5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
A slab of glass, an electrical insulator, is placed between the plates of an isolated, charged parallel-plate capacitor without touching them. Which statement describes the response of the charges in the glass?
Answer and reasoning
ACharges in each molecule shift slightly, leaving opposite charges on the slab's two faces.Correct In a dielectric the charges are bound to their molecules. The field pushes the positive and negative charges of each molecule slightly in opposite directions, so the glass becomes polarized: a thin layer of negative charge appears on the face nearer the positive plate and positive charge on the other face, while the slab stays neutral.
BElectrons flow freely through the glass until the field inside the glass is zero. A student who treats the glass like a conductor picks this. Glass has no free electrons; its charges shift only slightly, so the field inside it is reduced but not zero.
CNothing happens, because the glass contains no charges that the field is able to move. A student who thinks an insulator has no charges that respond to a field picks this. Every molecule of glass contains electrons and nuclei; they cannot flow through the glass, but the field shifts them slightly.
DThe glass gains a net negative charge, drawn out of the positive plate of the capacitor. A student who thinks the slab exchanges charge with the plates picks this. The glass does not touch the plates and is an insulator; polarization only separates its own charges, and its net charge stays zero.
A material has a dielectric constant κ = 4.0. Which statement about the material is correct?
Answer and reasoning
AIt lets charge pass through it 4.0 times as easily as air does. A student who reads 'permittivity' as how easily a material permits charge to flow picks this. Permittivity describes polarization, not conduction; a material with κ = 4.0 used in a capacitor is an insulator.
BAny field inside it is 4.0 times as strong as in a vacuum. A student who thinks a dielectric strengthens a field picks this. For the same charges, the polarized material reduces the field inside it, by the factor κ for a capacitor filled with it.
CIts value would double if a slab of it were twice as thick. A student who treats κ as depending on the size of the sample picks this. κ = ε/ε₀ is a property of the material, the same for a thin or a thick slab.
DIts permittivity is 4.0 times the permittivity of free space, ε₀.Correct The dielectric constant relates a material's permittivity to that of free space: κ = ε/ε₀. So κ = 4.0 means ε = 4.0ε₀ = 3.5 × 10⁻¹¹ C²/(N·m²).
Working κ = ε/ε₀, so ε = κε₀ = 4.0 × 8.85 × 10⁻¹² = 3.5 × 10⁻¹¹ C²/(N·m²). κ is a material property (independent of sample size) and describes polarization, not conduction; a filling dielectric reduces the field by κ.
The diagram shows an edge view of an isolated parallel-plate capacitor with a dielectric slab between its plates. Point P is inside the slab. Which describes the electric field produced at P by the polarized dielectric itself?
Answer and reasoning
ATo the left, and equal in magnitude to the field of the plates A student who thinks a dielectric cancels the field inside it, as a conductor does, picks this. The direction is correct, but the charges in a dielectric only shift slightly, so the dielectric's field is smaller than the plates' field and the net field is E₀/κ, not zero.
BTo the left, and smaller in magnitude than the field of the platesCorrect The plates' field points from the +Q plate to the −Q plate, to the right. The dielectric polarizes with negative charge on its left face and positive charge on its right face; these layers produce a field inside the slab pointing to the left, opposite to the plates' field. The bound charges shift only slightly, so this field is smaller than the plates' field and the net field in the slab is reduced, not zero.
CTo the right, which is the same direction as the field of the plates A student who thinks a dielectric strengthens the field picks this. The negative layer forms on the face nearer the +Q plate and the positive layer on the other face, so their field inside the slab points left, back toward the +Q plate.
DZero, because the slab has no net charge to produce a field A student who thinks a neutral object produces no field picks this. The slab is neutral overall, but its charge is separated into layers on its two faces, which produce a field between them, as a capacitor's plates do.
An isolated parallel-plate capacitor has plates of area 0.020 m² carrying charges of +4.0 nC and −4.0 nC. The space between the plates is filled with a material of dielectric constant κ = 3.0. What is the magnitude of the electric field inside the material? Use ε₀ = 8.85 × 10⁻¹² C²/(N·m²).
