Physics 2 Quiz: Charge Conservation And Conductors

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Charge Conservation And ConductorsQuestion 1 of 9

Two conducting spheres are connected by a long, thin conducting wire. Sphere 1 has radius $r$ and sphere 2 has radius $4r$ . A total charge $Q_{\text{tot}}$ is placed on the system. The system reaches electrostatic equilibrium.

At equilibrium, which statement about the surface charge densities $\sigma_1$ and $\sigma_2$ on the two spheres is correct, and what is the ratio $\sigma_1 / \sigma_2$ ?

\sigma_1 > \sigma_2

and

\sigma_1/\sigma_2 = 16

, because connected conductors equalize potential, which for spheres means surface charge density is inversely proportional to the square of the radius.

\sigma_1 = \sigma_2

and

\sigma_1/\sigma_2 = 1

, because charge distributes uniformly across the total surface area of connected conductors to minimize energy.

\sigma_1 < \sigma_2

and

\sigma_1/\sigma_2 = 1/4

, because the larger sphere has a stronger electric field at its surface and therefore attracts more charge per unit area.

\sigma_1 > \sigma_2

and

\sigma_1/\sigma_2 = 4

, because connected conductors equalize potential, which for spheres means surface charge density is inversely proportional to radius.

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Physics 2 Quiz

Physics 2 Quiz: Charge Conservation And Conductors

Practice Charge Conservation And Conductors 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 Charge Conservation And Conductors, 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

At equilibrium, which statement about the surface charge densities $\sigma_1$ and $\sigma_2$ on the two spheres is correct, and what is the ratio $\sigma_1 / \sigma_2$ ?

$\sigma_1 > \sigma_2$ and $\sigma_1/\sigma_2 = 16$ , because connected conductors equalize potential, which for spheres means surface charge density is inversely proportional to the square of the radius.
$\sigma_1 = \sigma_2$ and $\sigma_1/\sigma_2 = 1$ , because charge distributes uniformly across the total surface area of connected conductors to minimize energy.
$\sigma_1 < \sigma_2$ and $\sigma_1/\sigma_2 = 1/4$ , because the larger sphere has a stronger electric field at its surface and therefore attracts more charge per unit area.
$\sigma_1 > \sigma_2$ and $\sigma_1/\sigma_2 = 4$ , because connected conductors equalize potential, which for spheres means surface charge density is inversely proportional to radius. (correct answer)

Explanation: When two conductors are connected by a wire, charge redistributes until both are at the same electric potential. For an isolated conducting sphere of radius

R

carrying charge

Q

, the surface potential is

V = kQ/R

. Setting the potentials equal gives you the key relationship to work with. For the two spheres:

V_1 = V_2

means

\frac{kQ_1}{r} = \frac{kQ_2}{4r}

, so

Q_1/Q_2 = 1/4

. The larger sphere holds four times more charge. Now convert to surface charge density using

\sigma = Q/(4\pi R^2)

\sigma_1 = \frac{Q_1}{4\pi r^2}, \quad \sigma_2 = \frac{Q_2}{4\pi (4r)^2} = \frac{Q_2}{64\pi r^2}

Taking the ratio:

\frac{\sigma_1}{\sigma_2} = \frac{Q_1}{4\pi r^2} \cdot \frac{64\pi r^2}{Q_2} = 16 \cdot \frac{Q_1}{Q_2} = 16 \cdot \frac{1}{4} = 4

\sigma_1 > \sigma_2

and

\sigma_1/\sigma_2 = 4

, confirming D. Surface charge density is inversely proportional to radius, not radius squared. A gets the charge ratio right but forgets to account for the area factor when converting from charge to surface charge density — it stops one step too early. B is wrong because charge does not spread uniformly over total surface area; it distributes to equalize potential. C has the inequality completely backwards — the smaller sphere actually has the higher surface charge density and stronger surface field, not the larger one. A helpful pattern: equal potential → charge proportional to radius → surface charge density inversely proportional to radius. Don't confuse the charge ratio with the density ratio; that's the trap in A.

Question 2

A neutral conducting cube is placed in an initially uniform external electric field directed in the $+x$ direction. After electrostatic equilibrium is established inside the cube, the external field source is removed. Which of the following correctly describes what happens to the charge distribution on the cube's surfaces immediately after the field is removed?

