Conductor vs Dielectric: The Difference

In a conductor the charge carriers move freely, so in electrostatic equilibrium they arrange themselves until the field inside the material is zero. In a dielectric the charges are bound and can only shift slightly in place, so the field inside is reduced by the factor kappa rather than cancelled.

AP Physics: Unit 10 (topics 10.1 Electrostatics with Conductors, 10.4 Dielectrics, 10.6 Capacitors). This pair is a whole unit in AP Physics C: Electricity and Magnetism. Unit 10, Conductors and Capacitors, is weighted at 10 to 15 percent of the multiple-choice section over a suggested 8 to 16 class periods, running Topic 10.1 Electrostatics with Conductors, 10.2 Redistribution of Charge between Conductors, 10.3 Capacitors and 10.4 Dielectrics. On the conductor side, EK 10.1.A.1 defines an ideal conductor as a material in which electrons are able to move freely; EK 10.1.A.2 puts excess charge entirely on the surface through mutual repulsion; EK 10.1.A.3.i says the time to reach electrostatic equilibrium is so short as to be negligible; EK 10.1.A.3.ii makes the whole surface an equipotential; EK 10.1.A.4 states that the field is zero within the conductor; EK 10.1.A.5 puts the field perpendicular to the outer surface; and EK 10.1.A.6 notes that a conductor can itself be polarized in an external field. On the dielectric side, EK 10.4.A.1 says charges in a dielectric are not as free to move as in a conductor and the material polarizes instead; EK 10.4.A.2 defines the dielectric constant as the ratio of the material's permittivity to that of free space, kappa = epsilon over epsilon nought, which is printed on the C: E&M sheet; EK 10.4.A.3 puts the polarized dielectric's field opposite to the external field; EK 10.4.A.4 gives the derived form kappa = E nought over E for an isolated parallel-plate capacitor; and EK 10.4.A.5 gives C = kappa times C nought. AP Physics 2 compresses this into Topic 10.6 Capacitors inside Unit 10 at 15 to 18 percent, where EK 10.6.A.2.ii carries C = kappa epsilon nought A over d, EK 10.6.A.3.i carries E_C = Q over kappa epsilon nought A, and EK 10.6.A.6 says adding a dielectric changes the capacitance and induces a field in the dielectric opposite to the field between the plates. Its boundary statement requires only parallel-plate capacitors and ignores edge effects unless stated. The conductor results appear earlier in AP Physics 2 at EK 10.3.B.1. Both sheets set kappa to 1.0 for air-filled capacitors in the exam conventions box. The capacitance of a gap containing a partial conducting slab is not a CED result; it follows from the zero-field statement together with the printed field equation.

The distinction, stated once

Both materials respond to an external electric field, and both respond by separating charge. The difference is how far the charge can go.

In a conductor the carriers are free. AP Physics C: Electricity and Magnetism EK 10.1.A.1 defines an ideal conductor as a material in which electrons are able to move freely. Put one in a field and the electrons keep drifting for as long as any field remains inside the material to push them, which means they stop only when the interior field has been driven to zero. EK 10.1.A.4 records the endpoint: all excess charges reside on the surface of a conductor, so there is no net charge in the interior, and the electric field is zero within the conductor.

In a dielectric the carriers are bound. EK 10.4.A.1 states it as the contrast: 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. Each molecule stretches a little, positive one way and negative the other, and then it stops. There is no drift and no arriving at zero.

So the polarized dielectric does produce a field of its own, opposing the external one, as EK 10.4.A.3 says, but it cannot make it big enough to cancel. What comes out is a reduction by a fixed factor, the dielectric constant $\kappa$ , and EK 10.4.A.4 gives that factor as

\kappa = \frac{E_0}{E}

Cancel completely, or reduce by $\kappa$ . That is the whole difference, and every consequence below is that sentence applied to a capacitor.

