ARCHETYPE: Dielectric Slabs — Battery Connected or Disconnected?

The NCERT exercise book asks it with a straight face: a capacitor is charged, then a mica sheet slides between the plates — find the new charge, voltage and energy (a) while the voltage supply remained connected, (b) after the supply was disconnected. One phrase changes; every answer changes with it. Stored energy rises in (a) and falls in (b), and thousands of students, every year, compute a perfectly correct answer to the part that was not asked. This is the archetype where examiners test not a formula but a decision. Before any algebra, ask the only question that matters: is the battery still attached? Attached, the battery is a tyrant about voltage — $V$ is pinned, charge flows to obey it. Detached, the capacitor is an island — $Q$ has nowhere to go and freezes, while voltage floats. Every quantity ( $C$ , $Q$ , $V$ , $E$ , $U$ ) then follows from two lines of algebra and one honest piece of bookkeeping about who paid. So this article drills the decision itself: the two-protocol ledger for sliding a slab in or out, for pulling plates apart, and for the chained tricks JEE loves — insert connected, then disconnect, then remove. Learn the ledger once and the whole family of "what changes?" questions becomes a table you fill in thirty seconds.

Key takeaways

The one decision: battery connected ⇒ $V$ fixed (charge flows to hold it); battery disconnected ⇒ $Q$ fixed (charge has no path; voltage floats).
Slab of dielectric constant $\kappa$ fills the gap: $C \to \kappa C$ always. Connected: $Q \to \kappa Q$ , $E$ unchanged, $U \to \kappa U$ . Disconnected: $V \to V/\kappa$ , $E \to E/\kappa$ , $U \to U/\kappa$ .
Pick the energy costume that matches the frozen quantity: $U = Q^2/2C$ when $Q$ is fixed, $U = \tfrac12 CV^2$ when $V$ is fixed — then only $C$ changes.
The books always balance: at fixed $V$ the battery pays $V\Delta Q$ (twice the stored gain when a slab enters); at fixed $Q$ any energy drop became work done by the field on the slab.
The slab is pulled inward in both protocols — systems slide toward higher capacitance.
Key idea: decide what is frozen ( $Q$ or $V$ ), write the matching energy form, and let $C \to \kappa C$ (or $C \to \varepsilon_0 A/d_{\text{new}}$ ) do the rest — the answer is never a new formula, only the right constraint.

The core insight

A connected battery pins the capacitor's voltage while charge flows freely; a disconnected capacitor pins its charge while voltage floats — and every "slab in, slab out, plates apart" question is solved by freezing the right quantity, updating $C$ , and balancing the energy ledger.

Where this lives today

The condenser microphone in every phone call is column two doing its day job: a charged capacitor with one springy plate, run essentially at fixed charge. Sound pressure wiggles the gap, $C$ wiggles, and since $Q$ cannot follow, $V = Q/C$ wiggles instead — that voltage is the audio signal. The connected column runs the other frontier: a transistor's gate is a capacitor held at fixed drive voltage, and the industry's swap of silicon dioxide ( $\kappa \approx 3.9$ ) for hafnium dielectrics ( $\kappa \approx 20$ ) is a pure "battery connected, boost $\kappa$ , harvest $\kappa$ -fold charge at the same $V$ " play, executed billions of times per chip. And the deliberately movable dielectric survives in sensors: fuel gauges read how far the fuel has "slid into" a capacitor by exactly the shift this article taught you to score.

The setup

The hardware is minimal: a parallel-plate capacitor $C_0 = \varepsilon_0 A/d$ , a battery of emf $V_0$ , a switch, and a slab of dielectric constant $\kappa$ that can slide into the gap. The physics forks at the switch.

