5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
An uncharged parallel-plate capacitor with air between its plates is connected by wires to a battery. Which statement describes the movement of charge while the capacitor charges?
Answer and reasoning
AElectrons cross the gap from one plate to the other, making the plates oppositely charged. A student who pictures the current flowing across the gap, as it flows around a loop, picks this. The air gap is an insulator; charge moves through the wires and battery, from one plate to the other, and never crosses the gap.
BThe battery adds electrons to both plates, giving the capacitor a net negative charge. A student who thinks a battery puts extra charge into a capacitor picks this. The battery only moves charge from one plate to the other; the plates carry +Q and −Q, so the net charge of the capacitor stays zero.
CA plate becomes positive by gaining protons, which the battery sends to it along the wires. A student who thinks positive charges move in metals picks this. The nuclei of a metal stay in place; a plate becomes positive by losing electrons, which are the charges that move through the wires and battery.
DElectrons move from one plate through the wires and battery to the other; none cross the gap.Correct The metal plates and wires contain free electrons. The battery moves electrons away from one plate, which becomes positive, and onto the other, which becomes negative, through the wires and the battery. The air gap is an insulator, so no charge crosses it, and the plates end up with charges of equal magnitude and opposite sign.
An air-filled parallel-plate capacitor is charged and then disconnected. Which of the following changes would change its capacitance?
Answer and reasoning
AAdding more charge until each plate carries twice its charge A student who thinks capacitance grows with the charge stored picks this. Doubling the charge doubles ΔV as well, so C = Q/ΔV is unchanged.
BFilling the gap between the plates with a sheet of glassCorrect Capacitance depends only on the physical properties of the capacitor: its shape, the area and separation of the plates, and the material between them. Replacing the air with glass changes that material, so it changes the capacitance.
CCharging the capacitor again to twice the potential difference A student who reads ΔV in C = Q/ΔV as controlling C picks this. At twice the potential difference the charge on each plate is also twice as large, so the ratio Q/ΔV is unchanged.
DReplacing the plates with thicker plates, with the same gap A student who thinks of capacitance as how much charge the plates can hold, like a container, picks this. The charge sits on the facing surfaces; with the same facing area and the same gap, the capacitance is the same however thick the plates are.
A proton enters the uniform electric field between two oppositely charged, horizontal parallel plates, moving parallel to the plates. Gravitational effects are negligible. Which statement describes the proton's motion while it is between the plates?
Answer and reasoning
AIt turns to follow a field line, then moves straight toward the negative plate. A student who thinks charged particles travel along field lines picks this. The field line gives the direction of the force and acceleration, not of the velocity; the proton keeps its velocity parallel to the plates while it gains a perpendicular component.
BIt follows a parabolic path, with constant acceleration toward the negative plate.Correct The field between the plates is uniform, so the force eE on the proton is constant in magnitude and direction, toward the negative plate. Its acceleration is constant, and with an initial velocity perpendicular to that acceleration it follows a parabola, as a horizontally launched projectile does.
CIt curves toward the negative plate with an acceleration that grows as it nears it. A student who thinks the field is stronger near a plate, as it is near a point charge, picks this. The field between the plates is uniform, so the proton's acceleration is the same everywhere between them.
DIt moves along a circular arc, since the force on it is perpendicular to its velocity. A student who applies circular motion because the force starts out perpendicular to the velocity picks this. The electric force keeps the same direction while the velocity turns, so the force does not stay perpendicular; the path is a parabola.
A capacitor is charged from zero by a battery. The electric potential energy stored in the charged capacitor is equal to which of the following?
Answer and reasoning
AThe work done by an external agent to move the charge from one plate to the otherCorrect Charging moves charge from one plate to the other against the electric force of the charges already separated. The external agent, here the battery, does positive work on the charge, and that work is stored as the electric potential energy of the separated charge.
BThe work done by the electric field between the plates as the charge builds up A student who confuses work done by the field with work done by the external agent picks this. The charge is moved against the electric force, so the field does negative work; the stored energy equals the agent's positive work.
CThe kinetic energy gained by charges that cross the gap between the plates A student who thinks charge crosses the gap while the capacitor charges picks this. No charge crosses the insulating gap; the energy is stored in the separated charges on the plates, not as kinetic energy.
DThe product of the final charge Q and the final potential difference ΔV A student who thinks every bit of charge is moved through the final ΔV picks this. The first charge is moved through almost no potential difference, and ΔV grows as charge is added, so the work is (1/2)QΔV, half of QΔV.
An air-filled parallel-plate capacitor stays connected to an ideal battery, which keeps the potential difference across it constant. The energy stored in the capacitor is U₀. The plates are moved apart until their separation is four times as large, still much smaller than the plate dimensions. What is the new stored energy U?
Answer and reasoning
AU = 4U₀ A student who treats the charge as fixed, as for an isolated capacitor, picks this: U = Q²/(2C) with C divided by 4. The battery keeps ΔV constant, so the charge falls to a quarter and U = (1/2)CΔV² falls to a quarter.
BU = U₀/4Correct The battery holds ΔV constant. Quadrupling the separation divides the capacitance by 4, C = ε₀A/d, so UC = (1/2)QΔV = (1/2)CΔV² is divided by 4 as well; charge flows back to the battery, and Q is also a quarter of its original value.