Side by side

	Conductor	Dielectric
Charge carriers	Free to move through the material	Bound, and can only shift within a molecule
Response to an external field	Charges migrate to the surfaces	Molecules polarize in place
Field inside the material	Zero, in electrostatic equilibrium	$E_0/\kappa$ , reduced but not zero
Where excess charge sits	Entirely on the surface	Throughout the interior as well as the surface
Is the whole object at one potential?	Yes, its surface is an equipotential	No
Effect on capacitance when it fills the gap	Shorts the plates together	Multiplies the capacitance by $\kappa$
The number that describes it	None; the field is simply zero	$\kappa$ , dimensionless and greater than 1
Also called	A metal, in most problems	An insulator, when it is between capacitor plates
Example materials	Copper, aluminium	Glass, plastic, oil, air

The field inside row is the definitional split. Everything else on the table follows from whether that field is zero or merely smaller.

The also called row is worth pausing on, because the vocabulary is what makes this pair confusing in the first place. A dielectric is not a third kind of material. It is an insulator, described by the job it is doing: sitting in an electric field and polarizing. The general insulator behaviour, including charging and grounding, belongs to conductor vs insulator. This page is about what happens once that insulator is between two plates.

The field inside: cancelled, or divided by kappa

Set up a uniform external field $E_0$ and slide each material into it.

The conductor. Free electrons pile up on the face nearest the positive source and leave a deficit on the far face. Those two induced surface layers make a field inside the metal pointing opposite to $E_0$ , and the electrons keep moving until that opposing field is exactly $E_0$ in size. Then there is nothing left to push them, and they stop. Interior field: zero.

The stopping condition is what forces the exact cancellation. Any leftover interior field would still be pushing free charge, and while charge is being pushed the conductor is not in electrostatic equilibrium. EK 10.1.A.3.i adds that the time interval over which charges reach that equilibrium is so short as to be negligible, so on an exam a conductor has already got there.

The dielectric. The same two induced surface layers appear, but they are made of the stretched ends of bound molecules rather than of migrated electrons, and there are only so many of them. The opposing field they make is real, and smaller than $E_0$ . Interior field: $E_0/\kappa$ , with $\kappa > 1$ .

$\kappa$ is greater than 1 for every real insulating material precisely because EK 10.4.A.3 puts the polarized material's field opposite in direction to the external one rather than reinforcing it. Air is close enough to 1 that both AP exams treat it as exactly 1, which the exam conventions box on both the AP Physics 2 and the AP Physics C: E&M sheets states outright: capacitors are air-filled, with $\kappa = 1.0$ , unless a question says otherwise.

Three readings of $\kappa$ are all the same number, and the CED gives all three. EK 10.4.A.2 defines it as a permittivity ratio, $\kappa = \varepsilon/\varepsilon_0$ , and this is the version printed on the C: E&M equation sheet. EK 10.4.A.4 gives it as a field reduction, $\kappa = E_0/E$ . EK 10.4.A.5 gives it as a capacitance multiplier, $C = \kappa C_0$ . The dielectric constant entry sets out all three.

What each does to a capacitor

Both AP sheets print the parallel-plate capacitance with $\kappa$ already in it:

C = \frac{\kappa \varepsilon_0 A}{d}

So the answer for a dielectric is written on the sheet. Fill the gap with a material of dielectric constant $\kappa$ and the capacitance is multiplied by $\kappa$ . EK 10.4.A.5 states the same thing as $C = \kappa C_0$ , and AP Physics 2 EK 10.6.A.6 says that adding a dielectric between the plates changes the capacitance and induces a field in the dielectric in the opposite direction to the field between the plates.

Why it helps is worth one line of reasoning rather than memorising. Capacitance is $C = Q/\Delta V$ . The dielectric weakens the field in the gap, and a weaker field across the same separation means a smaller potential difference for the same charge, so the ratio $Q/\Delta V$ goes up. The device holds more charge per volt, which is what a capacitor is for.

A conductor cannot do this job, and the reason is the zero-field result again. A conducting slab that fills the whole gap puts the two plates in electrical contact. Charge flows from one to the other until both are at the same potential, $\Delta V$ goes to zero, and there is no capacitor left. That is a short circuit, not an improved capacitor.

