d². +q repels +Q with 4kqQ/d², pushing it toward −3q; −3q attracts +Q with 12kqQ/d², pulling it toward −3q. Both forces point toward −3q, so they add: 16kqQ/d².
3. C4.00kqQ/d²
   A student who uses the separation of the fixed particles, d, as r for both forces picks this: kqQ/d² + 3kqQ/d². Each force depends on its own distance to +Q, which is d/2.
4. D32.0kqQ/d²
   A student who doubles each force because both particles of each pair exert a force picks this. The force +Q experiences from each fixed particle is a single force, k|q₁q₂|/r²; the reaction forces act on the fixed particles, not on +Q.

Working Each distance is d/2, so 1/r² = 4/d². From +q: repulsive, kqQ(4/d²) = 4kqQ/d², pointing away from +q, toward −3q. From −3q: attractive, 3kqQ(4/d²) = 12kqQ/d², pointing toward −3q. Same direction: net = 16kqQ/d². Distractors: forces subtracted, 12 − 4 = 8; distance d used, 1 + 3 = 4; pair forces doubled, 32.

CED 8.1.A.3.ii · Read this in Fix

1. Aa₁ = 16a₂
   A student who thinks the force on each particle depends on its own charge picks this: particle 1 would feel 4 times the force of particle 2 and has one-quarter of its mass. The forces on the two particles have equal magnitudes, so only the masses differ.
2. Ba₁ = a₂
   A student who thinks the larger charge exerts the larger force picks this: particle 2 would feel 4 times the force but, with 4 times the mass, the same acceleration. The forces are equal in magnitude, so the lighter particle 1 has 4 times the acceleration.
3. Ca₁ = a₂ = 0
   A student who thinks the equal and opposite forces cancel picks this. The two forces act on different particles; each particle feels one unbalanced force and accelerates toward the other.
4. Da₁ = 4a₂ Correct
   The particles attract each other with forces of equal magnitude, F = k(4q)(q)/r², by Newton's third law. Particle 2 has four times the mass, so its acceleration is one-quarter as large: a₁ = F/m = 4 × F/(4m) = 4a₂.

Working Third law: each exerts a force of the same magnitude F = k(4q)(q)/r² on the other (attractive). a₁ = F/m, a₂ = F/(4m), so a₁ = 4a₂. Distractors: force on each ∝ own charge (F₁ = 4F₂): a₁/a₂ = 4 × 4 = 16; larger charge exerts larger force (force on 2 is 4 times force on 1): a₂ = 4F/4m = a₁; forces taken to cancel: both zero.

CED 8.1.A.3.ii · Read this in Fix

1. ARegion II, between the two fixed particles
   A student who expects the balance point to lie between the two charges picks this. Between opposite charges, one repels and the other attracts the third particle toward the same side, so the forces add.
2. BRegion III, to the right of both fixed particles
   A student who expects the balance point to be nearer the larger charge picks this. In region III the particle is closer to −4q than to +q, so the larger charge exerts the larger force everywhere there and the forces cannot balance.
3. CRegion I, to the left of both fixed particles Correct
   In region I the two forces point in opposite directions, and the particle is nearer the smaller charge, +q, so +q's force can balance −4q's larger charge. The balance point is where k|q|/x² = 4k|q|/(x + d)², at x = d beyond +q, for a third charge of either sign.
4. DNo region; opposite charges' forces cannot balance
   A student who thinks forces from opposite charges cannot balance picks this. Outside the pair the forces point in opposite directions, and beyond the smaller charge, in region I, they balance at a distance d from +q.

Working Region II: both forces point the same way (toward −4q for a positive third charge), so they cannot cancel. Region III: the third particle is nearer the larger charge −4q, which exerts the larger force there, so no balance. Region I: forces opposite; balance where kq/x² = 4kq/(x + d)² ⇒ x = d, i.e., a distance d beyond +q.

CED 8.1.A.3.ii · Read this in Fix

1. A(2kq² − Gm²)/r², directed away from particle 1
   A student who doubles the electric force because each particle exerts one on the other picks this: 2kq²/r² − Gm²/r². Particle 2 feels one electric force, kq²/r², so the net force is (kq² − Gm²)/r².
2. B(kq² − Gm²)/r², directed away from particle 1 Correct
   Particle 1 repels particle 2 electrically with kq²/r² (like charges) and attracts it gravitationally with Gm²/r² (gravity always attracts). The forces are opposite, and the electric one is larger, so the net force is (kq² − Gm²)/r², away from particle 1.
3. C(kq² + Gm²)/r², directed toward particle 1
   A student who thinks electric forces between charged particles attract, as gravity does, picks this. Two particles with charge +q repel each other; only the gravitational force is attractive.
4. Dkq²/r², directed away from particle 1
   A student who thinks small particles exert no gravitational force on each other picks this. Any two masses attract with Gm²/r²; here it is smaller than the electric force but still reduces the net force.

