Dielectrics — polarization, dielectric constant, effect on capacitance: Visual walkthrough

We assume only this much from before: a field $E$ is an arrow at every point that says "which way, and how hard, a $+ 1$ charge is pushed" (see Electric Field of Parallel Plates), and a capacitor stores charge $Q$ at voltage $V$ with $C = Q / V$ (see Capacitance and Capacitors). Everything else, we grow from scratch.

Definition Symbol glossary (everything used on this page)

Skim once, refer back as needed. Symbols are defined again in the step where they first do work.

$Q$ — charge on a plate (coulombs, C).
$A$ — area of one plate (m²).
$d$ — the gap between the two plates (m). When the slab fully fills the gap, its thickness $t$ equals $d$ — we assume this full-fill case throughout (Step 9 handles partial fill).
$σ_{f}$ — free surface charge density on the plates, $= Q / A$ (C/m²).
$σ_{b}$ — bound surface charge density on the slab faces (C/m²).
$ε_{0}$ — permittivity of free space, $8.85 \times 1 0^{- 12}$ C²/(N·m²) — the "exchange rate" from charge-per-area to field.
$E_{0}$ — the vacuum field with no slab; $V_{0} = E_{0} d$ is the vacuum voltage; $C_{0} = Q / V_{0} = ε_{0} A / d$ is the vacuum capacitance.
$E, V, C$ — the same quantities with the slab in place.
$p, P$ — one molecule's dipole moment, and dipole moment per volume.
$χ_{e}$ — electric susceptibility (dimensionless); $κ = 1 + χ_{e}$ — dielectric constant (dimensionless).

Step 1 — Two bare plates: the field we start with

WHAT. Put charge $+ Q$ on the left plate and $- Q$ on the right, area $A$ each, separated by a gap $d$ . Spread out, the charge is a surface charge density $σ_{f} = \frac{Q}{A} .$

$σ$ (sigma) = charge packed per square metre, units $C/m^{2}$ .
The subscript $f$ means free charge — the charge a battery actually delivered onto the metal. We call it "free" because it can move through the wire; remember the word, a different kind of charge is coming.
$d$ = the gap the field crosses; we'll need it in Step 8 to turn field into voltage.

WHY start here. Before any insulator exists there is only this. Its field is our reference — the thing the dielectric will later fight.

PICTURE. Uniform blue arrows march straight from $+$ plate to $-$ plate. The field of one charged sheet is $σ /2 ε_{0}$ ; two opposite sheets add between the plates (see Electric Field of Parallel Plates), giving $E_{0} = \frac{σ _{f}}{ε _{0}} .$

$E_{0}$ = the vacuum field, subscript $0$ = "empty space, nothing inserted."
$ε_{0}$ (epsilon-nought) = a fixed constant of nature, $8.85 \times 1 0^{- 12}$ in SI units. It converts "charge per area" into "field strength." Think of it as the exchange rate between charge-crowding and push.

Figure — Dielectrics — polarization, dielectric constant, effect on capacitance

Step 2 — Slide in a slab of neutral matter

WHAT. We push an insulating slab (glass, plastic, water in a bag) fully between the plates, so its thickness $t$ equals the gap $d$ . It is electrically neutral everywhere — same amount of $+$ and $-$ inside, so at first it changes nothing.

WHY. We need to know what the external field $E_{0}$ does to matter. Nothing in the metal changes; the story is now entirely about what happens inside the slab.

PICTURE. Draw the slab as a translucent block. Inside it, sketch the molecules as tiny neutral circles: a $+$ core and a $-$ cloud sitting right on top of each other, so each little unit has zero net displacement — no arrows yet.

Step 3 — The field stretches each molecule into a dipole

WHAT. The blue field $E_{0}$ pushes every $+$ core a hair to the right and pulls every $-$ cloud a hair to the left. Each molecule becomes a dipole: a $+$ and $-$ separated by a tiny distance $δ$ .

