Dielectrics — polarization, dielectric constant, effect on capacitance: Worked examples

This page is a drill through every case the dielectric topic can hand you. Before we solve anything, we lay out a matrix of case classes — the different kinds of question that exist — and then solve examples until every cell is covered. If you can do all of these, no dielectric problem can surprise you.

Prerequisites we lean on: Capacitance and Capacitors, Electric Field of Parallel Plates, Energy Stored in a Capacitor, Electric Susceptibility, Gauss's Law and the D-field, and the parent Dielectrics topic note.

Definition The symbols we build everything on (

$C_{0}$ , $σ_{f}$ , $σ_{b}$ ) Three symbols recur below — pin each to a picture before we use it.

== $C_{0}$ = the vacuum (empty-gap) capacitance== — the capacitance the same plates ( $A$ , $d$ ) have with nothing but vacuum (or air) between them. It is the "before" value; adding a dielectric turns it into $C = κ C_{0}$ . So " $C_{0} = 4 μ F$ empty" just means the gadget measured $4 μ F$ before any slab went in.
== $σ_{f}$ = free-charge surface density== — the charge per unit area the battery physically delivered onto the metal plates (the "free" electrons that can move through wires). On the $+$ plate $σ_{f} > 0$ ; on the $-$ plate $σ_{f} < 0$ . Same magnitude, opposite sign on the two plates.
== $σ_{b}$ = bound-charge surface density== — the charge that appears on the two faces of the dielectric because the dipoles line up (it is "bound": it cannot leave its molecule). It lives on the inner surfaces of the slab, right against each plate.

Recall Three master formulas + the regime question (built in the parent — repeated so you never scroll)

Three formulas do all the heavy lifting (symbols defined just above):

Field falls: $E = \frac{E _{0}}{κ}$ , where $E_{0} = \frac{σ _{f}}{ε _{0}}$ is the vacuum field of the free charge alone.
Capacitance rises: $C = κ C_{0} = \frac{κ ε _{0} A}{d}$ , where $C_{0}$ is the empty-gap capacitance.
Bound surface charge (magnitude): $σ_{b} = σ_{f} (1 - \frac{1}{κ})$ , sitting opposite in sign to the free charge it faces.

And one question you must ask before applying them:

Which quantity is held fixed — $Q$ (battery off) or $V$ (battery on)?

The scenario matrix

Every dielectric exam question is one (or a combination) of these case classes. The rightmost column names the worked example that nails it.

#	Case class	What is being tested	Covered by
A	Plain $κ$ boost	$C = κ C_{0}$ , nothing subtle	Ex 1
B	Battery disconnected ( $Q$ fixed)	$E, V, U$ all drop by $κ$	Ex 2
C	Battery connected ( $V$ fixed)	$E$ fixed, $Q, U$ rise; where charge comes from	Ex 3
D	Bound charge & field numbers	$σ_{b}$ , $E_{0}$ , $E$ from $σ_{f}, κ$	Ex 4
E	Degenerate limits	$κ \to 1$ (vacuum) and $κ \to \infty$ (conductor)	Ex 5
F	Partial fill (slab thinner than gap)	series of two regions, geometry	Ex 6 (figure)
G	Two dielectrics side-by-side vs stacked	parallel vs series capacitors	Ex 7 (figure)
H	Real-world word problem	choose material for safety/size	Ex 8
I	Exam-style twist (force pulling slab in)	energy method, sign of force	Ex 9

Cells E covers the two limiting inputs; A–D cover the everyday signs and regimes; F, G cover geometry degeneracies (a slab that only fills part of the gap, or splits the gap). Let's go.

Worked examples

Worked example Ex 2 — Case B · Battery DISCONNECTED (

$Q$ fixed) Charge a $C_{0} = 2 μ F$ (empty-gap) capacitor to $V_{0} = 100 V$ , then unplug the battery. Now insert $κ = 4$ . Find new $Q$ , $V$ , $E$ -ratio, and energy $U$ .

Forecast: With the battery gone, which quantity is frozen? Once you know that, predict the rest.