Figure

The one decision: connected pins V, disconnected pins Q

Why each constraint holds is worth one honest paragraph, because the marks live there. Connected: a battery maintains a fixed potential difference between its terminals, and ideal wires extend those potentials to the plates; disturb the capacitor (a slab, a plate shift) and charge flows until $V$ matches $V_0$ again. Disconnected: the plates are marooned. Charge is conserved, countable stuff with no conducting path off the plates — whatever $\pm Q_0$ was there at disconnection stays, whatever you do to the geometry. Voltage is not stuff; it is just $Q/C$ , and it moves the instant $C$ does. One anchor before the ledger. The capacitance never cares about the switch: it is pure geometry and filling,

C = \frac{\kappa\,\varepsilon_0 A}{d} \qquad (1)

where: $A$ is the plate area, $d$ the gap, $\kappa$ the dielectric constant of the filling ( $\kappa = 1$ for air). A slab multiplies $C$ by $\kappa$ ; doubling the gap halves it — identically in both protocols, which disagree only about which of $Q$ and $V$ absorbs the change. (Why the slab boosts $C$ belongs to Dielectrics & Polarisation; the formulas $U = Q^2/2C = \tfrac12CV^2$ come from Energy Stored in a Capacitor. Here we spend both.)

The master ledger

Everything this archetype asks lives in one table: freeze the right quantity, update $C$ by equation (1), and compute the rest from $Q = CV$ , $E = V/d$ , $U = Q^2/2C = \tfrac12CV^2$ .

quantity	slab IN, battery connected	slab IN, battery disconnected
capacitance $C$	$\kappa C_0$ (geometry, always)	$\kappa C_0$ (geometry, always)
voltage $V$	$V_0$ — pinned	$V_0/\kappa$ — falls
charge $Q$	$\kappa Q_0$ — battery supplies more	$Q_0$ — frozen
field $E$	$V_0/d$ — unchanged	$E_0/\kappa$ — falls
energy $U$	$\kappa U_0$ — rises	$U_0/\kappa$ — falls
who pays	battery: $W = V_0\Delta Q = 2(\kappa - 1)U_0$	field: does $U_0(1 - 1/\kappa)$ of work on the slab

Read the two most-tested cells aloud. Connected, the field does not change: $E = V/d$ with both $V$ and $d$ pinned — even though charge poured in. Disconnected, the energy falls though nothing left the circuit — the deficit is the work the field did dragging the slab in (hold the slab and you feel it as a pull).

Derivation AThe Disconnected Ledger and Where the Energy Went

We derive the full disconnected column of the ledger — slab of constant $\kappa$ slid slowly into an isolated charged capacitor — and close the energy books. The strategy: charge conservation freezes $Q_0$ (the physical input); geometry updates $C$ ; every other row is arithmetic from the definitions, so the energy audit has no freedom and its deficit must be mechanical work. Watch that the slab moves slowly: otherwise the deficit splits between work and kinetic energy.

Step 1. Freeze the charge. With the switch open there is no conducting path, so $Q = Q_0 = C_0V_0$ throughout. State this in an exam answer — it is the physics mark.

Step 2. Update the capacitance from geometry: $C = \kappa C_0$ (equation 1 with the slab filling the gap).

Step 3. Voltage follows from the definition $V = Q/C$ :

V = \frac{Q_0}{\kappa C_0} = \frac{V_0}{\kappa}

Why the voltmeter reading drops: the slab's polarisation partially cancels the field between the plates, and the potential difference is field times gap. Same trapped charge, weaker interior field, smaller voltage.

Step 4. Field: $E = V/d = E_0/\kappa$ — falls by the same factor.

Step 5. Energy, in the fixed- $Q$ costume:

U = \frac{Q_0^2}{2\,\kappa C_0} = \frac{U_0}{\kappa}

where: $U_0 = Q_0^2/2C_0$ is the starting stock. The energy fell by $\Delta U = U_0\left(1 - \dfrac{1}{\kappa}\right)$ , with nothing connected to carry it away.