CU = U₀ A student who thinks the stored energy is set by the battery's potential difference alone picks this. With ΔV fixed, U = (1/2)CΔV² is proportional to C, which falls to a quarter.
DU = U₀/16 A student who takes the capacitance to be inversely proportional to the square of the separation picks this, dividing C, and so U, by 16. C = ε₀A/d is inversely proportional to d, so U falls to a quarter.
Working ΔV fixed by the battery. C = ε₀A/d → C₀/4. U = (1/2)CΔV² → U₀/4. (Q = CΔV also falls to Q₀/4.)
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
10.3.A.1 Parallel-plate capacitor Fix
Parallel-plate capacitor
Two parallel conducting plates separated by a gap of air, vacuum or an insulating material. When it is charged, one plate carries charge +Q and the other −Q, so the capacitor as a whole has zero net charge.
Students often think While a capacitor charges, charge crosses the gap from one plate to the other. In fact No. The gap is an insulator (air, vacuum or another material), and no charge crosses it. Charge moves through the wires and the battery, from the plate that becomes positive to the plate that becomes negative.
Students often think A battery puts charge into a capacitor, so a charged capacitor carries a net charge. In fact No. Charging moves charge from one plate to the other, so the plates carry +Q and −Q and the capacitor's net charge is zero.
10.3.A.2 Charge on a capacitor (Q) Fix
Charge on a capacitor (Q)
The magnitude of the charge on either plate, not the sum of the charges on both plates (which is zero). SI unit: coulomb (C).
Capacitance (C)
C = Q/ΔV: the magnitude of the charge on each plate divided by the potential difference between the plates. SI unit: farad (F); 1 F = 1 C/V.
Potential difference between the plates (ΔV)
The difference between the electric potentials of the two plates, created by the separation of charge between them. SI unit: volt (V).
Dependence of capacitance on physical properties
A capacitor's capacitance is set by its geometry (the shape, size and spacing of its conductors) and by the material between the conductors. It does not depend on the charge stored or on ΔV: if Q changes, ΔV changes in proportion and Q/ΔV stays the same.
Concentric spherical capacitor
A conducting sphere of radius a inside a concentric conducting spherical shell of radius b, carrying charges +Q and −Q. With vacuum or air between them, C = 4πε₀ab/(b − a).
Coaxial cylindrical capacitor
A conducting cylinder of radius a inside a coaxial conducting cylindrical shell of radius b, both of length L much greater than b, carrying charges +Q and −Q. With vacuum or air between them, C = 2πε₀L/ln(b/a).
Capacitance of a parallel-plate capacitor
C = κε₀A/d, where A is the area of one plate, d is the separation of the plates and κ is the dielectric constant of the material between them (κ = 1.0 for air). It applies when d is much smaller than the dimensions of the plates.
Electric permittivity of free space (ε₀)
The constant in Gauss's law, ε₀ = 8.85 × 10⁻¹² C²/(N·m²), with k = 1/(4πε₀). The product of ε₀ and a length (for example ε₀A/d) has units of farads.
Students often think Q in C = Q/ΔV is the sum of the magnitudes of the charges on the two plates. In fact No. Q is the magnitude of the charge on each plate. A capacitor with +6.0 μC on one plate and −6.0 μC on the other has Q = 6.0 μC.
Students often think The ΔV in C = Q/ΔV is the potential of the positive plate. In fact No. ΔV is the potential difference between the two plates, the potential of the positive plate minus that of the negative plate. The potential of one plate depends on where V = 0 is chosen.
10.3.A.3 Uniform field between the plates Fix
Uniform field between the plates
Between two oppositely charged parallel plates the electric field has the same magnitude and direction at every point, pointing from the positive plate toward the negative plate, except near the edges of the plates, where the field lines bulge outward and the field is no longer uniform in magnitude or direction.
Field of a parallel-plate capacitor
By Gauss's law, each large plate with surface charge density σ produces a field of magnitude σ/(2ε₀) on both sides of it. Between the plates the two fields point the same way and add, giving E = σ/ε₀ = Q/(ε₀A); outside the plates they point in opposite directions and cancel. SI unit: N/C (= V/m).
Surface charge density (σ)
The charge per unit area on a plate, σ = Q/A. SI unit: C/m². The field between the plates is E = σ/ε₀, so it is proportional to σ.
Potential difference in a uniform field
From ΔV = −∫E⃗·dr⃗, the potential changes linearly with position between the plates, and the magnitude of the potential difference across a separation d is ΔV = Ed. The magnitude of the slope of a graph of V against x is E.
Charged particle between the plates
A particle of charge q and mass m in the uniform field has a constant acceleration of magnitude |q|E/m, along the field for positive q and opposite to it for negative q. If its initial velocity is parallel to the plates, it follows a parabolic path, as a projectile does near Earth's surface.
Students often think The field of a charged plate weakens with distance from it, as the field of a point charge does (∝ 1/r²), so the field between two plates is strongest near each plate. In fact No. Near a large, uniformly charged plate the field is σ/(2ε₀) at any distance, as long as the distance is small compared with the plate's size. Between two such plates the field is the same everywhere away from the edges.
