Dielectrics & Polarisation

Open any real capacitor — the ceramic chip in your phone, the electrolytic can in a power supply, the film capacitor in a fan regulator — and you will never find what the textbook diagrams show: two plates with nothing between them. There is always something in the gap. Glass in the original Leyden jar, waxed paper in radios, barium titanate ceramic in the thousand tiny capacitors soldered around your phone's processor. Engineers did not put it there for mechanical support. Fill the gap of a capacitor with an insulating material and its capacitance multiplies — by 2.6 for polystyrene, about 5.4 for mica, roughly 80 if you could use water. Same plates, same separation, same battery: several times the stored charge. The material is called a dielectric, the multiplying factor is the dielectric constant $\kappa$ , and the mechanism behind it — billions of molecules leaning into the field like a crowd tilting in the wind, without a single charge actually travelling anywhere — is called polarisation. This module is the physics of that trick: what a field does to an insulator, why the tilted molecules weaken the field inside the material by exactly the factor $\kappa$ , and how that turns into $C = \kappa C_0$ . It closes the concept spine of this chapter — you already own capacitance, the network rules and the energy formulas — and it arms you for the chapter's favourite exam question, the dielectric slab that only partly fills the gap. Every board in this module's list asks it; JEE asks it with relish.

Key takeaways

A dielectric is an insulator: its charges are bound to atoms and molecules. An electric field cannot make them flow — but it can stretch and turn them, giving each molecule a tiny dipole moment. That collective leaning is polarisation.
The aligned dipoles deposit induced (bound) surface charge on the dielectric's faces — negative facing the positive plate — whose field opposes the applied one. The inside field survives, but weakened: $E = E_0/\kappa$ (isolated capacitor).
The dielectric constant $\kappa$ is a pure number, always greater than 1: air 1.0006, paper ~3.7, mica ~5.4, water ~80. Capacitance multiplies: $C = \kappa C_0 = \dfrac{\kappa\varepsilon_0 A}{d}$ .
A slab of thickness $t$ that only partly fills the gap gives $C = \dfrac{\varepsilon_0 A}{d - t + t/\kappa}$ — the JEE workhorse. A metal slab is the $\kappa \to \infty$ special case.
Every dielectric fails above its dielectric strength (about $3\ \mathrm{MV/m}$ for air): the insulation breaks down and a spark jumps.
Key idea: an insulator answers an electric field by polarising instead of conducting — the induced surface charge partially cancels the field inside, and that partial cancellation is exactly what multiplies the capacitance by $\kappa$ .

The core insight

A dielectric placed in an electric field polarises — its bound charges shift slightly to make aligned molecular dipoles — so induced surface charge weakens the interior field to $E_0/\kappa$ , and a capacitor filled with it holds $\kappa$ times the capacitance: $C = \kappa\varepsilon_0 A/d$ .

Where this lives today

The highest-stakes dielectric on Earth is about two nanometres thick. Every transistor gate in a modern processor is a capacitor, and by 2007 its silicon-dioxide insulation ( $\kappa \approx 3.9$ ) had been thinned so far that electrons tunnelled straight through. The fix was pure equation (2): switch to hafnium-based high-κ dielectrics ( $\kappa \approx 20$ ), which keep the gate capacitance high while letting the insulator be thicker — billions of capacitors per chip, each exploiting $C = \kappa\varepsilon_0A/d$ at the atomic scale. Meanwhile the ceramic chip capacitors surrounding that processor use barium-titanate dielectrics with $\kappa$ in the thousands, packing microfarads into millimetre packages. Your kitchen runs the polar-molecule half of the story every day. A microwave oven drives water's permanent dipoles back and forth $2.45$ billion times per second; the molecules' frantic re-alignment against their neighbours' drag is the heating. Food heats because it is wet — a dry ceramic plate ( $\kappa$ small, no rotating dipoles) stays cool while the soup on it boils. Debye's leaning molecules, at dinner.

