10.1 Electric Charge and Electric Force

1. AIt contains no charged particles, because it has no net charge.
   A student who thinks a neutral object contains no charge picks this. 'Neutral' means that the net charge is zero, not that charge is absent: the coin's atoms contain protons and electrons in equal numbers.
2. BIt contains more electrons than protons, as an electron has less charge.
   A student who thinks the lighter electron carries less charge than the proton picks this, reasoning that extra electrons are needed to balance the protons. The electron's charge (−e) and the proton's (+e) have equal magnitudes, so a neutral object has equal numbers of each.
3. CIt contains huge numbers of protons and electrons, in equal numbers. Correct
   Charge is a property of the particles that make up all matter. Every copper atom contains 29 protons (+e each) and, in a neutral atom, 29 electrons (−e each), so the coin holds an enormous number of charged particles. They are equal in number, so the net charge is zero.
4. DIt contains protons whose charge is balanced by its many neutrons.
   A student who thinks neutrons cancel the charge of protons picks this. A neutron has no charge, so it cannot balance anything; the positive charge of the protons is balanced by the negative charge of an equal number of electrons.

CED 10.1.A.1 · Read this in Fix

1. A0.94 N
   A student who uses the gap between the surfaces, 0.24 m, as r picks this: (9.0 × 10⁹)(6.0 × 10⁻¹²)/(0.24)² = 0.94 N. The charges are modeled as being at the centers, so r = 0.24 m + 0.030 m + 0.030 m = 0.30 m.
2. B0.60 N Correct
   The point charges are at the centers, so r is the gap plus one radius of each sphere: r = 24 cm + 3.0 cm + 3.0 cm = 30 cm = 0.30 m. |F_E| = k|q₁q₂|/r² = (9.0 × 10⁹)(2.0 × 10⁻⁶)(3.0 × 10⁻⁶)/(0.30)² N = 0.60 N.
3. C0.18 N
   A student who divides by r instead of r² picks this: (9.0 × 10⁹)(6.0 × 10⁻¹²)/0.30 = 0.18 N. The force is inversely proportional to the square of the distance, so divide by (0.30 m)² = 0.090 m².
4. D0.42 N
   A student who adds each sphere's full diameter to the gap picks this: r = 0.24 + 0.060 + 0.060 = 0.36 m, giving (9.0 × 10⁹)(6.0 × 10⁻¹²)/(0.36)² = 0.42 N. The center of each sphere is one radius, 3.0 cm, from its surface, so r = 0.30 m.

Working r = gap + R₁ + R₂ = 0.24 m + 0.030 m + 0.030 m = 0.30 m. |F_E| = k|q₁q₂|/r² = (9.0 × 10⁹ N·m²/C²)(2.0 × 10⁻⁶ C)(3.0 × 10⁻⁶ C)/(0.30 m)² = 0.60 N.

CED 10.1.A.2 · Read this in Fix

1. A0.56 N
   A student who adds the three magnitudes picks this: 0.225 + 0.225 + 0.1125 = 0.56 N. The forces from B and C are at right angles to each other, so they combine to √2 × 0.225 N = 0.318 N, not 0.450 N.
2. B0.54 N
   A student who takes the sphere at A to be one side length, 0.30 m, from D picks this: 0.318 + 0.225 = 0.54 N. A is at the opposite corner, √2 × 0.30 m away, so its force is 0.225/2 = 0.113 N.
3. C0.14 N
   A student who divides by r instead of r² picks this: forces of 0.0675 N from B and C and 0.0477 N from A give 0.14 N. Coulomb's law divides by r², giving 0.225 N and 0.113 N.
4. D0.43 N Correct
   Each force on D points along the line from the other sphere to D, away from it (all charges are positive). B and C are 0.30 m away: each exerts kq²/a² = (9.0 × 10⁹)(1.5 × 10⁻⁶)²/(0.30)² = 0.225 N, at right angles, giving √2 × 0.225 = 0.318 N along the diagonal. A is √2 × 0.30 m away and exerts 0.225/2 = 0.1125 N along the same diagonal. Net force = 0.318 + 0.113 = 0.43 N.

Working kq²/a² = (9.0 × 10⁹)(1.5 × 10⁻⁶ C)²/(0.30 m)² = 0.225 N (from B and from C, perpendicular). Their resultant: √2(0.225 N) = 0.318 N along the diagonal AD, away from A. From A: kq²/(√2 a)² = 0.1125 N along the same line. Net = 0.225(√2 + 1/2) = 0.43 N.

CED 10.1.A.3 · Read this in Fix

1. AElectric forces between the charged particles of the two surfaces push the book up. Correct
   The book presses the surface atoms of the table slightly closer together, and the electric forces between the electrons and nuclei in the two surfaces push the book upward. This combined electric push is what we call the normal force, a contact force.
2. BThe table is solid, so it blocks the book without exerting any force on the book.
   A student who thinks a support only gets in the way picks this. Without an upward force the book would accelerate downward; the table exerts that force, which comes from electric forces between the particles of the two surfaces.
3. CThe book's weight comes back up from the table as a reaction to Earth's gravity.
   A student who thinks the table's push is the book's weight returned as a reaction picks this. The weight is exerted on the book by Earth; the table's push is a separate force with an electric origin, and it is not the third-law partner of the weight.
4. DA fundamental contact force, unrelated to charge, acts wherever surfaces touch.
   A student who thinks contact forces are a separate fundamental force picks this. Contact forces are not fundamental: they are the combined effect of electric forces between enormous numbers of charged particles in the surfaces.