Students often think The fields of the two plates add as magnitudes, so the field outside the plates is as strong as the field between them. In fact No. Each plate's field points away from a positive plate and toward a negative plate, on both sides of it. Between the plates the two fields point the same way and add; outside they point in opposite directions and cancel.
10.3.A.4 Work done to charge a capacitor Fix
Work done to charge a capacitor
An external agent, such as a battery, moves each small charge dq from one plate to the other against the electric force, doing work dW = ΔV dq = (q/C) dq, where q is the charge already on the plates. The total work to reach charge Q is Q²/(2C), the area under a graph of ΔV against q; it is stored as electric potential energy.
Students often think The force on one plate is its charge times the full field between the plates, σ/ε₀. In fact No. A plate cannot exert a force on itself; it experiences only the field of the other plate, σ/(2ε₀). The force on it is Q·Q/(2ε₀A) = Q²/(2ε₀A), half of Q times the field between the plates.
Students often think The energy stored in a capacitor equals the work done by the electric field between the plates as the charge builds up. In fact No. As the capacitor charges, each charge is moved against the electric force, so the field does negative work. The stored energy equals the positive work done by the external agent (the battery), which is the negative of the field's work.
10.3.A.5 Electric potential energy stored in a capacitor (UC) Fix
Electric potential energy stored in a capacitor (UC)
UC = (1/2)QΔV, equivalently Q²/(2C) or (1/2)CΔV². SI unit: joule (J).
Students often think The energy stored in a capacitor is QΔV, since every charge is moved through the potential difference ΔV. In fact No. The early charge is moved through a small potential difference and only the last through the full ΔV. The work, and the stored energy, is the area under a graph of ΔV against q, (1/2)QΔV.
Students often think A capacitor's charge stays the same when its capacitance changes, even while it stays connected to a battery. In fact No. The battery holds ΔV fixed, so the charge Q = CΔV changes with C; charge flows between the battery and the plates. Only an isolated capacitor keeps its charge.
16 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 16
One plate of a parallel-plate capacitor carries a charge of +6.0 μC and the other carries −6.0 μC. The positive plate is at an electric potential of +2.5 V and the negative plate is at −1.5 V. What is the capacitance of the capacitor?
Answer and reasoning
A1.5 μFCorrect C = Q/ΔV, where Q is the magnitude of the charge on each plate, 6.0 μC, and ΔV is the potential difference between the plates: 2.5 V − (−1.5 V) = 4.0 V. So C = (6.0 μC)/(4.0 V) = 1.5 μF.
B3.0 μF A student who adds the charges on both plates, 6.0 μC + 6.0 μC = 12 μC, picks this: (12 μC)/(4.0 V). Q in C = Q/ΔV is the magnitude of the charge on each plate, 6.0 μC.
C2.4 μF A student who uses the potential of the positive plate, 2.5 V, in place of the potential difference picks this: (6.0 μC)/(2.5 V). The potential difference between the plates is 2.5 V − (−1.5 V) = 4.0 V.
D6.0 μF A student who subtracts the sizes of the potentials, 2.5 V − 1.5 V = 1.0 V, picks this. Subtracting a negative potential adds its magnitude: ΔV = 2.5 V − (−1.5 V) = 4.0 V.
Working Q is the magnitude of the charge on each plate: Q = 6.0 μC. ΔV = V₊ − V₋ = 2.5 V − (−1.5 V) = 4.0 V. C = Q/ΔV = (6.0 μC)/(4.0 V) = 1.5 μF.
A capacitor consists of a metal sphere of radius 2.0 cm inside a thin, concentric metal shell of radius 3.0 cm, with air between them. The sphere carries charge +Q and the shell carries −Q. What is the capacitance? Use k = 9.0 × 10⁹ N·m²/C².
Answer and reasoning
A1.3 × 10⁻¹² F A student who adds the potentials of the sphere and the shell as magnitudes picks this: ΔV = kQ(1/a + 1/b), giving C = ab/(k(a + b)). The shell's charge is negative, so its contribution, −kQ/b, is subtracted.
B4.4 × 10⁻¹² F A student who uses the parallel-plate formula with A = 4πa² and d = b − a picks this: ε₀(4πa²)/(b − a) = a²/(k(b − a)). The field between the spheres is not uniform, so C must be found from ΔV = ∫E dr.
C6.7 × 10⁻¹² FCorrect The potential of the sphere is kQ/a from its own charge plus −kQ/b from the shell (the shell's potential is the same everywhere inside it), and the shell's potential is kQ/b − kQ/b = 0. So ΔV = kQ(1/a − 1/b) and C = Q/ΔV = ab/(k(b − a)) = (0.020 m)(0.030 m)/[(9.0 × 10⁹)(0.010 m)] = 6.7 × 10⁻¹² F.
D2.2 × 10⁻¹² F A student who takes the shell's potential inside it to be zero, because its field is zero there, picks this: ΔV = kQ/a, so C = a/k. The shell's field is zero inside, but its potential there is −kQ/b, not zero.
Working Between the conductors (a < r < b) a Gaussian sphere encloses +Q, so E = kQ/r². ΔV = ∫ab kQ/r² dr = kQ(1/a − 1/b) = kQ(b − a)/(ab). C = Q/ΔV = ab/(k(b − a)) = (0.020 m)(0.030 m)/[(9.0 × 10⁹ N·m²/C²)(0.010 m)] = 6.7 × 10⁻¹² F.