WHY this idea, and why "dipole"? A field pushes opposite charges opposite ways — that is the definition of a field. When you separate $+ q$ and $- q$ by $δ$ , physicists bottle that into one number, the dipole moment $p = q δ,$ because the pair's whole effect on the outside world depends only on this product, not on $q$ and $δ$ separately. (This is exactly the object that gets torqued in Electric Dipoles and Torque.)

$q$ = the charge that shifted, $δ$ = how far and which way $+$ moved relative to $-$ (we use $δ$ , not $d$ , to keep it distinct from the plate gap).
$p$ points from $-$ to $+$ , i.e. along the field here.

PICTURE. Same molecules as Step 2, now stretched: red $+$ nudged right, blue $-$ nudged left, a little green arrow $p$ on each pointing right.

Step 4 — Add up the dipoles: polarization $P$

WHAT. Billions of these tiny $p$ vectors, all pointing the same way, are hard to track one by one. So we sum them and divide by the volume they fill. That density is the polarization: $P = \frac{\sum p _{i}}{Volume} .$

WHY divide by volume? Because a fat slab has more molecules than a thin one but is not "more polarized" per bit of material. Dividing by volume gives an intensive quantity — a property of the material-in-this-field, independent of how much slab you grabbed.

Units check (this is the crucial bridge). $p$ has units $C \cdot m$ ; volume is $m^{3}$ ; so $[P] = \frac{C \cdot m}{m ^{3}} = \frac{C}{m ^{2}} .$ That is a charge per area — the same units as $σ$ . This is not a coincidence; Step 5 shows why.

PICTURE. Zoom out: the individual arrows blur into one smooth green field $P$ filling the slab, pointing right, same units-badge $C/m^{2}$ shown in the corner.

Step 5 — The great cancellation: bound surface charge

WHAT. Line up all the stretched molecules in a row. The $+$ end of one molecule sits right next to the $-$ end of its neighbour — they cancel. Everywhere inside, plus meets minus and the net charge is zero. Only at the two faces is there no neighbour to cancel with: the left face is left with exposed $-$ , the right face with exposed $+$ . Call that leftover bound surface charge $σ_{b}$ .

Bound charge $σ_{b}$ : real charge, but stuck to the molecules — it cannot run off through a wire. Opposite of "free."

Sign vs magnitude — the clean rule. The full statement is the vector one, $σ_{b} = P \cdot \overset{n}{^},$ where $\overset{n}{^}$ is the outward-pointing unit arrow on a face. The dot product carries the sign: on the right face $P$ points out ( $P \cdot \overset{n}{^} = + P$ , positive charge); on the left face $P$ points in ( $P \cdot \overset{n}{^} = - P$ , negative charge). When we later just write " $σ_{b} = P$ " we mean the magnitude on a face where $P$ is perpendicular; the sign is set by which face. Keep the vector form as the truth and $σ_{b} = P$ as its magnitude on our two flat faces.

Why the interior stays neutral. The general rule for volume bound charge is $ρ_{b} = - \nabla \cdot P$ (how much $P$ "spreads out or piles up" at a point). Our $P$ is uniform inside the slab, so $\nabla \cdot P = 0$ and $ρ_{b} = 0$ — no bulk charge, exactly the pairwise cancellation we drew. All the leftover charge is squeezed onto the two faces.

WHY $σ_{b}$ equals $P$ in magnitude. Two honest ways to write the slab's total dipole moment must agree. Way 1 (density $\times$ volume): $P \cdot (A t)$ , where $A$ is face area, $t$ is thickness. Way 2 (charge $\times$ its lever arm): the exposed charge $Q_{b} = σ_{b} A$ on each face, separated by the slab thickness $t$ , is itself one giant dipole: $Q_{b} \cdot t = σ_{b} A t$ . Set them equal: $P (A t) = σ_{b} A t ⟹ σ_{b} = P .$ The $A$ and $t$ cancel — geometry does all the work.

PICTURE. A row of dipoles inside a rectangle; every interior $\pm$ pair struck through (cancelled) in grey; the leftmost $-$ and rightmost $+$ circled in red/blue — those are $σ_{b}$ .