CED 10.1.A.4 · Read this in Fix

1. AThey exert repulsive forces of both kinds, as their charges are alike.
   A student who thinks gravitational forces follow the signs of the charges picks this. Gravitational forces depend on mass, not charge, and are always attractive.
2. BEach electron repels the other electrically and attracts it gravitationally. Correct
   The electrons have charges of the same sign, so their electric forces are repulsive. They also have mass, and gravitational forces are always attractive. Both forces act, in opposite directions; the electric force is by far the larger.
3. CThey exert only repulsive electric forces, having too little mass for gravity.
   A student who thinks only large bodies exert gravitational forces picks this. Any two objects with mass attract each other gravitationally; between electrons the force is extremely small, but it acts.
4. DTheir gravitational attraction exceeds their electric repulsion.
   A student who thinks gravity is the stronger force picks this. For two electrons the electric repulsion is about 10⁴² times the gravitational attraction, so the electrons push each other apart.

CED 10.1.B.1 · Read this in Fix

1. A1.3 × 10²⁸
   A student who writes the products q₁q₂ and m₁m₂ as e and m_p, without squaring, picks this: ke/(Gm_p) = 1.3 × 10²⁸. For two protons q₁q₂ = e² and m₁m₂ = m_p².
2. B4.2 × 10⁴²
   A student who uses the electron's mass, since the electron's charge has the same magnitude e, picks this: ke²/(Gm_e²) = 4.2 × 10⁴². The particles are protons, so the mass is m_p = 1.67 × 10⁻²⁷ kg.
3. C1.2 × 10³⁶ Correct
   F_E/F_g = (ke²/r²)/(Gm_p²/r²) = ke²/(Gm_p²); r cancels. = (9.0 × 10⁹)(1.60 × 10⁻¹⁹)²/[(6.67 × 10⁻¹¹)(1.67 × 10⁻²⁷)²] = 1.2 × 10³⁶. The electric force is vastly greater.
4. D1.2 × 10¹⁵
   A student who uses ε₀ in place of k picks this: ε₀e²/(Gm_p²) = 1.2 × 10¹⁵. Coulomb's constant is k = 1/(4πε₀) = 9.0 × 10⁹ N·m²/C², not ε₀.

Working F_E/F_g = ke²/(Gm_p²) = (9.0 × 10⁹)(2.56 × 10⁻³⁸)/[(6.67 × 10⁻¹¹)(2.79 × 10⁻⁵⁴)] = 1.2 × 10³⁶ (r cancels).

CED 10.1.B.2 · Read this in Fix

1. AThe Sun and planets are almost exactly neutral, so their electric forces nearly cancel. Correct
   Large bodies contain almost equal amounts of positive and negative charge, so the electric forces of their particles nearly cancel. The gravitational forces of all their particles are attractive and add up, so gravity dominates at large scales even though it is the weaker force between particles.
2. BElectric forces die out over short distances, but gravitational forces do not.
   A student who thinks electric forces act only at short range picks this. Both forces decrease as 1/r² and neither dies out; the electric forces between the Sun and planets are small only because the bodies are nearly neutral.
3. CElectric forces cannot act across the vacuum of space that separates these bodies.
   A student who thinks electric forces need a material to pass through picks this. Electric forces act across empty space, as gravitational forces do; Coulomb's law with ε₀ describes charges in free space.
4. DTheir enormous masses make gravity stronger than any electric force they could have.
   A student who explains the dominance of gravity by mass alone picks this. If the Sun and planets carried even a tiny fraction of their particles' charge as net charge, the electric forces would far exceed gravity; gravity wins because they are almost neutral.

CED 10.1.B.3 · Read this in Fix

1. AX lets charge flow through it more easily than Y does in a circuit.
   A student who confuses permittivity with conduction picks this. How easily charge flows through a material is a different property; permittivity concerns how much the charges shift within the material when it is polarized.
2. BX lets more of the electric force between charges pass through it.
   A student who reads 'permittivity' as how much force a material permits picks this. Permittivity measures polarization; in Coulomb's law the permittivity of free space appears in the denominator, as 4πε₀.
3. CX can hold a greater amount of net charge on its surface than Y can.
   A student who thinks permittivity measures how much charge a material can store picks this. Permittivity describes how strongly a material, which may be neutral, is polarized; it says nothing about the net charge it can hold.
4. DX is more strongly polarized than Y in the same electric field. Correct
   Electric permittivity measures the degree to which a material is polarized in an electric field, that is, how much its positive and negative charges separate. A greater permittivity means stronger polarization in the same field.