The induced surface charges remain fixed in place, because charges on the outer surface of a conductor are held there by the surface boundary and have no internal path available to recombine once the external driving field is gone.
The induced surface charges gradually recombine over a timescale of microseconds to milliseconds, driven by mutual repulsion of the like-sign charges on each face, until the surface returns to a uniform neutral distribution.
The induced surface charges slowly leak off into the surrounding air over many hours, because a conductor cannot sustain any surface charge distribution without a continuous external field to maintain the separation against thermal fluctuations.
The induced surface charges almost instantaneously vanish and the cube's surface returns to a neutral distribution, because the cube was never given net charge — the separation was a polarization of free electrons that immediately revert to a uniform distribution when the driving field is removed. (correct answer)

Explanation: When you see a question about conductors in external electric fields, the key distinction to keep in mind is the difference between net charge and charge separation. A neutral conductor placed in an external field undergoes polarization — free electrons redistribute internally, creating surface charge on opposite faces — but the total charge remains exactly zero throughout. This is precisely why D is correct. The induced charges were never "new" charges added to the cube; they were simply free electrons that shifted position due to the external field. Once that field is removed, there is no longer any force maintaining that separation. The electrons, driven by the now-unbalanced electrostatic forces from the positive face, redistribute almost instantaneously — on a timescale comparable to the relaxation time of a good conductor, roughly

10^{-19}

seconds for copper. The surface returns to a uniform neutral distribution effectively immediately. Choice A is wrong because it imagines surface charges as if they were somehow trapped or pinned. Free electrons in a conductor are highly mobile and are not "held" anywhere — they respond immediately to internal electric forces. Choice B contains a grain of truth about mutual repulsion driving recombination, but the timescale is wildly incorrect. The process happens in femtoseconds, not microseconds to milliseconds, and recombination doesn't require like-charges repelling each other — it's the attraction between the separated positive and negative regions that drives neutralization. Choice C is wrong because it confuses a conductor with a leaky capacitor or dielectric; conductors don't require a continuous field to retain charge, and the mechanism described doesn't apply here. Your study tip: always ask yourself whether a conductor has been given net charge or merely polarized — the answer completely changes the physics.

Question 3

An experimenter rubs a glass rod with silk, transferring $1.0 \times 10^{10}$ electrons from the rod to the silk. The experimenter then touches the glass rod to one end of a long copper wire, and touches the other end of the wire to a large copper plate that is connected to ground. The glass rod is then removed from the wire.

The rod is neutral and the silk carries $-1.6 \times 10^{-9}\,\text{C}$ ; the rod was neutralized because electrons flowed from ground through the wire onto the positively charged rod, with the ground supplying the $1.0\times10^{10}$ electrons needed. The total charge of the universe is conserved.
The rod is neutral and the silk carries $-1.6 \times 10^{-9}\,\text{C}$ ; the rod's deficit of electrons was replenished by electrons flowing from ground onto the rod. The net charge of the (rod + silk) system is $-1.6\times10^{-9}\,\text{C}$ , balanced by a $+1.6\times10^{-9}\,\text{C}$ deficit left in the ground. (correct answer)
The rod remains positively charged with $+1.6 \times 10^{-9}\,\text{C}$ and the silk remains negatively charged with $-1.6 \times 10^{-9}\,\text{C}$ ; glass is an insulator so charge cannot leave the rod through the copper wire, even at the point of contact.
Both the rod and the silk are neutral after the procedure; grounding the rod also neutralizes the silk because electrons redistribute through the ambient air between the silk and the ground until both objects reach zero net charge.

Explanation: When a question involves grounding a charged object, your first instinct should be to track where charge flows and apply conservation carefully — not just assume everything neutralizes. Here's what happens step by step. Rubbing the glass rod transfers

1.0 \times 10^{10}

electrons to the silk, leaving the rod with a charge of

+(1.0 \times 10^{10})(1.6 \times 10^{-19}\,\text{C}) = +1.6 \times 10^{-9}\,\text{C}

, and the silk with

-1.6 \times 10^{-9}\,\text{C}

. When the rod touches the grounded copper wire, electrons flow from the ground through the wire onto the positively charged rod, neutralizing it. The rod ends up neutral. The silk, however, was never connected to anything — it remains at

