In short

Once a dielectric of constant \kappa fills an entire parallel-plate gap, the capacitance becomes C = \kappa C_{0}. But real problems usually involve partial filling — a slab that covers only half the gap, or two slabs stacked vertically. The trick for all of them: redraw the capacitor as a combination of simpler capacitors and use series/parallel rules.

  • Dielectric fills part of the area (full height, half the plate) → two capacitors in parallel. C = \varepsilon_{0}(A - A_{d})/d + \kappa\varepsilon_{0}A_{d}/d.
  • Dielectric fills part of the gap (full area, partial thickness) → two capacitors in series. 1/C = d_{1}/(\varepsilon_{0}A) + d_{2}/(\kappa\varepsilon_{0}A).

When a dielectric is inserted or removed, the capacitance changes, and the capacitor is either pulled toward the dielectric or pushes it out. The force is derived from energy: F = -(dU/dx)_{Q} if the battery is disconnected and F = +(dU/dx)_{V} if the battery is connected (opposite signs because the battery does work in the connected case).

The energy bookkeeping is completely different in the two cases:

After insertion Q constant (battery off) V constant (battery on)
New capacitance \kappa C_{0} \kappa C_{0}
New charge Q_{0} (same) \kappa Q_{0}
New voltage V_{0}/\kappa V_{0} (same)
New energy U_{0}/\kappa \kappa U_{0}
Work by external agent negative — work extracted positive — battery does work

Every dielectric has a breakdown field E_\text{max} beyond which it conducts like a short-circuit. Dry air breaks down at about 3 \times 10^{6} V/m; polypropylene film at about 300 \times 10^{6} V/m — a hundred times better. This is what lets a 1 μF motor-run capacitor sit comfortably at 440 V in an industrial motor drive.

Walk into the projection room of an old single-screen cinema in Parel — the kind that still runs Hindi films from reels — and you will find a metal cabinet with a bank of oil-filled paper capacitors the size of cricket balls. Each one has two rolled-up strips of aluminium foil separated by kraft paper soaked in mineral oil. The paper is the dielectric. The oil fills in every microscopic gap and raises the breakdown voltage from a few hundred volts (air) to about 50 kV (paper-in-oil). That is how a single cap stores enough energy to drive the projector's arc lamp through the flicker of a 24-frame-per-second reel.

Cross the street to a textile mill in the same neighbourhood and you will find a very different kind of dielectric capacitor — the motor-run cap that sits in series with the auxiliary winding of a single-phase induction motor. This one is metallised polypropylene film, rated at 440 V AC. Its job is not to store energy briefly; it is to provide a phase-shifted current that lets the motor start and run smoothly. The dielectric constant \kappa of polypropylene (\kappa \approx 2.2) is modest — lower than paper — but its dielectric strength is enormous, which is what matters for a cap running hot at line voltage ten hours a day.

Both cases come down to the same physics: a dielectric slab between two plates. The basic rule — capacitance multiplies by \kappa — sits in the previous chapter. This chapter does the harder work: partial filling, multi-slab stacks, forces, and the two very different energy accountings depending on whether the battery is still attached when the dielectric moves.

The two fundamental partial-fill geometries

Almost every advanced capacitor problem reduces to one of two cases. Learn these two and you can solve the rest by mixing them.

Geometry A: Dielectric covers part of the area, full height

Imagine a parallel-plate capacitor with plate area A and gap d. A slab of dielectric with constant \kappa fills half of the space between the plates, but it occupies the full height d — it spans from the top plate to the bottom plate, covering only the left half of the area.

Dielectric fills half the area, full height — parallel combinationTop and bottom plates with a dielectric slab of constant kappa filling the left half of the gap. The right half is vacuum. The geometry is equivalent to two side-by-side capacitors sharing the same plates — one filled with dielectric, one empty — connected in parallel.+ + + + + + top plate, area A, potential V− − − − − − bottom plate, potential 0dielectric κarea A/2vacuumarea A/2d
A dielectric slab ($\kappa$, area $A/2$) fills half the plate area and runs the full height of the gap. The bottom and top surfaces of the slab are in contact with the plates.

Claim: the system behaves exactly as two capacitors in parallel: a plain-vacuum capacitor of area A/2 and a dielectric-filled capacitor of area A/2, both of gap d, connected at the top by the top plate and at the bottom by the bottom plate.