Step 6. Balance the books. The only other actor is whatever held the slab: the field pulls it inward (the fringing region grips its polarised edges), doing positive work $U_0(1 - 1/\kappa)$ on it — precisely the stored-energy deficit. Insert slowly and your hand absorbs it; release the slab and it becomes kinetic energy, the slab oscillating until losses park it inside. Verified: $C_0 = 200\ \mathrm{pF}$ , $V_0 = 300\ \mathrm{V}$ , $\kappa = 6$ : $Q_0 = 60\ \mathrm{nC}$ frozen; $V = 50\ \mathrm{V}$ ; $U$ : $9\ \mu\mathrm{J} \to 1.5\ \mu\mathrm{J}$ ; work on slab $= 7.5\ \mu\mathrm{J}$ — the column of Example 1(b).

Go deeper · Derivation B · 1 minThe Connected Ledger and Who Pays Double

Derivation BThe Connected Ledger and Who Pays Double

Now the connected column: same slab, same slow insertion, battery attached. The strategy: the battery pins $V_0$ , geometry updates $C$ , and the physics moves to the charge flow and the audit — the battery pays exactly twice what the capacitor banks, the difference again tugging the slab. Watch the payment rule: a battery pays $V\,\Delta Q$ (its voltage is flat), never $\tfrac12 V\Delta Q$ .

Step 1. Pin the voltage: $V = V_0$ throughout — the battery restores it continuously as the slab advances.

Step 2. Update capacitance: $C = \kappa C_0$ .

Step 3. Charge must rise to satisfy $Q = CV$ :

Q = \kappa C_0 V_0 = \kappa Q_0, \qquad \Delta Q = (\kappa - 1)\,C_0 V_0

where: $\Delta Q$ physically flows through the battery during insertion. The field $E = V_0/d$ is unchanged — extra plate charge exactly compensates the slab's polarisation; a favourite one-marker.

Step 4. Stored energy, fixed- $V$ costume: $U = \tfrac12 \kappa C_0 V_0^2 = \kappa U_0$ — a gain of $(\kappa - 1)U_0$ .

Step 5. The battery's invoice. It pushed $\Delta Q$ through a constant potential difference $V_0$ :

W_{\text{batt}} = V_0\,\Delta Q = (\kappa - 1)\,C_0V_0^2 = 2\,(\kappa - 1)\,U_0

Step 6. Audit: the battery spent $2(\kappa-1)U_0$ ; storage grew by $(\kappa-1)U_0$ ; the missing $(\kappa-1)U_0$ is once more the field's work pulling the slab in. Half banked, half spent on the tug — whenever charge falls through a pinned voltage, the split is fifty–fifty. Why the pull cannot vanish in either protocol: both ledgers show the system's available energy decreasing as the slab advances, so the force on the slab points inward regardless of the switch. The protocols disagree only about the financing — the field's own stock pays when disconnected; the battery pays when connected. *Verified: $C_0 = 200\ \mathrm{pF}$ , $V_0 = 300\ \mathrm{V}$ , $\kappa = 6$ : $Q$ : $60 \to 360\ \mathrm{nC}$ ; $U$ : $9 \to 54\ \mu\mathrm{J}$ (gain $45\ \mu\mathrm{J}$ ); battery pays $90\ \mu\mathrm{J}$ — exactly double the gain.*

Figure

The chained protocol: track C, Q, V through each phase, flipping the frozen quantity at every switch event

The plate-separation variants

The same two-column discipline answers the "plates pulled apart" family — only the geometry line changes: $d \to d'$ gives $C \to C_0\,d/d'$ . Doubling the gap halves $C$ ; then:

Disconnected, gap doubled: $Q$ frozen; $V = Q/C$ doubles; $E = \sigma/\varepsilon_0$ is set by the trapped charge alone and does not change; $U = Q^2/2C$ doubles. Who paid? You did — pulling the mutually attracting plates apart. (Field constant while $V$ doubles is the disconnected twin of the connected slab's "E unchanged" surprise.)