Answer and reasoning
A7.5 × 10³ N/CCorrect Without the material the field between the plates would be E₀ = Q/(ε₀A) = (4.0 × 10⁻⁹ C)/[(8.85 × 10⁻¹²)(0.020 m²)] = 2.3 × 10⁴ N/C. The polarized dielectric's field opposes it, and κ = E₀/E, so E = E₀/3.0 = 7.5 × 10³ N/C.
B2.3 × 10⁴ N/C A student who thinks the neutral dielectric produces no field of its own picks this, the field without the material. The polarized material's field opposes the plates' field, reducing the field by the factor κ = 3.0.
C6.8 × 10⁴ N/C A student who thinks a dielectric strengthens the field picks this, multiplying by κ. The dielectric's field is opposite to the plates' field, so the field is divided by κ: E = E₀/κ.
D3.8 × 10³ N/C A student who uses the field of a single sheet, σ/(2ε₀), for the capacitor picks this. Between the plates both plates' fields add to σ/ε₀ before the dielectric reduces it by κ.
Working Without the material: E₀ = σ/ε₀ = Q/(ε₀A) = (4.0 × 10⁻⁹ C)/[(8.85 × 10⁻¹²)(0.020 m²)] = 2.26 × 10⁴ N/C. With it: E = E₀/κ = 7.5 × 10³ N/C.
An air-filled parallel-plate capacitor is charged to a potential difference ΔV₀ and then disconnected, so it is isolated. A dielectric with dielectric constant κ is then inserted, filling the space between the plates. What is the new potential difference ΔV across the capacitor?
Answer and reasoning
AΔV = ΔV₀/κCorrect The isolated capacitor keeps its charge Q. Filling the gap with the dielectric multiplies the capacitance by κ, so ΔV = Q/C = Q/(κC₀) = ΔV₀/κ. Equivalently, the field falls to E₀/κ and ΔV = Ed falls by the same factor.
BΔV = κΔV₀ A student who thinks the dielectric strengthens the field picks this, making ΔV = Ed larger. The polarized dielectric's field opposes the plates' field, so the field and ΔV both fall by the factor κ.
CΔV = ΔV₀ A student who thinks a capacitor's ΔV stays fixed, as it would with a battery connected, picks this. The capacitor is isolated, so Q stays fixed; with C larger by κ, ΔV = Q/C is smaller by κ.
DΔV = ΔV₀/κ² A student who thinks the charge induced on the dielectric cancels part of the plates' charge, reducing it to Q/κ, picks this: (Q/κ)/(κC₀). The plates' charge is unchanged; the induced charge belongs to the dielectric and reduces only the field.
Working Isolated: Q fixed. C = κC₀, so ΔV = Q/C = ΔV₀/κ. (Equivalently E = E₀/κ and ΔV = Ed.)
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
10.4.A.1 Dielectric Fix
Dielectric
An electrically insulating material. Its charges are bound to their atoms or molecules: they can shift slightly in an electric field but cannot move through the material as the free electrons in a conductor do.
Polarization
The response of a dielectric to an external electric field: the positive and negative charges in each molecule shift slightly in opposite directions, or polar molecules turn to line up with the field. In a slab this leaves a thin layer of charge on each face, negative on the face nearer the positive plate and positive on the other; the slab's net charge stays zero.
Students often think A dielectric placed in a field behaves like a conductor: its charges move freely until the field inside it is zero. In fact No. In a dielectric the charges are bound to their molecules and can shift only slightly. The polarized dielectric reduces the field inside it but does not make it zero, unlike a conductor in equilibrium.
Students often think An insulator has no charges that an electric field can move, so a field has no effect on it. In fact No. An insulator contains electrons and nuclei in every molecule. They cannot flow through the material, but a field shifts them slightly, polarizing the material.
10.4.A.2 Electric permittivity of a material (ε) Fix
Electric permittivity of a material (ε)
The constant that takes the place of ε₀ in the equations for fields and capacitance when a linear dielectric fills the space. SI unit: C²/(N·m²), the same as ε₀.