-1.6 \times 10^{-9}\,\text{C}

. Charge conservation is satisfied because the ground now has a deficit of

1.0 \times 10^{10}

electrons, giving it a net charge of

+1.6 \times 10^{-9}\,\text{C}

. The rod + silk system carries

-1.6 \times 10^{-9}\,\text{C}

, exactly balanced by the ground's deficit. This is B. A is nearly right in its physical description, but claims "the total charge of the universe is conserved" without accounting for the charge left in the ground — it implies the ground is irrelevant, which misrepresents conservation. C incorrectly treats glass as a perfect insulator even at the point of contact. At a contact point, charge can transfer between conductors and insulators; that's exactly how the rod was charged in the first place. D invents a mechanism — charge redistribution through ambient air — that doesn't occur under normal conditions. The silk is isolated and cannot be neutralized remotely. Your takeaway: grounding neutralizes only the object in direct contact. Always ask "what is actually connected?" before applying conservation.

Question 4

A thin, neutral, conducting spherical shell of inner radius $a$ and outer radius $b$ surrounds a point charge $+Q$ at its center. An additional charge $+Q$ is then deposited directly onto the conducting shell. After electrostatic equilibrium is reached, what are the charges on the inner and outer surfaces of the shell?

Inner surface: $-Q$ ; outer surface: $+Q$ , because the shell was neutral before the extra charge was deposited, and adding $+Q$ to a shell that already had $-Q$ on its inner surface means only $+Q$ remains for the outer surface.
Inner surface: $0$ ; outer surface: $+2Q$ , because the deposited charge distributes itself with the original $+Q$ from the point charge across the outer surface, leaving the inner surface uncharged.
Inner surface: $-Q$ ; outer surface: $+2Q$ , because Gauss's law requires $-Q$ on the inner surface to zero the field in the conductor, and charge conservation (shell had $+Q$ added to its original neutral state) then puts $+2Q$ on the outer surface. (correct answer)
Inner surface: $+Q$ ; outer surface: $+Q$ , because the central charge repels positive charge to both surfaces equally, and the deposited charge distributes between the two surfaces to maintain equilibrium.

Explanation: Whenever you see a conducting shell surrounding a charge, your two essential tools are Gauss's law and charge conservation — apply them in that order. Start inside the conductor. In electrostatic equilibrium, the electric field within the conducting material itself must be zero. Draw a Gaussian surface entirely inside the conducting material (between radius

a

and

b

). Since

\vec{E} = 0

there, the net enclosed charge must be zero. The point charge

+Q

at the center is enclosed, so the inner surface must carry

-Q

to make the total zero. This is non-negotiable — Gauss's law demands it regardless of what's happening on the outer surface. Now apply charge conservation to the shell itself. The shell started neutral, and you deposited an additional

+Q

onto it, so the shell's total charge is

+Q

. Since

-Q

is already committed to the inner surface, the remaining charge on the outer surface must be

+Q - (-Q) = +2Q

. That makes C correct: inner surface

-Q

, outer surface

+2Q

. A is tempting but wrong — it correctly identifies

-Q

on the inner surface yet forgets to account for the deposited

+Q

, leaving the outer surface with only

+Q

instead of

+2Q

. B incorrectly assumes the

+Q

point charge somehow "merges" with the deposited charge on the outer surface, ignoring the mandatory

-Q

induction on the inner surface. D invents a mechanism where positive charge splits symmetrically between surfaces, which contradicts both Gauss's law and how induction actually works. Your strategy: always determine inner surface charge with Gauss's law first, then use charge conservation to find the outer surface charge.

Question 5

Two identical conducting spheres, each of radius $R$ , are separated by a center-to-center distance $d \gg R$ . Sphere 1 carries charge $+3Q$ and sphere 2 carries charge $-Q$ . They are connected by a thin conducting wire and allowed to reach equilibrium, then the wire is cut. Subsequently, a third identical uncharged conducting sphere 3 is touched simultaneously to both sphere 1 and sphere 2 (i.e., sphere 3 bridges them) and then removed.