So the pair splits cleanly by function: a dielectric is what you deliberately put between the plates, and a conductor is what the plates are made of.

Battery connected or battery removed: the section the exam question lives in

Inserting a dielectric always multiplies $C$ by $\kappa$ . Everything else about the answer depends on one question, and a problem that does not answer it cannot be solved.

Is the battery still connected?

If it is, the battery holds $\Delta V$ fixed at its own value. If it has been disconnected first, the charge on the plates has nowhere to go, so $Q$ is fixed instead. Those are two different constraints and they give opposite answers for almost every quantity.

Quantity	Battery attached, $\Delta V$ fixed	Battery removed first, $Q$ fixed
Capacitance $C$	$\times \kappa$	$\times \kappa$
Potential difference $\Delta V$	Unchanged	$\div \kappa$
Charge $Q$	$\times \kappa$	Unchanged
Field in the gap $E$	Unchanged	$\div \kappa$
Stored energy $U_C = \frac{1}{2}Q\Delta V$	$\times \kappa$	$\div \kappa$

Read the last row. The stored energy goes up in one column and down in the other, from the same physical act of sliding the same slab into the same capacitor. That is the single most informative line in this topic, and it is why the question is asked.

The energy behaves differently because the battery is a source. With it attached, it pushes extra charge onto the plates and does work doing so, and the stored energy rises. With it removed, nothing can supply energy, the slab is actually pulled into the gap by the field, and the energy that leaves the capacitor is the work done on the slab.

The CED states the right-hand column and leaves you to reason the left. EK 10.4.A.4 says the electric field between the plates of an isolated parallel-plate capacitor decreases when a dielectric is placed between the plates. "Isolated" is the word doing the work: it means the charge is stuck, which is the fixed- $Q$ column. If a question instead says the capacitor remains connected to the battery, the field in the gap does not change at all, because $E = \Delta V/d$ and neither of those has moved.

What if you slide a conductor in instead?

This is the cleanest way to see the two behaviours side by side, and it follows in two lines from the zero-field result.

Slide a conducting slab of thickness $t$ into a gap of width $d$ , without letting it touch either plate. Inside the metal the field is zero, by EK 10.1.A.4. Outside it, in the two remaining vacuum gaps, the field is unchanged, because the charge on the plates has not changed and the field between plates depends on that charge, per the printed $E_C = Q/(\kappa \varepsilon_0 A)$ .

So the potential difference is the field multiplied by the distance over which it actually acts, which is now $d - t$ rather than $d$ :

\Delta V = E(d - t) \qquad \Rightarrow \qquad C = \frac{\varepsilon_0 A}{d - t}

A conducting slab therefore increases the capacitance too, but by the geometric factor $d/(d-t)$ rather than by $\kappa$ , and it does so by effectively shortening the gap rather than by weakening the field. Push $t$ toward $d$ and the capacitance runs away to infinity, which is the mathematical version of shorting the plates.

Two differences fall straight out of that.

A dielectric that fills the gap gives $\kappa C_0$ , a factor set by the material. A conductor gives $d/(d-t)$ , a factor set by the geometry, with no material property in it at all.
A dielectric changes the field everywhere in the gap. A conductor leaves the field alone where it is and removes it entirely where the metal is.

The CED does not pose this problem as an essential knowledge item, so treat the result as a derivation rather than a quotable fact. The two statements it rests on, zero field inside a conductor and $E_C = Q/(\kappa\varepsilon_0 A)$ , are both CED material.

When it costs a mark

Answering a dielectric question without asking about the battery. The most expensive error here, because it produces an answer that is right for the other half of the question. Read the stem for "remains connected" or "is disconnected" before writing anything down.

Saying the field inside a dielectric is zero. It is $E_0/\kappa$ . Zero belongs to the conductor. This is the definitional confusion the page exists for.

