10.1 Electrostatics with Conductors

1. AEqual for the two, since excess charge spreads evenly over any sphere
   A student who thinks charge spreads out on any object picks this. On an insulator the electrons cannot move, so the charge stays near the point that was touched, far from P; only on the metal does it spread over the surface.
2. BGreater for the metal sphere, whose excess electrons spread over its surface Correct
   In metal, electrons move freely: the excess electrons repel one another and spread over the whole surface, so outside the sphere they act as a point charge at its center and the field at P is k|q|/R², where R is the radius. In plastic the charge cannot move, so it stays near the touched point, a distance 2R from P, and its field at P is about k|q|/(2R)², a quarter as large.
3. CGreater for the plastic sphere, as the metal lets its excess charge flow away
   A student who thinks a metal cannot hold excess charge picks this. The metal sphere stands on an insulator, so its excess electrons have nowhere to go; they stay on the sphere, spread over its surface, and produce the larger field at P.
4. DZero for the metal sphere, as the field around a charged conductor is zero
   A student who applies 'the field is zero in a conductor' outside the metal picks this. The field is zero within the metal in equilibrium; just outside a charged conductor it is not zero, and here it is larger than for the plastic sphere.

CED 10.1.A.1 · Read this in Fix

1. AOn both the outer and the inner surfaces, at P and at R
   A student who thinks excess charge spreads over every surface of a hollow conductor picks this. A Gaussian surface inside the metal around the empty cavity has zero field on it, so it encloses no net charge: the cavity wall carries no charge, and the excess is all on the outer surface.
2. BThroughout the metal walls, at Q as well as P and R
   A student who pictures charge soaking into a conductor picks this. Excess charge inside the metal would produce a field there and push itself outward; in equilibrium there is no net charge anywhere inside the metal, at Q or elsewhere.
3. COn the outer surface, at points such as P as well as T Correct
   The excess electrons repel one another and move freely through the metal, so they get as far apart as they can. In equilibrium they are all on the outer surface, spread over the whole box, including P. There are none inside the metal at Q, and none on the wall of the empty cavity at R.
4. DClose to T, the point that the charged rod touched
   A student who thinks charge stays where it is put picks this. That happens on an insulator; in a metal the electrons move freely and, repelling one another, spread over the whole outer surface.

CED 10.1.A.2.i · Read this in Fix

1. ASpread over its whole surface, as in electrostatic equilibrium Correct
   The time a conductor takes to reach electrostatic equilibrium is so short that it is negligible. Long before 0.01 s has passed, the excess electrons have spread over the surface into their equilibrium arrangement.
2. BStill close to the touched point, as electrons drift very slowly
   A student who pictures each electron traveling slowly across the sphere picks this. No electron has to travel across the sphere on its own: all the free electrons respond to the field at once, and equilibrium is reached in a negligibly short time.
3. CFlowing around its surface as a steady current that does not stop
   A student who thinks the charge on a conductor keeps moving picks this. Once the charge reaches equilibrium, the field in the metal is zero and there is no field along the surface, so there is no net flow; the charge is at rest.
4. DSpread evenly through the whole volume of the metal sphere
   A student who thinks excess charge fills a conductor's volume picks this. The excess electrons repel one another out to the surface; in equilibrium, which is reached almost at once, there is no excess charge inside the metal.

CED 10.1.A.3.i · Read this in Fix

1. A9.9 × 10⁻⁹ C/m²
   A student who thinks a charge in the cavity induces nothing on the shell picks this, putting only the shell's own 5.0 × 10⁻⁹ C on the outer surface. The cavity charge draws −3.0 × 10⁻⁹ C onto the inner surface, so the outer surface must carry 8.0 × 10⁻⁹ C.
2. B4.0 × 10⁻⁹ C/m²
   A student who thinks the cavity charge induces charge of its own sign on the inner surface picks this: +3.0 × 10⁻⁹ C inside leaves 2.0 × 10⁻⁹ C for the outer surface. A positive charge attracts electrons, so the inner surface is negative, −3.0 × 10⁻⁹ C.
3. C1.1 × 10⁻⁸ C/m²
   A student who shares the shell's own charge equally between its two surfaces picks this: 2.5 × 10⁻⁹ C on each, plus the +3.0 × 10⁻⁹ C induced outside, gives 5.5 × 10⁻⁹ C on the outer surface. The inner surface carries only the induced −3.0 × 10⁻⁹ C, so the outer carries 8.0 × 10⁻⁹ C.
4. D1.6 × 10⁻⁸ C/m² Correct
   The field in the metal is zero, so a Gaussian sphere within the metal encloses no net charge: the inner surface carries −3.0 × 10⁻⁹ C. The shell's net charge is +5.0 × 10⁻⁹ C, so the outer surface carries +5.0 × 10⁻⁹ − (−3.0 × 10⁻⁹) = +8.0 × 10⁻⁹ C, spread uniformly: σ = (8.0 × 10⁻⁹ C)/(4π(0.20 m)²) = 1.6 × 10⁻⁸ C/m².