Why parallel? Every point on the top plate is at the same potential V (metal plates are equipotentials). Every point on the bottom plate is at 0. So both sub-capacitors see the same voltage V. That is the defining condition for parallel combination.

Compute:

C_{1} \;=\; \frac{\varepsilon_{0}(A/2)}{d} \;=\; \frac{\varepsilon_{0}A}{2d} \qquad\text{(vacuum half)}
C_{2} \;=\; \frac{\kappa\,\varepsilon_{0}(A/2)}{d} \;=\; \frac{\kappa\,\varepsilon_{0}A}{2d} \qquad\text{(dielectric half)}
C_\text{total} \;=\; C_{1} + C_{2} \;=\; \frac{\varepsilon_{0}A}{2d}(1 + \kappa) \;=\; \frac{1 + \kappa}{2}\cdot \underbrace{\frac{\varepsilon_{0}A}{d}}_{\textstyle C_{0}}. \tag{1}

Why: parallel capacitances add (same voltage, charges add, total charge per volt is the sum of the individual charge-per-volts). The factor C_{0} = \varepsilon_{0}A/d is the capacitance with no dielectric. With \kappa = 1 (no dielectric anywhere) the formula gives C = C_{0}; with \kappa \to \infty (effective conductor in half the area) it gives C = \kappa C_{0}/2, doubling for every doubling of \kappa — linear behaviour in \kappa.

Geometry B: Dielectric covers full area, part of the height

Now stack a dielectric slab of thickness t < d and constant \kappa on top of the bottom plate. The slab spans the full plate area A, but only fills t of the gap; the remaining d - t is vacuum.

Dielectric fills full area, part of the height — series combinationTop and bottom plates with a dielectric slab of constant kappa filling part of the gap. The slab spans the entire plate area A but only occupies the lower portion t of the gap; the remaining d minus t is vacuum. The geometry is equivalent to two capacitors stacked on top of each other, in series.top plate, area A, potential Vvacuum, thickness (d − t)dielectric κ, thickness tbottom plate, potential 0d − tt
A dielectric slab of thickness $t$ and constant $\kappa$ fills the bottom portion of the gap across the full plate area. The top $d - t$ is vacuum. The two regions are stacked — the geometry is two capacitors in series.

Claim: the system behaves as two capacitors in series — a plain-vacuum capacitor of thickness d - t and a dielectric-filled capacitor of thickness t, stacked.

Why series? The charge on the top plate is +Q. By Gauss's law, every horizontal cross-section through the gap (whether in vacuum or in dielectric) carries the same displacement — equivalently, the same free charge per area is "seen" through every slice. That means the two regions share the same charge Q. Same charge → series.

Compute:

\frac{1}{C_\text{total}} \;=\; \frac{d - t}{\varepsilon_{0}A} \;+\; \frac{t}{\kappa\,\varepsilon_{0}A} \;=\; \frac{1}{\varepsilon_{0}A}\left[(d - t) + \frac{t}{\kappa}\right].
C_\text{total} \;=\; \frac{\varepsilon_{0}A}{(d - t) + t/\kappa}. \tag{2}

Why: series capacitances add reciprocally (1/C_\text{total} = 1/C_{1} + 1/C_{2}), because the same charge Q on each piece produces a voltage Q/C_{i} across that piece, and the voltages add to give the total voltage across the stack. Inverting that gives C_\text{total}.

Notice what (2) says: when \kappa \to \infty (the slab becomes a conductor), the effective gap shrinks from d to d - t — because the conducting slab short-circuits its own thickness. So you get C = \varepsilon_{0}A/(d - t), which is the capacitance you would have if you had simply removed thickness t from the gap and pushed the plates closer together. That is the correct physical limit.

Why it is always series or parallel (mixed cases reduce to these)

Any multi-slab capacitor can be analysed by slicing it into these two elementary cases:

A capacitor with four quadrants, each with a different \kappa, is two parallel pairs in series, or two series pairs in parallel, depending on how you group them. Any grouping that respects the "same voltage" or "same charge" rule gives the right answer.

Interactive: move the dielectric slab sideways

The figure below shows a capacitor of plate area A = 10\ \text{cm}^{2}, gap d = 1 mm, with a dielectric slab of constant \kappa = 4 (polypropylene-like) filling a variable fraction x of the plate area (full height). Drag the red slider from x = 0 (no dielectric) to x = 1 (fully filled) and watch the capacitance slide from C_{0} up to 4 C_{0}. This is Geometry A with a variable fraction.