CED 10.1.C.1 · Read this in Fix

1. ADiagram 1
   A student who thinks a negative charge pulls everything toward it picks this, with the electron cloud shifted toward the sphere. The sphere repels the negative electrons, so the cloud shifts away from it.
2. BDiagram 2 Correct
   The negative sphere pushes the atom's electrons away and pulls its positive nucleus toward it, so the electron cloud shifts slightly to the right of the nucleus. The atom is polarized, with its negative side away from the sphere, while its net charge stays zero. Diagram 2 shows this.
3. CDiagram 3
   A student who thinks a neutral atom is unaffected by a nearby charge picks this. The sphere pushes on the electrons and pulls on the nucleus in opposite directions, so they separate slightly: the atom is polarized.
4. DDiagram 4
   A student who thinks charge jumps across the gap to a nearby neutral object picks this. The sphere does not touch the atom, and no charge is transferred; polarization is a rearrangement of the atom's own electrons.

Working Negative sphere on the left repels electrons (cloud shifts right, away) and attracts the nucleus (left); no charge crosses the gap. Diagram 2 shows the cloud shifted away from the sphere.

CED 10.1.C.2 · Read this in Fix

1. A3.5 × 10⁻³¹ N
   A student who uses ε₀ in place of k picks this: ε₀e²/r² = (8.85 × 10⁻¹²)(1.60 × 10⁻¹⁹)²/(8.0 × 10⁻¹⁰)² = 3.5 × 10⁻³¹ N. Coulomb's constant is k = 1/(4πε₀), so ε₀ belongs in the denominator with the factor 4π.
2. B4.4 × 10⁻³⁰ N
   A student who thinks permittivity multiplies the force picks this: 4πε₀e²/r² = 4.4 × 10⁻³⁰ N. In Coulomb's law ε₀ is in the denominator: |F_E| = e²/(4πε₀r²).
3. C3.6 × 10⁻¹⁰ N Correct
   Coulomb's law is |F_E| = (1/(4πε₀))|q₁q₂|/r², and 1/(4πε₀) = k. With q₁ = q₂ = e: |F_E| = (1.60 × 10⁻¹⁹ C)²/[4π(8.85 × 10⁻¹² C²/(N·m²))(8.0 × 10⁻¹⁰ m)²] = 3.6 × 10⁻¹⁰ N.
4. D1.4 × 10⁻¹⁹ N
   A student who evaluates r² as 2r picks this: e²/(4πε₀ × 2r) = (1.60 × 10⁻¹⁹)²/[4π(8.85 × 10⁻¹²)(1.6 × 10⁻⁹)] = 1.4 × 10⁻¹⁹ N. r² = r × r = (8.0 × 10⁻¹⁰ m)² = 6.4 × 10⁻¹⁹ m².

Working |F_E| = (1/(4πε₀))e²/r² = (1.60 × 10⁻¹⁹ C)²/[4π(8.85 × 10⁻¹² C²/(N·m²))(8.0 × 10⁻¹⁰ m)²] = 2.56 × 10⁻³⁸/(7.12 × 10⁻²⁹) N = 3.6 × 10⁻¹⁰ N. (m50, ε₀ used in place of k: ε₀e²/r² = 3.5 × 10⁻³¹ N; m46, permittivity multiplies the force: 4πε₀e²/r² = 4.4 × 10⁻³⁰ N; m17, r² evaluated as 2r: e²/(4πε₀·2r) = 1.4 × 10⁻¹⁹ N.)

CED 10.1.C.3 · Read this in Fix

1. AIt has 3.0 × 10² fewer electrons than protons.
   A student who reads 'negative' as 'short of electrons' picks this. An electron carries charge −e, so a bead with fewer electrons than protons has a positive net charge; a negative net charge means extra electrons.
2. BIt has 3.0 × 10² more electrons than protons. Correct
   The net charge is negative, so the bead has more electrons (−e each) than protons (+e each). The number of excess electrons is |q|/e = (4.8 × 10⁻¹⁷ C)/(1.60 × 10⁻¹⁹ C) = 3.0 × 10².
3. CIt has 3.0 × 10² electrons and no protons at all.
   A student who thinks only a charged object's excess charge is present picks this. The bead's atoms contain enormous numbers of protons and electrons; the net charge tells only the difference, 3.0 × 10² more electrons than protons.
4. DIt has 4.8 × 10⁻¹⁷ more electrons than protons.
   A student who reads the charge in coulombs as a number of particles picks this. The coulomb is a unit of charge; the number of excess electrons is the charge divided by the charge of one electron: (4.8 × 10⁻¹⁷ C)/(1.60 × 10⁻¹⁹ C) = 3.0 × 10². (A number of particles must also be a whole number.)

Working Negative net charge → excess electrons. N = |q|/e = (4.8 × 10⁻¹⁷ C)/(1.60 × 10⁻¹⁹ C) = 3.0 × 10² more electrons than protons.