Working Gaussian sphere in the metal: E = 0 ⇒ q_enc = 0 ⇒ q_inner = −3.0 × 10⁻⁹ C. q_outer = Q_shell − q_inner = 5.0 × 10⁻⁹ + 3.0 × 10⁻⁹ = 8.0 × 10⁻⁹ C. Outer surface is spherical with nothing outside, so σ = q_outer/(4πb²) = 8.0 × 10⁻⁹/(4π × 0.040) = 1.6 × 10⁻⁸ C/m². Distractors: 5.0 × 10⁻⁹/(4πb²) = 9.9 × 10⁻⁹; 2.0 × 10⁻⁹/(4πb²) = 4.0 × 10⁻⁹; 5.5 × 10⁻⁹/(4πb²) = 1.1 × 10⁻⁸ C/m².

CED 10.1.A.4 · Read this in Fix

1. AArrow 2 Correct
   Just outside a conductor in equilibrium the field is perpendicular to the surface; a component along the surface would move the free surface charges. The conductor is negative, so the field points toward it, into the surface. Arrow 2 is perpendicular to the flat surface at P and points into the conductor.
2. BArrow 1
   A student who takes the field's direction to be the way an electron would be pushed picks this. The field points in the direction of the force on a positive test charge, which the negative conductor attracts: toward the surface, not away from it.
3. CArrow 3
   A student who thinks the field of a charged object points along the line to its center picks this. That holds for a sphere; at P on the flat surface, the line to the center is not perpendicular to the surface, and the field is perpendicular to the surface.
4. DArrow 4
   A student who thinks the field just outside a conductor can run along its surface picks this. A field component along the surface would push the free surface charges along it, so the conductor would not be in equilibrium.

Working Just outside a conductor in equilibrium, E⃗ is perpendicular to the surface (no component along it). The conductor is negative, so E⃗ points toward the surface: the arrow perpendicular to the flat face at P, pointing into the conductor.

CED 10.1.A.5 · Read this in Fix

1. AL positive, R negative
   A student who thinks electrons are pushed along the field picks this: the field of +Q points away from it, so the electrons would be sent to the far side. The force on an electron is opposite to the field; electrons are attracted toward +Q, so L becomes negative.
2. BL positive, R positive
   A student who thinks a nearby charge gives a conductor charge of its own sign, as contact would, picks this. Without contact no charge moves onto the sphere; its free electrons only move within it, toward +Q, so L becomes negative and R positive.
3. CL negative, R negative
   A student who thinks a charged object draws charge of the opposite sign onto a nearby conductor picks this. Nothing touches the sphere, so its net charge stays zero: the electrons that gather at L leave an equal positive charge at R.
4. DL negative, R positive Correct
   The positive charge attracts free electrons in the metal, which move toward it until the field inside the metal is zero and the sphere is again an equipotential. L gains excess electrons and is negative; R is left short of electrons and is positive. No charge has entered or left the sphere, so the two induced charges are equal in magnitude.

CED 10.1.A.6 · Read this in Fix

1. AIt is zero, whatever the external field and the box's own charge Correct
   The box is a closed conducting shell. Its free charges arrange themselves on its outer surface so that the field is zero in the metal, and with no charge in the empty space inside, the field there is zero too. This holds however strong the external field is and whatever net charge the box carries: this is electrostatic shielding.
2. BIt is weaker than the external field, but it is not zero
   A student who thinks a metal shell only weakens an outside field, absorbing part of it, picks this. The charges induced on the box's surface produce a field that exactly cancels the external field throughout the enclosed region.
3. CIt is not zero, because the box is not connected to ground
   A student who thinks a shield works only when grounded picks this. Shielding depends on the free charges in the closed shell rearranging on its surface; an isolated box does this just as a grounded one does.
4. DIt is not zero, as the box's own charge produces a field inside
   A student who thinks the charge on a shell produces a field inside it picks this. The box's excess charge is on its outer surface, arranged so that the field inside the metal and in the empty space it encloses is zero.