Dielectric constant (κ)
κ = ε/ε₀, the ratio of a material's permittivity to the permittivity of free space. It is a property of the material, has no units, equals 1 for vacuum (and 1.0 for air to two significant figures), and is greater than 1 for any dielectric.
Students often think The permittivity ε can be used in place of κ in C = κε₀A/d, giving εε₀A/d. In fact No. κ = ε/ε₀ has no units; ε has the units of ε₀, C²/(N·m²). In C = κε₀A/d the product κε₀ equals ε, so either use κε₀ or use ε alone, never εε₀.
Students often think A material with a high permittivity lets charge pass through it more easily, as its name suggests. In fact No. Permittivity describes how a material polarizes in a field, not how it conducts. Dielectrics with large κ are insulators.
10.4.A.3 Field of a polarized dielectric Fix
Field of a polarized dielectric
The charges on the faces of a polarized dielectric produce a field inside the dielectric that points opposite to the external field. It is smaller in magnitude than the external field, so the net field in the dielectric is reduced but not zero.
Students often think A dielectric increases the field between the plates, because it increases the capacitance. In fact No. The polarized dielectric produces a field opposite to the external field, so the field inside the dielectric is smaller: E = E₀/κ for an isolated capacitor filled with dielectric.
Students often think A dielectric with no net charge produces no electric field, so it does not change the field between the plates. In fact No. A polarized dielectric has zero net charge but separated charge: a layer of negative charge on one face and positive charge on the other. Like a capacitor's plates, these layers produce a field between them, inside the dielectric.
10.4.A.4 Field in an isolated capacitor with a dielectric Fix
Field in an isolated capacitor with a dielectric
When a dielectric fills the space between the plates of an isolated capacitor, the field between the plates decreases from E₀ to E = E₀/κ, so κ = E₀/E. The charge on the plates does not change.
Field in an air gap beside a dielectric slab
In an isolated parallel-plate capacitor partly filled by a slab parallel to the plates, the fields of the slab's two oppositely charged faces cancel outside the slab, so the field in the air gaps stays E₀ = σ/ε₀ while the field inside the slab is E₀/κ.
Students often think A dielectric slab reduces the field by the factor κ everywhere between the plates, even where there is air. In fact No. The field is reduced only inside the slab. The fields of the slab's two charged faces cancel outside it, so in the air gaps the field stays E₀ = σ/ε₀.
Students often think The field between the plates of a capacitor is that of one charged sheet, σ/(2ε₀), reduced by κ when a dielectric fills the gap. In fact No. Both plates contribute σ/(2ε₀) between them, so the field without the dielectric is Q/(ε₀A); the dielectric reduces it to Q/(κε₀A).
10.4.A.5 Capacitance with a dielectric Fix
Capacitance with a dielectric
C = κC₀ when a dielectric fills the whole space between the conductors, where C₀ is the capacitance with vacuum (or air) between them. A slab that fills only part of the space increases the capacitance by a smaller factor.
Isolated versus connected capacitor
An isolated capacitor keeps its charge Q when a dielectric is inserted, so ΔV = Q/C falls; a capacitor that stays connected to a battery keeps its ΔV, so its charge Q = CΔV rises as charge flows from the battery.
Students often think The capacitance depends only on the conductors' size and spacing, so the material between them has no effect. In fact No. The material between the conductors matters: filling the space with a dielectric of dielectric constant κ multiplies the capacitance by κ.
Students often think A dielectric reduces the capacitance of a capacitor, because it weakens the field between the plates. In fact No. A dielectric reduces the field, and so ΔV, for a given charge; with a smaller ΔV for the same Q, C = Q/ΔV is larger. Filling the space multiplies C by κ.
9 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 9
Two identical isolated parallel-plate capacitors carry equal charges, and the field between the plates of each is E₀. A metal slab is placed in one and a glass slab of the same size in the other; each slab is parallel to the plates, thinner than the gap and touches neither plate. EM is the field magnitude inside the metal and EG the field magnitude inside the glass. Which relationship is correct?