A coaxial cylindrical capacitor of length L consists of a metal cylinder of radius a inside a thin metal cylindrical shell of radius b, with vacuum between them; L is much greater than b. Using Gauss's law, which expression gives the capacitance?
Answer and reasoning
A4πε₀abL/(b − a) A student who uses a 1/r² field, as for a point charge, picks this: E = λ/(4πε₀r²) integrates to λ(b − a)/(4πε₀ab). By Gauss's law the field of a long cylinder of charge falls off as 1/r, and its integral gives a logarithm.
B2πε₀L/ln(b/a)Correct A Gaussian cylinder of radius r between the conductors encloses only the inner cylinder's charge, so E(2πrℓ) = λℓ/ε₀ and E = λ/(2πε₀r), with λ = Q/L. Then ΔV = ∫ab E dr = (λ/(2πε₀))ln(b/a), and C = Q/ΔV = λL/ΔV = 2πε₀L/ln(b/a).
C2πε₀aL/(b − a) A student who uses the parallel-plate formula, with the area of the inner cylinder, 2πaL, and the gap b − a, picks this. The field between the cylinders is not uniform (it falls as 1/r), so the capacitance must come from ΔV = ∫E dr.
Dπε₀L/ln(b/a) A student who adds a field from the outer cylinder's charge to the field of the inner cylinder picks this: the field, and so ΔV, doubles and C halves. A Gaussian cylinder between the conductors does not enclose the outer cylinder's charge, which produces no field inside it.
Working Charge +Q on the inner cylinder, −Q on the shell; λ = Q/L. Gaussian cylinder of radius r (a < r < b), length ℓ: E(2πrℓ) = λℓ/ε₀, so E = λ/(2πε₀r) (the shell's charge is outside the surface). ΔV = ∫ab λ/(2πε₀r) dr = (λ/(2πε₀))ln(b/a). C = Q/ΔV = λL/ΔV = 2πε₀L/ln(b/a). Distractors: E ∝ 1/r² → ΔV = (λ/(4πε₀))(1/a − 1/b) → C = 4πε₀abL/(b − a); parallel-plate with A = 2πaL, d = b − a → 2πε₀aL/(b − a); both cylinders' fields added (2λ/(2πε₀r)) → πε₀L/ln(b/a). Checked with sympy.
An air-filled parallel-plate capacitor has capacitance C₀. It is rebuilt with plates of three times the area and a separation four times as large, still with air between the plates. What is its new capacitance C?
Answer and reasoning
AC = 3C₀/4Correct For a parallel-plate capacitor C = κε₀A/d: C is proportional to the plate area and inversely proportional to the separation. Tripling the area multiplies C by 3; quadrupling the separation divides it by 4; so C = 3C₀/4.
BC = 12C₀ A student who thinks charge is stored in the space between the plates, so that C grows with the volume A·d, picks this: 3 × 4 = 12. A larger gap gives a larger ΔV for the same charge, so C is inversely proportional to d.
CC = 3C₀/16 A student who uses an inverse-square dependence on separation, as for the Coulomb force, picks this: 3/4² = 3/16. The field between large plates does not depend on d, so ΔV = Ed is proportional to d and C to 1/d.
DC = 3C₀ A student who thinks capacitance is set by the size of the plates alone picks this, multiplying by 3 for the area and ignoring the separation. For the same charge, a separation four times as large gives four times the ΔV, so C is divided by 4.
Working C = ε₀A/d, so C ∝ A/d. New C = ε₀(3A)/(4d) = (3/4)C₀.
The diagram shows three air-filled parallel-plate capacitors, X, Y and Z (not to scale), with the plate area, the plate separation and the charge on each plate labeled. Which ranking of their capacitances is correct?
Answer and reasoning
AY > Z > X A student who thinks C grows with the volume between the plates, A·d, picks this: X has 1, Y has 6 and Z has 3. A larger gap gives a larger ΔV for the same charge, so a larger separation lowers C.
BZ > X > Y A student who ranks by the charge on the plates, 3Q, 2Q and Q, picks this. The charge on a capacitor does not set its capacitance; ΔV changes in proportion to Q, so C = Q/ΔV depends only on A and d.
CX > Y > Z A student who takes C to be proportional to A/d², as in the inverse-square law, picks this: X has 1, Y has 3/4 and Z has 1/9. C is inversely proportional to d, not d², which puts Y above X.
DY > X > ZCorrect Capacitance depends only on the capacitor's construction: C = ε₀A/d, whatever charge it carries. In units of ε₀A/d: X has 1, Y has 3/2 and Z has 1/3. So Y > X > Z; the labeled charges do not affect the ranking.
Working C ∝ A/d, independent of Q. X: A/d = 1; Y: 3A/2d = 1.5; Z: A/3d = 0.33. So Y > X > Z. (A·d: 1, 6, 3 → Y > Z > X; by charge: 3, 2, 1 → Z > X > Y; A/d²: 1, 0.75, 0.11 → X > Y > Z.)
