AP Physics C: E&M · Topic 10.4

Topic 10.4: Dielectrics

Unit 10: Conductors and Capacitors10-15% of the multiple-choice section

A dielectric is an insulating material that polarizes in an external field instead of conducting. Placed between capacitor plates it sets up a field opposing the original one, which lowers the field in an isolated capacitor and raises the capacitance by a factor of the dielectric constant.

AP Physics: Unit 10 (topics 10.4 Dielectrics). Topic 10.4 of the current AP Physics C: Electricity and Magnetism course and exam description, inside Unit 10, weighted 10 to 15% of the multiple-choice section at about 8 to 16 class periods. One learning objective, 10.4.A, using the task verb describe, with five essential-knowledge statements: 10.4.A.1 defines a dielectric as a material in which charges are not as free to move as in a conductor and which instead polarizes in an external field; 10.4.A.2 gives the dielectric constant as kappa = epsilon over epsilon-zero, a relevant equation printed on the equation sheet; 10.4.A.3 states that the field created by a polarized dielectric opposes the external field; 10.4.A.4 states that the field between the plates of an isolated parallel-plate capacitor decreases when a dielectric is placed between them, with the derived equation kappa = E-zero over E; and 10.4.A.5 states that inserting a dielectric may change the capacitance, with the derived equation C = kappa C-zero. Topic 10.4 prints no boundary statement. Two of its three equations carry the Derived label and are absent from the equation sheet. Suggested skills are 1.B, 2.B, 2.D, 3.A and 3.B, five in all, the most of any Unit 10 topic, and 1.B and 3.A appear nowhere else in the unit. The exam conventions printed with the equation sheet state that capacitors are air-filled with kappa = 1.0. There is no AP Physics 2 topic of this name; the algebra-based course compresses the content into statement 10.6.A.6 inside its capacitors topic, while the underlying permittivity and polarization statements match in both courses, appearing as AP Physics C learning objective 8.1.C and AP Physics 2 learning objective 10.1.C with the same six statements in the same order.

What Topic 10.4 requires

Topic 10.4 has one learning objective and five essential-knowledge statements. Two of its three equations are labelled Derived, more than any other topic in Unit 10.

10.4.A, describe how a dielectric inserted between the plates of a capacitor changes the properties of the capacitor.

10.4.A.1 states that in a dielectric material, electric charges are not as free to move as they are in a conductor, and that instead the material becomes polarized in the presence of an external electric field.
10.4.A.2 states that the dielectric constant of a material relates the electric permittivity of that material to the permittivity of free space, with the relevant equation $\kappa = \dfrac{\varepsilon}{\varepsilon_0}$ .
10.4.A.3 states that the electric field created by a polarized dielectric is opposite in direction to the external field.
10.4.A.4 states that the electric field between the plates of an isolated parallel-plate capacitor decreases when a dielectric is placed between the plates, with the derived equation $\kappa = \dfrac{E_0}{E}$ .
10.4.A.5 states that the insertion of a dielectric into a capacitor may change the capacitance of the capacitor, with the derived equation $C = \kappa C_0$ .

Topic 10.4 prints no boundary statement. Unit 10's only boundary statement sits at the end of Topic 10.3 and names three capacitor shapes.

Suggested skills, five of them, the most of any Unit 10 topic: 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Skills 1.B and 3.A appear nowhere else in Unit 10. That combination, plotting data and designing a procedure, tells you what kind of question this topic supports: a laboratory one, where the dielectric constant is the thing being measured. The second worked example below is that experiment.

Unit 10 is weighted 10 to 15% of the multiple-choice section over a suggested 8 to 16 class periods.

A dielectric is not a conductor, and the CED says so by degree

Compare the two opening definitions in this unit, word for word.

Statement 10.1.A.1: "An ideal conductor is a material in which electrons are able to move freely."

Statement 10.4.A.1: "In a dielectric material, electric charges are not as free to move as they are in a conductor. Instead, the material becomes polarized in the presence of an external electric field."

Not "cannot move". Not as free to move. The framework is describing a spectrum rather than a switch, and that choice of words is what makes the rest of the topic work.