The physical picture

Put a slab of copper between charged capacitor plates and its free electrons surge sideways until the field inside the metal is exactly zero — that is the conductor doctrine from Electrostatics of Conductors & Shielding. Now swap the copper for glass. Glass has no free electrons; every charge is locked into a molecule. The field cannot start a current. What it can do is distort: in each molecule, the electron cloud is dragged a fraction of an atomic radius toward the positive plate while the nuclei lean the other way. No charge travels even a millionth of a millimetre — yet every molecule now has a slightly separated plus-end and minus-end. It has become a tiny electric dipole, the object you met in The Electric Dipole & Dipole Moment. Molecules come in two temperaments, and the field treats them differently. Non-polar molecules ( $\mathrm{O_2}$ , $\mathrm{N_2}$ , $\mathrm{CH_4}$ ) have no built-in dipole; the field must create one by stretching the charge cloud — this is induced polarisation. Polar molecules ( $\mathrm{H_2O}$ , $\mathrm{HCl}$ ) carry a permanent dipole moment from their lopsided chemistry; the field merely has to turn them, fighting the thermal jostling that wants to keep the orientations random. Either way the outcome is the same: dipole moments, on average, aligned with the field.

Figure

Polarisation of a dielectric slab — aligned molecular dipoles build bound surface charge that opposes the field

Now zoom out and look at the slab as a whole. Deep inside, the shifted plus-end of every molecule sits right next to the shifted minus-end of its neighbour: the interior cancels out, still electrically neutral on average. Only at the two faces does the pattern break. The face toward the positive plate ends in a layer of exposed minus-ends; the opposite face ends in exposed plus-ends. The slab has grown two thin sheets of induced surface charge — physicists call it bound charge, because unlike the charge on the metal plates it can never be skimmed off or conducted away; it exists only while the molecules stay stretched. Those two bound sheets are a charge pair just like the plates themselves — and their field points the opposite way to the plates' field. Here is the crucial difference from a conductor: the bound charge cannot grow without limit, because each molecule can only stretch so far. So the cancellation stalls partway. The interior field is not zero, just weakened — and the weakening factor is the material's personal number, $\kappa$ . A conductor is simply the runaway limit of this story: $\kappa \to \infty$ , cancellation complete, $E = 0$ inside. Insulator and conductor are two rungs of one ladder.

The dielectric constant, and what it does to a capacitor

The weakening of the field is the definition. For a capacitor holding fixed charge (battery disconnected), insert the dielectric so it fills the gap and:

E = \frac{E_0}{\kappa} \qquad (1)

where: $E_0$ — the field before insertion (made by the plate charge alone); $E$ — the net field inside the dielectric; $\kappa$ — the dielectric constant (also written $\varepsilon_r$ , the relative permittivity), a pure number ≥ 1 that belongs to the material, not to the capacitor. Air barely responds ( $\kappa = 1.0006$ — which is why "air capacitor" formulas are just the vacuum ones); paper manages ~3.7; water's strongly polar molecules reach ~80. Because $V = Ed$ across a uniform gap, the voltage falls by the same factor at fixed charge: $V = V_0/\kappa$ . That last line is the whole capacitance story. The plates hold the same $Q$ , but the potential difference has dropped to $V_0/\kappa$ , so:

C = \frac{Q}{V} = \frac{Q}{V_0/\kappa} = \kappa\,C_0, \qquad C = \frac{\kappa\,\varepsilon_0 A}{d} \qquad (2)

where: $C_0 = \varepsilon_0 A/d$ — the parallel-plate result from Capacitance & the Parallel-Plate Capacitor; $A$ , $d$ — plate area and separation. Read it as a bargain: the dielectric absorbs part of the field, the capacitor "feels" less voltage per unit of charge, so its appetite for charge per volt goes up. With the battery connected instead, $V$ is pinned; then it is the charge that multiplies, $Q = \kappa Q_0$ — either way, $C = \kappa C_0$ always. (Which quantity stays fixed — $Q$ or $V$ — decides everything else about energy; that full ledger is the archetype Dielectric Slabs: Battery Connected or Disconnected, coming in the E2 gym.)

material	$\kappa$ (room temperature)	dielectric strength (MV/m)
vacuum	exactly 1	—
air	1.0006	3
paper	~3.7	16
mica	~5.4	100
water	~80	(conducts too well to use)

Derivation AHow induced surface charge divides the field by kappa

We want equation (1) — the interior field weakened to $E_0/\kappa$ — built from nothing but sheet fields and the picture of bound charge, with the bonus of finding out exactly how much bound charge the faces carry. The strategy: treat the system as four charged sheets (two plate sheets, two bound sheets), use the sheet-field results you already own from Gauss's Law at Work, and let the measured $\kappa$ do the rest. Watch that we never need to know how the molecules produce the bound charge — the argument runs on superposition alone, which is what makes it exam-proof.

Step 1. The plates carry surface charge $\pm\sigma$ (free charge, delivered by the battery). Alone, they make the uniform interior field $E_0 = \sigma/\varepsilon_0$ — the parallel-plate composite from Gauss's Law at Work.