A student changes the separation d of the plates of an air-filled parallel-plate capacitor and measures its capacitance C each time. The graph shows C plotted against 1/d. Which conclusion about C is supported by the graph, with valid reasoning?
Answer and reasoning
AC is proportional to d, because C increases steadily moving to the right along the horizontal axis. A student who reads the horizontal axis as d rather than 1/d picks this. Moving right along the axis means 1/d increases, so d decreases; C increases as the plates get closer.
BC decreases linearly as d increases, because all of the data points lie on a single straight line. A student who takes any straight-line graph to mean a linear relationship with d picks this. The line is straight against 1/d, so C ∝ 1/d; against d itself the graph would be a curve, with C falling more and more slowly.
CC is inversely proportional to d, because the graph is a straight line through the origin.Correct A straight line through the origin on a graph of C against 1/d means C = (constant) × (1/d), so C is inversely proportional to d, as C = ε₀A/d predicts. The slope, about 1.8 × 10⁻¹³ F·m, is ε₀A.
DC is equal to the slope of the line, so it has one value at every separation d. A student who confuses the slope of a graph with the quantity plotted picks this. The plotted C rises from about 35 pF to about 177 pF as 1/d increases; the constant slope is ε₀A, in F·m, not a capacitance.
Working The data lie on a straight line through the origin of C against 1/d, so C = (slope)(1/d): C ∝ 1/d. Slope = 177 pF/(1000 m⁻¹) = 1.77 × 10⁻¹³ F·m = ε₀A.
The diagram shows an edge view of two large, parallel plates with charges +Q and −Q, far from other objects (not to scale). Points P, R and T are marked on the diagram; all three are far from the edges of the plates. Which ranking of the magnitudes of the electric field at the three points is correct?
Answer and reasoning
AER > EP > ET A student who thinks a plate's field weakens with distance, like a point charge's, picks this: R is very close to the negative plate. A large plate's field does not depend on the distance from it, so the field is the same at P and R.
BEP = ER > ETCorrect Between two large, oppositely charged plates the field is uniform: each plate contributes σ/(2ε₀) at any distance, and between the plates the two contributions point the same way, giving σ/ε₀ at both P and R. At T the two contributions point in opposite directions and cancel, so the field there is nearly zero.
CEP = ER = ET A student who adds the two plates' field magnitudes everywhere picks this. Outside the plates the positive plate's field points away from it and the negative plate's field points toward the negative plate, which is the opposite direction, so they cancel at T.
DET > EP = ER A student who thinks the fields of opposite charges cancel between them picks this. Between the plates both fields point from the positive plate toward the negative plate and add; they cancel outside.
Working Each plate's field has magnitude σ/(2ε₀), independent of distance. Between the plates both point from + to − and add: EP = ER = σ/ε₀. At T (outside) they point opposite ways and cancel: ET ≈ 0.
Two parallel metal plates, each of area A, carry charges +Q and −Q spread uniformly over their facing surfaces. The separation of the plates is much smaller than their dimensions. Using Gauss's law and the principle of superposition, which gives the magnitudes of the electric field between the plates and outside them, away from the edges?
Answer and reasoning
AQ/(ε₀A) both between and outside the plates A student who adds the magnitudes of the two plates' fields in every region picks this. Outside the plates the two fields point in opposite directions, so superposition gives zero there.
BZero between the plates, and Q/(ε₀A) outside A student who thinks the fields of opposite charges cancel where the charges face each other picks this. Between the plates both fields point from the positive plate toward the negative plate and add; they cancel outside.
C2Q/(ε₀A) between the plates, zero outside them A student who gives each plate the conductor-surface field σ/ε₀ and then adds the two picks this, counting each plate's contribution twice. Each plate alone produces σ/(2ε₀); the total between the plates is σ/ε₀ = Q/(ε₀A).
DQ/(ε₀A) between the plates, and zero outside themCorrect Gauss's law with a pillbox that straddles one plate gives that plate's field, σ/(2ε₀) = Q/(2ε₀A), on both sides of it. Between the plates the positive plate's field (away from it) and the negative plate's field (toward it) point the same way and add to Q/(ε₀A); outside they point in opposite directions and cancel.
Working One large sheet, σ = Q/A: pillbox with faces on both sides, 2EAp = σAp/ε₀, E = σ/(2ε₀) = Q/(2ε₀A), directed away from + and toward −. Between: both fields point from + to −: E = 2 × Q/(2ε₀A) = Q/(ε₀A). Outside: opposite directions, E = 0.
The graph shows the electric potential V as a function of position x between the plates of an isolated, air-filled parallel-plate capacitor; the plates are at the two ends of the graph. What is the magnitude of the surface charge density on each plate? Use ε₀ = 8.85 × 10⁻¹² C²/(N·m²).
Answer and reasoning
A4.4 × 10⁻⁷ C/m²Correct The field magnitude is the slope of V against x: E = (150 V)/(3.0 × 10⁻³ m) = 5.0 × 10⁴ V/m. Between the plates E = σ/ε₀, so σ = ε₀E = (8.85 × 10⁻¹²)(5.0 × 10⁴) = 4.4 × 10⁻⁷ C/m².
B8.9 × 10⁻⁷ C/m² A student who uses the field of a single sheet, E = σ/(2ε₀), picks this: σ = 2ε₀E. Between the plates both plates' fields add, so E = σ/ε₀ and σ = ε₀E.