Interactive: partial horizontal fill — capacitance linear in fractionStraight line showing C over C zero equals one plus kappa minus one times x, where x is the fraction of the plate area covered by a dielectric of constant four. A draggable red point on the x axis sets the fraction; a tracking dot rides the line. A readout panel shows the numerical values for C, charge, and stored energy.fraction of area covered by dielectric, xC / C₀0123450.20.40.60.81.0C/C₀ = 1 + (κ−1) xκ = 4 (polypropylene-like)drag the red dot along the x axis
As the dielectric fraction $x$ grows from 0 to 1, the capacitance rises linearly from $C_{0}$ to $\kappa C_{0}$. The interactive uses $\kappa = 4$ and $C_{0} \approx 100$ pF (taking $A = 10$ cm², $d = 1$ mm, $\varepsilon_{0} = 8.85\times 10^{-12}$). Drag the red point to set the fraction.

The linearity is instructive. Partial-area filling behaves like a linear interpolation between the empty and full-fill values. Partial-thickness filling (Geometry B), by contrast, is a reciprocal function of the fraction — so the two geometries have very different shapes of response.

Force on a dielectric being inserted

Hold a slab of dielectric at the edge of a charged parallel-plate capacitor. Release it. The slab accelerates into the gap — the capacitor pulls it in.

Why? Thermodynamics: the system reduces its potential energy by letting the slab slide in (at constant Q or constant V, depending on what's external). A reduction in potential energy means there is a net force in the direction of motion. That force is what pulls the slab in.

The cleanest way to compute the force is from energy. Two cases, with opposite signs.

Case A: Battery disconnected (Q constant)

After the capacitor is charged to Q_{0}, the battery is disconnected. The charge Q_{0} is frozen on the plates and cannot change. Now let the dielectric slide in by dx (so x grows). The capacitance C(x) grows — from (1) with variable fraction x,

C(x) \;=\; \frac{\varepsilon_{0}A}{d}\bigl(1 + (\kappa - 1)x\bigr) \;=\; C_{0}\bigl(1 + (\kappa - 1)x\bigr).

The energy, at constant Q = Q_{0}, is U(x) = Q_{0}^{2}/(2C(x)). As C grows, U decreases. The change:

\frac{dU}{dx}\bigg|_{Q} \;=\; -\frac{Q_{0}^{2}}{2\,C(x)^{2}}\cdot\frac{dC}{dx} \;=\; -\frac{Q_{0}^{2}}{2\,C(x)^{2}}\cdot C_{0}(\kappa - 1).

Because U decreases with x, the derivative is negative. The energy-into-force formula for a mechanically isolated system is

F \;=\; -\frac{dU}{dx}\bigg|_{Q} \;=\; +\frac{Q_{0}^{2}\,C_{0}(\kappa - 1)}{2\,C(x)^{2}}. \tag{3}

Why: the minus sign in "F = -dU/dx" is the usual rule for conservative forces (force points in the direction of decreasing potential energy). Since dU/dx is negative here, F is positive — the force points in the direction of increasing x, i.e., it pulls the dielectric into the capacitor. This matches the physical observation.

Case B: Battery still connected (V constant)

Now leave the battery attached, so it holds the plates at the fixed voltage V_{0}. As the dielectric slides in, more charge flows from the battery onto the plates to maintain the voltage. The system is no longer energy-isolated.

The capacitor's own energy at constant V = V_{0} is U_\text{cap}(x) = \tfrac{1}{2}C(x)V_{0}^{2}, which increases with x. So

\frac{dU_\text{cap}}{dx}\bigg|_{V} \;=\; \tfrac{1}{2}V_{0}^{2}\,\frac{dC}{dx} \;=\; \tfrac{1}{2}V_{0}^{2}\,C_{0}(\kappa - 1) \;>\; 0.

But while the capacitor's energy is rising, so is the battery's output. The battery supplies charge dQ = V_{0}\,dC at voltage V_{0}, doing work dW_\text{bat} = V_{0}\,dQ = V_{0}^{2}\,dC. This is twice the change in capacitor energy — so half of the battery's work goes into the capacitor, and the other half is available for mechanical work on the dielectric.