Answer and reasoning
AEM = EG < E₀ A student who thinks the charges in glass move freely, as in a metal, picks this, making the field zero inside both slabs. The charges in a dielectric can only shift slightly, so the field inside the glass is reduced, not removed.
BEM < EG < E₀Correct The metal has free electrons, which move until the field inside the metal is zero: EM = 0. The glass has only bound charges, which shift slightly; the polarized glass reduces the field inside it to E₀/κ, which is smaller than E₀ but not zero. So 0 = EM < EG < E₀.
CEM < E₀ < EG A student who thinks a dielectric strengthens the field picks this. The polarized glass produces a field opposite to the plates' field, so the field inside it is smaller than E₀.
DEM = EG = E₀ A student who thinks a slab with no net charge cannot change the field picks this. Both slabs are neutral, but each has separated charge on its faces, which reduces the field inside it: to zero in the metal and to E₀/κ in the glass.
A parallel-plate capacitor has plates of area 0.050 m² separated by 2.0 mm. The space between the plates is filled with a material whose permittivity is ε = 3.2 × 10⁻¹¹ C²/(N·m²). What is the capacitance? Use ε₀ = 8.85 × 10⁻¹² C²/(N·m²).
Answer and reasoning
A2.2 × 10⁻¹⁰ F A student who thinks the material between the plates does not affect the capacitance picks this, using ε₀A/d. The material has κ = 3.6, which multiplies the air-filled value by 3.6.
B6.1 × 10⁻¹¹ F A student who thinks the dielectric lowers the capacitance, because it lowers the field, picks this: (ε₀A/d)/κ. For the same charge the field and ΔV are smaller, so C = Q/ΔV is larger: κε₀A/d.
C8.0 × 10⁻¹⁰ FCorrect The dielectric constant is κ = ε/ε₀ = (3.2 × 10⁻¹¹)/(8.85 × 10⁻¹²) = 3.6, so C = κε₀A/d. Since κε₀ = ε, C = εA/d = (3.2 × 10⁻¹¹)(0.050 m²)/(2.0 × 10⁻³ m) = 8.0 × 10⁻¹⁰ F.
D7.1 × 10⁻²¹ F A student who puts the permittivity ε in place of κ in C = κε₀A/d picks this, multiplying by ε₀ twice. κ = ε/ε₀ is the ratio of the two permittivities; κε₀ is ε itself.
Working κ = ε/ε₀ = (3.2 × 10⁻¹¹)/(8.85 × 10⁻¹²) = 3.6. C = κε₀A/d = εA/d = (3.2 × 10⁻¹¹)(0.050 m²)/(2.0 × 10⁻³ m) = 8.0 × 10⁻¹⁰ F.
A coaxial cylindrical capacitor of length L has an inner metal cylinder of radius a and a thin outer metal cylindrical shell of radius b, with L much greater than b. The space between the cylinders is completely filled with a material of permittivity ε. Which expression gives the capacitance?
Answer and reasoning
A2πε₀L/ln(b/a) A student who thinks the material between the conductors does not affect the capacitance picks this, the value with vacuum between the cylinders. The material reduces the field and ΔV by κ = ε/ε₀, so C is κ times larger.
B2πεε₀L/ln(b/a) A student who uses ε in place of κ, multiplying the vacuum value by ε, picks this. The factor is κ = ε/ε₀, so κ × 2πε₀L/ln(b/a) = 2πεL/ln(b/a); the expression with εε₀ does not even have the units of capacitance.
C2πεL/ln(b/a)Correct Gauss's law gives the field between the cylinders without the material, λ/(2πε₀r), where λ = Q/L; the filling dielectric reduces it by κ = ε/ε₀, to λ/(2πεr). Then ΔV = ∫ab E dr = (λ/(2πε))ln(b/a) and C = λL/ΔV = 2πεL/ln(b/a), κ times the value with vacuum.
DπεL/ln(b/a) A student who adds a field from the outer cylinder's charge picks this: the field, and so ΔV, doubles and C halves. A Gaussian cylinder between the conductors encloses only the inner cylinder's charge.