Working Electric: kq²/r², repulsive (away from particle 1). Gravitational: Gm²/r², attractive (toward particle 1). Opposite directions; electric larger: net = (kq² − Gm²)/r², away from particle 1. Distractors: electric force doubled (m11) → (2kq² − Gm²)/r² away; electric taken as attractive like gravity (m36) → (kq² + Gm²)/r² toward; gravity ignored between particles (m37) → kq²/r² away.

CED 8.1.B.1 · Read this in Fix

1. AR/144
   A student who thinks the electric force falls off with distance faster than gravity picks this, applying the factor 1/9 to the electric force only: R/16 × 1/9. Both forces vary as 1/r², so tripling r leaves the ratio unchanged.
2. BR/8
   A student who writes the gravitational force with one mass, Gm/r², picks this: doubling the masses then only doubles the gravitational force. The force contains both masses, so it grows by 2 × 2 = 4.
3. CR
   A student who thinks the electric force is a fixed multiple of the gravitational force picks this. The ratio k|q₁q₂|/(Gm₁m₂) depends on the charges and masses, which have both changed.
4. DR/16 Correct
   Both forces vary as 1/r², so the ratio k|q₁q₂|/(Gm₁m₂) does not depend on the separation. Halving each charge multiplies the numerator by 1/4, and doubling each mass multiplies the denominator by 4, so the ratio becomes R/16.

Working F_E/F_G = k|q₁q₂|/(Gm₁m₂); r cancels. Charges: ×(1/2)(1/2) = ×1/4. Masses: ×2 × 2 = ×4. New ratio = R × (1/4)/4 = R/16. Distractors: electric force reduced by 1/9 but not gravity → R/144; gravity with one mass (×2) → R/8; ratio fixed → R.

CED 8.1.B.2 · Read this in Fix

1. AElectric forces weaken with distance much faster than gravity does, so they vanish over the distance to the Moon.
   A student who thinks electric forces are short-range picks this. Both forces vary as 1/r², so distance alone cannot make one negligible compared with the other; the electric forces are small because the bodies are nearly neutral.
2. BElectric forces are unable to act across the empty space between Earth and the Moon, but gravity can.
   A student who thinks electric forces need a material to act through picks this. Electric forces act across a vacuum, as between the electron and proton in an atom; they are negligible here because the bodies are nearly neutral.
3. CEarth and the Moon each contain nearly equal amounts of positive and negative charge, so the electric forces nearly cancel. Correct
   Large bodies are very nearly electrically neutral. The electric attractions and repulsions between the enormous numbers of charged particles in Earth and the Moon almost exactly cancel, while every particle's gravitational attraction adds, so gravity dominates at this scale.
4. DThe electric forces are equal and opposite, so they cancel, while gravity acts on the Moon alone.
   A student who thinks the two equal and opposite forces of an interaction cancel picks this. Those forces act on different bodies, Earth and the Moon, so they cannot cancel. Gravity also acts as an equal and opposite pair, one force on each body. The electric forces are negligible because each body is very nearly neutral.

Working Ratio for two protons: ke²/(Gm_p²) = (9.0 × 10⁹)(1.60 × 10⁻¹⁹)²/[(6.67 × 10⁻¹¹)(1.67 × 10⁻²⁷)²] ≈ 1.2 × 10³⁶.

CED 8.1.B.3 · Read this in Fix

1. A0.80F Correct
   In the liquid, |F⃗_E| = |q₁q₂|/(4πεr²). Halving the separation multiplies the force by 4, and replacing ε₀ with 5.0ε₀ divides it by 5: 4/5 = 0.80, so the force is 0.80F.
2. B20.0F
   A student who thinks a larger permittivity increases the force picks this: 4 × 5. The permittivity is in the denominator of Coulomb's law, so a medium with ε = 5.0ε₀ reduces the force by a factor of 5.
3. C0.40F
   A student who takes the force to vary as 1/r picks this: halving r doubles the force, 2/5 = 0.40. With 1/r², halving r multiplies the force by 4.
4. D0.20F
   A student who thinks the force does not depend on how far apart the spheres are picks this: applying only the stated change of medium, ×1/5, gives 0.20F. Halving the separation also multiplies the force by 4, so it becomes 4/5 = 0.80F.