Equate: work done on the dielectric by the capacitor equals dW_\text{bat} - dU_\text{cap}:

F\,dx \;=\; V_{0}^{2}\,dC - \tfrac{1}{2}V_{0}^{2}\,dC \;=\; \tfrac{1}{2}V_{0}^{2}\,dC.
F \;=\; +\tfrac{1}{2}V_{0}^{2}\,\frac{dC}{dx} \;=\; +\tfrac{1}{2}V_{0}^{2}\,C_{0}(\kappa - 1). \tag{4}

Why: the sign is positive (pulling in) with a plus sign in front of dU/dx, opposite to the Q-constant case. The reason is the battery: it pumps in enough energy to both increase U_\text{cap} and do mechanical work on the slab. If you naively wrote F = -dU_\text{cap}/dx in this case, you would get the wrong sign — the slab would appear to be pushed out. The battery is the reason the sign flips.

Summary of the sign-flip rule

F \;=\; -\frac{dU}{dx}\bigg|_{Q} \qquad \text{or} \qquad F \;=\; +\frac{dU_\text{cap}}{dx}\bigg|_{V}.

Both give the same direction of force (into the capacitor) even though the signs look opposite — because U decreases with x at constant Q but increases with x at constant V.

Energy changes when a dielectric is inserted

Tabulate the before/after state for the two cases. Take a capacitor with vacuum capacitance C_{0} charged to V_{0} with charge Q_{0} = C_{0}V_{0} and energy U_{0} = \tfrac{1}{2}C_{0}V_{0}^{2}. Insert a slab of \kappa filling the full gap.

Quantity Before After (Q constant) After (V constant)
Capacitance C_{0} \kappa C_{0} \kappa C_{0}
Charge Q Q_{0} Q_{0} (battery off) \kappa Q_{0} (extra flows in)
Voltage V V_{0} V_{0}/\kappa (drops) V_{0} (held)
Field E in gap V_{0}/d V_{0}/(\kappa d) V_{0}/d (unchanged)
Energy U \tfrac{1}{2}C_{0}V_{0}^{2} \tfrac{1}{2}(C_{0}V_{0})^{2}/(\kappa C_{0}) = U_{0}/\kappa \tfrac{1}{2}(\kappa C_{0})V_{0}^{2} = \kappa U_{0}
Work by external agent inserting the slab — negative: slab is pulled in, system does work on your hand negative: still pulled in, but net energy also flows from battery
Work by battery — zero (disconnected) V_{0}\cdot \Delta Q = V_{0}(\kappa Q_{0} - Q_{0}) = (\kappa - 1) V_{0} Q_{0} = 2(\kappa - 1)U_{0}

Two facts to take away:

Worked examples

Example 1: Motor-run capacitor in a textile mill

A single-phase motor-run capacitor has parallel plates of area A = 80\ \text{cm}^{2} = 8\times 10^{-3}\ \text{m}^{2} separated by a d = 40\ \mum polypropylene film (\kappa = 2.2). The motor applies 440 V across the capacitor. Find (a) its capacitance, (b) the charge stored, (c) the stored energy at 440 V, and (d) the electric field inside the film. Check whether it is safely below the breakdown field of polypropylene, about 300\times 10^{6} V/m.

Motor-run capacitor geometrySchematic of a parallel-plate capacitor with area 80 square centimetres, gap 40 micrometres, filled with polypropylene film of dielectric constant 2.2, connected across 440 volts.aluminium plate, area 80 cm²polypropylene film, κ = 2.2thickness 40 μmaluminium plate440 V
Motor-run cap: large area, very thin polypropylene film, high-voltage rating. Industrial motor drives run this kind of cap at line voltage continuously.

Step 1. Capacitance with the dielectric filling the full gap.

C \;=\; \kappa\,\frac{\varepsilon_{0}A}{d} \;=\; 2.2\times\frac{(8.85\times 10^{-12})(8\times 10^{-3})}{40\times 10^{-6}}.

Numerator: (8.85\times 10^{-12})(8\times 10^{-3}) = 7.08\times 10^{-14}.

Divided by 4\times 10^{-5}: 7.08\times 10^{-14}/4\times 10^{-5} = 1.77\times 10^{-9}.

C \;=\; 2.2 \times 1.77\times 10^{-9} \;=\; 3.9\times 10^{-9}\ \text{F} \;=\; 3.9\ \text{nF}.

Why: plug into C = \kappa\varepsilon_{0}A/d. Working in pure SI units — plate area in m², thickness in m — eliminates the unit-conversion errors that plague these problems. 3.9 nF is a reasonable size for a small motor-start cap; commercial 440-V motor-run caps in larger motors are typically 5–50 μF.

