Physics 2 Quiz: Dielectrics

7 questions · exam conditions

0:00

DielectricsQuestion 1 of 7

In a polar dielectric, permanent dipole moments align with an applied electric field, increasing the polarization $\mathbf{P}$ . In a nonpolar dielectric, induced dipole moments arise from charge separation within molecules. For both types, the macroscopic relationship $\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} = \kappa\varepsilon_0\mathbf{E}$ holds. A student argues: 'Since polar dielectrics have larger permanent dipole moments, they always have a larger dielectric constant $\kappa$ than nonpolar dielectrics at room temperature.' Which of the following best evaluates this claim?

The claim is correct, because the permanent dipole moments of polar dielectrics contribute more strongly to

\mathbf{P}

than induced moments in nonpolar materials, so

\kappa_{\text{polar}} > \kappa_{\text{nonpolar}}

is a universal rule at room temperature.

The claim is incorrect only at low temperatures, where thermal agitation is minimal and nonpolar materials develop induced moments comparable in magnitude to the aligned permanent moments of polar dielectrics, equalizing

\kappa

The claim is incorrect, because

\kappa

depends on the density of dipoles and the degree of alignment, not just on the existence of permanent moments. High-symmetry polar molecules may have suppressed net polarization due to geometric cancellation, and some nonpolar materials with highly polarizable electron clouds can have larger

\kappa

values.

The claim is correct for liquids but incorrect for solids, because in liquid polar dielectrics the molecules are free to rotate and fully align with the field, whereas in solid polar dielectrics crystalline constraints prevent rotation and eliminate the permanent-dipole contribution to

\kappa

← Back to quizzes

Physics 2 Quiz

Physics 2 Quiz: Dielectrics

Practice Dielectrics in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Dielectrics, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The claim is correct, because the permanent dipole moments of polar dielectrics contribute more strongly to $\mathbf{P}$ than induced moments in nonpolar materials, so $\kappa_{\text{polar}} > \kappa_{\text{nonpolar}}$ is a universal rule at room temperature.
The claim is incorrect only at low temperatures, where thermal agitation is minimal and nonpolar materials develop induced moments comparable in magnitude to the aligned permanent moments of polar dielectrics, equalizing $\kappa$ .
The claim is incorrect, because $\kappa$ depends on the density of dipoles and the degree of alignment, not just on the existence of permanent moments. High-symmetry polar molecules may have suppressed net polarization due to geometric cancellation, and some nonpolar materials with highly polarizable electron clouds can have larger $\kappa$ values. (correct answer)
The claim is correct for liquids but incorrect for solids, because in liquid polar dielectrics the molecules are free to rotate and fully align with the field, whereas in solid polar dielectrics crystalline constraints prevent rotation and eliminate the permanent-dipole contribution to $\kappa$ .

Explanation: Whenever you see a claim about dielectric constants, resist the instinct to think in absolutes —

\kappa

is a macroscopic quantity that emerges from multiple competing factors, not a single molecular property. The macroscopic relationship

\mathbf{P} = \varepsilon_0(\kappa - 1)\mathbf{E}

tells you that

\kappa

reflects how strongly a material polarizes per unit field. That depends on dipole density, degree of alignment, and molecular polarizability — not simply whether permanent dipoles exist. High-symmetry polar molecules (like CCl₄) can have individual bond dipoles that cancel geometrically, yielding negligible net polarization. Meanwhile, nonpolar materials with large, loosely bound electron clouds (like silicon or certain aromatic compounds) can have enormous induced polarizability, producing

\kappa

values that rival or exceed many polar dielectrics. Answer C correctly captures all of this nuance — it's the right choice. Answer A fails because it treats "permanent dipole exists → larger

\kappa

" as a universal law, ignoring geometric cancellation and the role of electron cloud polarizability in nonpolar materials. This is a classic overgeneralization trap. Answer B gets the temperature logic backwards. At low temperatures, thermal agitation is reduced, which actually helps polar molecules stay aligned — it doesn't help nonpolar materials "catch up." The premise of B is physically inverted. Answer D contains a grain of truth (rotation does matter in liquids), but it's wrong to say crystalline constraints eliminate the permanent-dipole contribution entirely — partial alignment and other mechanisms still contribute to

\kappa

in polar solids. Your study tip: when evaluating "always" or "never" claims about material properties, immediately look for counterexamples rooted in molecular geometry or alternative physical mechanisms — exams love testing these edge cases.