Giving $\kappa$ a unit. It is a ratio of two permittivities, so it is a pure number. An answer of " $\kappa = 3.0 \ \mathrm{F/m}$ " is wrong on inspection.

Using $\kappa < 1$ . Every real insulating material has $\kappa > 1$ , because the polarized material's field opposes the external one and therefore reduces it. A dielectric never strengthens a field or reduces a capacitance.

Forgetting that air-filled means $\kappa = 1$ . A capacitor problem that never mentions a slab has already told you the value of $\kappa$ , by the exam conventions box. Writing $C = \varepsilon_0 A/d$ in that case is correct, and looking for a missing $\kappa$ is wasted time.

Assuming inserting a dielectric always increases the stored energy. It does with the battery attached and it does not with the battery removed. Same slab, same capacitor, opposite answers.

Treating a dielectric as a conductor because both polarize. They do both polarize, and AP Physics C: E&M says so for the conductor too at EK 10.1.A.6. Polarization is the shared behaviour, not the distinguishing one. What distinguishes them is whether the charge can then keep going.

Where they behave alike, and why that lulls you

Three genuine similarities keep this pair muddled.

Both polarize in an external field. EK 10.1.A.6 says a conductor can be polarized in the presence of an external electric field, and calls it a consequence of the conductor remaining an equipotential surface. EK 10.4.A.1 says a dielectric becomes polarized in the presence of an external field. Same word, same picture on a diagram, different mechanism: one moves charge across the whole object, the other stretches it within each molecule.

Both are attracted to a charged object. A charged rod attracts a scrap of aluminium and a scrap of paper alike, because in both cases the near face acquires the opposite sign and is therefore closer to the rod than the like-signed far face. The demonstration does not distinguish them, which is why it is not evidence for either.

Both increase the capacitance of a capacitor when inserted, provided they do not touch both plates. For different reasons and by different factors, as the section above works out, but a question that asks only "does $C$ go up?" gets the same yes.

The distinction bites in exactly three places, which is where the exam looks:

Anything that asks for the field inside the material. Zero, or $E_0/\kappa$ .
Anything that asks whether the object is all at one potential. A conductor is, per EK 10.1.A.3.ii, which makes its surface an equipotential surface. A dielectric is not.
Anything that lets the slab touch both plates. A dielectric still works; a conductor shorts the capacitor.

How the exam frames it, and what to read next

This pair is a whole unit in AP Physics C: Electricity and Magnetism. Unit 10, Conductors and Capacitors, is weighted at 10 to 15 percent of the multiple-choice section over a suggested 8 to 16 class periods, and its four topics run in the order the physics does: 10.1 Electrostatics with Conductors, 10.2 Redistribution of Charge between Conductors, 10.3 Capacitors, and 10.4 Dielectrics. The conductor results come first because the capacitor is built out of them.

AP Physics 2 compresses the same material into one topic, 10.6 Capacitors, inside Unit 10 at 15 to 18 percent, with the conductor results appearing earlier at EK 10.3.B.1. Its boundary statement is narrower: only parallel-plate capacitors are required, and edge effects are ignored unless a question says otherwise. AP Physics C adds concentric spherical and coaxial cylindrical geometries.

Both courses print $C = \kappa \varepsilon_0 A / d$ and $U_C = \frac{1}{2}Q\Delta V$ . The AP Physics 2 sheet also prints the field between the plates as $E_C = Q/(\kappa \varepsilon_0 A)$ ; the C: E&M sheet prints $\kappa = \varepsilon/\varepsilon_0$ instead, and expects you to reach the field with Gauss's law.

From here: conductor, dielectric and electrostatic equilibrium carry the definitions; capacitance and parallel-plate capacitor cover the device; and conductor vs insulator handles charging, grounding and the behaviour of insulators outside a capacitor.