Step 2. The polarised slab presents two bound sheets: $-\sigma_b$ on the face toward the positive plate, $+\sigma_b$ on the far face. As a pair of sheets they make their own uniform field $E_b = \sigma_b/\varepsilon_0$ between them — pointing against $E_0$ , because the negative bound sheet faces the positive plate. Why the bound field must oppose: the molecules were stretched by $E_0$ itself, negative ends pulled toward the positive plate. A response arranged by the field always acts to reduce its cause — if it reinforced it, the stretching would run away and matter would be unstable.

Step 3. Superpose. Inside the dielectric the net field is

E = E_0 - E_b = \frac{\sigma - \sigma_b}{\varepsilon_0}

Step 4. Bring in the material's measured number: the definition $E = E_0/\kappa$ forces

\frac{\sigma - \sigma_b}{\varepsilon_0} = \frac{\sigma}{\kappa\,\varepsilon_0} \;\;\Rightarrow\;\; \sigma_b = \sigma\left(1 - \frac{1}{\kappa}\right)

Why this closing formula is worth memorising as a reading, not a formula: for a weak dielectric ( $\kappa \approx 1$ ) almost no bound charge appears; for polystyrene ( $\kappa = 2.6$ ) the faces carry about 62% of the plate charge; as $\kappa \to \infty$ , $\sigma_b \to \sigma$ — total cancellation, which is precisely a conductor. The dielectric constant measures how far up the conductor ladder a material can climb. *Verified: $\kappa = 1$ gives $\sigma_b = 0$ and $E = E_0$ (vacuum ✓); $\kappa \to \infty$ gives $\sigma_b = \sigma$ , $E = 0$ (conductor doctrine of Electrostatics of Conductors & Shielding ✓); for $\kappa = 2.6$ , $1 - 1/2.6 = 0.615$ ✓.*

The partial slab — the question JEE actually asks

Boards usually fill the whole gap; JEE prefers to slide in a slab of thickness $t < d$ and ask for the new capacitance. No new physics is needed — just the observation that the gap is now two regions in series: an air region of thickness $d - t$ with field $E_{\text{air}} = \sigma/\varepsilon_0$ , and a dielectric region of thickness $t$ where the field is $\kappa$ times weaker. Adding up the voltage across both regions ( $V = E_{\text{air}}(d-t) + \tfrac{E_{\text{air}}}{\kappa}t$ ) and dividing into $Q$ :

C = \frac{\varepsilon_0 A}{d - t + \dfrac{t}{\kappa}} \qquad (3)

where: $t$ — slab thickness; $d - t$ — the surviving air gap. Read the denominator as an effective gap: the slab shrinks its own thickness from $t$ down to $t/\kappa$ , as if the dielectric "compresses" the distance the field must cross. Two limits check it instantly: $t = 0$ returns $C_0$ ; $t = d$ returns $\kappa\varepsilon_0 A/d$ . And setting $\kappa \to \infty$ answers the companion trick question — a metal slab of thickness $t$ gives $C = \varepsilon_0 A/(d - t)$ : the conductor simply deletes its own thickness from the gap. (You can also get equation (3) by treating the two regions as capacitors in series via Capacitors in Series & Parallel — same answer, one more line.)

Figure

The partial slab — an air region and a weakened-field region in series

Limiting cases

case	result	meaning
$\kappa = 1$	$C = C_0$ , $\sigma_b = 0$	vacuum (and, to 0.06%, air): nothing to polarise
$\kappa \to \infty$	$E \to 0$ inside; slab of thickness $t$ : $C = \dfrac{\varepsilon_0 A}{d-t}$	a conductor is the ultimate dielectric — complete screening
$t \to 0$	$C \to C_0$	vanishing slab changes nothing
$t \to d$	$C \to \kappa C_0$	full filling recovers equation (2)
$E$ exceeds dielectric strength	breakdown: insulator sparks over	polarisation is stretching; past the limit, charges tear free
polar material heated	$\kappa$ falls	thermal jostling fights the alignment of permanent dipoles

The breakdown row deserves one more sentence, because students routinely merge two very different numbers. The dielectric constant says how well a material polarises (a pure number); the dielectric strength says how much field it can survive before its insulation fails (units V/m). Air breaks down near $3 \times 10^6\ \mathrm{V/m}$ — that crackle when you pull off a synthetic sweater is air failing as an insulator. Mica polarises modestly ( $\kappa \approx 5.4$ ) but survives brutal fields ( $\sim 100\ \mathrm{MV/m}$ ), which is why it lines high-voltage capacitors.