Question 2

Two identical parallel-plate capacitors, each with capacitance $C_0$ and plate separation $d$ , are connected in series across a battery of voltage $V$ . After being fully charged, the battery is disconnected. A dielectric with dielectric constant $\kappa = 2$ is then inserted to completely fill only one of the two capacitors.

What is the new voltage across the capacitor that does not contain the dielectric?

$\dfrac{2V}{3}$ , because after the dielectric is inserted the new series capacitance is $\frac{2C_0}{3}$ , and the voltage across the unmodified capacitor is determined by the ratio of the new series capacitance to the original capacitance of that element.
$\dfrac{V}{2}$ , because the total charge is conserved and, in a series circuit with isolated plates, each capacitor retains the same charge $Q = C_0 V/2$ ; dividing by $C_0$ gives $V/2$ for the unmodified capacitor. (correct answer)
$\dfrac{V}{3}$ , because the dielectric doubles the capacitance of one capacitor, and in a series circuit with conserved total charge the voltage divides inversely with capacitance, giving the unmodified capacitor one-third of the original voltage.
$\dfrac{V}{4}$ , because inserting a dielectric of $\kappa = 2$ into one capacitor reduces the voltage across it by a factor of 2, and by symmetry the voltage across the other capacitor must also be halved, leaving $V/4$ across the unmodified one.

Explanation: When capacitors are connected in series and then isolated from the battery, the key constraint shifts from voltage to charge. Charge cannot flow onto or off the isolated inner plates, so the total charge on the series combination is fixed. Before disconnection, the series capacitance is

C_{series} = \dfrac{C_0}{2}

, so the stored charge is

Q = \dfrac{C_0}{2} \cdot V = \dfrac{C_0 V}{2}

. This charge is locked in — inserting the dielectric cannot change it. Since both capacitors share the same series charge, each still carries

Q = \dfrac{C_0 V}{2}

. For the unmodified capacitor (still capacitance

C_0

), the voltage is simply

V_{unmod} = \dfrac{Q}{C_0} = \dfrac{V}{2}

. That confirms B is correct. Choice A incorrectly applies the voltage-divider ratio using the new series capacitance divided by the original element capacitance — this formula doesn't apply here. Voltage dividers work when voltage is fixed (battery connected), not when charge is fixed. Choice C gets the charge-conservation logic partially right but makes an arithmetic error. The dielectric doubles one capacitor to

2C_0

, and the total voltage now splits as

\frac{1}{3}V

across the modified and

\frac{2}{3}V

across the unmodified — yet C assigns those backwards, giving

V/3

to the wrong capacitor. Choice D invents a false symmetry argument. Inserting a dielectric affects only one capacitor; the other is physically unchanged. Study tip: Whenever a battery is disconnected before something changes, freeze the charge — that's your conserved quantity. Ask "what's locked in?" before solving.

Question 3

A student claims: 'Inserting a dielectric always increases the energy stored in a capacitor, because the dielectric constant $\kappa > 1$ increases the capacitance, and larger capacitance means more stored energy.' Under which condition is this claim correct?

When the capacitor plates are connected to a constant-voltage source during insertion, so the voltage remains fixed and $U = \frac{1}{2}CV^2$ increases with $C$ . (correct answer)
When the capacitor is isolated (battery disconnected) before insertion, so the charge is fixed and $U = Q^2/(2C)$ increases because $C$ appears in the denominator.
The claim is always correct regardless of boundary conditions, because the dielectric's polarization energy adds to the stored energy in every case.
When the dielectric is inserted slowly and quasi-statically, so that no energy is lost to Joule heating or radiation, ensuring all work by the source converts directly to stored energy.