$+Q/2$ , because after the wire equalizes spheres 1 and 2 (each gets $+Q$ ), touching sphere 3 to sphere 1 first transfers half of sphere 1's charge to sphere 3, giving sphere 3 a charge of $+Q/2$ , and the subsequent contact with sphere 2 leaves sphere 3's charge unchanged since both are already at $+Q/2$ .
$+2Q/3$ , because after the wire connects spheres 1 and 2 they each carry $+Q$ , giving a total of $+2Q$ ; when sphere 3 bridges both simultaneously, all three identical spheres share the total charge equally, leaving sphere 3 with $+2Q/3$ . (correct answer)
$+Q/2$ , because sphere 3 touches sphere 1 ( $+Q$ ) and sphere 2 ( $+Q$ ) in two separate sequential contacts, and each contact splits the charge equally, resulting in $+Q/2$ on sphere 3.
$+Q$ , because sphere 3 simultaneously contacts both spheres 1 and 2 and absorbs a charge equal to the average of the two spheres' charges, which is $(+Q + Q)/2 = +Q$ .

Explanation: When conducting spheres touch, charge redistributes until all connected conductors reach the same electric potential. For identical spheres, equal potential means equal charge — so the key is always identifying how many identical spheres are sharing charge at the same moment. First, when spheres 1 (

+3Q

) and 2 (

-Q

) are connected by the wire, the total charge is

+3Q + (-Q) = +2Q

, split equally between two identical spheres, giving each sphere

+Q

. After the wire is cut, both spheres carry

+Q

. Now sphere 3 (uncharged) simultaneously bridges both spheres. This is the critical detail. At the moment of contact, all three identical spheres are electrically connected, so the total charge

+Q + Q + 0 = +2Q

distributes equally among all three. Each sphere ends up with

+2Q/3

. When sphere 3 is removed, it carries

+2Q/3

, confirming answer B. Answer A describes a sequential touching process — sphere 3 contacts sphere 1, then sphere 2 separately — which is not what the problem states. Simultaneous bridging is fundamentally different from two separate contacts. Answer C makes the same sequential-contact error as A, applying the "split in half" rule twice in succession rather than recognizing that three spheres share charge simultaneously. Answer D claims sphere 3 "absorbs the average" of the two sphere charges, arriving at

+Q

. This misapplies averaging — when three equal spheres share charge, the share per sphere is the total divided by three, not the average of two. Study tip: Always count how many identical conductors are in simultaneous contact — charge divides equally among all of them at once, so the divisor is the total number connected, not just two.

Question 6

Two identical small conducting spheres, X and Y, carry charges of $+8\,\mu\text{C}$ and $-2\,\mu\text{C}$ respectively. They are touched together briefly and then separated. A third identical conducting sphere Z, initially uncharged, is then touched to sphere Y only and removed. What is the final charge on sphere Z?

$+1.5\,\mu\text{C}$ , because after X and Y share charge equally the combined charge is $+3\,\mu\text{C}$ , Y holds $+3\,\mu\text{C}$ , and Z takes half of Y's charge when touched. (correct answer)
$+1.5\,\mu\text{C}$ , because the net charge of $+6\,\mu\text{C}$ is shared among all three spheres equally, giving each $+2\,\mu\text{C}$ , and Z acquires half of Y's final share.
$+2\,\mu\text{C}$ , because the total charge $+6\,\mu\text{C}$ is divided equally among three identical spheres when all three are touched simultaneously.
$+3\,\mu\text{C}$ , because touching X and Y gives each $+3\,\mu\text{C}$ , and that entire charge transfers to the initially uncharged Z when they are touched.

Explanation: When identical conducting spheres touch, charge redistributes equally between them — this is the core principle being tested. The key here is that the process happens in two separate steps, not all at once. Step 1: X (

+8\,\mu\text{C}

) and Y (

-2\,\mu\text{C}

) are touched together. Their net charge is

+8 + (-2) = +6\,\mu\text{C}

, which splits equally, giving each sphere

+3\,\mu\text{C}

. Step 2: Uncharged Z touches only Y (now

+3\,\mu\text{C}

). The combined charge on Y and Z is

+3\,\mu\text{C}

, which splits equally — so Z ends up with

+1.5\,\mu\text{C}

. This makes A the correct answer. Choice B reaches the right number through wrong reasoning — it incorrectly imagines all three spheres sharing the total

+6\,\mu\text{C}

simultaneously, then applies yet another halving step. That's a double error. Choice C correctly identifies the total charge as

+6\,\mu\text{C}

and divides by three, but this would only apply if all three spheres were touched at the same time. The problem explicitly describes two separate touchings, so you can't treat it as one combined event. Choice D makes a classic blunder: assuming the full charge of Y transfers entirely to Z. When two identical spheres touch, charge always shares equally — none stays behind only if one sphere was truly isolated and fully discharged, which isn't the case here. Study tip: Always track charge transfers sequentially, step by step. Draw a small diagram noting the charge on each sphere after each contact — it prevents you from accidentally merging separate events into one.