Connected, gap doubled: $V$ pinned; $C$ halves so $Q$ halves — charge is pushed back into the battery; $E = V/d$ halves; $U$ halves. The audit is worth owning: storage fell $U_0/2$ , the battery received $V\Delta Q = U_0$ , yet you still did $+U_0/2$ of work pulling the plates apart — the books close as $U_0 + U_0/2 = U_0/2 + U_0$ . A slab thinner than the gap changes nothing about the protocol logic — only the capacitance formula (the air-plus-slab result covered in Dielectrics & Polarisation). Decide the frozen quantity first; then use whatever $C_{\text{new}}$ the geometry gives.

Limiting cases

case	result	meaning
$\kappa \to 1$	both columns collapse to "no change"	air slab is no slab
$\kappa \to \infty$ (metal slab, disconnected)	$V, E, U \to 0$	a conductor swallows the field; all stored energy went into the pull
disconnected, then reconnect	charge flows until $V = V_0$ again	reconnection converts column 2 into column 1 (with a burst of heat)
gap $d \to 2d$ , disconnected	$E$ unchanged, $V$ , $U$ double	trapped charge sets $E$ ; you paid the energy gain
gap $d \to 2d$ , connected	$E$ , $Q$ , $U$ halve	battery reclaims charge and energy
insert then disconnect then remove	$V \to \kappa V_0 > V_0$	the chained protocol: your muscles charged it above the battery

Worked examples

Board level — the NCERT mica-sheet classic (CBSE pattern, 3 marks)

A $200\ \mathrm{pF}$ parallel-plate air capacitor is charged by a $300\ \mathrm{V}$ supply. A mica sheet ( $\kappa = 6$ ) is then slid in to fill the gap completely. Find the charge, voltage and stored energy (a) if the supply remains connected, (b) if the sheet is inserted after the supply is disconnected.

Step 1. Common start: $Q_0 = C_0V_0 = 60\ \mathrm{nC}$ ; $U_0 = \tfrac12 C_0V_0^2 = 9\ \mu\mathrm{J}$ . Either way $C = \kappa C_0 = 1200\ \mathrm{pF}$ .

Step 2. (a) Connected — $V$ pinned at $300\ \mathrm{V}$ : $Q = CV = 1200\ \mathrm{pF} \times 300\ \mathrm{V} = \boxed{360\ \mathrm{nC}}$ ; $U = \tfrac12 CV^2 = \boxed{54\ \mu\mathrm{J}}$ ( $= \kappa U_0$ ✓).

Step 3. (b) Disconnected — $Q$ frozen at $\boxed{60\ \mathrm{nC}}$ : $V = Q/C = 60\ \mathrm{nC}/1200\ \mathrm{pF} = \boxed{50\ \mathrm{V}}$ ( $= V_0/\kappa$ ✓); $U = Q^2/2C = \boxed{1.5\ \mu\mathrm{J}}$ ( $= U_0/\kappa$ ✓).

Step 4. One sentence each for the method marks: (a) the battery drove in extra charge to hold $V$ ; (b) the trapped charge's weakened field reads as lower voltage, and the missing $7.5\ \mu\mathrm{J}$ became the field's work on the incoming sheet.

Board level — plates pulled apart both ways (ISC pattern)

A $6\ \mu\mathrm{F}$ parallel-plate capacitor is charged to $10\ \mathrm{V}$ . The plate separation is then slowly doubled. Find the new voltage, charge and energy (a) with the battery disconnected first, (b) with the battery still connected.

Step 1. Start: $Q_0 = 60\ \mu\mathrm{C}$ , $U_0 = \tfrac12(6\ \mu\mathrm{F})(10)^2 = 300\ \mu\mathrm{J}$ . Doubling $d$ halves the capacitance: $C = 3\ \mu\mathrm{F}$ in both parts.

Step 2. (a) Disconnected — $Q$ frozen at $60\ \mu\mathrm{C}$ : $V = Q/C = 60/3 = \boxed{20\ \mathrm{V}}$ (doubles); $U = Q^2/2C = \dfrac{(60)^2}{2\times3} = \boxed{600\ \mu\mathrm{J}}$ . The extra $300\ \mu\mathrm{J}$ is the work you did against the plates' attraction.