CED 10.1.A.1.i · Read this in Fix

1. AEach is a whole-number multiple of 3.2 × 10⁻¹⁹ C.
   A student who takes the smallest measured charge as the basic unit picks this. The claim fails for two of the drops: 4.8/3.2 = 1.5 and 11.2/3.2 = 3.5, which are not whole numbers. The smallest drop here carries 2e, not e.
2. BEach result is one reading of a single charge, 7.0 × 10⁻¹⁹ C.
   A student who averages the five results picks this: (4.8 + 9.6 + 3.2 + 11.2 + 6.4)/5 ≈ 7.0. Each drop carries its own whole number of elementary charges, so the results are not repeated readings of one quantity, and their mean has no special meaning.
3. CThe charges follow no pattern, so charge can take any value.
   A student who thinks charge is continuous picks this, seeing only that the five values differ. They do follow a pattern: every one is a whole-number multiple of 1.6 × 10⁻¹⁹ C, as expected if charge comes in units of e.
4. DEach charge is a whole-number multiple of 1.6 × 10⁻¹⁹ C. Correct
   Dividing each result by 1.6: 4.8/1.6 = 3, 9.6/1.6 = 6, 3.2/1.6 = 2, 11.2/1.6 = 7 and 6.4/1.6 = 4, all whole numbers. The data are consistent with charge coming in whole-number multiples of the elementary charge, e = 1.6 × 10⁻¹⁹ C.

Working q/(1.6 × 10⁻¹⁹ C): 4.8 → 3, 9.6 → 6, 3.2 → 2, 11.2 → 7, 6.4 → 4, all whole numbers. With 3.2: 4.8/3.2 = 1.5 and 11.2/3.2 = 3.5 (not whole). Mean = 35.2/5 ≈ 7.0 (units 10⁻¹⁹ C).

CED 10.1.A.1.ii · Read this in Fix

1. A+3.2 × 10⁻¹⁹ C Correct
   Protons carry +e, electrons −e and neutrons no charge. Net charge = (20 − 18)e = +2e = +2(1.60 × 10⁻¹⁹ C) = +3.2 × 10⁻¹⁹ C. The ion has two fewer electrons than protons, so it is positive.
2. B−3.2 × 10⁻¹⁹ C
   A student who thinks a shortage of electrons makes an object negative picks this. Each electron carries −e, so having two fewer electrons than protons leaves a net charge of +2e, which is positive.
3. C−2.9 × 10⁻¹⁸ C
   A student who thinks the 20 neutrons cancel the charge of the 20 protons picks this, leaving only the 18 electrons: −18e = −2.9 × 10⁻¹⁸ C. Neutrons have no charge; the protons' +20e is balanced only by the electrons.
4. D+3.5 × 10⁻¹⁸ C
   A student who gives every particle in the nucleus a charge of +e picks this: (20 + 20 − 18)e = +22e = +3.5 × 10⁻¹⁸ C. Only the protons are charged; the neutrons add mass, not charge.

Working Net charge = (N_p − N_e)e = (20 − 18)(1.60 × 10⁻¹⁹ C) = +3.2 × 10⁻¹⁹ C; neutrons contribute 0.

CED 10.1.A.1.iii · Read this in Fix

1. AEqual in magnitude, and pointing in the same direction
   A student who thinks a positive charge pushes every other charge away picks this. The direction depends on both signs: the positive sphere repels the proton but attracts the electron, so the two forces point in opposite directions.
2. BSmaller in magnitude, as the electron carries less charge
   A student who thinks the lighter electron carries less charge than the proton picks this. The electron's charge, −e, has the same magnitude as the proton's, +e, so at the same distance the forces have equal magnitudes.
3. CLarger in magnitude, as the electron has much less mass
   A student who thinks a lighter particle feels a larger force picks this. Coulomb's law contains the charges and the distance, not the masses, so the forces have equal magnitudes; the electron's smaller mass gives it a larger acceleration, not a larger force.
4. DEqual in magnitude, and opposite in direction Correct
   The proton's charge is +e and the electron's is −e, so |q₁q₂| is the same and, at the same distance, Coulomb's law gives forces of equal magnitude. The proton (same sign as the sphere) is repelled and the electron (opposite sign) is attracted, so the forces point in opposite directions.

CED 10.1.A.1.iii · Read this in Fix

1. ABoth pairs, since each object carries only a small charge
   A student who thinks a point charge is an object with a small charge picks this. The model depends on size compared with distance, not on the amount of charge: the 10 cm spheres only 12 cm apart are not small compared with their separation.
2. BNeither pair, since only particles such as electrons are points
   A student who takes 'point' literally picks this. A point charge is a model: any object whose size is negligible compared with the relevant distances can be treated as one, as the 2.0 mm beads 1.0 m apart can.
3. CPair X only, since the beads are tiny next to their separation Correct
   A point charge is a model for an object whose size is negligible compared with the distances in the situation. The beads (2.0 mm) are 500 times smaller than their separation (1.0 m), so the model fits. The spheres (10 cm) are nearly as large as their separation (12 cm), so they cannot be treated as points.
4. DPair Y only, since beads 1.0 m apart exert no force at all
   A student who thinks electric forces stop acting beyond a short range picks this, reasoning that beads 1.0 m apart do not interact, so there is nothing to model. Coulomb's law gives a force at any separation; the beads still exert forces on each other, and they are the pair whose size is negligible compared with their separation.