The mechanism sits in Topic 8.1, whose statement 8.1.C.2 says electric polarization can be modeled as the induced rearrangement of electrons by an external electric field, resulting in a separation of positive and negative charges within a material or medium. In a conductor, electrons travel all the way to the surface and keep going until the interior field is exactly zero. In a dielectric, they shift a little within each molecule or region and then stop, because they are bound. So:

	Conductor	Dielectric
Charge mobility	Free (10.1.A.1)	Not as free (10.4.A.1)
Response to an external field	Charge migrates to the surface	Material polarizes in place
Interior field afterwards	Exactly zero (10.1.A.4)	Reduced, not zero (10.4.A.4)
Where the induced charge sits	Conductor's outer surface	Faces of the dielectric slab

The right way to read a dielectric is as a partial conductor, and the dielectric constant as the measure of how partial. Push $\kappa$ to infinity and you recover a conductor: the field inside goes to zero. Set $\kappa = 1$ and you have vacuum, or air to the accuracy this course works at, and nothing happens at all.

That also explains why the two closely related quantities in this topic behave the way they do. Statement 8.1.C.1 defines electric permittivity as a measurement of the degree to which a material or medium is polarized in the presence of an electric field, and statement 8.1.C.4.i says that in a given material, permittivity is determined by the ease with which electrons can change configurations within the material. A material whose electrons rearrange easily has high permittivity, therefore high $\kappa$ , therefore a strong opposing field. See conductor vs insulator for the two extremes.

Two definitions of kappa, and why they agree

The dielectric constant arrives twice in this topic, with two different definitions, and only one of them is printed on the equation sheet.

Statement 10.4.A.2, printed on the sheet:

\kappa = \frac{\varepsilon}{\varepsilon_0}

This is a material definition. It says the dielectric constant is a ratio of permittivities, a pure number with no units, telling you how much more readily this material polarizes than vacuum does. Since $\varepsilon > \varepsilon_0$ for any real matter, $\kappa > 1$ always.

Statement 10.4.A.4, labelled Derived and not on the sheet:

\kappa = \frac{E_0}{E}

This is an operational definition, and note the condition attached: statement 10.4.A.4 specifies the field between the plates of an isolated parallel-plate capacitor. It says you can measure $\kappa$ by measuring the field before and after inserting the slab, and it only reads correctly when the charge on the plates is held fixed.

The two agree, and the reason is the third statement. Statement 10.4.A.3 says the electric field created by a polarized dielectric is opposite in direction to the external field. So the net field between the plates is the plate field minus the dielectric's own field, and the more polarizable the material, the more it subtracts. Take an isolated capacitor with charge $Q$ fixed on its plates. The plate field alone is $E_0 = Q/(\varepsilon_0 A)$ from Topic 10.3, and with the material present the same charge produces $E = Q/(\varepsilon A)$ , since $\varepsilon$ is the permittivity of the medium the field is in. Divide:

\frac{E_0}{E} = \frac{\varepsilon}{\varepsilon_0} = \kappa

The two definitions are the same statement, one written about the material and one about a measurement you can make.

And that gives the third equation, statement 10.4.A.5's derived $C = \kappa C_0$ . With $Q$ fixed and the field cut by $\kappa$ , the potential difference $\Delta V = Ed$ is cut by $\kappa$ too, so $C = Q/\Delta V$ is multiplied by $\kappa$ . Equivalently, replace $\varepsilon_0$ with $\varepsilon = \kappa\varepsilon_0$ in the parallel-plate result and you land on the printed $C = \dfrac{\kappa\varepsilon_0 A}{d}$ , which the sheet does carry.

So of this topic's three equations, one is printed and two are yours to produce. The CED's Required Equations note explains why: derived equations "demonstrate the final results of derivations expected of students on the exam".

May change the capacitance: reading statement 10.4.A.5 carefully

Statement 10.4.A.5 is worded with a hedge that is easy to read past: the insertion of a dielectric into a capacitor may change the capacitance. Not "changes". The derived equation $C = \kappa C_0$ shows the factor, and $\kappa = 1$ for vacuum, and to the accuracy this course uses, for air.

That connects to a printed exam convention worth knowing by heart. The conventions box on the AP Physics C: E&M equation sheet states: "Capacitors are air-filled $(\kappa = 1.0)$ ." So unless a question names a dielectric, every capacitor on the exam has $\kappa = 1$ and the $\kappa$ in $C = \kappa\varepsilon_0 A/d$ is quietly doing nothing. It is still worth writing, because a question that later inserts a slab has already given you the factor.