CED 10.1.A.7 · Read this in Fix

1. AElectrons moved from the sphere onto the rod, leaving the sphere short of electrons Correct
   In a metal only electrons are free to move; the nuclei stay in place. Attracted by the positive rod, electrons moved from the sphere onto it. The sphere now has fewer electrons than protons, and this deficit, on its surface, can be modeled as positive charge spread over the surface.
2. BProtons moved from the rod onto the sphere and spread out over its surface
   A student who thinks positive charge moves as electrons do picks this. Protons are bound in nuclei, which stay in place in both the glass and the metal; the sphere became positive by losing electrons.
3. CPositive charge was created on the sphere at the point that the rod touched
   A student who thinks charge is created by contact picks this. Charge is conserved: the sphere's positive charge equals the negative charge carried off by the electrons that moved onto the rod, and it is spread over the whole surface.
4. DThe sphere's electrons moved into its interior, leaving its surface positive
   A student who confuses charging with rearranging charge picks this. Moving electrons within the sphere would leave its net charge zero; it is positive because electrons left it. Nor can excess charge stay inside the metal in equilibrium.

CED 10.1.A.2.ii · Read this in Fix

1. AZero at both times, as the field inside a conductor is zero
   A student who applies 'the field is zero in a conductor' at every instant picks this. The rule holds only in electrostatic equilibrium; before it is reached, the excess electrons in the metal produce a field, and that field is what drives them to the surface.
2. BChanged but nonzero, as the excess charge spreads through the metal
   A student who thinks excess charge spreads out through a conductor's volume picks this. The excess electrons repel one another and move freely, so they move out until they are all on the surface and the field inside is zero.
3. CUnchanged and nonzero, as the electrons stay where they stopped
   A student who thinks excess charge stays where it was put on a conductor picks this. In a metal the excess electrons repel one another and move freely, so they move out to the surface, and in equilibrium the field at P is zero.
4. DNonzero at first, then zero, as electrons move to the surface Correct
   Just after they arrive, the excess electrons are bunched together inside the metal; a Gaussian surface around them encloses net charge, so there is a field at P. That field pushes free charges, and the excess electrons repel one another out to the surface. In equilibrium the field everywhere within the metal, P included, is zero.

CED 10.1.A.3 · Read this in Fix

1. A2.7 × 10² V
   A student who thinks the charge on a hollow conductor is shared between its inner and outer surfaces picks this: half on each gives k(2.0 × 10⁻⁹ C)(1/0.10 m + 1/0.20 m) = 2.7 × 10² V. With the cavity empty, all the charge is on the outer surface, and V_P = kQ/(0.20 m).
2. B7.2 × 10² V
   A student who treats the sphere as a point charge at its center at every distance picks this: kQ/(0.050 m). That model holds only outside the charge. Inside, the field is zero, so the potential stays at its value at the outer surface, 1.8 × 10² V.
3. C9.0 × 10² V
   A student who calculates potential with the field expression picks this: kQ/(0.20 m)² = 9.0 × 10² N/C, a field, not a potential. The potential at the surface is kQ/r = 1.8 × 10² V, and it has the same value throughout the cavity.
4. D1.8 × 10² V Correct
   In equilibrium all the excess charge is on the outer surface, radius 0.20 m. Outside, the sphere acts as a point charge at its center, so at the surface V = kQ/(0.20 m) = (9.0 × 10⁹)(4.0 × 10⁻⁹)/(0.20) = 1.8 × 10² V. The field is zero in the metal and in the empty cavity, so ΔV = −∫E⃗·dr⃗ = 0 from the surface in to P: V_P = 1.8 × 10² V.

Working All the excess charge is on the outer surface (b = 0.20 m); a Gaussian surface in the metal encloses no charge because the cavity is empty. At the outer surface V = kQ/b = (9.0 × 10⁹ N·m²/C²)(4.0 × 10⁻⁹ C)/(0.20 m) = 1.8 × 10² V. E = 0 in the metal and in the empty cavity, so V is constant from the outer surface to P: V_P = 1.8 × 10² V. Distractors: half the charge on each surface, k(2.0 × 10⁻⁹)(1/0.10 + 1/0.20) = 2.7 × 10² V; point-charge model inside, kQ/0.050 = 7.2 × 10² V; kQ/b² = 9.0 × 10² (field expression).