Step 6 — The bound charge fights back: the net field shrinks

WHAT. Now two sheets of charge live at the slab's faces: bound $- σ_{b}$ hugging the $+$ plate, bound $+ σ_{b}$ hugging the $-$ plate. Each sheet makes its own field $E_{b} = σ_{b} / ε_{0}$ pointing opposite to $E_{0}$ (from the bound $+$ toward the bound $-$ , i.e. leftward). The real field inside is the tug-of-war winner: $E = E_{0} - E_{b} = \frac{σ _{f} - σ _{b}}{ε _{0}} .$

$E_{0}$ = original push (right), $E_{b}$ = the dielectric's counter-push (left).
Subtraction, not addition, because they point opposite ways — that single minus sign is the whole "dielectrics weaken the field" story.

WHY the bound sheet opposes. The $+$ plate attracted every molecule's $-$ end, so the face nearest the $+$ plate is negative. A negative sheet next to a positive plate makes a field pointing back toward the plate — against $E_{0}$ . It must oppose; that is geometry, not luck.

PICTURE. The strong blue $E_{0}$ arrows, a shorter red $E_{b}$ arrow pointing the other way at each spot, and the surviving green net arrow $E$ — visibly shorter than $E_{0}$ .

Step 7 — Close the loop: $E$ shrinks by exactly $κ$

WHAT. For our linear dielectric (see the assumption box), the harder you push, the more it polarizes — proportionally: $P = ε_{0} χ_{e} E .$

$χ_{e}$ (chi, "kai") = electric susceptibility — the material's eagerness to polarize (see Electric Susceptibility). It is dimensionless (units of $ε_{0} E$ already match $P$ 's C/m², so $χ_{e}$ carries no units). Big $χ_{e}$ = stretches a lot; $χ_{e} = 0$ = vacuum. It links to the more common permittivity language by $ε = ε_{0} (1 + χ_{e}) = ε_{0} κ$ , where $ε$ is the material's permittivity and $ε_{r} = ε / ε_{0} = κ$ is the "relative permittivity."

Feed $σ_{b} = P = ε_{0} χ_{e} E$ back into Step 6: $E = \frac{σ _{f} - ε _{0} χ _{e} E}{ε _{0}} = E_{0} - χ_{e} E .$ Now solve for $E$ . WHY the next line: $E$ appears on both sides, so we add $χ_{e} E$ to both sides to collect every $E$ on the left — the standard "get the unknown alone" move: $E + χ_{e} E = E_{0} \Rightarrow E (1 + χ_{e}) = E_{0} \Rightarrow E = \frac{E _{0}}{1 + χ _{e}} = \frac{E _{0}}{κ} .$

WHY this is the finish line. We define $κ = 1 + χ_{e},$ so the messy susceptibility collapses into one dimensionless number: the factor by which the field shrank. $κ = 1$ means "no eagerness, vacuum"; big $κ$ (water $\approx 80$ ) means "field almost wiped out."

PICTURE. Side-by-side: left panel tall blue arrow $E_{0}$ ; right panel the same arrow scaled down to $E_{0} / κ$ , with a $\div κ$ label bridging them.

Step 8 — From weaker field to bigger capacitance

WHAT. Voltage between plates is field $\times$ gap: $V = E d$ , where $d$ is the plate separation (from Electric Field of Parallel Plates). Before the slab, the vacuum voltage is $V_{0} = E_{0} d,$ and the vacuum capacitance is $C_{0} = \frac{Q}{V _{0}} = \frac{Q}{E _{0} d} = \frac{Q}{( σ _{f} / ε _{0} ) d} = \frac{ε _{0} A}{d} .$ Now insert the slab with the charge $Q$ held fixed (battery unplugged). Using $E = E_{0} / κ$ from Step 7: $V = E d = \frac{E _{0}}{κ} d = \frac{E _{0} d}{κ} = \frac{V _{0}}{κ} .$ Same charge, but the voltage across it is $κ$ times smaller. Capacitance is charge-per-volt: $C = \frac{Q}{V} = \frac{Q}{V _{0} / κ} = κ \frac{Q}{V _{0}} = κ C_{0} .$

WHY it rises. $C$ measures "charge you can store per volt of strain." The dielectric relieves the strain (lowers $V$ ) for free, so each volt now holds $κ$ times more charge.