Explanation: When analyzing dielectric insertion problems, the key is identifying the boundary condition: is voltage fixed (battery connected) or charge fixed (battery disconnected)? The student's claim hinges entirely on which formula applies. With a battery connected, voltage

V

stays constant. Since

U = \frac{1}{2}CV^2

, inserting a dielectric increases

C

by a factor of

\kappa

, which directly increases stored energy by the same factor. This makes A correct — the claim holds precisely when voltage is held fixed by an external source. B is actually backwards. When the capacitor is isolated, charge

Q

is fixed, so you must use

U = \frac{Q^2}{2C}

. Because

C

appears in the denominator, increasing

C

decreases stored energy — the opposite of the student's claim. The dielectric is actually pulled in by the electric field, and the system does work on the dielectric, reducing stored energy. C is wrong because the outcome is not universal — it depends entirely on boundary conditions, as shown above. Dielectric polarization doesn't simply "add" energy; the energy bookkeeping changes depending on whether a source is present. D conflates a thermodynamic process condition (quasi-static) with an energy outcome. Inserting slowly affects how energy is transferred, not whether energy stored increases. Even a quasi-static insertion with the battery disconnected still reduces stored energy. Study tip: Always identify the constraint first — constant

V

or constant

Q

— then choose the correct energy formula. These two cases give opposite results for energy when a dielectric is inserted, making this a favorite exam trap.

Question 4

A parallel-plate capacitor is connected to a constant-voltage source $V_0$ . A dielectric slab of dielectric constant $\kappa > 1$ is slowly inserted between the plates while the voltage source remains connected. Which of the following statements about the free charge on the plates and the work done by the voltage source is correct?

The free charge on the plates increases by a factor of $\kappa$ , and the work done by the voltage source equals the increase in stored energy, so no net energy is supplied beyond what is stored.
The free charge on the plates increases by a factor of $\kappa$ , and the work done by the voltage source is twice the increase in stored energy, with the excess energy accounted for by the mechanical work done on the dielectric slab. (correct answer)
The free charge on the plates remains constant because the voltage is fixed, and the work done by the voltage source is zero since $V_0$ does not change.
The free charge on the plates decreases by a factor of $\kappa$ because the bound charges from the dielectric partially neutralize the free charges, and the voltage source does negative work to maintain $V_0$ .

Explanation: When a capacitor stays connected to a voltage source while you insert a dielectric, the voltage is locked at

V_0

, but the capacitance changes — and that drives everything else. Here's the core logic: inserting a dielectric increases capacitance by a factor of

\kappa

, since

C = \kappa C_0

. Because

Q = CV

and

V_0

is fixed, the free charge on the plates must increase by the same factor:

Q = \kappa C_0 V_0

. This immediately rules out C (charge doesn't stay constant — capacitance changed, so charge must change to maintain the same voltage) and D (charge increases, not decreases; bound charges from the dielectric don't neutralize free charges — they actually cause the battery to supply more free charge to maintain

V_0

). Now for the energy accounting. The new stored energy is

U = \frac{1}{2}\kappa C_0 V_0^2

, an increase of

\Delta U = \frac{1}{2}(\kappa - 1)C_0 V_0^2

. The battery supplies charge

\Delta Q = (\kappa - 1)C_0 V_0

, doing work

W_{battery} = V_0 \cdot \Delta Q = (\kappa - 1)C_0 V_0^2 = 2\Delta U

. The battery does twice the work stored — the extra half goes into the mechanical work done on the slab (the dielectric is pulled in, so the system does negative mechanical work on whoever holds it, or equivalently the field does positive work on the dielectric). This confirms B is correct. A is wrong because it claims the battery's work equals the stored energy increase — it actually supplies twice that amount. Study tip: Whenever a capacitor problem involves a battery staying connected, remember the battery does work

W = V \cdot \Delta Q

, which always equals twice the energy stored from that extra charge. Energy conservation then tells you where the other half went.

Question 5

A capacitor is constructed with two different dielectric slabs of equal thickness $d/2$ and equal area $A/2$ each, placed side by side (i.e., each slab spans the full plate separation $d$ but only half the plate area), with dielectric constants $\kappa_1$ and $\kappa_2$ . Which expression gives the total capacitance, and which circuit analogy correctly describes this geometry?