Battery gone $\Rightarrow$ $Q$ cannot change (nowhere for charge to flow). $Q = C_{0} V_{0} = 2 \times 1 0^{- 6} \times 100 = 2 \times 1 0^{- 4} C = 200 μ C$ , and it stays $200 μ C$ . Why this step? Isolated plates conserve charge — this is the anchor of the whole regime.
New capacitance $C = κ C_{0} = 8 μ F$ . Why this step? Geometry unchanged, medium changed — Case A rule still applies.
New voltage $V = \frac{Q}{C} = \frac{200 μ C}{8 μ F} = 25 V$ . Why this step? $Q$ fixed, $C$ up by $4$ $\Rightarrow$ $V$ down by $4$ : $100/4 = 25$ ✔.
Field ratio $E / E_{0} = V / V_{0} = 25/100 = 1/4 = 1/ κ$ . Why this step? $E = V / d$ with $d$ fixed, so $E$ tracks $V$ exactly.
Energy $U = \frac{Q ^{2}}{2 C} = \frac{( 2 \times 1 0 ^{- 4} ) ^{2}}{2 \times 8 \times 1 0 ^{- 6}} = 2.5 \times 1 0^{- 3} J = 2.5 mJ$ (was $\frac{1}{2} C_{0} V_{0}^{2} = 10 mJ$ ). Why this step? $Q$ fixed form of energy; $C$ in the denominator, so energy falls.

Verify: Energy dropped $10 \to 2.5 mJ$ . Where did $7.5 mJ$ go? Into work done by the field as it sucked the slab in (Ex 9 makes this concrete). Consistency: $U = \frac{1}{2} Q V = \frac{1}{2} (200 μ C) (25 V) = 2.5 mJ$ ✔.

Worked example Ex 3 — Case C · Battery CONNECTED (

$V$ fixed) Same $C_{0} = 2 μ F$ empty-gap capacitor, but the $V = 100 V$ battery stays plugged in. Insert $κ = 3$ . Find $C$ , $Q$ , the extra charge from the battery, the field ratio, and $U$ .

Forecast: Now which quantity is frozen? Guess $Q$ 's direction and $E$ 's direction.

Battery clamps $V = 100 V$ fixed. Since $E = V / d$ and $d$ fixed, $E$ is unchanged by insertion. Why this step? This is the trap-avoider: with a battery you do not use $E = E_{0} / κ$ .
$C = κ C_{0} = 3 \times 2 = 6 μ F$ .
$Q = C V = 6 \times 1 0^{- 6} \times 100 = 6 \times 1 0^{- 4} C = 600 μ C$ (was $C_{0} V = 200 μ C$ ). Why this step? $V$ fixed, $C$ up by $3$ $\Rightarrow$ $Q$ up by $3$ .
Extra charge pumped by battery $= 600 - 200 = 400 μ C$ . Why this step? Conservation — the increase must come from somewhere; the battery is the source.
$U = \frac{1}{2} C V^{2} = \frac{1}{2} (6 \times 1 0^{- 6}) (100)^{2} = 3 \times 1 0^{- 2} J = 30 mJ$ (was $10 mJ$ ). Why this step? $V$ fixed form; $C$ in numerator, so energy rises.

Verify: $E$ unchanged but $C, Q, U$ all up by $κ = 3$ ( $6/2 = 3$ , $600/200 = 3$ , $30/10 = 3$ ) ✔. Contrast Ex 2: same $κ$ -ish insertion, opposite energy trend — because a different thing was held fixed.

Worked example Ex 4 — Case D · Bound charge & fields from scratch

Free-charge density $σ_{f} = 8.85 \times 1 0^{- 6} C/m^{2}$ on the plates (magnitude on the $+$ plate; the $-$ plate carries the same magnitude, opposite sign), dielectric $κ = 4$ . Find $E_{0}$ , $E$ , and the bound surface charge $σ_{b}$ .

Forecast: Will $σ_{b}$ be bigger or smaller than $σ_{f}$ ? (It's bound charge that only partly cancels — so...?)

Vacuum field $E_{0} = \frac{σ _{f}}{ε _{0}} = \frac{8.85 \times 1 0 ^{- 6}}{8.85 \times 1 0 ^{- 12}} = 1.0 \times 1 0^{6} V/m$ . Why this step? Field of the free charge alone, before the dielectric responds (Electric Field of Parallel Plates).
Net field $E = \frac{E _{0}}{κ} = \frac{1 0 ^{6}}{4} = 2.5 \times 1 0^{5} V/m$ . Why this step? No battery mentioned about $σ_{f}$ — $σ_{f}$ is the given free charge, so $E$ shrinks by $κ$ .
Bound charge $σ_{b} = σ_{f} (1 - \frac{1}{κ}) = 8.85 \times 1 0^{- 6} \times (1 - 0.25) = 6.64 \times 1 0^{- 6} C/m^{2}$ . Why this step? From $E = \frac{σ _{f} - σ _{b}}{ε _{0}} = \frac{σ _{f}}{κ ε _{0}}$ , solve for $σ_{b}$ . We subtract $σ_{b}$ because the bound face hugging the $+$ plate carries the opposite ( $-$ ) sign, partly cancelling it (see the sign-convention callout above).