Step 2. Charge on the plates at 440 V.

Q \;=\; CV \;=\; (3.9\times 10^{-9})(440) \;=\; 1.7\times 10^{-6}\ \text{C} \;=\; 1.7\ \mu\text{C}.

Step 3. Stored energy at the instantaneous peak voltage of 440 V.

U \;=\; \tfrac{1}{2}CV^{2} \;=\; \tfrac{1}{2}(3.9\times 10^{-9})(440)^{2} \;=\; \tfrac{1}{2}(3.9\times 10^{-9})(1.94\times 10^{5}).
U \;=\; \tfrac{1}{2}(7.55\times 10^{-4}) \;=\; 3.8\times 10^{-4}\ \text{J} \;=\; 0.38\ \text{mJ}.

Why: for a running AC capacitor, this is the peak stored energy; it oscillates between zero and U at 50 Hz. The capacitor is not storing energy to be delivered later — it is storing and returning energy every half-cycle as part of the reactive-power exchange with the motor.

Step 4. Electric field inside the film.

E \;=\; \frac{V}{d} \;=\; \frac{440}{40\times 10^{-6}} \;=\; 1.1\times 10^{7}\ \text{V/m}.

Step 5. Compare with breakdown strength of polypropylene, E_\text{max} \approx 300\times 10^{6} V/m.

\frac{E}{E_\text{max}} \;=\; \frac{1.1\times 10^{7}}{3.0\times 10^{8}} \;\approx\; 3.7\%.

Why: the capacitor is running at only 3.7% of the dielectric's breakdown field — a healthy safety margin. Textile mill engineers design for at least a 5:1 margin to handle switching surges, which can double or triple the peak voltage for microseconds.

Result: C = 3.9 nF, Q = 1.7\ \muC, U = 0.38 mJ, E = 1.1\times 10^{7} V/m (well below breakdown).

What this shows: A dielectric does two things simultaneously — it raises the capacitance by the factor \kappa (from 1.8 nF to 3.9 nF) and it allows a much higher operating voltage than vacuum would (because air breaks down at 3 MV/m, only a thirty-seventh of the polypropylene limit). For industrial AC capacitors, the dielectric strength, not the \kappa value, is the headline specification.

Example 2: A capacitor with two dielectric slabs stacked

A parallel-plate capacitor has plate area A = 100\ \text{cm}^{2} = 10^{-2}\ \text{m}^{2} and gap d = 2 mm. The gap is filled by two slabs of equal thickness t_{1} = t_{2} = 1 mm, stacked vertically — the top slab is mica (\kappa_{1} = 6) and the bottom slab is bakelite (\kappa_{2} = 4). A battery holds the plates at V = 100 V. Find the capacitance, the charge, and the electric field in each slab.

Two stacked dielectric slabs in seriesParallel plate capacitor with the gap filled by two horizontal slabs. The top slab is mica with dielectric constant six and thickness one millimetre. The bottom slab is bakelite with dielectric constant four and thickness one millimetre. One hundred volts across the plates.top plate, V = 100 Vmica κ₁ = 6 t₁ = 1 mmbakelite κ₂ = 4 t₂ = 1 mmbottom plate, 0 V
Two dielectric slabs stacked: mica on top, bakelite below. The geometry is series — the two slabs share the same charge, and the voltages across them add to 100 V.

Step 1. Recognise the geometry as series.

The two slabs are stacked, both span the full plate area. Using Geometry B for each slab gives two series capacitors.

Step 2. Compute each sub-capacitance.

Vacuum-equivalent capacitance for each slab (treating each slab as if it were a full-area capacitor at its own thickness):

C_{1} \;=\; \frac{\kappa_{1}\,\varepsilon_{0}A}{t_{1}} \;=\; \frac{6\times(8.85\times 10^{-12})\times 10^{-2}}{10^{-3}} \;=\; 5.31\times 10^{-10}\ \text{F} \;=\; 531\ \text{pF}.
C_{2} \;=\; \frac{\kappa_{2}\,\varepsilon_{0}A}{t_{2}} \;=\; \frac{4\times(8.85\times 10^{-12})\times 10^{-2}}{10^{-3}} \;=\; 3.54\times 10^{-10}\ \text{F} \;=\; 354\ \text{pF}.