C2.2 × 10⁻⁷ C/m² A student who gives each plate a field of σ/ε₀ and adds the two, E = 2σ/ε₀, picks this: σ = ε₀E/2. Each plate contributes σ/(2ε₀), for a total of σ/ε₀.
D1.3 × 10⁻⁹ C/m² A student who uses the potential, 150 V, in place of the field picks this: ε₀ × 150. The field is the slope of the graph, (150 V)/(3.0 × 10⁻³ m); the units show the difference: ε₀ × (V/m) gives C/m², ε₀ × V does not.
Working From the graph: V falls from 150 V at x = 0 to 0 at x = 3.0 mm. E = |ΔV/Δx| = 150 V/(3.0 × 10⁻³ m) = 5.0 × 10⁴ V/m. E = σ/ε₀ ⇒ σ = ε₀E = (8.85 × 10⁻¹²)(5.0 × 10⁴) = 4.4 × 10⁻⁷ C/m².
An air-filled parallel-plate capacitor is charged and then disconnected from everything, so it is isolated. The magnitude of the field between its plates is 6.0 × 10⁴ N/C. The plates are then pulled apart until their separation is four times as large; the separation is still much smaller than the plate dimensions. What is the new magnitude of the field between the plates?
Answer and reasoning
A2.4 × 10⁵ N/C A student who takes the field to be proportional to the potential difference, whatever the separation, picks this: ΔV becomes four times as large. The field is ΔV/d, and d is also four times as large, so the field is unchanged.
B1.5 × 10⁴ N/C A student who thinks ΔV stays fixed picks this: E = ΔV/d with d four times as large. The capacitor is isolated, so its charge, not its ΔV, stays fixed; σ and so the field are unchanged.
C6.0 × 10⁴ N/CCorrect The isolated capacitor keeps its charge, and its plate area is unchanged, so the surface charge density σ = Q/A is unchanged. The field between the plates is E = σ/ε₀, proportional to σ and independent of the separation, so it stays 6.0 × 10⁴ N/C; it is ΔV = Ed that increases, to four times its value.
D3.8 × 10³ N/C A student who thinks the field between the plates weakens as the inverse square of their separation, as for point charges, picks this: 6.0 × 10⁴ N/C divided by 16. The field of a large plate does not depend on distance; E = σ/ε₀ is unchanged.
Working Isolated: Q fixed, A fixed ⇒ σ = Q/A fixed. E = σ/ε₀ does not depend on d, so E = 6.0 × 10⁴ N/C (unchanged); ΔV = Ed becomes four times as large. Distractors: E ∝ ΔV → ×4 = 2.4 × 10⁵; ΔV fixed, E = ΔV/d → ÷4 = 1.5 × 10⁴; E ∝ 1/d² → ÷16 = 3.8 × 10³ N/C.
An electron (charge magnitude e, mass m) enters the region between two parallel plates with speed v₀ parallel to the plates, midway between them, as shown in the diagram. The potential difference between the plates is ΔV, and the electron leaves the region without striking a plate. Gravitational effects are negligible. Which expression gives the magnitude of the electron's velocity component perpendicular to the plates as it leaves the region?
Answer and reasoning
AeΔVL/(mv₀) A student who uses the potential difference as the field, E = ΔV, picks this. The field is the potential difference per unit distance, E = ΔV/d; the expression also has the wrong units for a speed.
B√(2eΔV/m) A student who thinks the electron gains kinetic energy eΔV, the full potential difference, picks this. The electron starts midway and does not reach a plate, so it moves through only part of ΔV; the perpendicular velocity must come from a = eΔV/(md) acting for t = L/v₀.
CeΔVL/(2mdv₀) A student who takes the final velocity to be half of at picks this, confusing it with the average velocity. Starting from zero with constant acceleration, v = at; at/2 is the average perpendicular velocity.
DeΔVL/(mdv₀)Correct The field between the plates is uniform, E = ΔV/d, so the electron has constant acceleration a = eE/m = eΔV/(md) perpendicular to the plates, like a projectile in a gravitational field. Its velocity component parallel to the plates stays v₀, so it is between the plates for t = L/v₀. Starting from zero, the perpendicular component becomes at = eΔVL/(mdv₀).
Working E = ΔV/d; a = eE/m = eΔV/(md), perpendicular to the plates and constant. Parallel component constant: time between plates t = L/v₀. v⊥ = at = eΔVL/(mdv₀). Distractors: E = ΔV → eΔVL/(mv₀); ΔK = eΔV for full pd → √(2eΔV/m); v = at/2 → eΔVL/(2mdv₀). Checked with sympy.
An isolated parallel-plate capacitor with plate area A has charges +Q and −Q on its plates, which are separated by a distance d much smaller than the plate dimensions; there is vacuum between them. An external agent slowly pulls the plates apart until their separation is 2d. Which expression gives the work done by the external agent?
Answer and reasoning
AQ²d/(ε₀A) A student who takes the force on a plate as its charge times the full field between the plates, Q × Q/(ε₀A), picks this. A plate is acted on only by the other plate's field, σ/(2ε₀), so the force, and the work, are half as large.