Question 7

A student claims: 'The interior of a conductor in electrostatic equilibrium must be free of net charge. Therefore, if I place a small charged insulating bead inside a hollow conducting shell, the interior of the conductor itself still has no net charge, so the shell's outer surface also remains uncharged.'

Which of the following correctly evaluates the student's argument?

The argument is partially correct: the conductor material itself has no net charge in its bulk, but the inner surface of the shell acquires an induced charge equal and opposite to the bead's charge, and charge conservation then requires the outer surface to carry a charge equal in sign and magnitude to the bead's charge. (correct answer)
The argument is fully correct: the shell's conductor material has zero net charge in equilibrium, so neither the inner nor the outer surface can develop any net charge regardless of what is placed inside the cavity.
The argument is incorrect because the electric field of the bead penetrates the conductor and distributes the bead's charge uniformly over both surfaces of the shell, making each surface carry half the bead's charge.
The argument is correct for the inner surface but incorrect for the outer surface: the inner surface acquires an induced charge equal in sign to the bead, and the outer surface acquires the opposite sign, as required by charge conservation.

Explanation: When a charged object is placed inside a hollow conducting shell, you need to apply two principles together: Gauss's Law and charge conservation. The student's mistake is conflating "no net charge in the bulk conductor material" with "no net charge on the surfaces." Here's the correct reasoning behind answer A. The conductor's free electrons rearrange until the electric field inside the conductor material itself is zero — this is electrostatic equilibrium. To cancel the field from the bead (charge

+q

), the inner surface of the shell must accumulate

-q

. Since the conductor's bulk remains neutral, those electrons drawn to the inner surface must come from somewhere: charge conservation demands the outer surface is left with

+q

. The bead doesn't "give" its charge to the shell — it induces a redistribution. The outer surface then radiates field lines outward exactly as if the bead's charge sat at the center. Answer B is wrong because it ignores the induced surface charges entirely. Yes, the bulk material has zero net charge, but the surfaces absolutely do develop charge — the inner and outer surface charges just cancel each other, preserving the conductor's overall neutrality. Answer C is wrong on two counts: the field does NOT penetrate the conductor (that's the whole point of shielding), and the charge doesn't split "half-half" — it's

-q

on the inner surface and

+q

on the outer. Answer D reverses the signs. The inner surface acquires the opposite sign to the bead (

-q

for a

+q

bead), not the same sign. Study tip: Always apply charge conservation as a check — if the inner surface has

-q

, the outer surface must have

+q

to keep the shell's total charge at zero.

Question 8

A neutral conducting sphere A is mounted on an insulating stand. A negatively charged rod is brought near (but not touching) the left side of sphere A. While the rod is held in place, sphere A is briefly connected by a conducting wire to a distant, grounded conducting sphere B, and then the wire is removed. Finally, the rod is withdrawn.

After the entire procedure is complete, what is the net charge state of sphere A, and what is the primary mechanism responsible for that outcome?

Sphere A is positively charged, because electrons flowed from A to ground through sphere B while the rod was present, leaving A with a net deficit of electrons. (correct answer)
Sphere A is negatively charged, because the rod deposited electrons onto A by direct conduction when it was brought close, and those electrons remained on A after the rod was withdrawn.
Sphere A remains neutral, because the charges redistributed equally between A and B during contact, and when the wire was removed the charge distribution on A returned symmetrically to neutral.
Sphere A is positively charged, because conventional current carried positive charges from the ground onto A through the wire, directly neutralizing the electrons repelled to the far side of A and leaving a net positive charge.