CED 10.1.A.1.iv · Read this in Fix

1. A0.22F Correct
   |F_E| = k|q₁q₂|/r² is proportional to each charge and to 1/r². Doubling q₂ multiplies the force by 2; tripling r multiplies it by 1/3² = 1/9. The new force is (2/9)F ≈ 0.22F.
2. B0.67F
   A student who takes the force to be inversely proportional to r picks this: 2 × (1/3) = 2/3 ≈ 0.67. The force depends on 1/r², so tripling r divides it by 9, giving (2/9)F ≈ 0.22F.
3. C0.11F
   A student who thinks the force on sphere 2 depends only on the charge of sphere 1, which exerts it, picks this and applies only the distance change: 1/9 ≈ 0.11. The force depends on the product q₁q₂, so doubling sphere 2's own charge doubles the force on it too.
4. D0.44F
   A student who takes the force to be proportional to the square of the changed charge picks this: 2² × (1/9) = 4/9 ≈ 0.44. The force is proportional to q₁ times q₂, so doubling one charge doubles the force.

Working F ∝ q₁q₂/r². New/old = (1)(2)/(3)² = 2/9 ≈ 0.22, so F_new ≈ 0.22F. (INVR: 2/3 ≈ 0.67; SOURCEONLY: 1/9 ≈ 0.11; QSQUARED: 4/9 ≈ 0.44.)

CED 10.1.A.2 · Read this in Fix

1. A0.067 m
   A student who takes the force to be inversely proportional to r picks this: 0.60 m/9 = 0.067 m. The force depends on 1/r², so a force 9 times as large needs a separation 3 times smaller, 0.20 m.
2. B0.20 m Correct
   The force is proportional to 1/r², so F₂/F₁ = (r₁/r₂)². Here (r₁/r₂)² = 9.0 N/1.0 N = 9, so r₁/r₂ = 3 and r₂ = 0.60 m/3 = 0.20 m.
3. C0.0074 m
   A student who squares the factor instead of taking its square root picks this: 0.60 m/9² = 0.0074 m. Since F ∝ 1/r², r changes by the square root of the force factor: √9 = 3.
4. D1.8 m
   A student who puts r² in the numerator, so that the force grows with distance, picks this: 0.60 m × √9 = 1.8 m. The electric force decreases as the separation increases, so a larger force needs the spheres closer together.

Working F ∝ 1/r²: (r₁/r₂)² = F₂/F₁ = 9.0/1.0 = 9 → r₂ = r₁/3 = 0.60 m/3 = 0.20 m.

CED 10.1.A.2 · Read this in Fix

1. Ax = (1/2)d
   A student who puts the balance point midway between the charges picks this. At x = d/2 the +4q sphere, being 4 times the charge at the same distance, exerts a force 4 times as large, so the net force is not zero.
2. Bx = (4/5)d
   A student who treats the point like a center of mass, nearer the larger charge, picks this. Nearer the +4q sphere its force is even larger than the force from +q; the forces balance nearer the smaller charge, at x = d/3.
3. Cx = (1/5)d
   A student who takes the force to be inversely proportional to distance picks this: q/x = 4q/(d − x) gives x = d/5. With the correct 1/r² dependence, (d − x)² = 4x², so x = d/3.
4. Dx = (1/3)d Correct
   Let the third sphere (charge q₃) be at x. The forces from +q and +4q are in opposite directions, so the net force is zero where k|q q₃|/x² = k|4q q₃|/(d − x)². Then (d − x)² = 4x², so d − x = 2x and x = d/3, nearer the smaller charge.

Working kq q₃/x² = k(4q)q₃/(d − x)² → (d − x)² = 4x² → d − x = 2x (taking the root between the charges) → x = d/3. (1/r belief: d − x = 4x → x = d/5.)

CED 10.1.A.2 · Read this in Fix

1. A1.00h
   A student who thinks the force on the bead depends only on the charge +Q of the sphere that exerts it picks this: nothing that matters seems to change, so the height stays h. The force kQq/h² is proportional to the bead's own charge too, so with charge 4q the force at any height is 4 times as large and the bead rests higher, at 2.00h.
2. B4.00h
   A student who takes the electric force to be inversely proportional to h picks this: kQ(4q)/h′ = kQq/h gives h′ = 4h. The force varies as 1/h², so h² becomes 4 times as large and h′ = 2.00h.
3. C2.00h Correct
   At rest, the upward electric force balances the weight: kQq/h² = mg, so h² = kQq/(mg). The weight is unchanged, and with charge 4q, h² must be 4 times as large. The new height is √4 × h = 2.00h.
4. D16.0h
   A student who squares the factor again when working back from h² picks this: h² must become 4 times as large, and squaring the 4 gives 16. The height changes by the square root of that factor, √4 = 2, so h′ = 2.00h.

Working At rest, the upward electric force balances the weight: kQq/h² = mg, so h² = kQq/(mg) and h² ∝ q. The weight is unchanged, so with charge 4q, h′² = 4h² and h′ = √4·h = 2.00h. (m20, force independent of the bead's own charge: h′ = 1.00h; m16, F ∝ 1/h: kQ(4q)/h′ = kQq/h gives h′ = 4.00h; m21, factor squared again instead of square-rooted: 4² = 16, h′ = 16.0h.)