B−Q²d/(4ε₀A) A student who keeps ΔV fixed while the plates separate picks this: U = (1/2)CΔV² would fall to half its value. The capacitor is isolated, so Q stays fixed; ΔV rises and the stored energy increases, so the agent does positive work.
CQ²d/(2ε₀A)Correct The work done by the external agent equals the increase in stored energy. With Q fixed, U = Q²/(2C) = Q²x/(2ε₀A) at separation x, so W = Q²(2d)/(2ε₀A) − Q²d/(2ε₀A) = Q²d/(2ε₀A). The same result follows from the constant force Q²/(2ε₀A) that one plate exerts on the other, acting over the distance d.
DkQ²/(2d) A student who treats the plates as point charges attracting with force kQ²/x² picks this, integrating from d to 2d. The field of a large plate does not fall off with distance, so the attraction is constant, Q²/(2ε₀A).
Working Isolated: Q fixed. U = Q²/(2C) = Q²x/(2ε₀A). W = U(2d) − U(d) = Q²(2d − d)/(2ε₀A) = Q²d/(2ε₀A). Check by force: each plate is in the other's field σ/(2ε₀), so F = Q²/(2ε₀A), constant; W = Fd = Q²d/(2ε₀A). Distractors: full field σ/ε₀ → F = Q²/(ε₀A), W = Q²d/(ε₀A); ΔV held fixed → U = (1/2)CΔV² halves, ΔU = −Q²d/(4ε₀A); Coulomb point charges → ∫d2d kQ²/x² dx = kQ²/(2d). Checked with sympy.
The graph shows the potential difference ΔV across a capacitor as a function of the charge Q on its plates. How much work must be done by an external agent to increase the charge on the capacitor from 16 μC to 32 μC?
Answer and reasoning
A3.2 × 10⁻⁵ J A student who applies (1/2)QΔV to the changes, (1/2)(16 μC)(4.0 V), picks this. That is only the triangle above the 4.0 V level; the charge added is also moved through the 4.0 V already there, which adds a rectangle of 64 μJ.
B1.9 × 10⁻⁴ J A student who takes the stored energy as QΔV picks this: (32 μC)(8.0 V) − (16 μC)(4.0 V). The energy stored at each charge is (1/2)QΔV, the area of the triangle under the graph.
C1.3 × 10⁻⁴ J A student who takes the work to be the energy stored at the end, (1/2)(32 μC)(8.0 V), picks this. The capacitor already stored (1/2)(16 μC)(4.0 V) = 32 μJ; the agent supplies only the increase, 96 μJ.
D9.6 × 10⁻⁵ JCorrect Moving a small charge dQ onto the plates takes work ΔV dQ, so the work is the area under the graph between 16 μC and 32 μC. From the graph ΔV is 4.0 V at 16 μC and 8.0 V at 32 μC, so the area is (1/2)(32 μC)(8.0 V) − (1/2)(16 μC)(4.0 V) = 96 μJ = 9.6 × 10⁻⁵ J.
Working From the graph ΔV = 4.0 V at 16 μC and 8.0 V at 32 μC (C = 4.0 μF). W = area under ΔV against Q between the two charges = (1/2)(32 μC)(8.0 V) − (1/2)(16 μC)(4.0 V) = 128 μJ − 32 μJ = 96 μJ = 9.6 × 10⁻⁵ J.
The uniform electric field between the plates of an air-filled parallel-plate capacitor has magnitude E. Each plate has area A, and the plates are a distance d apart. Which expression gives the electric potential energy stored in the capacitor?
Answer and reasoning
Aε₀E²Ad/2Correct The field between the plates is E = Q/(ε₀A), so the charge on each plate is Q = ε₀EA, and the potential difference across the uniform field is ΔV = Ed. Then UC = (1/2)QΔV = (1/2)(ε₀EA)(Ed) = ε₀E²Ad/2.
Bε₀E²Ad A student who uses U = QΔV picks this. As the capacitor charges, ΔV rises from zero, so the average potential difference through which the charge is moved is ΔV/2 and UC = (1/2)QΔV.
Cε₀E²A/(2d³) A student who writes the potential difference as ΔV = E/d and uses U = (1/2)CΔV² with C = ε₀A/d picks this: (1/2)(ε₀A/d)(E/d)². The field is the potential difference per unit length, so ΔV = Ed; the units check, (V/m) × m = V, shows this.
Dε₀E²A/(2d) A student who uses the field itself as the potential difference in U = (1/2)CΔV², with C = ε₀A/d, picks this: (1/2)(ε₀A/d)E². The potential difference across the gap is Ed; a field in V/m cannot stand in for a potential difference in V.
Working E = σ/ε₀ = Q/(ε₀A) ⇒ Q = ε₀EA. ΔV = Ed. UC = (1/2)QΔV = (1/2)(ε₀EA)(Ed) = ε₀E²Ad/2. Distractors: QΔV → ε₀E²Ad; ΔV = E/d in UC = (1/2)CΔV² with C = ε₀A/d → (1/2)(ε₀A/d)(E/d)² = ε₀E²A/(2d³); ΔV = E in UC = (1/2)CΔV² → (1/2)(ε₀A/d)E² = ε₀E²A/(2d). Checked with sympy.