Two other conditions on $C = \kappa C_0$ are worth being explicit about, since neither is printed.

The slab fills the gap. The factor is exactly $\kappa$ when the dielectric occupies the whole space between the plates. Topic 10.4 prints nothing about a partially filled gap, and neither does Topic 10.3.

The geometry is unchanged. $C = \kappa C_0$ compares the same capacitor with and without the material. It is not a statement about two different capacitors.

Statement 10.4.A.4 carries its own condition, and this one is printed: the field decrease is stated for an isolated parallel-plate capacitor. Isolated means the charge is trapped, with no battery attached. That single word is the difference between the two cases in the next section, and it is the most commonly missed word in Unit 10.

The two circuit cases, which decide every dielectric question

Insert the same slab into the same capacitor in two different circuits and almost everything comes out differently. What is pinned in each case is the whole story, and it follows from Topic 10.3's constant-charge and constant-voltage distinction.

Quantity	Isolated: battery disconnected, $Q$ fixed	Connected: battery attached, $\Delta V$ fixed
Capacitance $C$	Multiplied by $\kappa$	Multiplied by $\kappa$
Charge $Q$	Unchanged	Multiplied by $\kappa$
Potential difference $\Delta V$	Divided by $\kappa$	Unchanged
Field between plates $E$	Divided by $\kappa$	Unchanged
Stored energy $U_C$	Divided by $\kappa$	Multiplied by $\kappa$

The capacitance row is the same in both, because statement 10.3.A.2.i says capacitance depends only on the physical properties of the capacitor, and inserting the slab changed one of those. Everything else depends on the circuit.

Note which row statement 10.4.A.4 applies to. It says the field decreases for an isolated capacitor, and the table shows why the qualifier is there: with a battery connected, the field does not decrease at all. $\Delta V$ is pinned by the battery and $d$ is pinned by the geometry, so $E = \Delta V/d$ cannot move. The dielectric still polarizes and still produces its opposing field, but the battery pushes extra charge onto the plates to compensate, exactly enough to keep the net field where it was.

The energy rows are the ones to reason out rather than memorise. Pick the energy form whose two quantities are pinned:

Isolated, $Q$ fixed: use $U = \dfrac{Q^2}{2C}$ . Raising $C$ by $\kappa$ divides $U$ by $\kappa$ . The energy fell, and it fell because the field did work pulling the slab in. Let go of a slab near the edge of an isolated charged capacitor and it gets sucked in.
Connected, $\Delta V$ fixed: use $U = \frac{1}{2}C(\Delta V)^2$ . Raising $C$ by $\kappa$ multiplies $U$ by $\kappa$ . The energy rose, and the battery supplied it, along with the work done on the slab.

Both cases have the slab pulled in. What differs is who pays.

Where the algebra-based course puts dielectrics

There is no AP Physics 2 topic called Dielectrics. Both the underlying idea and one consequence of it are in that course, so here is the precise map.

Permittivity and polarization are identical in the two courses. AP Physics C's Topic 8.1 carries learning objective 8.1.C, describe the electric permittivity of a material or medium, with statements 8.1.C.1 through 8.1.C.4.ii. AP Physics 2 carries learning objective 10.1.C with the same title and the same six statements, numbered 10.1.C.1 through 10.1.C.4.ii, saying the same things in the same order: permittivity as a measure of the degree of polarization, polarization as induced rearrangement of electrons, the constant $\varepsilon_0$ of free space, the permittivity of matter differing from it because of composition and arrangement, permittivity determined by the ease with which electrons change configuration, and the conductor-versus-insulator distinction. Nothing at all changes there.

The capacitor consequence is one sentence in AP Physics 2 and five statements in AP Physics C. AP Physics 2's statement 10.6.A.6, inside its capacitors topic, reads: "Adding a dielectric between two plates of a capacitor changes the capacitance of the capacitor and induces an electric field in the dielectric in the opposite direction to the field between the plates." That is Physics C's 10.4.A.3 and 10.4.A.5 compressed into one line.