CED 10.1.A.2 · Read this in Fix

1. AF is inversely proportional to r², since doubling r divides F by 4 Correct
   Reading the graph, F = 4.0 N at r = 0.10 m and 1.0 N at r = 0.20 m, and 1.0 N at 0.20 m becomes 0.25 N at 0.40 m: each doubling of r divides F by 4, as Coulomb's law predicts for F ∝ 1/r².
2. BF is inversely proportional to r, since doubling r divides F by 2
   A student who expects simple inverse proportion picks this. The graph shows that doubling r from 0.10 m to 0.20 m reduces F from 4.0 N to 1.0 N, a factor of 4, not 2.
3. CF decreases linearly with r, by an equal amount for each step
   A student who reads 'decreases as r increases' as a straight line picks this. Equal steps in r give unequal drops: from 0.10 m to 0.20 m F falls by 3.0 N, from 0.20 m to 0.30 m by only about 0.56 N. The curve flattens out.
4. DF falls to zero once the separation exceeds about 0.30 m
   A student who thinks electric forces act only at short range picks this. The data point at 0.40 m shows F = 0.25 N, which is small but not zero; Coulomb's law gives a force at every separation.

Working From the graph: F(0.10 m) = 4.0 N, F(0.20 m) = 1.0 N, F(0.40 m) = 0.25 N. Doubling r divides F by 4 each time, so F ∝ 1/r². (Linear test: drops of 3.0 N from 0.10 to 0.20 m but about 0.56 N from 0.20 to 0.30 m.)

CED 10.1.A.2 · Read this in Fix

1. A3.6 × 10⁻⁶ N
   A student who divides by r instead of r² picks this: (9.0 × 10⁹)(1.2 × 10⁻¹⁷)/0.030 = 3.6 × 10⁻⁶ N. The force depends on 1/r²: divide by (0.030 m)² = 9.0 × 10⁻⁴ m².
2. B1.8 × 10⁻⁶ N
   A student who evaluates r² as 2r picks this: (9.0 × 10⁹)(1.2 × 10⁻¹⁷)/0.060 = 1.8 × 10⁻⁶ N. r² = r × r = (0.030 m)² = 9.0 × 10⁻⁴ m².
3. C1.2 × 10⁻⁴ N Correct
   Convert to SI units: q₁ = 2.0 × 10⁻⁹ C, q₂ = 6.0 × 10⁻⁹ C (magnitude), r = 0.030 m. |F_E| = k|q₁q₂|/r² = (9.0 × 10⁹)(2.0 × 10⁻⁹)(6.0 × 10⁻⁹)/(0.030)² N = 1.2 × 10⁻⁴ N.
4. D1.2 × 10⁻⁸ N
   A student who substitutes 3.0 for r without converting centimeters to meters picks this: (9.0 × 10⁹)(1.2 × 10⁻¹⁷)/3.0² = 1.2 × 10⁻⁸ N. With k in N·m²/C², r must be in meters: 0.030 m.

Working |F_E| = k|q₁q₂|/r² = (9.0 × 10⁹ N·m²/C²)(2.0 × 10⁻⁹ C)(6.0 × 10⁻⁹ C)/(0.030 m)² = 1.2 × 10⁻⁴ N.

CED 10.1.A.2 · Read this in Fix

1. ARow 1
   A student who thinks the larger charge exerts the larger force picks this row, with the longer arrow on Q, the sphere with the smaller charge. The forces are a pair with equal magnitudes: both depend on the same product q_P q_Q.
2. BRow 3 Correct
   Both charges are negative, so the spheres repel: the force on each points directly away from the other. By Coulomb's law the magnitude k|q_P q_Q|/r² is the same for both forces, so the two arrows have equal lengths. Only row 3 shows this.
3. CRow 2
   A student who thinks a negative charge pulls every other charge toward it picks this row. Whether charges attract or repel depends on both signs: two negative charges repel.
4. DRow 4
   A student who thinks only the larger charge exerts a force picks this row, with a force on Q only. Each sphere exerts a force on the other: P is pushed away from Q just as Q is pushed away from P.

Working Both charges negative → repulsion: each force points away from the other sphere. Magnitudes k|q_P q_Q|/r² are equal (Newton's third law pair). Only row 3 has equal arrows pointing apart.

CED 10.1.A.3.i · Read this in Fix

1. AIt points directly toward A, along the line from B to A. Correct
   A and B carry charges of opposite sign, so they attract. The electrostatic force is directed along the line joining the two spheres, so the force on B points directly toward A, whatever B's velocity.
2. BIt points directly away from A, along the line from A to B.
   A student who thinks a positive charge pushes every other charge away picks this. The direction depends on both signs: A is positive and B is negative, so A pulls B toward it.
3. CIt points straight upward, in the same direction as B's velocity.
   A student who thinks a moving object has a force in its direction of motion picks this. The electric force depends only on the positions and signs of the charges: it lies along the line joining A and B, and B's upward velocity plays no part in it.
4. DThere is none, since A's charge has the smaller magnitude.
   A student who thinks a charge of smaller magnitude cannot exert a force on one of larger magnitude picks this. Every pair of charged objects interacts: A exerts a force on B of magnitude k|q_A q_B|/r², equal to the force B exerts on A.