Step 3. (b) Connected — $V$ pinned at $10\ \mathrm{V}$ : $Q = CV = \boxed{30\ \mu\mathrm{C}}$ (half returns to the battery); $U = \tfrac12(3)(10)^2 = \boxed{150\ \mu\mathrm{J}}$ .

Step 4. Audit of (b), the probed part: storage fell $150\ \mu\mathrm{J}$ while the battery absorbed $V\Delta Q = 300\ \mu\mathrm{J}$ ; the balance, $+150\ \mu\mathrm{J}$ , is your work: $300 + 150 = 150 + 300$ ✓.

Competitive level — the chained protocol (JEE-Main pattern, ~3 minutes)

A $2\ \mu\mathrm{F}$ capacitor is connected to a $100\ \mathrm{V}$ battery, and a slab of $\kappa = 3$ is inserted to fill the gap. The battery is then disconnected, and the slab is slowly pulled out. Find (a) the final voltage across the capacitor, (b) the final stored energy, (c) the work done in pulling the slab out.

Step 1. Phase one (connected insertion): $C_1 = \kappa C_0 = 6\ \mu\mathrm{F}$ , $V = 100\ \mathrm{V}$ pinned, so $Q_1 = 600\ \mu\mathrm{C}$ , $U_1 = \tfrac12(6\ \mu\mathrm{F})(100)^2 = 30\ \mathrm{mJ}$ .

Step 2. Phase two — disconnect. From this instant $Q = 600\ \mu\mathrm{C}$ is frozen. The decision has flipped mid-problem; re-state it.

Step 3. Phase three (slab out at fixed $Q$ ): $C \to C_0 = 2\ \mu\mathrm{F}$ , so $V = Q/C_0 = 600/2 = \boxed{300\ \mathrm{V}}$ — three times the battery that charged it.

Step 4. Energy: $U = \dfrac{Q^2}{2C_0} = \dfrac{(600\ \mu\mathrm{C})^2}{2 \times 2\ \mu\mathrm{F}} = \boxed{90\ \mathrm{mJ}}$ .

Step 5. Work done by you $= U_{\text{final}} - U_{\text{after disconnect}} = 90 - 30 = \boxed{60\ \mathrm{mJ}}$ — pulling the slab out against the inward pull is how the capacitor got richer than its battery ever made it. Timing note: chained protocols are bookkeeping — track $(C, Q, V)$ per phase in a margin table, flipping the frozen quantity at each switch event. Budget ~3 min; the classic error is freezing the original $Q_0 = 200\ \mu\mathrm{C}$ instead of the connected-phase $Q_1 = 600\ \mu\mathrm{C}$ .

Competitive level — two capacitors, one switch difference (NEET pattern, ~2 minutes)

Two identical $4\ \mu\mathrm{F}$ capacitors A and B are each charged to $12\ \mathrm{V}$ . A remains connected to its battery; B is disconnected. A slab of $\kappa = 2$ is inserted into both. Find the ratio of the final stored energies $U_A : U_B$ and the final charge on each.

Step 1. Start (both): $Q_0 = 48\ \mu\mathrm{C}$ , $U_0 = \tfrac12(4\ \mu\mathrm{F})(12)^2 = 288\ \mu\mathrm{J}$ .

Step 2. A (connected): $U_A = \kappa U_0 = 2 \times 288 = 576\ \mu\mathrm{J}$ ; $Q_A = \kappa Q_0 = \boxed{96\ \mu\mathrm{C}}$ .

Step 3. B (disconnected): $U_B = U_0/\kappa = 288/2 = 144\ \mu\mathrm{J}$ ; $Q_B = \boxed{48\ \mu\mathrm{C}}$ (frozen), with $V_B = 6\ \mathrm{V}$ .