Common trap“A dielectric increases everything about the capacitor”

It reliably increases exactly one thing: the capacitance, always, by $\kappa$ . Everything else depends on what is held fixed. Battery disconnected (charge $Q$ frozen): the field, the voltage and the stored energy all fall — $E \to E_0/\kappa$ , $V \to V_0/\kappa$ , $U \to U_0/\kappa$ . Battery connected (voltage pinned): the charge and stored energy rise instead, and the field between the plates returns to its old value $V/d$ . If an option claims "inserting a dielectric increases the potential difference," it is wrong in both protocols. Train the reflex: before touching any dielectric question, write down which of $Q$ or $V$ is fixed — the full bookkeeping is the coming E2-gym archetype, but the reflex starts here.

Worked examples

Board level — filling the gap (CBSE/ISC pattern, 3 marks)

A parallel-plate capacitor has square plates of side $8.0\ \mathrm{cm}$ separated by $2.0\ \mathrm{mm}$ of air. It is charged to $100\ \mathrm{V}$ and the battery is removed. A polystyrene slab ( $\kappa = 2.6$ ) is then inserted, filling the gap. Find (a) the capacitance before and after, (b) the charge, (c) the voltage and field inside the slab after insertion. ( $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2/N\,m^2}$ .)

Step 1. Before: $A = (0.080)^2 = 6.4 \times 10^{-3}\ \mathrm{m^2}$ , so $C_0 = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85\times10^{-12})(6.4\times10^{-3})}{2.0\times10^{-3}} = \boxed{28.3\ \mathrm{pF}}$ .

Step 2. After: $C = \kappa C_0 = 2.6 \times 28.3 = \boxed{73.6\ \mathrm{pF}}$ — geometry unchanged, material does the multiplying.

Step 3. Charge: fixed the moment the battery left. $Q = C_0 V_0 = 28.3\ \mathrm{pF} \times 100\ \mathrm{V} = \boxed{2.83\ \mathrm{nC}}$ — and it stays $2.83$ nC after insertion (nowhere to go).

Step 4. New voltage: $V = Q/C = V_0/\kappa = 100/2.6 = \boxed{38.5\ \mathrm{V}}$ . Field inside the slab: $E = V/d = 38.5/(2.0\times10^{-3}) = 1.92\times10^{4}\ \mathrm{V/m}$ — exactly the old field $5.0\times10^4\ \mathrm{V/m}$ divided by $2.6$ ✓. The marks-scheme sentence worth writing: "With the battery disconnected the charge is constant, so V and E fall by the factor κ while C rises by κ."

Competitive level — the partial slab (JEE-Main pattern, ~3 minutes)

A parallel-plate capacitor has plate area $200\ \mathrm{cm^2}$ and separation $3.0\ \mathrm{mm}$ . It is charged to $300\ \mathrm{V}$ and disconnected from the battery. A dielectric slab of thickness $2.0\ \mathrm{mm}$ and $\kappa = 4.0$ is slid in, faces parallel to the plates. Find (a) the new capacitance, (b) the new potential difference, (c) the field in the air gap and in the slab.

Step 1. Before: $C_0 = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85\times10^{-12})(200\times10^{-4})}{3.0\times10^{-3}} = 59.0\ \mathrm{pF}$ ; charge banked: $Q = C_0 V_0 = 59.0\ \mathrm{pF}\times300\ \mathrm{V} = 17.7\ \mathrm{nC}$ , frozen from here on.

Step 2. New capacitance by the effective gap, equation (3): $d - t + t/\kappa = 3.0 - 2.0 + 2.0/4.0 = 1.5\ \mathrm{mm}$ , so $C = \dfrac{\varepsilon_0 A}{1.5\ \mathrm{mm}} = \boxed{118\ \mathrm{pF}}$ — exactly double, because the effective gap halved.

Step 3. New voltage: $V = Q/C = 17.7\ \mathrm{nC}/118\ \mathrm{pF} = \boxed{150\ \mathrm{V}}$ .