Verify: $σ_{b} < σ_{f}$ (partial cancellation, never total) ✔. Check the net: $σ_{f} - σ_{b} = 8.85 - 6.64 = 2.21 \times 1 0^{- 6}$ , and $E = \frac{2.21 \times 1 0 ^{- 6}}{8.85 \times 1 0 ^{- 12}} = 2.5 \times 1 0^{5} V/m$ — matches step 2 ✔.

Worked example Ex 5 — Case E · The two DEGENERATE limits

A $C_{0} = 10 μ F$ vacuum (empty-gap) capacitor. (a) Slide in "vacuum" ( $κ = 1$ ). (b) Slide in a perfect conductor filling the gap — treat as $κ \to \infty$ . Discuss $C$ , $E$ , and why.

Forecast: For (b), what happens to the voltage across a conductor? What does that force $C$ to do?

(a) $κ = 1 \Rightarrow C = 1 \cdot C_{0} = 10 μ F$ . Nothing changes. Why this step? $κ = 1$ means zero susceptibility — the "dielectric" polarizes not at all, i.e. it is vacuum. Sanity floor of the formula.
(b) $κ \to \infty \Rightarrow C = κ C_{0} \to \infty$ . Why this step? A conductor cannot sustain an internal field, so $E \to 0$ inside it; $V = E d \to 0$ , and $C = Q / V \to \infty$ for finite $Q$ .
Field: (a) unchanged $E = E_{0}$ ; (b) $E = E_{0} / κ \to 0$ . Why this step? Consistent with $κ = E_{0} / E$ : infinite $κ$ means the medium fully cancels the field.

Verify: Both limits are monotone-consistent with $C = κ C_{0}$ : as $κ$ climbs $1 \to \infty$ , $C$ climbs $C_{0} \to \infty$ and $E$ falls $E_{0} \to 0$ . No formula breaks at the endpoints ✔. (A real conductor slab of thickness $t < d$ gives a finite boost — that's the partial-fill logic of Ex 6.)

Worked example Ex 6 — Case F · Partial fill (slab thinner than the gap)

Plate separation $d = 3 mm$ , area $A$ giving $C_{0} = \frac{ε _{0} A}{d} = 6 μ F$ empty (empty-gap capacitance). A dielectric slab $κ = 3$ , thickness $t = 1 mm$ , sits against one plate; the remaining $2 mm$ is air. Find the new capacitance.

Figure — what it shows: the two horizontal black bars are the plates ( $+ Q$ on top, $- Q$ on bottom). The red block hugging the bottom plate is the $1 mm$ dielectric slab; the white space above it (labelled "air") is the remaining $2 mm$ . The double-headed arrows on the right mark the two thicknesses $t = 1 mm$ and $d - t = 2 mm$ ; the arrow on the left marks the full gap $d = 3 mm$ . The picture is telling you the gap is split into two stacked regions.

Forecast: Is a partial slab more or less effective than a full slab of the same $κ$ ? Predict a number between $6$ and $18 μ F$ .

The gap is now two regions in series: a dielectric layer (thickness $t = 1 mm$ ) and an air layer (thickness $d - t = 2 mm$ ). As the figure shows, the red slab sits below the air region, stacked one on top of the other. Why this step? The same charge $Q$ crosses both layers; voltages add. That is the definition of series (Capacitance and Capacitors).
Region capacitances (same area $A$ ): dielectric $C_{1} = \frac{κ ε _{0} A}{t}$ , air $C_{2} = \frac{ε _{0} A}{d - t}$ . Why this step? Each layer is its own parallel-plate capacitor with its own thickness.
Series combine: $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} = \frac{t}{κ ε _{0} A} + \frac{d - t}{ε _{0} A} = \frac{1}{ε _{0} A} (\frac{t}{κ} + d - t)$ . Why this step? Series capacitors add reciprocals (voltages add for shared $Q$ ).
So $C = \frac{ε _{0} A}{\frac{t}{κ} + d - t}$ . Plug in with $ε_{0} A = C_{0} d = 6 μ F \times 3 mm = 18 μ F\cdotmm$ : $C = \frac{18 μ F\cdotmm}{\frac{1}{3} + ( 3 - 1 ) mm} = \frac{18}{\frac{1}{3} + 2} = \frac{18}{7/3} = \frac{54}{7} \approx 7.71 μ F .$ Why this step? $\frac{t}{κ} = \frac{1}{3} mm$ is the "electrical thickness" of the slab — it looks $κ$ times thinner to the field.