Two large, parallel metal plates, each of area A, are far from other objects, and their separation is much smaller than their dimensions. One plate carries a net charge +4Q and the other carries a net charge −Q. Which expression gives the magnitude of the electric field between the plates, away from their edges?
Answer and reasoning
A5.0 Q/(ε₀A) A student who gives each plate a field of σ/ε₀ picks this: 4Q/(ε₀A) + Q/(ε₀A). σ/ε₀ is the total field just outside a conductor; each plate on its own, as a large sheet, contributes σ/(2ε₀).
B1.5 Q/(ε₀A) A student who thinks the fields of a positive and a negative plate cancel between the plates picks this: 2.0 Q/(ε₀A) − 0.50 Q/(ε₀A). A positive test charge between the plates is pushed away from the positive plate and pulled toward the negative plate, both in the same direction, so the fields add.
C1.0 Q/(ε₀A) A student who pairs off +Q with −Q, treats the extra +3Q as having no effect between the plates, and uses E = Q/(ε₀A) for a capacitor picks this. The field of the extra charge does reach the gap: superposing the two plates' fields gives 2.5 Q/(ε₀A).
D2.5 Q/(ε₀A)Correct Each plate acts as a large sheet with field (its charge)/(2ε₀A): 2.0 Q/(ε₀A) from the +4Q plate and 0.50 Q/(ε₀A) from the −Q plate. Between the plates both fields point from the positive plate toward the negative plate, so they add to 2.5 Q/(ε₀A).
Working Gauss's law for one large sheet with total charge qs (pillbox through the sheet, flux through both ends): 2EAp = (qs/A)Ap/ε₀, so E = qs/(2ε₀A) on both sides, pointing away from a positive sheet and toward a negative one. Plate with +4Q: 4Q/(2ε₀A) = 2.0 Q/(ε₀A), pointing away from it. Plate with −Q: 0.50 Q/(ε₀A), pointing toward it. Between the plates both point from the +4Q plate toward the −Q plate, so they add: E = 2.5 Q/(ε₀A) (= 5Q/(2ε₀A)). (Check: E = 0 in each metal plate requires outer faces +1.5Q each and inner faces +2.5Q and −2.5Q; E = σinner/ε₀ = 2.5 Q/(ε₀A).) Distractors: each plate's field taken as σ/ε₀ → 4Q/(ε₀A) + Q/(ε₀A) = 5.0 Q/(ε₀A); fields of the + and − plates taken to cancel between them → 2.0 − 0.50 = 1.5 Q/(ε₀A); only the matched ±Q taken to act across the gap, with E = Q/(ε₀A) → 1.0 Q/(ε₀A). Checked with sympy.
Two large, parallel plates are a distance d apart, much smaller than their dimensions, and the potential difference between them is ΔV. A particle with charge +q and mass m is released from rest between the plates, a distance x from the positive plate. Gravitational effects are negligible. Which expression gives the time the particle takes to reach the negative plate?
Answer and reasoning
A√(2m(d − x)/(qdΔV)) A student who rearranges ΔV = E/d, so E = ΔVd, picks this: the acceleration becomes qΔVd/m. Between parallel plates ΔV = Ed, so E = ΔV/d; checking units shows this expression is not a time.
B√(2md(d − x)/(qΔV))Correct The field between the plates is uniform, E = ΔV/d, so the particle has constant acceleration qΔV/(md). Starting from rest, it covers d − x = (1/2)at², which gives t = √(2md(d − x)/(qΔV)).
C√(md(d − x)/(2qΔV)) A student who finds the final speed v correctly but then divides the distance d − x by v, as if the particle moved at that speed all the way, picks this. The particle starts from rest, so its average speed is v/2 and the time is twice as long.
D√(2m(d − x)²/(qΔV)) A student who thinks the particle gains kinetic energy qΔV, the whole potential difference between the plates, picks this: v = √(2qΔV/m) at the negative plate, and with average speed v/2, t = 2(d − x)/v. Starting a distance x from the positive plate, the particle moves through only the fraction (d − x)/d of ΔV.
Working Away from the edges the field between the plates is constant in magnitude and direction: E = ΔV/d, from the positive plate toward the negative plate. The force qE is constant, so the acceleration is constant: a = qΔV/(md). From rest, the distance d − x is covered in time t with d − x = (1/2)at², so t = √(2(d − x)/a) = √(2md(d − x)/(qΔV)). Distractors: ΔV = E/d, so E = ΔVd and a = qΔVd/m → √(2m(d − x)/(qdΔV)); final speed v = √(2qΔV(d − x)/(md)) used as if constant, t = (d − x)/v → √(md(d − x)/(2qΔV)); kinetic energy qΔV (the whole potential difference) at the negative plate, v = √(2qΔV/m), with average speed v/2, t = 2(d − x)/v → √(2m(d − x)²/(qΔV)). The key, the constant-speed and the full-ΔV options have units of time; the ΔV = E/d option does not. Checked with sympy (proton, ΔV = 100 V, d = 0.020 m, x = 0.005 m: key 2.5 × 10⁻⁷ s; distractors 1.3 × 10⁻⁵, 1.3 × 10⁻⁷, 2.2 × 10⁻⁷).
Compiled from the AP Physics C: Electricity and Magnetism Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