The field inside each material

A uniform external electric field of $E_0 = 1.2 \times 10^4 \ \mathrm{N/C}$ fills a region. A slab of metal and a slab of dielectric with $\kappa = 3.0$ are each placed in it, one at a time. (a) Find the field inside the metal, once electrostatic equilibrium is reached. (b) Find the field inside the dielectric. (c) Find the field produced by the induced charges in each case.

(a) The metal is a conductor, so its free electrons move until nothing is left to push them. EK 10.1.A.4 gives the result directly: the electric field is zero within the conductor. $E_{\mathrm{inside}} = 0$ .
(b) The dielectric constant is defined as the field reduction factor, $\kappa = E_0/E$ , from EK 10.4.A.4. Rearranging, $E = E_0/\kappa = (1.2 \times 10^4)/3.0 = 4.0 \times 10^3 \ \mathrm{N/C}$ .
(c) In the conductor, the induced surface charges must produce a field that exactly cancels the external one, so their field is $1.2 \times 10^4 \ \mathrm{N/C}$ pointing opposite to $E_0$ . Nothing less would stop the electrons moving.
In the dielectric, the polarized molecules produce a field of $1.2 \times 10^4 - 4.0 \times 10^3 = 8.0 \times 10^3 \ \mathrm{N/C}$ , also opposite to $E_0$ , per EK 10.4.A.3. That is two thirds of the external field, which is $(1 - 1/\kappa)$ of it.
Sanity check on the general form: the induced field is $E_0(1 - 1/\kappa)$ . Setting $\kappa = 3.0$ gives $E_0 \times 2/3 = 8.0 \times 10^3 \ \mathrm{N/C}$ , matching. Letting $\kappa$ grow without limit sends the induced field to $E_0$ and the interior field to zero, which is the conductor. A conductor behaves like a dielectric with an unlimited $\kappa$ .

Zero inside the metal; $4.0 \times 10^3 \ \mathrm{N/C}$ inside the dielectric. The induced charges make $1.2 \times 10^4 \ \mathrm{N/C}$ in the conductor and $8.0 \times 10^3 \ \mathrm{N/C}$ in the dielectric, both opposing the external field.

Battery attached, then battery removed

A parallel-plate capacitor has plate area $A = 0.020 \ \mathrm{m^2}$ and separation $d = 1.0 \ \mathrm{mm}$ , air-filled, and is charged by a $12 \ \mathrm{V}$ battery. A dielectric with $\kappa = 3.0$ is then inserted to fill the gap. Find $C$ , $Q$ , $\Delta V$ , the field in the gap and the stored energy, before and after, (a) with the battery still connected and (b) with the battery disconnected first. Use $\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C^2/(N \cdot m^2)}$ .

Start with the air-filled capacitor, where the conventions box sets $\kappa = 1.0$ . $C_0 = \varepsilon_0 A/d = (8.85 \times 10^{-12})(0.020)/(1.0 \times 10^{-3})$ . Numerator: $1.77 \times 10^{-13}$ . So $C_0 = 1.77 \times 10^{-10} \ \mathrm{F} = 177 \ \mathrm{pF}$ .
$Q_0 = C_0 \Delta V = (1.77 \times 10^{-10})(12) = 2.124 \times 10^{-9} \ \mathrm{C}$ . Field: $E_0 = \Delta V/d = 12/(1.0 \times 10^{-3}) = 1.2 \times 10^4 \ \mathrm{V/m}$ . Energy: $U_0 = \frac{1}{2}Q\Delta V = \frac{1}{2}(2.124 \times 10^{-9})(12) = 1.274 \times 10^{-8} \ \mathrm{J}$ .
The new capacitance is the same in both parts: $C = \kappa C_0 = 3.0 \times 1.77 \times 10^{-10} = 5.31 \times 10^{-10} \ \mathrm{F} = 531 \ \mathrm{pF}$ .
(a) Battery connected. It holds $\Delta V = 12 \ \mathrm{V}$ . Then $Q = C\Delta V = (5.31 \times 10^{-10})(12) = 6.372 \times 10^{-9} \ \mathrm{C}$ , three times the original: the battery pushed more charge on.
Field: $E = \Delta V/d = 12/(1.0 \times 10^{-3}) = 1.2 \times 10^4 \ \mathrm{V/m}$ , unchanged, because neither the voltage nor the separation moved. Energy: $U = \frac{1}{2}(6.372 \times 10^{-9})(12) = 3.823 \times 10^{-8} \ \mathrm{J}$ , three times the original.
(b) Battery removed first. The charge is trapped at $Q = 2.124 \times 10^{-9} \ \mathrm{C}$ . Then $\Delta V = Q/C = (2.124 \times 10^{-9})/(5.31 \times 10^{-10}) = 4.0 \ \mathrm{V}$ , one third of the original.
Field: $E = 4.0/(1.0 \times 10^{-3}) = 4.0 \times 10^3 \ \mathrm{V/m}$ , one third of the original, which is EK 10.4.A.4 for an isolated capacitor. Energy: $U = \frac{1}{2}(2.124 \times 10^{-9})(4.0) = 4.248 \times 10^{-9} \ \mathrm{J}$ , one third of the original.
Compare the two energy answers: $3.823 \times 10^{-8} \ \mathrm{J}$ against $4.248 \times 10^{-9} \ \mathrm{J}$ , a factor of nine apart from the same slab in the same capacitor. With the battery attached the battery supplied the extra energy; with it removed, the field pulled the slab in and the capacitor gave energy up.