Why: each slab on its own, with the bottom and top metal plates as its plates, would have this capacitance. In the full capacitor, these two sub-capacitances are in series.

Step 3. Combine in series.

\frac{1}{C} \;=\; \frac{1}{C_{1}} + \frac{1}{C_{2}} \;=\; \frac{1}{531} + \frac{1}{354}\ \text{pF}^{-1} \;=\; 0.001884 + 0.002825 \;=\; 0.004709\ \text{pF}^{-1}.
C \;=\; \frac{1}{0.004709} \;\approx\; 212\ \text{pF}.

Why: for series, reciprocals add. The total capacitance is always less than either individual capacitance — here, less than 354 pF. Adding more dielectrics in series reduces the total capacitance, a useful guardrail for spotting arithmetic errors.

Step 4. Charge on the plates.

Q \;=\; CV \;=\; (2.12\times 10^{-10})(100) \;=\; 2.12\times 10^{-8}\ \text{C} \;=\; 21.2\ \text{nC}.

Step 5. Voltage across each slab. In series, the same Q sits on each sub-capacitor, and the voltage across each sub-capacitor is Q/C_{i}.

V_{1} \;=\; \frac{Q}{C_{1}} \;=\; \frac{2.12\times 10^{-8}}{5.31\times 10^{-10}} \;=\; 39.9\ \text{V}.
V_{2} \;=\; \frac{Q}{C_{2}} \;=\; \frac{2.12\times 10^{-8}}{3.54\times 10^{-10}} \;=\; 60.0\ \text{V}.

Check: V_{1} + V_{2} = 39.9 + 60.0 \approx 100\ \text{V}. ✓

Why: in series, the bigger-capacitance slab (mica) gets the smaller voltage, and the smaller-capacitance slab (bakelite) gets the bigger voltage. The bakelite carries 60 V across 1 mm while the mica carries only 40 V across 1 mm.

Step 6. Field in each slab.

E_{1} \;=\; \frac{V_{1}}{t_{1}} \;=\; \frac{39.9}{10^{-3}} \;=\; 3.99\times 10^{4}\ \text{V/m}.
E_{2} \;=\; \frac{V_{2}}{t_{2}} \;=\; \frac{60.0}{10^{-3}} \;=\; 6.00\times 10^{4}\ \text{V/m}.

Why: inside each slab the field is uniform (same free charge Q on each cross-section gives the same free surface charge density, and the dielectric just reduces the field by its own \kappa). The bakelite, with the smaller \kappa, carries the larger field — because E in a dielectric is inversely proportional to \kappa for fixed free-charge density.

Result: C = 212 pF, Q = 21.2 nC, V_{1} = 39.9 V in mica, V_{2} = 60.0 V in bakelite, E_{2}/E_{1} = 1.5 = \kappa_{1}/\kappa_{2}.

What this shows: In a stack of dielectrics, the slab with the smaller \kappa always carries the larger field. If you are designing a high-voltage capacitor with two dielectric layers, the weaker-\kappa layer is the one that will break down first. This is why engineers put the stronger (higher breakdown) dielectric on the outside of a multi-layer stack — so that the unavoidable higher-field layer is also the more breakdown-resistant one.

Example 3: Energy change when a dielectric is inserted — both scenarios

A parallel-plate capacitor with C_{0} = 20\ \muF is charged to V_{0} = 100 V. A glass slab (\kappa = 5) is then slid in to completely fill the gap. Compute the charge, voltage, and stored energy both before and after the insertion, for two cases: (a) the battery stays connected during insertion; (b) the battery is disconnected before insertion.

Step 1. Before the insertion (same for both cases).

Q_{0} \;=\; C_{0}V_{0} \;=\; (20\times 10^{-6})(100) \;=\; 2\times 10^{-3}\ \text{C} \;=\; 2\ \text{mC}.
U_{0} \;=\; \tfrac{1}{2}C_{0}V_{0}^{2} \;=\; \tfrac{1}{2}(20\times 10^{-6})(100)^{2} \;=\; \tfrac{1}{2}(20\times 10^{-6})(10^{4}) \;=\; 0.1\ \text{J}.

Step 2. After insertion, case (a) — battery connected, V held at 100 V.

New capacitance: C' = \kappa C_{0} = 5\times 20\ \mu\text{F} = 100\ \mu\text{F}.

New charge: Q' = C'V_{0} = (100\times 10^{-6})(100) = 10\ \text{mC}.