Working Charges +Q, −Q; λ = Q/L. Without the material, Gauss's law gives E₀ = λ/(2πε₀r) (the outer shell's charge is not enclosed). The filling dielectric reduces the field by κ = ε/ε₀: E = λ/(2πκε₀r) = λ/(2πεr). ΔV = ∫ab E dr = (λ/(2πε))ln(b/a). C = λL/ΔV = 2πεL/ln(b/a) = κ × 2πε₀L/ln(b/a). Distractors: material ignored → 2πε₀L/ln(b/a); ε used as κ → 2πεε₀L/ln(b/a); outer shell's field added → πεL/ln(b/a). Checked with sympy.
The diagram shows an edge view of an isolated parallel-plate capacitor with charges +Q and −Q on its plates. Before the dielectric slab was inserted, the field between the plates had magnitude E₀. The slab, which touches neither plate, is now in place. Points A and B are marked on the diagram. Which relationship between the field magnitudes EA and EB is correct?
Answer and reasoning
AEA = EB < E₀ A student who thinks the slab reduces the field everywhere between the plates picks this. The fields of the slab's two charged faces cancel outside the slab, so the reduction happens only inside it; at A the field is still E₀.
BEB > EA = E₀ A student who thinks a dielectric strengthens the field picks this. The induced charge on each face of the slab is opposite to the charge on the nearer plate, so the slab's field inside it opposes the plates' field.
CEA = EB = E₀ A student who thinks the neutral slab produces no field picks this. The slab's faces carry separated charge, which produces a field inside the slab opposite to E₀, reducing the field at B.
DEB < EA = E₀Correct The isolated plates keep their charge, so their own field is still E₀. The polarized slab has charge layers of opposite sign on its two faces; like a pair of plates, they produce a field only between them, inside the slab, opposite to E₀. So the field in the air gap stays E₀, and the field in the slab is reduced to E₀/κ.
Working Isolated: σ on the plates unchanged, so the plates alone give E₀. Slab faces carry −σi (left) and +σi (right): their field is zero outside the slab and σi/ε₀ opposite to E₀ inside. EA = E₀; EB = E₀/κ < E₀.
An isolated parallel-plate capacitor had a field E₀ between its plates before a dielectric slab, parallel to the plates, was inserted into the shaded region. The graph shows the electric potential V as a function of position x between the plates after the slab is in place; the plates are at the two ends of the graph. What is the dielectric constant of the slab?
Answer and reasoning
A4.0 A student who compares potential differences instead of fields picks this: 80 V across the air gaps against 20 V across the slab. The field is the drop per unit width; the air gaps total 4 mm and the slab 3 mm, so the slopes must be compared.
B2.1 A student who thinks the isolated capacitor's potential difference did not change when the slab went in takes the field before insertion as E₀ = 100 V/7 mm = 14 V/mm and divides by the slab's 6.7 V/mm. The capacitor is isolated, so its charge stays fixed and ΔV falls when the slab is inserted; E₀ is the field that remains in the air gaps, 20 V/mm.
C1.4 A student who reads E₀ from the air gaps, 20 V/mm, but then applies C = κC₀ to the partly filled capacitor, so that κ = C/C₀ = ΔV₀/ΔV, picks this: ΔV₀ = (20 V/mm)(7 mm) = 140 V and 140/100 = 1.4. C = κC₀ holds only when the dielectric fills the whole gap; here the capacitance rose by less than κ, and κ is the ratio of the fields, 20/6.7 = 3.0.
D3.0Correct The field in each region is the magnitude of the slope of V against x. In the air gaps it is (100 − 60) V/(2 mm) = 20 V/mm; because the plates' charge is unchanged and the fields of the slab's charged faces cancel outside the slab, this is E₀. In the slab it is (60 − 40) V/(3 mm) = 6.7 V/mm. So κ = E₀/E = 20/6.7 = 3.0.
