10.6 Capacitors

1. AElectrons move from one plate, through the wires and battery, to the other plate. Correct
   Charging moves charge; it does not create it. Electrons are pulled off one plate, leaving it positive, pass through the wires and the battery, and are pushed onto the other plate, making it negative. The plates end with equal amounts of charge of opposite sign, the capacitor's net charge stays zero, and no charge crosses the insulating gap.
2. BElectrons cross the gap between the two plates, from one plate to the other.
   A student who pictures charge flowing straight through the capacitor picks this. The plates are separated by an insulator, so no charge crosses the gap; electrons leave one plate and reach the other only by going around the circuit, through the battery.
3. CThe battery creates new charge and places some on each of the two plates.
   A student who thinks a battery is a source of charge picks this. A battery does not create charge; it moves charge that is already in the plates and wires. The electrons that end up on the negative plate came from the positive plate.
4. DThe battery adds charge to one plate, and the other plate remains neutral.
   A student who thinks charging a capacitor gives it extra charge picks this. If only one plate gained charge, the capacitor would have a net charge. In fact the plates gain charges of equal magnitude and opposite sign, +Q and −Q, and the net charge stays zero.

CED 10.6.A.1 · Read this in Fix

1. A3.6 × 10⁻⁵ F
   A student who thinks capacitance is the amount of charge a capacitor holds picks this: the charge, 3.6 × 10⁻⁵ C, written in farads. Capacitance is the charge per volt of potential difference, so the charge must be divided by 12 V.
2. B3.0 × 10⁻⁶ F Correct
   C = Q/ΔV, where Q is the magnitude of the charge on each plate and ΔV is the potential difference across the capacitor: C = (3.6 × 10⁻⁵ C)/(12 V) = 3.0 × 10⁻⁶ F.
3. C6.0 × 10⁻⁶ F
   A student who takes Q to be the total charge on both plates picks this: (2 × 3.6 × 10⁻⁵ C)/(12 V). Q is the magnitude of the charge on ONE plate; the other plate carries −3.6 × 10⁻⁵ C.
4. D2.2 × 10⁻⁴ F
   A student who thinks capacitance measures the energy stored picks this: (1/2)QΔV = (1/2)(3.6 × 10⁻⁵ C)(12 V) = 2.2 × 10⁻⁴, which is the stored energy in joules. Capacitance is the ratio Q/ΔV, in farads.

Working C = Q/ΔV with Q the magnitude of the charge on each plate: C = (3.6 × 10⁻⁵ C)/(12 V) = 3.0 × 10⁻⁶ F.

CED 10.6.A.2 · Read this in Fix

1. AThe field has the same magnitude at points 1, 2 and 3. Correct
   Between two oppositely charged parallel plates the field is uniform: the same in magnitude and direction everywhere except near the edges. Points 1, 2 and 3 are all between the plates and far from the edges, so the field has the same magnitude at each, whatever their distances from the plates.
2. BThe field is strongest at points 1 and 3, closest to the plates.
   A student who treats each plate like a point charge, whose field weakens with distance, picks this. Large charged plates do not behave like point charges: between them the field is the same at every point away from the edges.
3. CThe field weakens steadily from point 1 toward point 3.
   A student who pictures the field flowing out of the positive plate and weakening as it goes picks this. Field lines are not a flow that gets used up; between the plates they are straight, parallel and evenly spaced, showing a uniform field.
4. DThe field is zero at all three points, as the fields cancel.
   A student who thinks the fields of opposite charges cancel picks this. Between the plates both fields point from the positive plate toward the negative plate, so they add; they cancel only outside the plates.

Working Large parallel plates, points far from the edges: the field between the plates is uniform, so E₁ = E₂ = E₃ (nonzero, directed from the positive plate to the negative plate).