New energy: U' = \tfrac{1}{2}C'V_{0}^{2} = \tfrac{1}{2}(100\times 10^{-6})(10^{4}) = 0.5\ \text{J}.

Change in energy: \Delta U_\text{cap} = 0.5 - 0.1 = 0.4\ \text{J}.

Why: voltage is held constant by the battery, so V does not change. The capacitance goes up by \kappa, so Q = CV goes up by \kappa too (from 2 mC to 10 mC). The extra 8 mC of charge had to come from somewhere — it came from the battery. The stored energy also rises by \kappa, from 0.1 J to 0.5 J.

Work done by the battery during insertion:

W_\text{bat} \;=\; V_{0}\,\Delta Q \;=\; (100)(8\times 10^{-3}) \;=\; 0.8\ \text{J}.

Of this 0.8 J, only 0.4 J went into the capacitor as stored energy. The other 0.4 J did mechanical work on the glass slab as it was pulled in (and, if you held the slab back, dissipated as friction or muscle effort on your side).

Step 3. After insertion, case (b) — battery disconnected, Q held at Q_{0}.

New capacitance (same): C' = 5\times 20 = 100\ \muF.

Charge unchanged: Q' = 2\ \text{mC}.

New voltage: V' = Q'/C' = 2\times 10^{-3}/100\times 10^{-6} = 20\ \text{V}.

New energy:

U' \;=\; \frac{Q'^{2}}{2C'} \;=\; \frac{(2\times 10^{-3})^{2}}{2\times 100\times 10^{-6}} \;=\; \frac{4\times 10^{-6}}{2\times 10^{-4}} \;=\; 0.02\ \text{J}.

Change in energy: \Delta U_\text{cap} = 0.02 - 0.1 = -0.08\ \text{J}.

Why: charge is frozen on the plates (the battery wire has been cut), so Q = Q_{0}. The capacitance goes up by \kappa, so the voltage drops by \kappa (from 100 V to 20 V). The energy U = Q^{2}/(2C) drops by \kappa (from 0.1 J to 0.02 J). The missing 0.08 J went into the mechanical work of pulling the slab in — the capacitor did work on the slab, accelerating it to some speed (or, if you held the slab back, the slab did work on your hand).

Step 4. Summary table.

Quantity Before Case a (V const) Case b (Q const)
C 20 μF 100 μF 100 μF
Q 2 mC 10 mC 2 mC
V 100 V 100 V 20 V
U 0.1 J 0.5 J 0.02 J
\Delta U_\text{cap} — +0.4 J -0.08 J

Result: With the battery connected, inserting the dielectric quintuples the stored energy (from 0.1 to 0.5 J). With the battery disconnected, inserting the dielectric reduces the stored energy to a fifth (0.1 to 0.02 J). The capacitor pulls the dielectric in both times, but the energy bookkeeping differs because the battery is a variable energy source.

What this shows: Whether the battery is attached is the single most important detail in any dielectric-insertion problem. Memorising "U = \tfrac{1}{2}CV^{2}" alone is not enough — you have to ask which variable is held fixed. The trap that most JEE students fall into is computing U' = \tfrac{1}{2}C'V_{0}^{2} in case (b), where V is no longer V_{0} after the insertion. Always ask: what's conserved?

Common confusions

If you came to understand partial-fill geometries, the force on a dielectric, and the two energy bookkeepings, you have what you need. What follows is for readers who want to see the breakdown-voltage scaling argument, the derivation of the force formula from first principles, and the compound-dielectric edge cases that show up on JEE Advanced.

Dielectric strength and the scaling argument

The breakdown field E_\text{max} of a dielectric is the field at which it stops insulating and starts conducting — typically because free electrons, accelerated by the field, kick bound electrons loose and the result is a cascading avalanche. For a dielectric of thickness d, the breakdown voltage is V_\text{max} = E_\text{max}\,d. So to hold more volts, make the dielectric thicker. But making the dielectric thicker reduces capacitance as 1/d. The design trade-off is:

C \;=\; \frac{\kappa\varepsilon_{0}A}{d}, \qquad V_\text{max} \;=\; E_\text{max}\,d, \qquad U_\text{max} \;=\; \tfrac{1}{2}CV_\text{max}^{2} \;=\; \tfrac{1}{2}\kappa\varepsilon_{0}A\,d\,E_\text{max}^{2}.