Verify: Bounds check — full slab would give $κ C_{0} = 18 μ F$ , no slab gives $6 μ F$ ; our $7.71$ sits between and near the low end because only $1/3$ of the gap is filled ✔.

Worked example Ex 7 — Case G · Two dielectrics: side-by-side vs stacked

Same plates, $C_{0} = 6 μ F$ empty-gap capacitance. Two materials $κ_{1} = 2$ , $κ_{2} = 4$ , each filling half the space. Do it two ways: (a) side-by-side (each covers half the area), (b) stacked (each fills half the thickness).

Figure — what it shows: two mini-capacitors drawn side by side. Left diagram ("side-by-side = PARALLEL"): the gap is split into a left white column ( $κ_{1} = 2$ ) and a right red column ( $κ_{2} = 4$ ), each spanning the full plate separation but only half the area. Right diagram ("stacked = SERIES"): the gap is split into a lower white layer ( $κ_{1} = 2$ ) and an upper red layer ( $κ_{2} = 4$ ), each spanning the full area but only half the thickness. Left = same voltage across both halves (parallel); right = same charge through both layers (series).

Forecast: Which arrangement gives the bigger capacitance — sharing area, or sharing thickness? Guess before computing.

(a) Side-by-side = parallel. Each half has area $A /2$ , full gap $d$ . $C_{a} = \frac{κ _{1} ε _{0} ( A /2 )}{d} + \frac{κ _{2} ε _{0} ( A /2 )}{d} = \frac{( κ _{1} + κ _{2} )}{2} C_{0}$ . Why this step? Same voltage across both halves, charges add — parallel. This is the left diagram.
$C_{a} = \frac{2 + 4}{2} \times 6 = 3 \times 6 = 18 μ F$ . (This is $C_{0}$ times the average $κ = 3$ .) Why this step? Parallel with equal areas literally averages the two $κ$ 's.
(b) Stacked = series. Each layer thickness $d /2$ , full area $A$ . $C_{i} = \frac{κ _{i} ε _{0} A}{d /2} = 2 κ_{i} C_{0}$ . Series: $\frac{1}{C _{b}} = \frac{1}{2 κ _{1} C _{0}} + \frac{1}{2 κ _{2} C _{0}}$ . Why this step? Same charge through both stacked layers, voltages add — series. This is the right diagram.
$\frac{1}{C _{b}} = \frac{1}{2 C _{0}} (\frac{1}{κ _{1}} + \frac{1}{κ _{2}})$ . The clean closed form of a two-capacitor series is "product over sum," so $C_{b} = 2 C_{0} \cdot \frac{κ _{1} κ _{2}}{κ _{1} + κ _{2}} = 2 (6) \cdot \frac{2 \cdot 4}{2 + 4} = 12 \cdot \frac{8}{6} = 16 μ F .$ Why this step? Series of two capacitors is "product over sum"; the effective $κ$ is the harmonic-type combination $\frac{2 κ _{1} κ _{2}}{κ _{1} + κ _{2}} = \frac{16}{6} \approx 2.67$ , and $C_{b} = κ_{eff} C_{0} = 2.67 \times 6 = 16 μ F$ .

Verify: Side-by-side ( $C_{a} = 18 μ F$ ) beats stacked ( $C_{b} = 16 μ F$ ). General truth: parallel (arithmetic mean of $κ$ , here $3$ ) $\geq$ series (harmonic-type mean, here $2.67$ ) ✔. Both exceed $C_{0} = 6 μ F$ and both lie below the all- $κ_{2}$ value $4 \times 6 = 24 μ F$ ✔. Physical read: putting the weaker dielectric in series drags the whole stack down toward its low value, which is why stacked loses to side-by-side.

Worked example Ex 8 — Case H · Real-world word problem

You must build a $C = 100 pF$ capacitor that holds off $V = 2000 V$ without sparking. Plate area is fixed at $A = 1.0 \times 1 0^{- 3} m^{2}$ . Air breaks down at $3 \times 1 0^{6} V/m$ ; a plastic with $κ = 3$ breaks down at $20 \times 1 0^{6} V/m$ . Can air do it? Find the plate gap $d$ with the plastic.