Explanation: When you see a question involving a charged rod brought near (but not touching) a conductor, you're being tested on charging by induction — a fundamentally different process from direct conduction. Keep that distinction front and center. Here's what happens step by step: The negative rod repels free electrons in sphere A, pushing them away from the left side toward the right. When the grounding wire connects A to distant sphere B, those repelled electrons have a path to escape — they flow from A through the wire into B (and effectively into Earth's infinite reservoir). This leaves sphere A with a deficit of electrons, meaning a net positive charge. When the wire is removed, that positive charge is locked onto A. Finally, when the rod is withdrawn, there's no longer an external field to cause redistribution — A simply remains positively charged. Answer A correctly captures this entire chain of reasoning. Answer B is wrong because the rod never touched sphere A — no direct conduction occurred. Proximity alone cannot transfer charge; it only induces redistribution of existing charges. Answer C is wrong because it imagines the charges "returning to neutral" after the wire is removed, but electrons that already flowed out through the wire are gone — they cannot spontaneously return. Answer D describes "conventional current carrying positive charges from ground," which is a conceptual error. Ground doesn't supply positive charges; rather, excess electrons leave A toward ground. The result (A becoming positive) is the same as A describes, but D's mechanism is physically incorrect. Your study tip: Always trace electron flow, not conventional current, when analyzing induction problems. Ask yourself: which way do electrons move, and what gets left behind?

Question 9

A physicist has two objects: Object P is a block of silicon doped with a very small impurity concentration (making it a semiconductor with resistivity $\sim 10^3\,\Omega\cdot\text{m}$ ), and Object Q is a block of sulfur (resistivity $\sim 10^{15}\,\Omega\cdot\text{m}$ ). A charge is deposited on the surface of each object. The physicist wants to classify each as a 'conductor' or 'insulator' for the purpose of predicting whether the deposited charge will remain localized or spread over the surface within a few seconds.

Which classification and prediction is most physically justified?

P behaves as an insulator and Q as a conductor: sulfur's molecular structure permits surface charge to migrate freely along its surface, while silicon's strong covalent bonds prevent charge from moving even when a conductor is touched to it.
Both P and Q behave as insulators on the timescale of seconds: neither material has free electrons like a metal, so the deposited surface charge cannot migrate in either material regardless of their large difference in resistivity.
P behaves as a conductor and Q as an insulator: the charge relaxation time $\tau = \varepsilon/\sigma$ is negligibly short for P (nanoseconds), so charge on P redistributes nearly instantly, while $\tau$ for Q is many hours or longer, so charge on Q remains essentially localized indefinitely on any human timescale. (correct answer)
Both P and Q behave as conductors on the timescale of seconds: any material with nonzero mobile carrier density will redistribute surface charge given enough time, and both materials have measurable carrier concentrations at room temperature.

Explanation: When classifying materials as conductors or insulators, the key framework isn't a binary "metals vs. everything else" — it's the charge relaxation time

\tau = \varepsilon/\sigma

, where

\varepsilon

is permittivity and

\sigma

is conductivity. This timescale tells you how quickly a deposited charge redistributes itself. If

\tau

is much shorter than your observation window (seconds, in this case), the material behaves as a conductor. If

\tau

is far longer, it behaves as an insulator. For Object P (silicon,

\sigma \approx 10^{-3}\,\text{S/m}

\tau = \varepsilon/\sigma \approx (10^{-11})/(10^{-3}) \sim 10^{-8}\,\text{s}

— nanoseconds. Any deposited charge redistributes almost instantaneously relative to a human timescale: P behaves as a conductor. For Object Q (sulfur,

\sigma \approx 10^{-15}\,\text{S/m}

\tau \sim 10^{4}\,\text{s}

— several hours. Charge stays localized for far longer than seconds: Q behaves as an insulator. This makes C the correct answer. A is physically backwards and invents a fictional mechanism for sulfur — its molecular structure does not allow free surface migration. B commits the classic trap of conflating "not a metal" with "insulator." Semiconductors have real, measurable conductivity; the relaxation-time calculation directly refutes this. D overcorrects in the opposite direction: while both materials have nonzero carrier density, the timescale matters enormously. Q's relaxation time of hours means it is functionally an insulator on a seconds-long observation. Your study tip: whenever a question involves charge redistribution, immediately think

\tau = \varepsilon/\sigma

. The absolute resistivity value only matters relative to your timescale of interest — that ratio is what defines "conductor" vs. "insulator" in practice.