Since A\,d is the volume \mathcal{V} of the dielectric, the maximum energy per unit volume is

u_\text{max} \;=\; \tfrac{1}{2}\kappa\varepsilon_{0}E_\text{max}^{2}.

This is the figure of merit for choosing a dielectric: more \kappa is good, but much more breakdown field is better (it enters squared). Polypropylene at \kappa = 2.2, E_\text{max} = 3\times 10^{8} V/m gives u_\text{max} = \tfrac{1}{2}(2.2)(8.85\times 10^{-12})(9\times 10^{16}) \approx 880 kJ/m³. Ceramic at \kappa = 3000, E_\text{max} = 10^{7} V/m gives u_\text{max} = \tfrac{1}{2}(3000)(8.85\times 10^{-12})(10^{14}) \approx 1.3 MJ/m³. Ceramic wins on energy density, but polypropylene wins on reliability because \kappa = 2.2 is a more linear, more temperature-stable response.

The force from first principles

The Q-constant case in the main text derives

F \;=\; \frac{Q^{2}}{2C^{2}}\cdot\frac{dC}{dx}.

For the horizontal-slide geometry with a slab of width L sliding sideways into a capacitor of plate width L (so the fraction x is the dimensionless slab position from 0 to 1), A_\text{in} = xL^{2} and

C(x) \;=\; \frac{\varepsilon_{0}L^{2}}{d}\bigl(1 + (\kappa - 1)x\bigr), \qquad \frac{dC}{dx} \;=\; \frac{\varepsilon_{0}L^{2}(\kappa - 1)}{d}.

So at slab position x,

F(x) \;=\; \frac{Q^{2}}{2C(x)^{2}}\cdot\frac{\varepsilon_{0}L^{2}(\kappa - 1)}{d}.

The force decreases as the slab moves in, because C grows in the denominator squared. This explains why a dielectric-insertion is a damped, not accelerating, motion — the pull weakens as the dielectric takes up more of the plate area.

This is also one of the surprisingly elegant features of the energy method: we never had to compute the actual forces on individual surface charges or polarisation charges in the fringing field. The energy bookkeeping handles all of that automatically. You get the net result without parsing the microscopic geometry.

Compound dielectric with different κ values in two halves — Geometry A generalised

If instead of "dielectric fills half, vacuum fills half", you have "dielectric \kappa_{1} fills the left half, dielectric \kappa_{2} fills the right half", Geometry A gives

C \;=\; \frac{\varepsilon_{0}A}{2d}(\kappa_{1} + \kappa_{2}).

This is just the parallel sum with both halves filled. Use this when two dielectrics are side-by-side, sharing the same plate-to-plate distance.

The asymmetric force when the plates are not symmetric

A subtle issue: what if one plate is larger than the other? Then the system no longer has the left-right symmetry of the standard partial-fill problem, and the dielectric can experience a torque as well as a linear force — it tends to rotate to align with the larger plate's edge. This is the origin of the alignment forces in liquid-crystal displays, where the dielectric anisotropy of the liquid crystal molecules interacts with the electric field of the pixel electrodes.

For the simple symmetric problems in this chapter, only the linear force matters. The symmetric formulas (3) and (4) are the full answer.

Partial-fill with vertical displacement in Geometry B

If instead of pulling the dielectric out sideways, you lift it off the bottom plate by a gap g (so g of vacuum, t of dielectric, d - t - g of vacuum from top), the three-series-capacitor formula gives

\frac{1}{C} \;=\; \frac{g}{\varepsilon_{0}A} + \frac{t}{\kappa\varepsilon_{0}A} + \frac{d - t - g}{\varepsilon_{0}A} \;=\; \frac{(d - t) + t/\kappa}{\varepsilon_{0}A}.

The g drops out — shifting the slab up or down within the gap does not change the capacitance, as long as the slab stays between the plates. Only the thickness matters, not the position. The reader can verify this by sliding a slab vertically in a textbook problem and watching C stay constant. This is the "vertical-position independence" theorem for a single slab between parallel plates.

When the field is strong enough to change \kappa

Real dielectrics are not perfectly linear — \kappa depends weakly on E at low fields and strongly on E as you approach breakdown. The linear approximation P = \varepsilon_{0}(\kappa - 1)E is the first term of a Taylor series; the next term (quadratic in E) becomes important in nonlinear-optics devices like frequency-doubling crystals. For JEE Advanced and standard capacitor problems, the linear approximation is always sufficient.

Where this leads next