CED 10.6.A.3 · Read this in Fix

1. AU_C decreases, since C decreases while the potential difference stays the same.
   A student who assumes the potential difference stays fixed picks this: with ΔV fixed, a smaller C would mean a smaller Q = CΔV and a smaller U_C = (1/2)QΔV. The capacitor is disconnected, so Q is fixed; ΔV increases as C decreases, and the stored energy increases.
2. BU_C increases, since the student does positive work pulling apart plates that attract. Correct
   The plates carry opposite charges and attract each other, so the student pulls each plate in the direction it moves and does positive work on the system. The energy stored in a capacitor equals the work done by an external force to separate the charge, so U_C increases. (Check: Q is fixed and ΔV = Q/C rises as C falls, so U_C = (1/2)QΔV increases.)
3. CU_C stays the same, since the charge on each of the plates does not change.
   A student who thinks the stored energy depends only on the charge picks this. The charge is unchanged, but the student does work separating the plates, and U_C = (1/2)QΔV increases because ΔV increases.
4. DU_C increases, since the electric field between the plates becomes stronger.
   A student who takes the field to be set by the potential difference picks this: ΔV increases, so the field seems to increase. The claim is right but the reason is wrong: E_C = Q/(κε₀A) is unchanged. U_C increases because of the work done separating the plates.

CED 10.6.A.4 · Read this in Fix

1. A1.8 × 10⁻⁴ J
   A student who counts the charge on both plates uses Q = 2 × 30 μC = 60 μC and picks this: (1/2)(60 × 10⁻⁶ C)(6.0 V). Q is the charge on each plate, 30 μC.
2. B5.0 × 10⁻⁶ J
   A student who takes the slope of the graph as the stored energy picks this: (60 μC)/(12 V) = 5.0 μF. The slope is the capacitance; the energy is the area under the line up to 6.0 V.
3. C1.5 × 10⁻⁵ J
   A student who writes the stored energy as (1/2)CΔV, taking C = 5.0 × 10⁻⁶ F from the slope and not squaring ΔV, picks this: (1/2)(5.0 × 10⁻⁶ F)(6.0 V). That product is in coulombs, not joules; U_C = (1/2)QΔV = (1/2)(30 × 10⁻⁶ C)(6.0 V).
4. D9.0 × 10⁻⁵ J Correct
   From the graph, Q = 30 μC when ΔV = 6.0 V. U_C = (1/2)QΔV = (1/2)(30 × 10⁻⁶ C)(6.0 V) = 9.0 × 10⁻⁵ J, which is the triangular area under the line from 0 to 6.0 V.

Working From the graph, at ΔV = 6.0 V, Q = 30 μC (slope = 60 μC/12 V = 5.0 μF). U_C = (1/2)QΔV = (1/2)(30 × 10⁻⁶ C)(6.0 V) = 9.0 × 10⁻⁵ J, the area of the triangle under the line from 0 to 6.0 V.

CED 10.6.A.5 · Read this in Fix

1. AIt points downward, along the plates' field, so that the net field is increased.
   A student who thinks a dielectric strengthens the field picks this. Each induced surface charge is opposite in sign to the charge on the plate it faces, so the induced field opposes the plates' field and the net field is reduced.
2. BIt is zero, since the dielectric has no free charges that are able to move.
   A student who thinks an insulator cannot respond to a field because its charges cannot flow picks this. Charges in an insulator cannot move through it, but they can shift slightly within each molecule. That polarization produces the induced surface charges and the opposing field.
3. CIt points upward, opposite to the plates' field, so the net field is reduced. Correct
   The plates' field points downward, from the positive top plate to the negative bottom plate. It polarizes the dielectric: within its molecules, electrons shift slightly toward the positive plate, leaving the top surface of the slab negative and the bottom surface positive. These induced charges produce an upward field, opposite to the plates' field, so the net field in the dielectric is weaker, by the factor κ, than without it.
4. DIt points upward and cancels the plates' field, so that the net field is zero.
   A student who treats the dielectric like a conductor picks this. In a conductor, free charges move until the field inside is zero. In a dielectric the charges shift only slightly, so the induced field is smaller than the plates' field and the net field is reduced, not zero.