Step 4. Ratio: $U_A : U_B = \kappa U_0 : U_0/\kappa = \kappa^2 : 1 = \boxed{4 : 1}$ — the general result: the two protocols differ by a clean factor $\kappa^2$ , a favourite NEET one-liner. Timing note: recognise "same slab, different switch" and jump straight to $\kappa U_0$ vs $U_0/\kappa$ — the $\kappa^2$ ratio needs no intermediate numbers. Under a minute; the distractor $\kappa : 1$ punishes those who remember only one column.

Exam traps

Answering the wrong column. Underline "supply remains connected" or "after disconnecting" before writing anything. NCERT-style both-part questions award the second part's marks for changed reasoning, not changed arithmetic.

Applying $\kappa$ to the wrong quantity. Connected: $Q$ , $U$ multiply by $\kappa$ ; $V$ , $E$ hold. Disconnected: $V$ , $E$ , $U$ divide by $\kappa$ ; $Q$ holds. $C$ multiplies in both. An answer where everything got multiplied is a ledger failure.

The battery pays $V\Delta Q$ , not $\tfrac12 V\Delta Q$ . The half belongs to the capacitor's triangle, because its own voltage grew from zero; the battery's voltage never budged, so its invoice is the full rectangle. Mixing the two is the classic energy-audit error, and it is exactly a factor of two.

Mid-problem switch flips. JEE chains protocols (insert connected → disconnect → remove). At each switch event freeze the current value of the newly conserved quantity, never the original. A margin table of $(C, Q, V)$ per phase prevents the standard disaster.

"E must change if Q changed." At fixed $V$ , $E = V/d$ is pinned however much charge flows; at fixed $Q$ with the gap changed, $E = \sigma/\varepsilon_0$ is pinned whatever $V$ does. The field follows its own controller; memorise both surprises.

Sign of your work. Pulling a slab out or plates apart at fixed $Q$ , you do positive work (the system resists) and the stored energy rises; letting a slab in, the field does the work. State who works on what — ISC and CBSE schemes reserve a mark for it.

Constants and units. $\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{C^2/N\,m^2}$ where needed; stay in pF–V–nC or $\mu$ F–V– $\mu$ C throughout. The only "constant" that matters is the one you underline in the stem: connected or not.

Origins · 1 minMichael Faraday and how this was discovered

Michael Faraday

1791–1867 · Royal Institution, London

Faraday earns his place in this ledger for the 1837 experiments that put the filling of a capacitor on physics' books. He built twin spherical capacitors, charged one, and shared its charge with the other — first air-gapped, then with shellac, glass or sulphur filling one. The filled sphere always held more charge at the same potential, by a factor depending only on the material. He named it specific inductive capacity — the $\kappa$ you have been multiplying by all article — and read it as evidence that "induction" acts through the particles of the dielectric, not across empty distance: the polarisation picture, decades early. Characteristically, his was a disconnected, charge-conserving setup — Faraday's shared charge was the frozen $Q$ of your column two. The farad honours the man who showed that what stands between the plates is part of the capacitor.

History

1745

— The Leyden jar stores charge in glass — a dielectric capacitor before either word exists, carried about disconnected

1837

— Faraday's concentric-sphere experiments define specific inductive capacity: the dielectric constant enters physics via a fixed-charge protocol

1881

— The farad is adopted as the international unit of capacitance, fixing Faraday's name to the subject

1920

s — Variable air capacitors — plate overlap changed at fixed connection — tune the first generation of home radios

2007

— Intel ships hafnium-based high-κ transistor gates: dielectric choice at

\kappa \approx 20

becomes the lever keeping Moore's law alive

Going further

Texts. NCERT Physics Class 12, Ch. 2 (Exercise 2.9 is Example 1's ancestor) · HC Verma, Concepts of Physics Vol. 2, Ch. 31 · DC Pandey, Electricity and Magnetism, capacitor chapter (chained-protocol bank).