Before: $C_0 = 177 \ \mathrm{pF}$ , $Q_0 = 2.124 \ \mathrm{nC}$ , $12 \ \mathrm{V}$ , $1.2 \times 10^4 \ \mathrm{V/m}$ , $1.274 \times 10^{-8} \ \mathrm{J}$ . Battery attached: $531 \ \mathrm{pF}$ , $6.372 \ \mathrm{nC}$ , $12 \ \mathrm{V}$ , $1.2 \times 10^4 \ \mathrm{V/m}$ , $3.823 \times 10^{-8} \ \mathrm{J}$ . Battery removed: $531 \ \mathrm{pF}$ , $2.124 \ \mathrm{nC}$ , $4.0 \ \mathrm{V}$ , $4.0 \times 10^3 \ \mathrm{V/m}$ , $4.248 \times 10^{-9} \ \mathrm{J}$ .

A conducting slab instead of a dielectric

The same air-filled capacitor, $A = 0.020 \ \mathrm{m^2}$ and $d = 1.0 \ \mathrm{mm}$ , has a conducting slab of thickness $0.40 \ \mathrm{mm}$ slid into the gap without touching either plate. (a) Find the new capacitance. (b) Compare it with the $\kappa = 3.0$ dielectric that fills the whole gap. (c) Say what happens if the conducting slab is made thick enough to touch both plates.

(a) The field inside the conducting slab is zero, so the potential difference is built up only across the two remaining air gaps, whose widths add to $d - t = 1.0 - 0.40 = 0.60 \ \mathrm{mm}$ .
The field in those gaps is unchanged, since it is set by the charge on the plates: $E = Q/(\varepsilon_0 A)$ with $\kappa = 1$ . So $\Delta V = E(d-t)$ and $C = Q/\Delta V = \varepsilon_0 A/(d-t)$ .
$C = (8.85 \times 10^{-12})(0.020)/(0.60 \times 10^{-3}) = (1.77 \times 10^{-13})/(6.0 \times 10^{-4}) = 2.95 \times 10^{-10} \ \mathrm{F} = 295 \ \mathrm{pF}$ .
Check the factor: $C/C_0 = d/(d-t) = 1.0/0.60 = 5/3$ , and $177 \times 5/3 = 295 \ \mathrm{pF}$ . The factor contains only lengths, with no material property in it.
(b) The dielectric gave $\kappa C_0 = 3.0 \times 177 = 531 \ \mathrm{pF}$ , a factor of $3.0$ set by the material. The conducting slab gave $295 \ \mathrm{pF}$ , a factor of $1.67$ set by its thickness. Two different mechanisms, two different-sized answers.
(c) If the slab reaches both plates, the two plates are joined by a conductor. Charge flows until they are at the same potential, so $\Delta V = 0$ and $C = Q/\Delta V$ has no finite value. The formula says the same thing: $d - t \to 0$ sends $C$ to infinity. In practice the capacitor is shorted and stores nothing.
A full-gap dielectric does not do this, because its charges are bound and none of them can cross from one plate to the other. That is the difference between the two materials in a single sentence.