Step 4. Fields region by region. The plate charge is unchanged, so the air-gap field is the original $E_{\text{air}} = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A) = 1.0\times10^{5}\ \mathrm{V/m}$ ; inside the slab $E_{\text{slab}} = E_{\text{air}}/\kappa = 2.5\times10^{4}\ \mathrm{V/m}$ . Audit the voltage: $10^5 \times 1.0\ \mathrm{mm} + 2.5\times10^4 \times 2.0\ \mathrm{mm} = 100 + 50 = 150\ \mathrm{V}$ ✓. Timing note: the entire question is the effective-gap denominator — write $d - t + t/\kappa$ in the first ten seconds and the rest is arithmetic. Budget 3 minutes; if the slab were metal, skip even that: $C = \varepsilon_0 A/(d-t) = 177\ \mathrm{pF}$ .

Exam traps

The effective-gap formula, mangled. The denominator is $d - t + \dfrac{t}{\kappa}$ — subtract the whole slab, add back its compressed thickness. Common corruptions: $d - t(1 - \kappa)$ (sign error, gives $C < C_0$ ... or worse) and $d - t/\kappa$ (forgets to remove the slab first). Check any candidate formula with $t = 0$ (must give $C_0$ ) and $t = d$ (must give $\kappa C_0$ ) — five seconds, kills three options.

$E = E_0/\kappa$ used with the battery connected. The division by $\kappa$ is a fixed-charge statement. With the battery attached, $V$ is pinned, so the full-gap field ends up back at $V/d$ — the battery ships in extra charge to rebuild it. State which quantity is frozen before writing anything.

Dielectric constant vs dielectric strength. $\kappa$ is a pure number (how well it polarises); strength is a field in V/m (when it sparks). "Mica ( $\kappa = 5.4$ ) withstands $100\ \mathrm{MV/m}$ " uses both correctly. Papers plant options that swap them.

"The dielectric blocks the field completely." That is a conductor. A dielectric only reduces the interior field, by $\kappa$ ; bound charges cannot flow, so cancellation stalls. Writing $E = 0$ inside a dielectric slab costs the whole part.

Bound charge treated as free. The induced surface charge $\sigma_b = \sigma(1 - 1/\kappa)$ cannot be conducted away, does not add to $Q$ on the plates, and disappears when the field does. It alters fields, never the capacitor's charge inventory.

Unit slips in $\varepsilon_0 A/d$ . Area in $\mathrm{cm^2}$ and gap in mm must both go to SI before the formula — the $10^{-4}$ and $10^{-3}$ conversions are where pF answers silently become nF. All boards here expect $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^2/N\,m^2}$ ; JEE/NEET occasionally hand you $1/4\pi\varepsilon_0 = 9\times10^9$ instead — convert, don't guess.

Origins · 1 minPeter Debye and how this was discovered

Peter Debye

1884–1966 · Zurich, Leipzig & Cornell

Debye is the physicist who worked out what this article's "leaning molecules" really are. In 1912, while others treated insulators as passive filler, he proposed that many molecules carry permanent electric dipole moments — water chief among them — and showed how an applied field partially aligns them against thermal chaos, predicting how a material's dielectric constant should fall as temperature rises. The idea explained why water's $\kappa \approx 80$ towers over glass's ~5, turned dielectric measurements into a tool for reading molecular shapes (a bent water molecule polarises; a symmetric $\mathrm{CO_2}$ barely does), and earned him the 1936 Nobel Prize in Chemistry. The unit of molecular dipole moment is named the debye in his honour — fittingly, since the polarisation physics of every capacitor dielectric, every microwave oven, and every solvent in a chemistry lab runs on the mechanism he identified.

History

1745

— The Leyden jar stores the first "electric fire" in glass — a dielectric capacitor before either word exists.

1837

— Faraday fills twin capacitors with shellac and glass and discovers specific inductive capacity: the dielectric constant enters physics.

1912

— Debye proposes permanent molecular dipole moments and explains how fields align polar molecules against heat.

1936

— Debye receives the Nobel Prize in Chemistry for dipole moments and molecular structure.

2007

— Intel replaces silicon dioxide with hafnium-based high-κ dielectrics in processor gates — polarisation physics rescues Moore's law.

Going further

Texts. NCERT Physics Class 12, Ch. 2 (Electrostatic Potential and Capacitance, §2.10–2.14) · HC Verma, Concepts of Physics Vol. 2, Ch. 31 · Halliday, Resnick & Walker, Ch. 25.5–25.8.