Provenance. All results are standard. A verifier should recompute: Example 1's $60\ \mathrm{nC}/9\ \mu\mathrm{J}$ start, (a) $360\ \mathrm{nC}$ , $54\ \mu\mathrm{J}$ , (b) $50\ \mathrm{V}$ , $1.5\ \mu\mathrm{J}$ ; Example 2's $600\ \mu\mathrm{J}$ vs $150\ \mu\mathrm{J}$ with the $300 + 150 = 150 + 300\ \mu\mathrm{J}$ audit; Example 3's $300\ \mathrm{V}$ , $90\ \mathrm{mJ}$ , $60\ \mathrm{mJ}$ ; Example 4's $\kappa^2 = 4:1$ ratio, $96$ vs $48\ \mu\mathrm{C}$ ; both derivations' $\kappa$ -ledgers at $\kappa = 6$ .

Prerequisites. Capacitance & the Parallel-Plate Capacitor · Energy Stored in a Capacitor · Dielectrics & Polarisation. Leads to. Charge Sharing & the Lost Energy · Sign Discipline in V, U & W · Capacitor Energy and Dielectrics (UG deep dive).

Why exactly does disconnecting freeze the charge and not the voltage?

Charge is conserved stuff — countable electrons — and with the switch open it has no path off the plates, so the number is locked. Voltage is not stuff: it is the ratio $Q/C$ and changes the instant $C$ does. Connected, the logic flips because the battery actively pumps charge whenever the plate voltage drifts from its emf: $V$ is enforced, $Q$ becomes adjustable. One quantity conserved by nature, one enforced by machinery — that asymmetry is the whole archetype.

How can the voltage end up higher than the battery that charged it, as in the chained example?

Because after disconnection the battery is out of the story, and you become the energy source. Pulling the slab out at fixed charge forces the same $Q$ onto a smaller capacitance, so $V = Q/C$ climbs — financed by the work your hand does against the inward pull. Nothing is free: the energy rose from 30 to 90 mJ in the example, every joule through your muscles. This genuine voltage-amplification trick is how electrophorus demonstrations and some high-voltage generators actually work.

Where does the lost energy physically go in the disconnected case?

Into mechanical work on the slab. The fringing field at the capacitor's mouth grips the slab's polarised edges and pulls it inward; lower it slowly and your restraining hand absorbs the energy. Release it and the slab accelerates, overshoots, and oscillates — the energy briefly kinetic — until losses convert it to heat and the slab settles inside. No resistor is needed: the "circuit" spent its energy on motion.

Why does the battery pay twice the stored gain during connected insertion?

The battery moves charge $\Delta Q$ through its own fixed voltage $V_0$ , so its outlay is the full rectangle $V_0\Delta Q$ — but storage grows by only half that; the other half is the field's work tugging the slab in. It is the same fifty–fifty split you meet when a battery charges an empty capacitor through a wire: whenever charge falls through a pinned potential difference, half the invoice escapes storage. At fixed $V$ , the escape route is the mechanical pull.

Does the slab really get pulled in, or do I always have to push it?

Pulled, in both protocols — surprising, since disconnected energy falls while connected energy rises. What sets the force is not the stored energy alone but the whole system's available energy (capacitor plus battery, when connected), and both bookkeepings show it decreasing as the slab advances: systems slide toward higher capacitance. The gripping happens at the edges, where the fringing field is non-uniform — the uniform interior field, perpendicular to the slab's motion, cannot pull it sideways.

What changes if the slab is thinner than the gap?

Only the capacitance formula. A slab of thickness $t < d$ leaves an air layer, and the stack gives $C = \varepsilon_0 A/(d - t + t/\kappa)$ — derived in Dielectrics & Polarisation. The protocol logic is untouched: decide connected (pin $V$ ) or disconnected (pin $Q$ ), then use the new $C$ in the same ledger. Neat check: $t = d$ recovers $\kappa\varepsilon_0A/d$ ; $t = 0$ recovers air.