PICTURE. Two bar-meters: $V$ bar drops to $1/ κ$ its height; the $C$ bar grows to $κ$ its height. Same $Q$ label sits on both.

Step 9 — The edge cases you must not skip

Every scenario the reader could hit, shown so none is a surprise.

Vacuum / air ( $χ_{e} = 0$ ): no molecules stretch, $σ_{b} = 0$ , $E = E_{0}$ , $κ = 1$ , $C = C_{0}$ . The whole derivation quietly reduces to the bare-plate case — good, it must.
Battery kept connected ( $V$ fixed): now $V$ , hence $E = V / d$ , is forced constant by the battery. The field does not drop. Instead $Q = C V = κ C_{0} V$ rises by $κ$ ; the battery pumps extra free charge to feed the newly-hungry capacitor (energy story in Energy Stored in a Capacitor).
Can $κ < 1$ ? No. $χ_{e} \geq 0$ for ordinary matter (dipoles align with the field, never against), so $κ = 1 + χ_{e} \geq 1$ . The field can only weaken, never strengthen.
Slab thinner than the gap (partial fill, $t < d$ ): only the filled fraction gets $E_{0} / κ$ ; the empty part keeps $E_{0}$ . The capacitor behaves like two capacitors in series — beyond this page, but note it is not just $κ C_{0}$ . This is why we assumed the full-fill case $t = d$ everywhere above.

The one-picture summary

The full causal chain, left to right: free charge makes $E_{0}$ → $E_{0}$ stretches molecules into $p$ → summed, that is $P$ → $P$ leaves bound $σ_{b}$ on the faces → $σ_{b}$ opposes $E_{0}$ , shrinking $E$ by $κ$ → smaller $E$ means smaller $V$ means bigger $C$ .

Recall Feynman retelling of the whole walkthrough

Start with two charged plates a gap $d$ apart: arrows shoot straight across (Step 1). Now slide in a neutral block that just fills the gap (Step 2). The arrows grab every molecule and stretch it — plus end this way, minus end that way — making millions of tiny see-saws all tipped the same direction (Step 3). Blur those tiny tips together and you get one smooth "amount-of-tipping-per-cubic-metre" field called $P$ (Step 4). Inside the block, every plus touches a neighbour's minus and they cancel; only the two outer faces are left with naked charge (Step 5). Those naked face-charges sit with the wrong sign next to each plate, so they push back and partly cancel the original arrows (Step 6). The more you push, the more they tip and push back, and when you solve that tug-of-war the arrows come out exactly $κ$ times shorter (Step 7). Shorter arrows means less voltage strain across the same gap, so each volt now holds $κ$ times more charge — the capacitance $C = Q / V$ is $κ$ times bigger (Step 8). Turn the eagerness off ( $κ = 1$ ) and everything snaps back to bare plates; keep the battery plugged in and the field stays put while extra charge floods in instead (Step 9). One block of glass, one number $κ$ , and it all follows.

Recall

Where does the minus sign in $E = E_{0} - E_{b}$ come from? ::: The bound charge sheet sits with the opposite sign next to each plate, so its field points against $E_{0}$ . Why does $σ_{b} = P$ (in magnitude) exactly? ::: Equating "dipole density × volume" with "exposed face charge × slab thickness" — the area and thickness cancel. Why is there no bound charge in the bulk? ::: $ρ_{b} = - \nabla \cdot P = 0$ because $P$ is uniform inside the slab. Why does $C$ rise even though we added an insulator? ::: The dielectric lowers the voltage strain for the same charge, so each volt stores $κ$ times more charge. Battery connected — what changes instead of $E$ ? ::: $Q$ (and stored energy) rises by $κ$ ; $E$ and $V$ are pinned by the battery. What is $C_{0}$ for a parallel plate? ::: $C_{0} = ε_{0} A / d$ .

Connections

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