CED 10.1.A.3.ii · Read this in Fix

1. AV_A > V_B > V_C
   A student who thinks the potential is highest where the charge density is highest picks this. The charge density is greatest at the pointed end, but the potential is the same everywhere on and in the conductor: a potential difference along the surface would drive charge along it.
2. BV_A = V_B = V_C Correct
   In electrostatic equilibrium there is no field inside the metal and no field component along its surface, so no work is done moving a charge between any two points of the conductor: ΔV = −∫E⃗·dr⃗ = 0. The whole conductor, surface and interior, is at one potential, even though the charge density and the field just outside differ from place to place.
3. CV_A = V_B > V_C
   A student who thinks zero field means zero potential picks this. The field inside the metal is zero, so the potential there does not change from its value at the surface: V_C equals V_A and V_B, which are not zero for a positively charged conductor.
4. DV_C > V_B > V_A
   A student who treats the conductor as a point charge at its center, even inside it, picks this, ranking the points by their distance from the center. The charge is on the surface and the field inside the metal is zero, so C is at the potential of the surface, the same as at A and B.

Working E = 0 in the metal and E⃗ ⊥ surface just outside, so ΔV = −∫E⃗·dr⃗ = 0 along any path within the metal or along its surface: V_A = V_B = V_C.

CED 10.1.A.3.ii · Read this in Fix

1. A−Q/(5πε₀R)
   A student who thinks the potential increases along the field picks this: the magnitude is right, but the field points outward from the positive sphere, so the potential falls from the center (at the surface's value) to P. V_center − V_P is positive.
2. B−Q/(20πε₀R)
   A student who thinks zero field means zero potential picks this, taking V_center = 0 and so V_center − V_P = −Q/(4πε₀(5R)). Where E = 0 the potential is constant, not zero: the center is at the surface's potential, Q/(4πε₀R).
3. CQ/(5πε₀R) Correct
   At and outside the surface the sphere acts as a point charge: V_surface = Q/(4πε₀R) and V_P = Q/(4πε₀(5R)). Inside the metal E = 0, so the center is at the surface's potential. V_center − V_P = (Q/(4πε₀R))(1 − 1/5) = Q/(5πε₀R); it is positive, as the potential falls along the outward field.
4. D13Q/(40πε₀R)
   A student who spreads the charge through the sphere's volume picks this, using the uniformly charged insulating sphere's potential at its center, 3Q/(8πε₀R), and subtracting Q/(20πε₀R). On a conductor the charge is on the surface, so the potential inside equals the surface value, Q/(4πε₀R).

Working Charge on surface; outside, V(r) = Q/(4πε₀r), so V(R) = Q/(4πε₀R) and V(5R) = Q/(20πε₀R). E = 0 inside ⇒ V_center = V(R). V_center − V_P = Q/(4πε₀R) − Q/(20πε₀R) = 4Q/(20πε₀R) = Q/(5πε₀R). Distractors: sign reversed, −Q/(5πε₀R); V_center taken as 0, −Q/(20πε₀R); uniform volume charge, V_center = 3Q/(8πε₀R), so 3Q/(8πε₀R) − Q/(20πε₀R) = 13Q/(40πε₀R).

CED 10.1.A.3.ii · Read this in Fix

1. AE_X = E_Y > E_Z
   A student who thinks charge spreads evenly over any conductor picks this. That is true of an isolated sphere, but on this shape the charge density, and so the field just outside, is greater at the pointed end than at the rounded end.
2. BE_X > E_Y > E_Z Correct
   The surface charge density is greatest where the surface is most sharply curved, at the pointed end, and smallest on the broad rounded end. Just outside a conductor the field is perpendicular to the surface with magnitude σ/ε₀, so it is largest at X. Inside the metal, at Z, it is zero.
3. CE_Y > E_X > E_Z
   A student who mixes up total charge with charge density picks this, reasoning that the broad end has more surface to hold charge. The field just outside depends on the charge per unit area, σ, not on how much charge a whole region holds, and σ is greatest at the pointed end.
4. DE_X = E_Y = E_Z
   A student who thinks points at the same potential have the same field picks this. The conductor is at one potential, but the field depends on how the potential changes with position: it is zero inside the metal, and just outside it is largest where the charge density is greatest.