Forecast: The plastic has higher $κ$ and higher breakdown strength — which one actually saves you here?

Minimum gap so the field stays below air's limit: $d_{m i n} = \frac{V}{E _{max,air}} = \frac{2000}{3 \times 1 0 ^{6}} = 6.7 \times 1 0^{- 4} m$ . Why this step? $E = V / d$ ; to keep $E$ under breakdown you need $d$ large enough.
Capacitance air would give at that gap: $C_{air} = \frac{ε _{0} A}{d _{m i n}} = \frac{8.85 \times 1 0 ^{- 12} \times 1 0 ^{- 3}}{6.7 \times 1 0 ^{- 4}} = 1.33 \times 1 0^{- 11} F = 13.3 pF$ . Why this step? At the safe gap air can only reach $13.3 pF$ — far short of $100 pF$ . So plain air fails the spec.
With plastic, breakdown allows a smaller gap: $d = \frac{V}{E _{max,plastic}} = \frac{2000}{20 \times 1 0 ^{6}} = 1.0 \times 1 0^{- 4} m$ . Why this step? A higher breakdown field lets the field be more intense safely, so the plates may sit closer — a smaller $d$ means a bigger $C$ .
Resulting capacitance: $C = \frac{κ ε _{0} A}{d} = \frac{3 \times 8.85 \times 1 0 ^{- 12} \times 1 0 ^{- 3}}{1.0 \times 1 0 ^{- 4}} = 2.66 \times 1 0^{- 10} F = 265 pF \geq 100 pF$ ✔. Why this step? Both effects help — $κ$ multiplies $C$ by $3$ , and the smaller safe gap multiplies it again — and together they clear the target.

Verify: Air maxes at $13.3 pF$ (fails); plastic reaches $265 pF$ (passes the $100 pF$ target with margin) ✔. Lesson: it wasn't only $κ$ — the higher breakdown field was the decisive property. Units of step 4: $\frac{(F/m) \cdot m ^{2}}{m} = F$ ✔.

Worked example Ex 9 — Case I · Exam twist: the force that pulls the slab in

Disconnected capacitor of Ex 2 ( $Q = 200 μ C$ , $C_{0} = 2 μ F$ , $κ = 4$ ). We found $U$ dropped from $10 mJ$ to $2.5 mJ$ . Show this drop equals work done by the field, and hence the field pulls the slab in (not out).

Forecast: Systems fall toward lower energy. If inserting the slab lowers $U$ , which direction is the slab pulled?

Energy change on full insertion: $Δ U = U_{final} - U_{initial} = 2.5 - 10 = - 7.5 mJ$ . Why this step? We already computed both ends in Ex 2; the sign is the physics.
Work-energy: with $Q$ fixed and no battery, energy conservation gives $W_{by field on slab} = - Δ U = + 7.5 mJ > 0$ . Why this step? The stored energy that vanished went into mechanical work on the slab.
Positive work done by the field along the direction of motion ⇒ the force acts inward (dragging the slab into the gap). Why this step? $W = \int F d x > 0$ over inward motion means $F$ points inward.

Verify: Direction check via energy landscape — deeper insertion means higher $κ_{eff}$ , higher $C$ , lower $U = Q^{2} /2 C$ ; the system always slides toward more overlap ✔. Magnitude of released energy $7.5 mJ = 10 - 2.5$ ✔, consistent with Ex 2.

Recall Matrix self-test (cover the answers)

Battery ON, insert $κ$ : what is frozen? ::: The voltage $V$ (and hence $E = V / d$ ); $Q$ and $U$ rise by $κ$ . Battery OFF, insert $κ$ : what is frozen? ::: The charge $Q$ ; $V$ , $E$ , and $U$ fall (last one by factor $κ$ ). Slab fills only part of the gap — series or parallel? ::: Series (dielectric layer + air layer share the same $Q$ ). Two dielectrics splitting the area — series or parallel? ::: Parallel (same $V$ , charges add). Two dielectrics splitting the thickness — series or parallel? ::: Series (same $Q$ , voltages add). Limit $κ \to \infty$ models what? ::: A conductor filling the gap: $E \to 0$ , $C \to \infty$ . Disconnected slab insertion lowers $U$ — force direction? ::: Inward; the field pulls the slab in.

Connections

Capacitance and Capacitors
Electric Field of Parallel Plates
Energy Stored in a Capacitor
Electric Susceptibility
Gauss's Law and the D-field
Electric Dipoles and Torque