What AP Physics C adds beyond it:

$\kappa = \varepsilon/\varepsilon_0$ as an explicit relevant equation (10.4.A.2), tying the dielectric constant back to the permittivity statements in Unit 8.
The comparison of dielectric to conductor by degree of charge mobility (10.4.A.1).
$\kappa = E_0/E$ for an isolated capacitor as a derived result (10.4.A.4), which is the version you can actually measure.
$C = \kappa C_0$ as a derived result (10.4.A.5), where Physics 2 says only that the capacitance changes.
Five suggested skills including experimental design, where Physics 2's Topic 10.6 does not treat dielectrics as a separate measurable at all.

Both courses print $C = \kappa\varepsilon_0 A/d$ on their equation sheets, and both print $\kappa = \varepsilon/\varepsilon_0$ on the AP Physics C sheet only. AP Physics 2's sheet instead prints $E_C = \dfrac{Q}{\kappa\varepsilon_0 A}$ , which carries the dielectric constant inside the plate-field equation; the AP Physics C sheet prints no plate-field equation at all, dielectric or otherwise.

If you want the algebra-based treatment, the AP Physics 2 capacitors topic has it in a sentence. This page is for the calculus-based course, where dielectrics carry a topic, two derived equations and a laboratory skill.

Traps

Assuming the field always drops when a dielectric goes in. Only for an isolated capacitor, which is the word statement 10.4.A.4 uses. With a battery connected, the field is unchanged.

Assuming the energy always drops. Isolated, it drops by $\kappa$ . Connected, it rises by $\kappa$ . Decide which quantity is pinned before you decide the direction.

Taking $\kappa$ to have units. It is a ratio of two permittivities, so it is dimensionless. If your $\kappa$ has units, you have divided the wrong pair of things.

Using $\kappa < 1$ . Real materials polarize more than vacuum, so $\kappa > 1$ always, and $\kappa = 1$ is the vacuum or air case that the exam conventions assume by default.

Confusing $\varepsilon$ with $\varepsilon_0$ in a formula. $\varepsilon$ is the permittivity of the material the field is actually in. The equation sheet's constants table gives $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$ for free space only.

Treating the dielectric's induced surface charge as free charge. It is bound charge, produced by polarization within the material, so it does not move onto the plates and it does not appear in the charge on the capacitor. That is the difference from the induced charge on a conductor in Topic 10.1.

Forgetting the $\kappa$ on the sheet's capacitance equation. $C = \kappa\varepsilon_0 A/d$ is printed with $\kappa$ in it. Leaving it out works only because the printed conventions set it to $1.0$ for the air-filled default.

Applying $C = \kappa C_0$ to a slab that does not fill the gap. The factor is exactly $\kappa$ for a filled gap. Topic 10.4 prints nothing about partial filling.

How Topic 10.4 is assessed

The AP Physics C: Electricity and Magnetism exam is 3 hours long, with 42 multiple-choice questions worth 50% in 85 minutes and 4 free-response questions worth 50% in 95 minutes. The free-response questions always appear in the same order: Mathematical Routines (10 points), Translation Between Representations (12 points), Experimental Design and Analysis (10 points), and Qualitative/Quantitative Translation (8 points).

Topic 10.4 is the only topic in Unit 10 that carries skill 3.A, create experimental procedures that are appropriate for a given scientific question, and the only one carrying skill 1.B, create quantitative graphs with appropriate scales and units, including plotting data. Together those point at the third free-response question, Experimental Design and Analysis, which the CED describes as expecting a scientifically sound method that varies a single parameter and measures how that change affects a single characteristic, using equipment realistically available in a high school laboratory, and then asks students to plot given data on a graph whose slope or intercept determines a physical quantity.

The CED suggests two activities for this topic, and both are recognisably that question in classroom form. In one, students justify with evidence why pure water has such a high dielectric constant, that is, the polarity of the molecules, and also justify why impure water gives leaky capacitors, that is, the impurities conduct current. In the other, students build a parallel-plate capacitor from two pieces of foil with a piece of paper between them, measure the capacitance, then increase the number of sheets of paper and record data with the purpose of finding the dielectric constant of the paper.

That second one is the archetypal Topic 10.4 laboratory question, and the second worked example below runs it end to end: the linearisation, the graph, the slope, and the dielectric constant that comes out of it.