Working σ is largest at the most sharply curved part (the point); just outside, E = σ/ε₀ ⇒ E_X > E_Y; inside the metal E_Z = 0.

CED 10.1.A.3.iii · Read this in Fix

1. AGraph (1) Correct
   Within the metal the field is zero in equilibrium, because all the excess charge is on the surface and a Gaussian sphere of radius r < R encloses none. At r = R the field jumps to its largest value; outside, the sphere acts as a point charge at its center, E = kQ/r². The correct graph is zero up to R, jumps at R and then falls off as 1/r².
2. BGraph (2)
   A student who thinks the excess charge is spread through the metal picks this graph, which rises in proportion to r inside, as for a uniformly charged insulating sphere. On a conductor the charge is all on the surface, so the field inside is zero.
3. CGraph (3)
   A student who treats the sphere as a point charge at its center everywhere picks this graph, which grows without limit toward r = 0. The 1/r² field holds only outside the charge; within the metal the field is zero.
4. DGraph (4)
   A student who thinks that points at the same potential have the same field picks this graph, which is constant and nonzero inside. The metal is at one potential, and that means the field inside is zero: E_r = −dV/dr, and V does not change with r there.

Working Conductor in equilibrium: all excess charge on the surface. Gaussian sphere of radius r < R encloses no charge ⇒ E = 0 for r < R. For r ≥ R, E = kQ/r², largest at r = R. The graph is zero inside, jumps at R, then falls off as 1/r².

CED 10.1.A.4 · Read this in Fix

1. AIt is stronger on the side of the shell nearer to the moved charge
   A student who thinks the charge's field passes through the shell as if the metal were not there picks this. The inner surface charge cancels the field of +q throughout the metal and beyond; the field outside comes from the uniform +q on the outer surface, which does not move.
2. BIt is weaker on that side, where the inner wall's negative charge gathers
   A student who thinks the charge on the inner surface affects the field outside picks this. The inner surface charge does gather near +q, but together with +q it produces zero field everywhere beyond the cavity wall, so the field outside is unchanged.
3. CIt is unchanged, as the outer surface's charge stays spread uniformly Correct
   Wherever +q is, the inner surface carries −q, arranged so that it and +q together produce zero field in the metal and everywhere beyond. The outer surface carries +q; nothing inside affects it, so it spreads uniformly over the outer sphere. The field outside is kq/r² from the shell's center, before and after the move.
4. DIt stays zero, since the shell blocks the field of any charge inside
   A student who thinks a closed shell shields the outside from charges inside it picks this. The charge inside induces +q on the outer surface, so there is a field kq/r² outside the shell, before and after the move; it does not change when the small sphere moves.

CED 10.1.A.4 · Read this in Fix

1. Aσ/(2ε₀)
   A student who treats the conductor's surface like a thin charged sheet, with field on both sides, picks this: flux EA through each end face gives 2EA = σA/ε₀. Inside the metal the field is zero, so flux leaves only through the outer face, and E = σ/ε₀.
2. B4πkσ Correct
   Inside the metal E = 0, so no flux passes through the inner end face. Just outside, the field is perpendicular to the surface: it is parallel to the pillbox's sides, so no flux passes through them, and perpendicular to the outer face, giving Φ_E = EA. The pillbox encloses charge σA, so EA = σA/ε₀ and E = σ/ε₀ = 4πkσ.
3. Ckσ
   A student who writes Gauss's law with k in place of 1/ε₀ picks this: EA = kσA gives E = kσ. Gauss's law is Φ_E = q_enc/ε₀, and 1/ε₀ = 4πk, not k.
4. DσA/ε₀
   A student who takes the flux through the pillbox to be the field picks this. σA/ε₀ is the flux, in N·m²/C; dividing it by the area of the face it passes through, A, gives the field, σ/ε₀.

Working Gauss's law on the pillbox: inner face in metal (E = 0): no flux; sides parallel to E⃗ (E⃗ ⊥ surface): no flux; outer face: Φ = EA. q_enc = σA. EA = σA/ε₀ ⇒ E = σ/ε₀ = 4πkσ. Distractors: flux through both faces, 2EA = σA/ε₀ ⇒ σ/(2ε₀); Φ = kq ⇒ kσ; flux reported as field ⇒ σA/ε₀.