CED 10.6.A.6 · Read this in Fix

1. AC_Y = 2C_X, since Y holds twice as much charge on its plates as X does.
   A student who thinks capacitance is the charge a capacitor holds picks this. Y does hold twice the charge, but it also has twice the potential difference, so Q/ΔV, the capacitance, is the same as for X.
2. BC_X = 2C_Y, since C = Q/ΔV and ΔV across Y is twice that across X.
   A student who reads C = Q/ΔV as saying that C falls when ΔV rises picks this. That would follow only if Q stayed fixed; here Q doubles along with ΔV, so their ratio is unchanged. Capacitance is set by the plates and the gap, not by ΔV.
3. CC_X = C_Y, since charging to 12 V changes Q but not the plates or gap. Correct
   Capacitance depends only on the physical properties of a capacitor: its plate area, plate separation and the material between the plates. These are the same for X and Y, so C_X = C_Y. Y's larger potential difference gives it twice the charge (Q = CΔV), not a larger capacitance.
4. DC_Y = 4C_X, since Y stores four times as much energy as X does.
   A student who thinks capacitance measures the energy stored picks this. Y does store four times the energy (U_C = (1/2)QΔV with Q and ΔV both doubled), but capacitance is Q/ΔV, which is the same for both.

Working C = κε₀(A/d): A, d and κ are the same, so C_X = C_Y. Then Q_Y = CΔV_Y = 2Q_X and U_Y = (1/2)Q_YΔV_Y = 4U_X, while Q_Y/ΔV_Y = Q_X/ΔV_X = C.

CED 10.6.A.2.i · Read this in Fix

1. A96 pF
   A student who thinks a wider gap lets a capacitor store more picks this: (16 pF)(3)(2) = 96 pF. Capacitance is inversely proportional to d, so moving the plates apart lowers it.
2. B12 pF
   A student who treats the plates like point charges, with an inverse-square dependence on distance, picks this: (16 pF)(3)/2² = 12 pF. For parallel plates C is inversely proportional to d, not to d².
3. C48 pF
   A student who thinks only the plate area matters picks this: (16 pF)(3) = 48 pF. The separation matters too: C is proportional to 1/d, so doubling d halves the capacitance.
4. D24 pF Correct
   C = κε₀(A/d): capacitance is proportional to the plate area and inversely proportional to the plate separation. Tripling A multiplies C by 3 and doubling d divides it by 2, so C = (16 pF)(3/2) = 24 pF.

Working C = κε₀(A/d), so with κ unchanged C ∝ A/d: C₂ = (16 pF)(3)/(2) = 24 pF.

CED 10.6.A.2.ii · Read this in Fix

1. A1.2 × 10⁻¹⁰ F
   A student who thinks the material between the plates does not affect the capacitance leaves out κ and picks this: ε₀A/d. The dielectric multiplies the capacitance by κ = 3.0.
2. B3.5 × 10⁻¹⁰ F Correct
   C = κε₀(A/d) = (3.0)(8.85 × 10⁻¹² C²/(N·m²))(0.020 m²)/(1.5 × 10⁻³ m) = 3.5 × 10⁻¹⁰ F.
3. C3.9 × 10⁻¹¹ F
   A student who thinks an insulating material between the plates reduces the capacitance divides by κ and picks this. A dielectric increases the capacitance: C = κε₀(A/d).
4. D8.0 × 10⁻¹⁶ F
   A student who thinks a wider gap increases the capacitance multiplies by d instead of dividing and picks this. C is inversely proportional to the plate separation.

Working C = κε₀(A/d) = (3.0)(8.85 × 10⁻¹² C²/(N·m²))(0.020 m²)/(1.5 × 10⁻³ m) = 3.54 × 10⁻¹⁰ F ≈ 3.5 × 10⁻¹⁰ F.

CED 10.6.A.2.ii · Read this in Fix

1. AHalved
   A student who assumes the potential difference stays the same uses E = ΔV/d with ΔV fixed and picks this. The capacitor is disconnected, so it is the charge that stays fixed; ΔV doubles along with d, and E is unchanged.
2. BDoubled
   A student who takes the field to be set by the potential difference alone picks this: ΔV doubles, so the field seems to double. The field is the potential difference per unit distance, ΔV/d, and both ΔV and d double.
3. CQuartered
   A student who treats the plates like point charges, with a field that falls off as 1/d², picks this. Between large parallel plates the field does not depend on distance: E_C = Q/(κε₀A).
4. DThe same Correct
   E_C = Q/(κε₀A) depends on the charge per unit area of the plates, not on their separation. Q, A and κ are all unchanged, so the field stays the same. (The potential difference doubles, because the same field now extends across twice the distance.)