Skill 2.D, predicting factors of change from functional dependence, is the multiple-choice version. A question gives you an insertion and asks what happens to four quantities. The table two sections up is the answer to all of them, once you have decided whether the battery is connected.

The Unit 10 Progress Check is about 18 multiple-choice questions and 4 free-response questions.

Inserting a dielectric with the battery still connected

An air-filled parallel-plate capacitor of capacitance $C_0 = 120$ pF is connected across a $9.0$ V battery and fully charged. With the battery still connected, a slab of dielectric constant $\kappa = 3.4$ is slid in to fill the gap completely. Find the charge, stored energy and field before and after, and account for the energy the battery supplied.

Identify what is pinned. The battery stays connected, so $\Delta V = 9.0$ V throughout. The charge is free to change.
Before, charge and energy. $Q_0 = C_0\Delta V = (120 \times 10^{-12})(9.0) = 1.08 \times 10^{-9}$ C, that is $1.08$ nC. $U_0 = \frac{1}{2}Q_0\Delta V = \frac{1}{2}(1.08 \times 10^{-9})(9.0) = 4.86 \times 10^{-9}$ J, that is $4.86$ nJ.
After, capacitance. By the derived equation of statement 10.4.A.5, $C = \kappa C_0 = (3.4)(120) = 408$ pF.
After, charge. $Q = C\Delta V = (408 \times 10^{-12})(9.0) = 3.67 \times 10^{-9}$ C, that is $3.67$ nC, larger by exactly $\kappa$ .
After, field. Unchanged. $E = \Delta V/d$ , and neither the battery voltage nor the plate separation moved. This is why statement 10.4.A.4's field decrease is stated only for an isolated capacitor. The slab still polarizes and still produces an opposing field, but the battery drives extra charge onto the plates to cancel that effect exactly.
After, energy. With $\Delta V$ pinned, use $U = \frac{1}{2}C(\Delta V)^2$ : multiplying $C$ by $3.4$ multiplies $U$ by $3.4$ , so $U = (3.4)(4.86) = 16.5$ nJ. Check directly: $\frac{1}{2}(3.67 \times 10^{-9})(9.0) = 1.65 \times 10^{-8}$ J.
Energy audit. The battery pushed an extra $\Delta Q = 3.67 - 1.08 = 2.59$ nC through a fixed $9.0$ V, so it delivered $W_{\text{batt}} = \Delta Q\,\Delta V = (2.59 \times 10^{-9})(9.0) = 23.3$ nJ. The stored energy rose by only $16.5 - 4.86 = 11.7$ nJ, so $11.7$ nJ went elsewhere: into work done on the slab as it was pulled in, and into whatever the circuit dissipated.
Compare with the isolated case, to see how much the circuit matters. Disconnect the battery first and $Q$ stays at $1.08$ nC, $\Delta V$ falls to $9.0/3.4 = 2.65$ V, the field falls by $3.4$ , and the energy falls to $4.86/3.4 = 1.43$ nJ. Same slab, same capacitor, opposite direction for three of the four quantities.

Battery connected: $C$ rises from $120$ pF to $408$ pF, $Q$ rises from $1.08$ nC to $3.67$ nC, $\Delta V$ and $E$ are unchanged, and $U_C$ rises from $4.86$ nJ to $16.5$ nJ. The battery supplied $23.3$ nJ, of which $11.7$ nJ became stored energy. Isolated instead, the charge would have held and the potential difference, field and energy would each have fallen by a factor of $3.4$ .

Measuring the dielectric constant of paper from a graph

Following the CED's suggested activity, a student builds a parallel-plate capacitor from two pieces of aluminium foil of overlapping area $A = 0.020\ \mathrm{m^2}$ , separated by $N$ sheets of paper each of thickness $t = 0.10$ mm. A capacitance meter gives $6.18$ nF for one sheet, $3.14$ nF for two, $2.04$ nF for three and $1.56$ nF for four. Determine the dielectric constant of the paper, and state what you would plot.