Provenance. Downshifted from the UG modules "Capacitor Energy and Dielectrics" and "Dielectric Materials and Insulators" (HS rewrite: sheet-superposition argument in place of the P and D machinery; all numbers recomputed). A verifier should recompute: $C_0 = 28.3$ pF and $\kappa C_0 = 73.6$ pF for $8\times8$ cm at 2.0 mm; $V = 100/2.6 = 38.5$ V; the partial-slab example $C_0 = 59.0$ pF, effective gap 1.5 mm, $C = 118$ pF, $V = 150$ V, fields $1.0\times10^5$ and $2.5\times10^4$ V/m (voltage audit $100 + 50$ V); $\sigma_b/\sigma = 1 - 1/2.6 = 0.615$ .

Prerequisites. Capacitance & the Parallel-Plate Capacitor · Energy Stored in a Capacitor · The Electric Dipole & Dipole Moment. Leads to. Dielectric Slabs: Battery Connected or Disconnected (coming in the E2 gym) · Dielectric Materials and Insulators (UG deep dive).

Why is the dielectric constant always greater than 1 — could a material have κ less than 1?

Not in ordinary electrostatics. The bound charge is arranged by the applied field itself: negative molecular ends get pulled toward the positive plate, so the induced field always opposes the applied one, and opposition means reduction — $E < E_0$ , hence $\kappa = E_0/E > 1$ . A material with $\kappa < 1$ would need its molecules to lean against the force acting on them, amplifying whatever field appears; any stray charge would then trigger runaway growth, and the material would be unstable. Vacuum's $\kappa = 1$ is the floor: nothing there to polarise at all.

If water has κ = 80, why don't we build capacitors filled with water?

Two spoilers. First, real water is never pure: dissolved ions make it conduct, and a conducting gap is a shorted capacitor, not a better one. Second, even ultrapure water self-ionises enough to leak charge over seconds. Water's giant $\kappa$ still matters enormously — it is why salt dissolves (the polarised water swarm weakens the ionic attraction by ~80, an idea you can now read directly from equation 1) — but as a capacitor filling, engineers use polar solids like ceramics: high κ, no free ions.

Does the induced bound charge reduce the charge stored on the plates?

No — it lives on the dielectric's faces, not on the plates, and it cannot flow anywhere. With the battery disconnected the plate charge is untouched by insertion; with the battery connected the plate charge actually grows (the battery ships more in to hold V). What the bound charge changes is the field: its sheet field subtracts from the plates' sheet field inside the material. Think of it as a partial counter-capacitor that the material builds internally, never as a withdrawal from the plates' account.

What is actually different between polar and non-polar molecules here?

A polar molecule (water, HCl) is permanently lopsided: its chemistry puts the electron cloud off-centre, so it carries a built-in dipole moment even with no field. A non-polar molecule (O₂, CH₄) is symmetric, dipole-free until a field stretches it. Both polarise under a field — non-polar by stretching (weak, temperature-blind), polar by rotating into line (strong, and weakened by heating since thermal jostling randomises orientations). That is why polar materials have the big κ values, and why κ for water falls as it warms.

Is the field inside the dielectric E₀/κ even when the battery stays connected?

Careful — that formula is the fixed-charge statement. Battery connected means V is pinned; for a fully filled gap the field settles at $E = V/d$ , the same value as before insertion, because the battery pumps in κ times the charge to rebuild it. The reliable order of operations: decide what is frozen (Q or V), get the new C from geometry-plus-κ (that part never changes), then recover everything else from $Q = CV$ . The complete energy bookkeeping is the E2-gym archetype coming up.

What physically happens during dielectric breakdown?

Polarisation is stretching, and every stretch has a limit. Past the material's dielectric strength, stray free electrons (there are always a few, from cosmic rays or impurities) are accelerated hard enough between collisions to knock new electrons free; each of those knocks free more — an avalanche. In a microsecond the insulator hosts a conducting plasma channel: a spark. Air fails near 3 MV/m — which sets the roughly 30,000 V-per-centimetre rule for sparks, and is exactly why high-voltage capacitors use mica or oil rather than trusting air gaps.

My textbook writes ε_r and yours writes κ — are they different quantities?

Same number, two costumes. NCERT and most Indian boards write $\varepsilon_r$ (relative permittivity) and define the absolute permittivity $\varepsilon = \varepsilon_r\varepsilon_0$ ; many international texts and AP write κ (dielectric constant). All formulas convert by pure substitution: $C = \kappa\varepsilon_0 A/d = \varepsilon_r\varepsilon_0 A/d = \varepsilon A/d$ . The one habit to avoid is mixing conventions mid-solution — pick the paper's symbol and stay with it.