$C = \dfrac{2\varepsilon_0 A}{(1/\kappa_1+1/\kappa_2)d}$ ; the two sections act as capacitors in parallel, because each dielectric region independently stores charge and the capacitances add harmonically.
$C = \dfrac{2\kappa_1\kappa_2\varepsilon_0 A}{(\kappa_1+\kappa_2)d}$ ; the two sections act as capacitors in series, since the electric field must pass through both dielectrics sequentially from one plate to the other.
$C = \dfrac{\varepsilon_0 A(\kappa_1 + \kappa_2)}{2d}$ ; the two sections act as capacitors in series, because the total charge must divide between the two dielectric regions to satisfy the boundary conditions at the interface.
$C = \dfrac{\varepsilon_0 A(\kappa_1 + \kappa_2)}{2d}$ ; the two sections act as capacitors in parallel, since both share the same voltage (connected to the same plates) and their charges add. (correct answer)

Explanation: When a capacitor has dielectric slabs placed side by side (each occupying half the area but the full gap), the key question is: do they share the same voltage, or the same charge? Since both regions are bounded by the same two conducting plates, they share identical voltage

V

across them. This is the defining condition for a parallel combination. For each half, you have a standard capacitor with area

A/2

and gap

d

C_1 = \frac{\kappa_1 \varepsilon_0 (A/2)}{d}, \quad C_2 = \frac{\kappa_2 \varepsilon_0 (A/2)}{d}

In parallel, capacitances add directly:

C = C_1 + C_2 = \frac{\varepsilon_0 A}{2d}(\kappa_1 + \kappa_2)

This confirms D is correct — parallel combination, charges add, voltages are equal. A is wrong on two counts: it gives a harmonic (series-like) formula and incorrectly claims "charges add harmonically." That formula would apply to stacked dielectrics, not side-by-side ones. B describes a series combination and uses the harmonic mean of

\kappa_1

and

\kappa_2

. Series applies when the electric field passes sequentially through both materials — that's the stacked geometry, not this one. C gets the correct formula but misidentifies the circuit analogy as series. The math is right but the physical reasoning is backward — this is a classic trap designed to test whether you truly understand why the formula works. Study tip: Always ask yourself first — do the dielectric regions share the same voltage or the same charge? Same voltage → parallel. Same charge → series. The geometry tells you everything.

Question 6

An air-filled parallel-plate capacitor (capacitance $C_0$ , plate separation $d$ ) is connected to a battery and charged to voltage $V_0$ . The battery is then disconnected. Next, a technician partially inserts a dielectric slab ( $\kappa = 4$ ) so that it covers exactly half the plate area while leaving the other half as vacuum, filling the full gap $d$ throughout. The technician observes that the voltage across the capacitor changes.

What is the new voltage across the capacitor after the partial insertion?

$\dfrac{4V_0}{5}$ , because the dielectric slab covering half the area reduces the effective electric field in that region by a factor of $\kappa = 4$ , so the average field across the full plate area decreases by the weighted factor $\frac{1}{2}\left(1 + \frac{1}{4}\right) = \frac{5}{8}$ , giving a voltage of $\frac{4}{5}V_0$ .
$\dfrac{V_0}{4}$ , because inserting a dielectric over half the area effectively reduces the electric field by $\kappa = 4$ everywhere between the plates, cutting the voltage to one-quarter of its original value.
$\dfrac{2V_0}{5}$ , because the partial insertion creates a parallel combination with total capacitance $C_{\kappa} + C_{\text{vac}} = 2C_0 + \frac{C_0}{2} = \frac{5}{2}C_0$ , and with fixed charge $Q = C_0 V_0$ , the new voltage is $Q/C_{\text{new}} = \frac{2V_0}{5}$ . (correct answer)
$\dfrac{V_0}{2}$ , because only half the plate area has the dielectric, so the capacitance effectively doubles (from $\kappa = 4$ acting on half the area, contributing $2C_0$ , while the vacuum half contributes nothing new) and the voltage halves with fixed charge.