Write the model. With the paper filling the gap, $C = \dfrac{\kappa\varepsilon_0 A}{d}$ and $d = Nt$ , so $C = \dfrac{\kappa\varepsilon_0 A}{t}\cdot\dfrac{1}{N}$ .
Linearise, which is skill 1.B. $C$ against $N$ is a curve. $C$ against $1/N$ is a straight line through the origin with slope $\dfrac{\kappa\varepsilon_0 A}{t}$ . Plot $C$ on the vertical axis and $1/N$ on the horizontal.
Tabulate. $1/N$ takes the values $1$ , $0.500$ , $0.333$ and $0.250$ , against $C$ of $6.18$ , $3.14$ , $2.04$ and $1.56$ nF.
Get the slope through the origin. Using $\text{slope} = \dfrac{\sum x_i y_i}{\sum x_i^2}$ with $x = 1/N$ and $y = C$ in nanofarads: $\sum x_i y_i = 6.18 + 1.570 + 0.680 + 0.390 = 8.82$ and $\sum x_i^2 = 1 + 0.250 + 0.1111 + 0.0625 = 1.4236$ . So the slope is $\dfrac{8.82}{1.4236} = 6.20$ nF.
Solve for $\kappa$ . From $\text{slope} = \dfrac{\kappa\varepsilon_0 A}{t}$ , $\kappa = \dfrac{(\text{slope})\,t}{\varepsilon_0 A} = \dfrac{(6.20 \times 10^{-9})(1.0 \times 10^{-4})}{(8.85 \times 10^{-12})(0.020)}$ .
Evaluate. Numerator $= 6.20 \times 10^{-13}$ ; denominator $= 1.77 \times 10^{-13}$ . So $\kappa = 3.50$ , dimensionless as it must be.
Sanity check the value. It is greater than $1$ , which statement 10.4.A.2 requires for any real material since $\varepsilon > \varepsilon_0$ . It is nowhere near infinite, which is what a conductor would give. A paper capacitor behaving as a partial conductor with $\kappa$ of about $3.5$ is a sensible result.
Name the single varied parameter. The number of paper sheets, and therefore the plate separation. Everything else, the overlap area and the material, is held fixed. That is the requirement the Experimental Design and Analysis question states.

Plot capacitance against $1/N$ , which is a straight line through the origin of slope $\kappa\varepsilon_0 A/t$ . The slope is $6.20$ nF, giving $\kappa = \dfrac{(\text{slope})t}{\varepsilon_0 A} = 3.50$ . The dielectric constant is dimensionless and greater than one, as it must be for any real material.

Getting kappa backwards, from two measured fields

A parallel-plate capacitor with plate separation $d = 2.0$ mm is charged and then isolated from its circuit. With vacuum between the plates the field is measured as $E_0 = 7.5 \times 10^4$ V/m. A slab is slid in to fill the gap and the field falls to $E = 2.5 \times 10^4$ V/m. Find the dielectric constant, the permittivity of the slab, the potential difference before and after, and the factor by which the stored energy changed.

Check the condition on the equation before using it. Statement 10.4.A.4 gives $\kappa = E_0/E$ for an isolated parallel-plate capacitor. The problem says the capacitor was isolated from its circuit, so the charge is fixed and the relation applies.
Dielectric constant. $\kappa = \dfrac{E_0}{E} = \dfrac{7.5 \times 10^4}{2.5 \times 10^4} = 3.0$ , dimensionless.
Permittivity of the material, from statement 10.4.A.2 rearranged: $\varepsilon = \kappa\varepsilon_0 = (3.0)(8.85 \times 10^{-12}) = 2.66 \times 10^{-11}\ \mathrm{C^2/(N \cdot m^2)}$ .
Potential difference before. The field is uniform, so $\Delta V_0 = E_0 d = (7.5 \times 10^4)(2.0 \times 10^{-3}) = 150$ V.
Potential difference after. $\Delta V = E d = (2.5 \times 10^4)(2.0 \times 10^{-3}) = 50$ V, lower by the same factor of $3.0$ , as it must be since $d$ did not change.
Energy factor. The charge is fixed, so use $U = \frac{1}{2}Q\Delta V$ : with $Q$ unchanged and $\Delta V$ divided by $3.0$ , the stored energy is divided by $3.0$ . Equivalently $U = Q^2/2C$ with $C$ multiplied by $\kappa$ .
Where the energy went. It was not destroyed. The field pulled the slab into the gap, so the capacitor did work on the slab as it entered. Release a slab at the edge of an isolated charged capacitor and it accelerates inward for exactly this reason.
Cross-check the polarization picture. Statement 10.4.A.3 says the polarized dielectric's own field opposes the external one. Here the plate field alone is still $7.5 \times 10^4$ V/m, because the plate charge never changed, and the slab contributes $5.0 \times 10^4$ V/m in the opposite direction, leaving the measured $2.5 \times 10^4$ V/m.