CED 10.1.A.5 · Read this in Fix

1. A(1/2)ε₀E₀
   A student who takes the field of each charged face on its own to be σ/ε₀, the result for a conductor's surface, picks this: 2σ/ε₀ = E₀. One charged plane on its own produces σ/(2ε₀); the two faces together give σ/ε₀ inside the metal, so σ = ε₀E₀.
2. B2ε₀E₀
   A student who thinks the metal blocks the field of the charge on its far face picks this, counting only the nearer face: σ/(2ε₀) = E₀. The fields of both faces act at every point inside the metal, and together, σ/ε₀, they cancel E₀.
3. Cε₀E₀ Correct
   In equilibrium the field in the metal is zero. The induced charge is −σ on the face where the external field enters and +σ on the face where it leaves. Each large plane produces a field of magnitude σ/(2ε₀); inside the metal the two add to σ/ε₀, directed opposite to E₀. For zero net field, σ/ε₀ = E₀, so σ = ε₀E₀. (Outside the slab the two induced fields cancel, leaving E₀.)
4. D4πε₀E₀
   A student who uses k in place of 1/ε₀ picks this, writing the two faces' field in the metal as kσ = E₀, so σ = E₀/k = 4πε₀E₀. The two planes give σ/ε₀ in the metal, and σ/ε₀ = E₀ gives σ = ε₀E₀.

Working Induced −σ (entry face) and +σ (exit face). Each plane: σ/(2ε₀). Inside the metal the two fields point the same way, opposite E₀: total σ/ε₀. Equilibrium: E₀ − σ/ε₀ = 0 ⇒ σ = ε₀E₀. Distractors: each face σ/ε₀ ⇒ 2σ/ε₀ = E₀ ⇒ (1/2)ε₀E₀; one face only ⇒ σ/(2ε₀) = E₀ ⇒ 2ε₀E₀; kσ = E₀ ⇒ σ = E₀/k = 4πε₀E₀.

CED 10.1.A.6 · Read this in Fix

1. A5.0 × 10² N/C
   A student who thinks the shell shields the outside from the charge inside it picks this, using only the shell's −5.0 × 10⁻⁹ C. The charge inside induces charge on the shell's surfaces, and outside the field depends on all the enclosed charge, −2.0 × 10⁻⁹ C.
2. B2.0 × 10² N/C Correct
   The field in the metal is zero, so the inner surface carries −3.0 × 10⁻⁹ C and the outer surface carries −5.0 × 10⁻⁹ − (−3.0 × 10⁻⁹) = −2.0 × 10⁻⁹ C. Outside, the net enclosed charge, −2.0 × 10⁻⁹ C, acts from the center: E = k(2.0 × 10⁻⁹ C)/(0.30 m)² = 2.0 × 10² N/C, directed toward the center.
3. C8.0 × 10² N/C
   A student who adds the magnitudes of the charges picks this: 3.0 × 10⁻⁹ + 5.0 × 10⁻⁹ = 8.0 × 10⁻⁹ C. With signs, the enclosed charge is +3.0 × 10⁻⁹ − 5.0 × 10⁻⁹ = −2.0 × 10⁻⁹ C.
4. D1.8 × 10³ N/C
   A student who measures the distance from the shell's outer surface picks this: k(2.0 × 10⁻⁹ C)/(0.10 m)². Outside a spherically symmetric distribution, the charge acts from the center, so r = 0.30 m.

Working Gaussian sphere in the metal: q_inner = −3.0 × 10⁻⁹ C; q_outer = −5.0 × 10⁻⁹ − (−3.0 × 10⁻⁹) = −2.0 × 10⁻⁹ C, uniform. At r = 0.30 m: q_enc = +3.0 − 5.0 = −2.0 × 10⁻⁹ C, E = (9.0 × 10⁹)(2.0 × 10⁻⁹)/(0.30)² = 2.0 × 10² N/C (inward). Distractors: shell charge only 5.0 × 10⁻⁹ ⇒ 5.0 × 10²; magnitudes added 8.0 × 10⁻⁹ ⇒ 8.0 × 10²; r from surface 0.10 m ⇒ 1.8 × 10³ N/C.