Working E_C = Q/(κε₀A): Q, κ and A unchanged, so E_C is unchanged. Consistency check: C = κε₀A/d halves, so ΔV = Q/C doubles, and E = ΔV/d = (2ΔV₀)/(2d₀) = ΔV₀/d₀.

CED 10.6.A.3.i · Read this in Fix

1. AE = ΔV/(κd)
   A student who thinks a dielectric always reduces the field by the factor κ picks this. That is true when the charge is fixed. Here the battery fixes ΔV, so the charge on the plates increases by κ and the field is ΔV/d, as it would be without the dielectric.
2. BE = ΔV/d²
   A student who carries the r² of the point-charge field E = kq/r² into the plate relationship, writing E = ΔV/d², picks this. Substituting Q = κε₀AΔV/d into E_C = Q/(κε₀A) gives d to the first power: E = ΔV/d.
3. CE = ΔV/d Correct
   Q = CΔV = κε₀AΔV/d, and E_C = Q/(κε₀A). Substituting, κ, ε₀ and A cancel: E = ΔV/d. With the battery holding ΔV fixed, the dielectric increases the charge on the plates by the factor κ, which exactly offsets the κ in E_C.
4. DE = κ·ΔV/d
   A student who thinks a dielectric strengthens the field picks this: κ times the field there would be without it. The induced charges on the dielectric oppose the plates' field; with ΔV fixed, the extra charge drawn onto the plates exactly makes up for this, leaving E = ΔV/d.

Working C = κε₀A/d, so Q = CΔV = κε₀AΔV/d. E_C = Q/(κε₀A) = (κε₀AΔV/d)/(κε₀A) = ΔV/d.

CED 10.6.A.3.i · Read this in Fix

1. APath 1 Correct
   The field points from the positive top plate to the negative bottom plate, so the force on the negative electron is upward, toward the positive plate, and constant. As for a projectile in a uniform gravitational field, the electron keeps its horizontal velocity while its upward velocity grows at a steady rate, so its path is a parabola that curves toward the positive plate from the moment it enters.
2. BPath 2
   A student who takes the electric force on any charge to be in the direction of the field picks this, the parabola curving toward the negative plate. The force on a negative charge is opposite to the field, so the electron curves toward the positive plate.
3. CPath 3
   A student who thinks a charged particle moves along the field lines picks this, the path that turns at once and runs straight up across the gap. The force is along the field lines, but the electron keeps its horizontal velocity; the constant force bends the path gradually into a parabola, as gravity does for a ball thrown horizontally.
4. DPath 4
   A student who thinks the electron's initial motion carries it straight ahead for a while before the force takes effect picks this, the path that runs straight and only then bends. The force acts from the moment the electron enters the field, so the path starts to curve at once.

Working F⃗_E = qE⃗ with q = −e: E⃗ points down (from + to −), so the force on the electron is up and constant, a = eE/m_e. Horizontal velocity constant, vertical velocity increases steadily from zero: a parabola curving toward the positive (top) plate from the entry point.

CED 10.6.A.3.ii · Read this in Fix

1. A2.0 V Correct
   The capacitor is disconnected, so the charge on its plates stays the same. The dielectric multiplies the capacitance by κ = 2.0, so ΔV = Q/C = (4.0 V)/2.0 = 2.0 V. Equivalently, the dielectric halves the field between the plates, and ΔV = Ed.
2. B8.0 V
   A student who thinks a dielectric reduces the capacitance divides C by κ and gets ΔV = Q/(C₀/2.0) = 2.0 × 4.0 V = 8.0 V. A dielectric increases the capacitance, so the same charge gives a smaller potential difference.
3. C4.0 V
   A student who thinks a capacitor keeps the potential difference it was charged to picks this. Once disconnected, the capacitor keeps its charge, not its potential difference; with a larger capacitance, ΔV = Q/C falls.
4. D0.0 V
   A student who treats the dielectric as a conductor, which would cancel the field between the plates completely, picks this. A dielectric only partly cancels the field, reducing it by the factor κ, so ΔV falls to 2.0 V, not to zero.