Working |F⃗_E| = |q₁q₂|/(4πεr²). Halving r: ×4. ε = 5.0ε₀: ×1/5. New force = 4/5 F = 0.80F. Distractors: ε in the numerator → 4 × 5 = 20F; 1/r with ε → 2/5 = 0.40F; separation change ignored → 1/5 = 0.20F.

CED 8.1.C.4 · Read this in Fix

1. Aε₀ > ε_Z > ε_Y > ε_X
   A student who thinks permittivity measures how easily a field passes through a material picks this, expecting the field to pass most easily through empty space and least easily through X. Permittivity measures polarization, which is greatest in X.
2. Bε_X = ε_Y = ε_Z = ε₀
   A student who thinks every material has the permittivity ε₀ picks this. ε₀ is the permittivity of free space; a material's permittivity differs from it and depends on how easily its electrons change configuration.
3. Cε_Z > ε_Y > ε_X > ε₀
   A student who expects the material with the most tightly held electrons, the 'best' insulator, to have the largest permittivity picks this. Permittivity grows with the ease of electron rearrangement, so X, not Z, has the largest.
4. Dε_X > ε_Y > ε_Z > ε₀ Correct
   In a given material, permittivity is determined by how easily its electrons change configuration: most easily in X, least in Z, so ε_X > ε_Y > ε_Z. Free space has no electrons to rearrange and is not polarized, so its permittivity, ε₀, is the smallest.

Working Permittivity is determined by the ease with which electrons change configuration: X most, then Y, then Z. Free space contains no electrons to rearrange, so it has the smallest permittivity, ε₀.

CED 8.1.C.4.i · Read this in Fix

1. ACopper is made of charged particles, while glass is made up entirely of uncharged particles.
   A student who thinks insulators contain no charged particles picks this. Glass, like copper, is made of atoms containing protons and electrons; the difference is how easily charge carriers move through it.
2. BSome electrons in copper move freely through the metal; glass has very few such mobile charges. Correct
   Copper is a conductor: some of its electrons are free to move through the metal. Glass is an insulator: its electrons are bound to atoms and molecules, so charge carriers cannot move easily through it.
3. CCopper can be given a net charge, while glass cannot be given any net charge at all.
   A student who thinks insulators cannot be charged picks this. Glass can be given a net charge, for example by rubbing; the charge simply cannot move easily through it.
4. DProtons carry charge through the copper, while in the glass the protons are held in place.
   A student who thinks positive charge moves by protons moving picks this. In copper, as in glass, the protons stay in the nuclei; the mobile charge carriers in copper are electrons.

CED 8.1.C.4.ii · Read this in Fix

1. A−0.67q
   A student who adds the magnitudes of the two repulsive forces on a corner particle, 2kq²/a², picks this. The two forces are 60° apart; their components across the line to the center cancel and only the components along it add, giving √3kq²/a², so |Q| = q/√3, not 2q/3.
2. B−1.73q
   A student who uses the side a as the distance between the center particle and a corner picks this. In Coulomb's law r is the distance between the two charges, here a/√3, so the center particle's force on a corner is 3kq|Q|/a², and |Q| = q/√3.
3. C−0.58q Correct
   Each corner particle is pushed away from the center by the other two: two forces of kq²/a² at 30° either side of the line to the center, with resultant √3kq²/a². The center particle, a/√3 from each corner, must pull each corner inward with 3kq|Q|/a², so Q is negative and |Q| = q/√3 ≈ 0.58q. By symmetry the three forces on the center particle cancel.
4. D−1.00q
   A student who takes the electric force to fall off as 1/r picks this: √3kq²/a = kq|Q|/(a/√3) gives |Q| = q. Coulomb's law has 1/r²; with the center a/√3 from each corner, the center particle's force is 3kq|Q|/a², so |Q| = q/√3.

Working Each corner particle is repelled by the other two with forces of kq²/a², each at 30° to the line from that corner to the center, so their resultant is 2(kq²/a²)cos 30° = √3kq²/a², directed away from the center. The center is a/√3 from each corner, so the center particle must pull each corner toward the center with kq|Q|/(a/√3)² = 3kq|Q|/a²: Q is negative and 3|Q| = √3q, so Q = −q/√3 = −0.577q ≈ −0.58q. The center particle is then in equilibrium by symmetry (three equal forces 120° apart). Distractors (sympy-checked): the two repulsive forces added as magnitudes, 2kq²/a² = 3kq|Q|/a² ⇒ −0.67q; the side a used as the center-to-corner distance, kq|Q|/a² = √3kq²/a² ⇒ −1.73q; force taken ∝ 1/r, √3kq²/a = kq|Q|/(a/√3) ⇒ −1.00q.

CED 8.1.A.3 · Read this in Fix