CED 10.1.A.3.ii · Read this in Fix

1. AIt is positive, since only objects with opposite charges attract each other
   A student who thinks attraction always means opposite charges picks this. A neutral ball is also attracted, because it is polarized by the rod; attraction therefore shows only that the ball is not negative.
2. BIt could have any charge, as a negative rod pulls in every nearby object
   A student who thinks a negative charge pulls every other charge toward it picks this. The direction of the force depends on both signs: a negatively charged ball would be repelled by the rod, so the swing toward the rod rules it out.
3. CIt is neutral, and gravity between the rod and ball pulls the ball over
   A student who thinks nearby everyday objects attract each other noticeably by gravity picks this. For everyday masses the gravitational force is tiny: for a 100 g rod and a 1 g ball 3 cm apart, Gm₁m₂/r² is less than 10⁻¹¹ N, far too small to swing the ball; the attraction is electric.
4. DIt is positive or neutral; a neutral ball is polarized and attracted Correct
   Opposite charges attract, so a positive ball would swing toward the rod. A neutral ball would too: the rod pushes the ball's electrons slightly away, leaving the near side slightly positive, and the nearer positive charge is attracted more strongly than the farther negative charge is repelled. Only a negative ball is ruled out, since it would be repelled.

CED 10.1.A.3.ii · Read this in Fix

1. AThe forces of the neutral surfaces' positive and negative charges add up to zero.
   A student who thinks neutral objects can exert no electric forces on each other picks this. At the atomic distances where surfaces touch, the charges of neighboring atoms are at different distances, so the electric forces do not cancel; they produce friction and the normal force.
2. BElectric forces between particles are negligible compared with their gravity.
   A student who thinks gravity is the stronger force between particles picks this. Between charged particles the electric force is enormously larger than the gravitational force (about 10³⁶ times for two protons).
3. CAdding up the forces of so many particle interactions is impractical. Correct
   Contact forces arise from electric forces between the charged particles of the surfaces, but the surfaces contain an enormous number of particles, all interacting. Treating their combined effect as one contact force, such as friction or a normal force, is far more convenient than summing every interaction.
4. DContact forces are a separate fundamental interaction, not an electric one.
   A student who thinks contact forces are a fundamental force of their own picks this. Contact forces are nonfundamental: they are the combined effect of electric forces, described as contact forces only for convenience.

CED 10.1.A.4 · Read this in Fix

1. AGlass conducts charge from one side to the other, unlike empty free space.
   A student who thinks permittivity measures conduction picks this. Glass is an insulator: charge does not flow through it easily. Its permittivity comes from polarization, the slight shifting of charges within it.
2. BGlass blocks electric forces completely, as every solid material is able to do.
   A student who thinks solid matter blocks electric forces picks this. Electric forces act through glass; the glass responds by becoming polarized, which is what gives it a permittivity different from ε₀.
3. CGlass holds a net charge on its surface, while free space holds none.
   A student who thinks permittivity measures stored charge picks this. Neutral glass has a permittivity too; permittivity describes how strongly the glass is polarized, not how much net charge it carries.
4. DGlass contains electrons whose arrangement changes in an electric field. Correct
   Free space contains no matter to polarize. Glass is made of atoms whose electrons shift slightly in an electric field, so glass is polarized; its permittivity, which measures that polarization, depends on its composition and arrangement and differs from ε₀.

CED 10.1.C.4 · Read this in Fix

1. AX's is smaller, since its electrons rearrange less easily in a field Correct
   The permittivity of a material is determined by how easily its electrons can change their configuration. Y's loosely held electrons shift more readily, so Y is more strongly polarized in a given field and has the larger permittivity; X's is smaller.
2. BX's is larger, since its tightly held electrons resist the field more
   A student who pictures permittivity as the strength with which a material resists a field picks this. Permittivity measures how much a material is polarized: tightly held electrons shift less, so X's permittivity is the smaller.
3. CThey are equal, since every material has the same permittivity, ε₀
   A student who thinks ε₀ applies to every material picks this. ε₀ is the permittivity of free space; the permittivity of a material differs from it and depends on how easily its electrons rearrange.
4. DBoth are zero, since charge cannot flow easily through either insulator
   A student who thinks permittivity measures how easily charge flows picks this. Insulators do not conduct, but their electrons can still shift slightly within their atoms, so they are polarized and have permittivities greater than ε₀.