CED 10.1.A.7 · Read this in Fix

1. AElectric fields cannot pass through solid material, so the metal wall of the sphere keeps the external field from reaching the cavity.
   A student who thinks electric fields cannot pass through matter picks this. Fields pass through glass or plastic, for example; the cavity is field-free because the charges induced on the metal's outer surface produce a field that cancels the external field there.
2. BThe metal sphere is at zero potential, and at any point where the potential is zero, the electric field is zero as well.
   A student who links zero potential with zero field picks this. The field is the rate of change of potential with distance, not its value; zero potential at a point says nothing about the field there, and the sphere's potential need not be zero.
3. CNo field lines are drawn anywhere in the cavity, and an electric field is present just where the field lines are drawn.
   A student who thinks the field exists only along the drawn lines picks this. A field line diagram shows a sample of lines, and the field exists between them. The empty cavity in the diagram is what the argument must explain, not evidence for it.
4. DThe lines end on the left of the outer surface and restart on the right, at induced charges whose field cancels the external field in the sphere. Correct
   Field lines end on negative charge and begin on positive charge, so the diagram shows negative charge induced on the left of the outer surface and positive charge induced on the right. These charges arrange themselves so that their field exactly cancels the external field throughout the region the outer surface encloses, the metal and the empty cavity alike. This is electrostatic shielding, and it makes the field at P zero.

CED 10.1.A.7 · Read this in Fix

1. A3.0 kq/a
   A student who thinks the small sphere induces no charge on the shell, so that the shell's own charge 2q sits on its outer surface unaffected, picks this: 2kq/a + k(2q)/(2a). A Gaussian surface in the metal must enclose zero charge, so the cavity wall carries −q and the outer surface 3q.
2. B6.0 kq/a
   A student who finds the surface charges correctly but treats each charged surface as a point charge at the center, even at points inside it, picks this: k(q − q + 3q)/(a/2). Inside a uniformly charged spherical surface its potential is constant, k(charge)/(radius), not k(charge)/r.
3. C1.0 kq/a
   A student who takes the shell to be at zero potential because the field in its metal is zero picks this: integrating the field kq/r² inward from the cavity wall gives 2kq/a − kq/a. The metal's potential is constant, not zero; it equals the potential of its outer surface, 3kq/(2a).
4. D2.5 kq/a Correct
   The field in the metal is zero, so the cavity wall carries −q and, by conservation of charge, the outer surface carries 3q. Inside each charged spherical surface its potential is constant, k(charge)/(radius). At P: 2kq/a − kq/a + 3kq/(2a) = 2.5 kq/a.

Working E = 0 in the metal, so a Gaussian sphere inside the metal encloses zero charge: the cavity wall carries −q. Charge conservation: the outer surface carries 2q + q = 3q. At points inside a uniformly charged spherical surface, its potential is constant, k(charge)/(its radius); the small sphere gives kq/r. At P (r = a/2): V = kq/(a/2) + k(−q)/a + k(3q)/(2a) = (2 − 1 + 1.5)kq/a = 2.5 kq/a. Check by integration: outside, E = k(3q)/r², so V(2a) = 3kq/(2a); the metal is an equipotential, so V(a) = 1.5 kq/a; in the cavity E = kq/r², so V(a/2) = V(a) + kq/(a/2) − kq/a = 2.5 kq/a. Distractors: no induced charge, shell's 2q on its outer surface → kq/(a/2) + k(2q)/(2a) = 3.0 kq/a; each surface treated as a point charge at the center → k(q − q + 3q)/(a/2) = 6.0 kq/a; metal taken to be at V = 0 because E = 0 there → kq/(a/2) − kq/a = 1.0 kq/a. Checked with sympy.

CED 10.1.A.4 · Read this in Fix

1. AQ/(ε₀A)
   A student who gives each charged face, on its own, the field σ/ε₀ picks this: outside the plate both faces' fields point away from it, giving 2σ/ε₀ = Q/(ε₀A). σ/ε₀ is the total field just outside a conductor, from all of its charge; each face alone produces σ/(2ε₀).
2. BQ/(2ε₀A) Correct
   With zero field in the metal, the charge divides equally between the two faces, σ = Q/(2A) on each. A pillbox with one end in the metal gives E = σ/ε₀ = Q/(2ε₀A) just outside either face.
3. CQ/(8πε₀A)
   A student who writes Gauss's law as Φ_E = kq_enc picks this: the pillbox gives E = kσ = kQ/(2A) = Q/(8πε₀A). Gauss's law is Φ_E = q_enc/ε₀, and 1/ε₀ = 4πk.
4. D0
   A student who applies 'E = 0 for a conductor' outside the metal as well picks this. The field is zero only within the metal; just outside a charged conductor it is σ/ε₀, perpendicular to the surface.