$295 \ \mathrm{pF}$ with the $0.40 \ \mathrm{mm}$ conducting slab, against $531 \ \mathrm{pF}$ with the full $\kappa = 3.0$ dielectric. A conductor spanning the whole gap shorts the capacitor; a dielectric spanning it is exactly what a capacitor is designed for.

Frequently asked questions

What is the difference between a conductor and a dielectric?

In a conductor the charge carriers move freely through the material, so they keep drifting until the electric field inside the conducting material is zero. In a dielectric the charges are bound and can only shift slightly within each molecule, so the material polarizes and produces an opposing field that reduces the external field by the factor kappa without cancelling it. Cancelled to zero, or divided by kappa: that is the whole difference, and every capacitor consequence follows from it.

Is the electric field inside a dielectric zero?

No. That is the conductor result. Inside a dielectric the field is E0 divided by kappa, where E0 is the field that was there before and kappa is the dielectric constant. AP Physics C: E&M EK 10.4.A.4 gives exactly this as the definition of kappa in its derived form. A field of 1.2 times ten to the fourth newtons per coulomb becomes 4.0 times ten to the third inside a dielectric with kappa equal to 3.0, and zero inside a metal.

Why does inserting a dielectric increase capacitance?

Because capacitance is charge divided by potential difference, and the dielectric weakens the field in the gap. A weaker field over the same plate separation means a smaller potential difference for the same stored charge, so the ratio goes up. The factor is exactly kappa, which both AP sheets build into the printed formula C = kappa times epsilon nought times A over d, and which the CED states separately as C = kappa times C nought.

What happens to a capacitor when a dielectric is inserted with the battery still connected?

The battery holds the potential difference fixed, so the voltage does not change and neither does the field in the gap, since the field is the voltage divided by the separation. The capacitance is multiplied by kappa, and because C = Q over delta V with delta V fixed, the charge is multiplied by kappa too, with the battery supplying the extra. The stored energy, one half Q delta V, is therefore also multiplied by kappa. Every one of these answers changes if the battery is disconnected first.

What happens if the battery is disconnected before the dielectric is inserted?

Then the charge is trapped on the plates and cannot change. The capacitance is still multiplied by kappa, so the potential difference is divided by kappa, and the field in the gap is divided by kappa too. This is the case AP Physics C: E&M EK 10.4.A.4 describes when it says the field between the plates of an isolated parallel-plate capacitor decreases when a dielectric is inserted. The stored energy is divided by kappa, so it falls, in contrast with the connected case where it rises.

Can you use a conductor instead of a dielectric in a capacitor?

Not if it fills the gap. A conducting slab touching both plates joins them electrically, charge flows until they reach the same potential, the potential difference goes to zero and the capacitor is shorted. A conducting slab of thickness t that does not touch either plate does increase the capacitance, to epsilon nought times A over the quantity d minus t, because the field inside the slab is zero and so the potential difference is built up over a shorter distance. That factor is set by the geometry rather than by any property of the metal.

Is the dielectric constant ever less than 1?

Not for any real insulating material. It is 1 for vacuum by construction and greater than 1 otherwise, because the polarized material produces a field opposite in direction to the external one, per EK 10.4.A.3, and an opposing field can only reduce the total. Kappa is also dimensionless, since it is the ratio of the material's permittivity to the permittivity of free space, so an answer carrying a unit is wrong on inspection. Both AP equation sheets set kappa to 1.0 for air-filled capacitors unless the question says otherwise.