Working Air gaps: slope magnitude (100 − 60) V/2 mm = 20 V/mm (also (40 − 0)/2); the air-gap field equals E₀, because the fields of the slab's charged faces cancel outside the slab and the plates' charge is unchanged. Slab: (60 − 40) V/3 mm = 6.7 V/mm. κ = E₀/E = 20/6.67 = 3.0. Distractors: ratio of drops 80/20 = 4.0; ΔV taken as unchanged by insertion, so E₀ = 100 V/7 mm, (100/7)/(6.67) = 2.1; ΔV₀/ΔV = (20 × 7)/100 = 1.4.
An air-filled parallel-plate capacitor has capacitance C₀. A slab of dielectric with κ = 4.0, with the same area as the plates and half as thick as the gap, is placed flat against one plate. What is the new capacitance?
Answer and reasoning
A4.0C₀ A student who applies C = κC₀ to a slab that fills only half the gap picks this. The field is reduced only inside the slab; the air half still has the field E₀, so C increases by less than κ.
B1.6C₀Correct For charge Q on the plates, the field is E₀ = Q/(ε₀A) in the air half and E₀/κ in the slab half, so ΔV = E₀(d/2) + (E₀/4.0)(d/2) = 0.625E₀d. Without the slab ΔV₀ = E₀d, so C = Q/ΔV = C₀/0.625 = 1.6C₀.
C2.0C₀ A student who treats the slab like a conductor, with zero field inside, picks this: only the air half would contribute to ΔV. The field inside a dielectric is E₀/κ, not zero, so the slab half adds to ΔV.
D2.5C₀ A student who averages the dielectric constants of the two halves, (1 + 4.0)/2, picks this. The slab and the air lie one after the other along the field, so their potential differences add: ΔV = E₀(d/2)(1 + 1/4.0).
Working Charge Q on the plates: field E₀ = Q/(ε₀A) in the air half, E₀/κ in the slab half. ΔV = E₀(d/2) + (E₀/κ)(d/2) = (E₀d/2)(1 + 1/κ). C = Q/ΔV = (ε₀A/d) × 2κ/(κ + 1) = (8/5)C₀ = 1.6C₀. Distractors: κC₀ = 4.0C₀; slab as conductor → ΔV = E₀d/2 → 2.0C₀; averaged κ (1 + 4.0)/2 → 2.5C₀.
An air-filled parallel-plate capacitor is connected to an ideal battery, and the charge on its positive plate is measured as 24 μC. A dielectric slab that fills the space between the plates is then inserted while the battery stays connected, and the charge on the positive plate becomes 72 μC. Which claim is supported by these measurements, with valid reasoning?
Answer and reasoning
AThe field between the plates tripled, because the field is proportional to the plates' charge. A student who links the field to the plates' charge alone picks this. ΔV and d are both unchanged, so E = ΔV/d is unchanged; the extra charge on the plates is matched by the opposing field of the polarized dielectric.
BThe stored energy did not change, because the battery kept ΔV between the plates fixed. A student who thinks the stored energy is set by the battery's ΔV alone picks this. U = (1/2)QΔV, and Q tripled while ΔV stayed the same, so the stored energy tripled.
CThe slab's dielectric constant is 3.0, as ΔV stayed fixed while C tripled.Correct The battery holds ΔV constant, so the capacitance is proportional to the charge: C rose by the factor 72/24 = 3.0. For a dielectric that fills the gap, C = κC₀, so κ = 3.0.
DThe extra 48 μC came from the slab, whose charges moved across onto the plates. A student who thinks the dielectric exchanges charge with the plates picks this. The slab is an insulator and stays neutral; the extra 48 μC flowed from the battery through the wires.
Working Battery connected: ΔV fixed. C = Q/ΔV rises by 72/24 = 3.0, and a filling dielectric gives C = κC₀, so κ = 3.0. E = ΔV/d unchanged; U = (1/2)QΔV tripled; the extra 48 μC came through the battery.
An isolated parallel-plate capacitor has a uniform electric field of magnitude E₀ between its plates. A dielectric of dielectric constant κ is then inserted, completely filling the space between the plates. Which expression gives the magnitude of the electric field produced inside the dielectric by the polarized dielectric alone?