CED 10.1.C.4.i · Read this in Fix

1. ADiagram 1
   A student who thinks static charge stays where it is placed on any material picks this. That is true for the plastic rod, but in the copper rod the charge carriers move easily, and the repelling charges spread out.
2. BDiagram 2 Correct
   Copper is a conductor: its charge carriers move easily, so the like charges repel one another and spread out along the rod. Plastic is an insulator: its charge carriers cannot move easily, so the excess charge stays near the end where it was placed. Diagram 2 shows this.
3. CDiagram 3
   A student who thinks excess charge spreads over any object picks this. Repulsion pushes the charges apart, but in the plastic rod they cannot move easily, so they stay near the end where they were placed.
4. DDiagram 4
   A student who thinks an insulator cannot hold charge picks this. The plastic rod keeps its excess charge; because the charge cannot move easily, it stays at the end where it was placed.

Working Copper (conductor): charge carriers move easily, so the repelling excess charges spread along the rod. Plastic (insulator): the charge stays at the left end. Diagram 2 shows this.

CED 10.1.C.4.ii · Read this in Fix

1. Aq = √(2mgL sin θ tan θ/k)
   A student who takes the electric force to be proportional to 1/r picks this: kq²/(2L sin θ) = mg tan θ. Coulomb's law has r² in the denominator, kq²/(2L sin θ)² = mg tan θ, so q = 2L sin θ √(mg tan θ/k).
2. Bq = 4mgL² sin²θ tan θ/k
   A student who writes the product of the two equal charges, q × q, as a single q picks this: kq/(2L sin θ)² = mg tan θ. Each sphere has charge q, so the product is q², and solving kq²/(2L sin θ)² = mg tan θ needs a square root: q = 2L sin θ √(mg tan θ/k).
3. Cq = 2L sin θ √(mg tan θ/k) Correct
   For each sphere, the vertical component of the tension balances the weight, T cos θ = mg, and the horizontal component 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, which gives q = 2L sin θ √(mg tan θ/k).
4. Dq = 2L sin θ √(mg/(k tan θ))
   A student who takes the vertical component of the tension to be T sin θ and the horizontal component to be T cos θ picks this: the electric force then comes out as mg/tan θ. The angle is measured from the vertical, so T cos θ = mg and T sin θ = F_E, which give F_E = mg tan θ and q = 2L sin θ √(mg tan θ/k).

Working Each sphere is in equilibrium under its weight mg (down), the electric force F_E (horizontal, away from the other sphere) and the tension T along its thread. Vertical: T cos θ = mg. Horizontal: T sin θ = F_E. Dividing: F_E = mg tan θ. Each sphere is L sin θ from the vertical through the support, so the separation is r = 2L sin θ. Coulomb's law: kq²/r² = mg tan θ, so q² = mg tan θ (2L sin θ)²/k and q = 2L sin θ √(mg tan θ/k). Errors: F ∝ 1/r (m16): kq²/(2L sin θ) = mg tan θ gives q = √(2mgL sin θ tan θ/k); q·q written as q (m41): kq/r² = mg tan θ gives q = 4mgL² sin²θ tan θ/k; components swapped (T sin θ = mg, T cos θ = F_E): F_E = mg/tan θ gives q = 2L sin θ √(mg/(k tan θ)).

CED 10.1.A.2 · Read this in Fix

1. A8.50kq²/d²
   A student who adds the magnitudes of the two forces, whatever their directions, picks this: 8.00kq²/d² + 0.50kq²/d². The +q sphere repels the +2q sphere toward +x while the −4q sphere attracts it toward −x, so the forces partly cancel and the net force is 7.50kq²/d².
2. B8.00kq²/d²
   A student who thinks the +q sphere, having the smaller charge, exerts no force on the +2q sphere picks this, counting only the force of the −4q sphere. Every charge exerts a force on every other: the +q sphere pushes the +2q sphere with 0.50kq²/d² toward +x, so the net force is 7.50kq²/d².
3. C6.00kq²/d²
   A student who takes the +q sphere to be one spacing d from the +2q sphere picks this: its force would then be k(q)(2q)/d² = 2.00kq²/d², and 8.00 − 2.00 = 6.00. The +q sphere is at x = 0, a distance 2d away, so its force is k(q)(2q)/(2d)² = 0.50kq²/d² and the net force is 7.50kq²/d².
4. D7.50kq²/d² Correct
   The +q sphere, 2d away, repels the +2q sphere in the +x direction with a force k(q)(2q)/(2d)² = 0.50kq²/d². The −4q sphere, d away, attracts it in the −x direction with a force k(4q)(2q)/d² = 8.00kq²/d². The two forces are along the same line in opposite directions, so the net force is 8.00kq²/d² − 0.50kq²/d² = 7.50kq²/d², toward −x.

Working Force of the +q sphere on the +2q sphere: separation 2d, like charges, so repulsive, directed in +x (away from x = 0): k(q)(2q)/(2d)² = 0.50kq²/d². Force of the −4q sphere: separation d, opposite charges, so attractive, directed in −x (toward x = d): k(4q)(2q)/d² = 8.00kq²/d². The forces are opposite: net = 8.00 − 0.50 = 7.50kq²/d², in the −x direction. Errors: magnitudes added (m27): 8.00 + 0.50 = 8.50kq²/d²; smaller +q exerts no force on the larger +2q (m29): 8.00kq²/d²; +q taken to be a distance d away (new): 8.00 − 2.00 = 6.00kq²/d².

CED 10.1.A.3 · Read this in Fix