Working Excess charge resides on the surface, on the two faces (edges negligible). Inside the metal E = 0; treating the faces as large planes with densities σ₁ and σ₂, the field in the metal is (σ₁ − σ₂)/(2ε₀) = 0, so σ₁ = σ₂ = Q/(2A). Pillbox across one face (inner end in the metal): EA_p = σA_p/ε₀, so E = σ/ε₀ = Q/(2ε₀A). (Equivalently, the plate as one thin sheet of total density Q/A gives Q/(2ε₀A) on each side.) Distractors: each face's field on its own taken as σ/ε₀, both adding outside → 2 × Q/(2ε₀A) = Q/(ε₀A); Gauss's law as Φ = kq_enc → E = kQ/(2A) = Q/(8πε₀A); 'E = 0 around a conductor' → 0. Checked with sympy.

CED 10.1.A.2 · Read this in Fix

1. A1.4 × 10⁴ N/C
   A student who thinks joined conductors share their charge equally picks this: 4.0 × 10⁻⁹ C on the smaller sphere gives (9.0 × 10⁹)(4.0 × 10⁻⁹)/(0.050)². Equal charges would put the smaller sphere at three times the potential of the larger, so charge would flow until the potentials are equal, leaving 2.0 × 10⁻⁹ C on the smaller sphere.
2. B2.9 × 10³ N/C
   A student who thinks the charge spreads uniformly over the whole surface, whatever its shape, picks this: one σ = Q/(4π(r₁² + r₂²)) on both spheres, and E = σ/ε₀. The surface charge density is greater on the more sharply curved, smaller sphere; equal potentials give Q ∝ r and so σ ∝ 1/r.
3. C7.2 × 10³ N/C Correct
   The wire makes the two spheres one conductor, so in equilibrium they are at the same potential: kQ₁/r₁ = kQ₂/r₂, and the charge divides in proportion to the radii, 1 : 3. The smaller sphere holds 2.0 × 10⁻⁹ C, and just outside it E = kQ₁/r₁² = (9.0 × 10⁹)(2.0 × 10⁻⁹)/(0.050)² = 7.2 × 10³ N/C.
4. D2.2 × 10⁴ N/C
   A student who thinks the smaller, more sharply curved sphere holds the larger share of the charge, in inverse proportion to its radius, picks this: 6.0 × 10⁻⁹ C on the smaller sphere. It is the charge density that is greater on the smaller sphere; the charge itself divides in proportion to the radii, so the smaller sphere holds only 2.0 × 10⁻⁹ C.

Working The spheres and wire form one conductor, so in equilibrium the spheres are at the same potential: kQ₁/r₁ = kQ₂/r₂ ⇒ Q₁/Q₂ = r₁/r₂ = 0.050/0.15 = 1/3. With Q₁ + Q₂ = 8.0 × 10⁻⁹ C: Q₁ = 2.0 × 10⁻⁹ C, Q₂ = 6.0 × 10⁻⁹ C. Each isolated sphere's charge is uniform, so just outside the smaller sphere E = kQ₁/r₁² = (9.0 × 10⁹)(2.0 × 10⁻⁹)/(0.050)² = 7.2 × 10³ N/C (equivalently σ₁/ε₀ with σ₁ = Q₁/(4πr₁²)). The larger sphere: E = kQ₂/r₂² = 2.4 × 10³ N/C, smaller, as 10.1.A.3.iii states for the more gently curved surface. Distractors: equal shares, 4.0 × 10⁻⁹ C each → 1.4 × 10⁴ N/C; one uniform σ over both surfaces, σ = Q/(4π(r₁² + r₂²)), E = σ/ε₀ = kQ/(r₁² + r₂²) = 2.9 × 10³ N/C; charge in inverse proportion to radius, Q₁ = 6.0 × 10⁻⁹ C → 2.2 × 10⁴ N/C.

CED 10.1.A.3.iii · Read this in Fix