Explanation: Whenever a dielectric is partially inserted into a capacitor, the key is recognizing the correct circuit model. The two halves — one with dielectric, one without — share the same voltage and act as capacitors in parallel, not in series. Here's the reasoning for choice C. Before disconnection, the charge stored is

Q = C_0 V_0

. After the battery is disconnected, this charge is fixed. The partial insertion creates two parallel capacitors: the dielectric half has capacitance

C_\kappa = \kappa \cdot \frac{C_0}{2} = 4 \cdot \frac{C_0}{2} = 2C_0

, and the vacuum half has

C_{\text{vac}} = \frac{C_0}{2}

. Their total is

C_{\text{new}} = 2C_0 + \frac{C_0}{2} = \frac{5C_0}{2}

. With fixed charge, the new voltage is

V = \frac{Q}{C_{\text{new}}} = \frac{C_0 V_0}{\frac{5C_0}{2}} = \frac{2V_0}{5}

. That confirms C. Choice A incorrectly averages the electric fields across regions, treating the geometry as a series-like problem. The voltage isn't a weighted average of field reductions — it comes from

Q/C_{\text{new}}

. Choice B wrongly applies

\kappa = 4

to the entire gap, as if the dielectric were fully inserted. Only half the area is covered, so only half the capacitance is scaled by

\kappa

. Choice D claims the vacuum half "contributes nothing," ignoring that it still stores charge and adds

\frac{C_0}{2}

to the total capacitance. Study tip: When a dielectric covers part of the plate area, always split the capacitor into two parallel capacitors. Parallel means same voltage, additive capacitance — never skip the vacuum half.

Question 7

A parallel-plate capacitor with plate area $A$ and separation $d$ is connected to a battery of EMF $\mathcal{E}$ and fully charged. The battery is then disconnected. A dielectric slab with dielectric constant $\kappa = 3$ is subsequently inserted to completely fill the gap between the plates.

After the dielectric is inserted (with the battery disconnected), which of the following correctly describes what happens to the electric field between the plates and the energy stored in the capacitor?

The electric field decreases by a factor of 3, and the stored energy decreases by a factor of 3, because the charge on the plates remains fixed while the capacitance triples. (correct answer)
The electric field remains unchanged, and the stored energy increases by a factor of 3, because the dielectric introduces additional polarization energy that adds to the original stored energy.
The electric field decreases by a factor of 3, and the stored energy increases by a factor of 3, because the dielectric increases the capacitance while holding the voltage constant across the plates.
The electric field remains unchanged, and the stored energy decreases by a factor of 3, because the induced surface charges on the dielectric reduce the effective plate separation without altering the field.

Explanation: Whenever a capacitor question involves a dielectric, your first move should be identifying whether the battery is connected or disconnected — this single detail controls everything. If the battery is disconnected, charge is conserved (

Q = \text{const}

); if it stays connected, voltage is conserved. Here, the battery is disconnected before the dielectric is inserted, so the charge

Q

on the plates cannot change. Inserting a dielectric with

\kappa = 3

triples the capacitance:

C' = \kappa C_0 = 3C_0

. Since

Q

is fixed but

C

triples, the voltage drops by a factor of 3:

V' = Q/C' = V_0/3

. The electric field, which equals

V/d

, therefore also drops by a factor of 3. The stored energy

U = Q^2/(2C)

then decreases by the same factor of 3, giving

U' = U_0/3

. This confirms answer A is correct. B is wrong because the electric field does change — it's tied to voltage, which drops when capacitance increases at constant charge. There is no "polarization energy" added; in fact, energy is released (it goes into the mechanical work of pulling the dielectric in). C describes the scenario where the battery remains connected (constant voltage), not the disconnected case. With constant voltage, energy would actually increase by

\kappa

, but the field would stay the same — C incorrectly mixes both scenarios. D is wrong on both counts: the field does change, and the idea that the dielectric reduces "effective plate separation" is not a real mechanism here. Study tip: Memorize these two cases as a pair — battery connected means constant

V

; battery disconnected means constant

Q

. Every dielectric insertion problem hinges on that distinction.