Answer and reasoning
AE₀ A student who thinks the dielectric's charges move freely, as in a conductor, until the field inside it is zero picks this: the dielectric's field would then cancel E₀. In a dielectric the charges are bound and shift only slightly, so the net field inside is reduced to E₀/κ, not to zero.
B0 A student who thinks a dielectric with no net charge produces no field picks this. The polarized dielectric has separated layers of charge on its two faces, which produce a field between them, as a capacitor's plates do.
C(κ − 1)E₀ A student who thinks the dielectric increases the field to κE₀ picks this, taking the dielectric's own field as the increase, κE₀ − E₀. The induced charges produce a field opposite to the plates' field, so the net field inside falls to E₀/κ.
D(1 − 1/κ)E₀Correct The plates' charges still produce E₀, and the net field inside the dielectric is E₀/κ. The polarized dielectric's own field, opposite to the plates' field, makes up the difference: E₀ − E₀/κ = (1 − 1/κ)E₀.
Working The capacitor is isolated, so the plates' charges, and the field E₀ they produce, do not change. With the dielectric filling the gap, the net field inside it is E = E₀/κ. The net field is the plates' field plus the dielectric's own field, which is opposite to the plates' field: E₀/κ = E₀ − Ed, so Ed = E₀ − E₀/κ = (1 − 1/κ)E₀. Distractors: dielectric treated as a conductor, cancelling E₀ → E₀; neutral dielectric taken to produce no field → 0; dielectric taken to raise the net field to κE₀, so its own field is κE₀ − E₀ → (κ − 1)E₀. Checked with sympy (κ = 3: key 0.67E₀; distractors E₀, 0, 2E₀).
An air-filled parallel-plate capacitor has capacitance C₀. A slab of dielectric with dielectric constant κ, exactly as thick as the gap between the plates, is inserted so that it fills the space between the plates over half of the plate area; the other half of the space between the plates is still air. Edge effects are negligible. Which expression gives the new capacitance?
Answer and reasoning
A(κ + 1)C₀/2Correct The plates are equipotentials, so both halves have the same field ΔV/d. The charge density is then ε₀ΔV/d on the air half and κε₀ΔV/d on the dielectric half, so Q = (A/2)(κ + 1)ε₀ΔV/d and C = Q/ΔV = (κ + 1)C₀/2.
BκC₀ A student who multiplies the capacitance by κ whenever a dielectric is inserted, however much of the space it fills, picks this. C = κC₀ holds only when the dielectric fills all of the space between the plates; here half of it is still air.
C2κC₀/(κ + 1) A student who treats the slab as if it lay flat against a plate, filling half of the gap, picks this: the air and dielectric layers would then carry the same charge density and their potential differences would add. Side by side, the two halves share the potential difference instead, so their charges add.
D(κ + 1)C₀ A student who takes the field between the plates to be σ/(2ε₀), the field of one sheet, reduced by κ in the dielectric, picks this: each half's charge density, and so the capacitance, is doubled. Both plates contribute to the field between them, so it is σ/ε₀ in air and σ/(κε₀) in the dielectric.
Working Let the plates have area A and separation d, so C₀ = ε₀A/d. Each plate is a conductor and so an equipotential: the potential difference ΔV is the same across the air half and the dielectric half, and so is the separation, so the field is ΔV/d in both halves. Air half: E = σair/ε₀, so σair = ε₀ΔV/d. Dielectric half: the field is reduced by κ from the value the free charge alone would produce, E = σκ/(κε₀), so σκ = κε₀ΔV/d. Charge on the positive plate: Q = (A/2)(σair + σκ) = (κ + 1)ε₀AΔV/(2d). C = Q/ΔV = (κ + 1)ε₀A/(2d) = (κ + 1)C₀/2. Distractors: capacitance multiplied by κ even though only half is filled → κC₀; slab treated as stacked against a plate, filling half the gap (ΔV = (σ/ε₀)(d/2) + (σ/(κε₀))(d/2)) → 2κC₀/(κ + 1); field in each half taken as σ/(2ε₀), reduced by κ in the dielectric → σair = 2ε₀ΔV/d, σκ = 2κε₀ΔV/d → (κ + 1)C₀. Checked with sympy.
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