Working Disconnected, so Q is fixed. The capacitance becomes κC₀ = 2.0C₀. ΔV = Q/C = Q/(2.0C₀) = (4.0 V)/2.0 = 2.0 V.

CED 10.6.A.6 · Read this in Fix

1. A4.00
   A student who carries the r² of the point-charge field E = kq/r² into the plate relationship, writing E = ΔV/d², picks this: doubling d then quarters the field, so the charge must be 4 times as large. Between large parallel plates E = ΔV/d, so doubling d halves the field and the charge must double.
2. B1.00
   A student who takes the field to be set by the potential difference alone, E = ΔV, picks this: with ΔV unchanged, the field and the charge needed seem unchanged. The field is the potential difference per unit distance, ΔV/d, so doubling d halves it and the charge must double.
3. C0.50
   A student who multiplies instead of dividing, E = ΔV·d, picks this: doubling d then seems to double the field, so half the charge would do. ΔV = Ed, so E = ΔV/d; plates farther apart give a weaker field and need a larger charge, twice as large.
4. D2.00 Correct
   Between the plates E = ΔV/d, so doubling d with ΔV fixed halves the field. The drop is at rest, so |q|E = mg; with mg unchanged and E halved, |q| must double: the factor is 2.00.

Working Between the plates E = ΔV/d. At rest, |q|E = mg, so |q| = mgd/ΔV ∝ d with m, g and ΔV fixed. Doubling d doubles |q|: factor 2.00. Errors: E = ΔV/d² → |q| ∝ d², factor 4.00; E = ΔV (d left out) → factor 1.00; E = ΔV·d → |q| ∝ 1/d, factor 0.50.

CED 10.6.A.3 · Read this in Fix

1. Av₀√(m/(qΔV))
   A student who writes the field as E = ΔV·d picks this: the acceleration becomes qΔVd/m and t² = m/(qΔV). Since ΔV = Ed, the field is E = ΔV/d, which gives a = qΔV/(md) and a distance v₀·d√(m/(qΔV)).
2. Bv₀√(md/(qΔV))
   A student who takes the field to be set by the potential difference alone, E = ΔV, picks this: the acceleration becomes qΔV/m and t² = md/(qΔV). The same ΔV across plates farther apart gives a weaker field, E = ΔV/d, so a = qΔV/(md) and the distance is v₀·d√(m/(qΔV)).
3. Cv₀·d√(m/(2qΔV))
   A student who takes the vertical distance to be the final vertical speed times the time, d/2 = (at)t, picks this: t² = md²/(2qΔV). The vertical velocity grows steadily from zero, so the average is half the final value and d/2 = (1/2)at², which gives t² = md²/(qΔV) and a distance v₀·d√(m/(qΔV)).
4. Dv₀·d√(m/(qΔV)) Correct
   The field between the plates is E = ΔV/d, so the particle's vertical acceleration is a = qΔV/(md). Starting with no vertical velocity, it moves d/2 vertically in a time t given by d/2 = (1/2)at², so t = d√(m/(qΔV)). As for a projectile, the horizontal velocity stays v₀, so the horizontal distance is v₀t = v₀·d√(m/(qΔV)).

Working The field between the plates is uniform, E = ΔV/d, so the particle has a constant vertical acceleration a = qE/m = qΔV/(md), toward the negative plate, and no horizontal acceleration. It starts with no vertical velocity and must move d/2 vertically: d/2 = (1/2)at², so t² = d/a = md²/(qΔV) and t = d√(m/(qΔV)). The horizontal velocity stays v₀, so x = v₀t = v₀·d√(m/(qΔV)). Errors: E = ΔV·d (m27): a = qΔVd/m, t² = m/(qΔV), x = v₀√(m/(qΔV)); E = ΔV, d left out (m16): a = qΔV/m, t² = md/(qΔV), x = v₀√(md/(qΔV)); distance = final speed × time, d/2 = at² (new): t² = md²/(2qΔV), x = v₀·d√(m/(2qΔV)).

CED 10.6.A.3.ii · Read this in Fix