$\kappa = E_0/E = 3.0$ , so $\varepsilon = \kappa\varepsilon_0 = 2.66 \times 10^{-11}\ \mathrm{C^2/(N \cdot m^2)}$ . The potential difference falls from $150$ V to $50$ V, and with the charge fixed the stored energy falls by a factor of $3.0$ , the missing energy having gone into work done on the slab as the field drew it in.

Frequently asked questions

What is a dielectric in AP Physics C?

CED statement 10.4.A.1 defines it by contrast with a conductor: in a dielectric material, electric charges are not as free to move as they are in a conductor, and instead the material becomes polarized in the presence of an external electric field. The wording is deliberate. A conductor's charges travel all the way to its surface until the interior field is exactly zero; a dielectric's charges shift a little in place and stop, so the field inside is reduced but not eliminated. The dielectric constant measures how far along that spectrum a material sits.

What is the dielectric constant?

It is the ratio of a material's electric permittivity to the permittivity of free space, kappa equals epsilon over epsilon-zero, which is CED statement 10.4.A.2 and is printed on the AP Physics C: Electricity and Magnetism equation sheet. It is dimensionless and greater than one for any real material, since matter polarizes more readily than vacuum. Statement 10.4.A.4 gives a second, equivalent definition for an isolated parallel-plate capacitor, kappa equals E-zero over E, the ratio of the field without the slab to the field with it.

Does inserting a dielectric always reduce the electric field in a capacitor?

Only for an isolated capacitor, which is the condition CED statement 10.4.A.4 attaches. Isolated means the battery is disconnected and the charge on the plates is fixed, and then the field falls by a factor of kappa. With a battery still connected, the potential difference and the plate separation are both pinned, so E equals delta V over d cannot change; the slab still polarizes, but the battery drives extra charge onto the plates that exactly compensates. That one word, isolated, is the most commonly missed condition in the unit.

What happens to the energy stored when a dielectric is inserted?

It depends on the circuit. With the battery disconnected the charge is fixed, so use U equals Q squared over 2C: raising the capacitance by kappa divides the stored energy by kappa, and the missing energy went into work done on the slab as the field pulled it in. With the battery connected the potential difference is fixed, so use U equals one half C delta V squared: the stored energy is multiplied by kappa, and the battery supplied more than that, the excess going into the slab and dissipation.

Why does the capacitance increase by a factor of kappa?

Take an isolated capacitor, so the plate charge is fixed. The polarized dielectric produces a field opposing the plate field, per statement 10.4.A.3, which reduces the net field by kappa and therefore the potential difference by kappa, since delta V equals Ed. Then C equals Q over delta V goes up by kappa, giving the derived equation C equals kappa C-zero. Equivalently, replace epsilon-zero with the material's permittivity in the parallel-plate result to get the printed C equals kappa epsilon-zero A over d.

Which dielectric equations are on the AP Physics C equation sheet?

Two of the three relevant ones. The sheet prints kappa equals epsilon over epsilon-zero from statement 10.4.A.2, and C equals kappa epsilon-zero A over d from Topic 10.3. It does not print kappa equals E-zero over E or C equals kappa C-zero, both of which the CED labels Derived equations. The framework's Required Equations note explains that label: derived equations demonstrate the final results of derivations expected of students on the exam, so you are meant to produce them rather than look them up.

Does AP Physics 2 cover dielectrics?

In one sentence rather than a topic. AP Physics 2's statement 10.6.A.6 says adding a dielectric between two plates of a capacitor changes the capacitance and induces an electric field in the dielectric in the opposite direction to the field between the plates. The permittivity groundwork matches in both courses, appearing as AP Physics C's learning objective 8.1.C and AP Physics 2's 10.1.C with the same six statements in the same order. What AP Physics C adds is the dielectric constant equation, the field-ratio and capacitance-ratio derived equations, and a laboratory skill for measuring kappa.