Dielectrics — polarization, dielectric constant, effect on capacitance: Question bank

Quick symbol reminder so nothing here is used unexplained:

$σ_{f}$ = free surface charge (what the battery puts on the metal plates).
$σ_{b}$ = bound surface charge (the exposed, uncancelled charge on the dielectric's faces).
$E_{0} = σ_{f} / ε_{0}$ = field from free charge alone (vacuum value).
$E$ = actual net field inside the dielectric; $κ = E_{0} / E \geq 1$ is the dielectric constant.
" $Q$ fixed" ⇔ battery disconnected; " $V$ fixed" ⇔ battery still connected.

True or false — justify

Bound charge on a dielectric face is always opposite in sign to the nearby plate's free charge.

True — the plate's field pulls each molecule's opposite-sign end toward it, so the face facing the

+

plate exposes

-

charge. That opposition is exactly why the net field shrinks.

A perfect vacuum has

κ = 1

True — with no matter there are no dipoles, so

χ_{e} = 0

and

κ = 1 + χ_{e} = 1

. This is the floor value; nothing ordinary drops below it.

Inserting a dielectric can make

κ < 1

if the material resists polarizing.

False — resisting polarization just means small

χ_{e}

, and

χ_{e} \geq 0

, so

κ = 1 + χ_{e} \geq 1

. "Resistance" pushes

κ

toward

1

, never below it.

With the battery disconnected, inserting a dielectric lowers the stored energy.

True —

Q

is fixed and

U = Q^{2} /2 C

; since

C

rises by

κ

U

falls by

κ

. The missing energy became work done by the field pulling the slab in.

With the battery connected, inserting a dielectric lowers the stored energy.

False —

V

is fixed and

U = \frac{1}{2} C V^{2}

;

C

rises by

κ

U

rises by

κ

. The battery supplies the extra energy (plus more that goes to heat/work).

The field inside the dielectric is unchanged when the battery stays connected.

True — the battery pins

V

, and

E = V / d

with

d

fixed, so

E

can't move; instead

Q

climbs to

κ Q_{0}

Bound charge contributes to Gauss's law for

E

True — Gauss's law for

E

counts all charge, free and bound. The convenience of

D = ε_{0} κ E

is that

D

responds to free charge only, letting you skip

σ_{b}

— see Gauss's Law and the D-field.

Polarization

P

has the same units as a surface charge density.

True — dipole moment per volume is

(C\cdotm) / m^{3} = C / m^{2}

, which is why aligned dipoles literally look like a sheet of charge

σ_{b} = P

on each face.

A polar molecule like water polarizes by stretching, just like a non-polar one.

False — water already has a permanent dipole; the field mainly rotates it into partial alignment (torque, see Electric Dipoles and Torque), whereas non-polar molecules gain an induced dipole by charge-cloud stretching.

Spot the error

"I added a dielectric, so I added matter, so the field inside must be stronger."

The added matter contributes bound charge that opposes the free charge, so the net field drops by

κ

. More matter here means more opposition, not more field.

"With the battery connected I used

E = E_{0} / κ

to get the new field."

Wrong regime — the

E_{0} / κ

rule assumes

Q

(hence

σ_{f}

) is fixed. With a battery,

V

and therefore

E = V / d

are pinned; it is

Q

(and

σ_{f}

) that grows by

κ

"Bound charge is real charge on the surface, so I can wire it up and drain it."

Bound charge cannot flow — it is the exposed end of locked-in molecular dipoles, not free carriers. You cannot conduct it away; remove the field and it vanishes.

σ_{b} = σ_{f}

because bound charge cancels the free charge."

It only partially cancels:

σ_{b} = σ_{f} (1 - 1/ κ)

, which is always less than

σ_{f}

. If they were equal the net field would be zero, which happens only in the

κ \to \infty

(conductor) limit.

"Energy went down when I inserted the slab, so energy was destroyed."

Energy is conserved — the drop

Δ U

equals the mechanical work the field did sucking the slab in. It converted stored field energy into kinetic/mechanical energy, it didn't vanish.

κ = 1 + χ_{e}

, so a bigger field gives a bigger

κ

For a linear dielectric

χ_{e}

(and hence

κ

) is a material property independent of field strength.

P

scales with

E

, but their ratio

χ_{e}

stays fixed.

"I'll compute the capacitor's field using the total charge

σ_{f} + σ_{b}

and

ε_{0}

The net field is

E = (σ_{f} - σ_{b}) / ε_{0}

— bound and free charge on the two faces have opposite signs, so they subtract, not add. Adding them double-counts and reverses the physics.

Why questions

Why does inserting a dielectric let a capacitor store more charge at the same voltage?

The bound charge cancels part of the field, so for a given

V = E d

you can now pile on

κ

times more free charge before the field (and voltage) reach the same value — that extra packing is higher capacitance, per Capacitance and Capacitors.

Why is the relation written

σ_{b} = P \cdot \overset{n}{^}

rather than just

σ_{b} = P

The exposed charge depends on how the polarization pokes through the face; only the component along the outward normal

\overset{n}{^}

leaves charge on that face. On a slab whose faces are perpendicular to

P

the dot product reduces to

\pm P

Why does the net field drop by exactly the factor

κ

and not some other number?

Because

P = ε_{0} χ_{e} E

makes the opposition proportional to the field itself; solving

E = E_{0} - χ_{e} E

gives

E (1 + χ_{e}) = E_{0}

, and

1 + χ_{e}

is defined as

κ

. The self-consistency of "field causes polarization which reduces field" bakes in the single factor.

Why can we treat the aligned dipoles' interior as electrically neutral?

Inside the slab, the

+

end of each molecule sits against the

-

end of its neighbour, so interior charges cancel pairwise. Only the two outer faces are left with uncancelled charge.

Why does the parallel-plate formula gain a

κ

but keep

A

and

d

Geometry (

A

d

) is untouched by filling the gap; only the medium changed, and the medium enters solely through

κ

, giving

C = κ ε_{0} A / d

(compare the vacuum case in Electric Field of Parallel Plates).

Why does a strong dielectric like water (

κ \approx 80

) barely leave any field inside?

A large

χ_{e}

means the dipoles align strongly and produce nearly as much bound charge as the free charge, so

E = E_{0} / κ

is tiny — most of the free-charge field is neutralised, per Electric Susceptibility.

Edge cases

What happens to

κ

and to the field as the material approaches a perfect conductor (

χ_{e} \to \infty

κ \to \infty

, so

E = E_{0} / κ \to 0

— the bound charge fully cancels the free charge, exactly like the zero interior field of a conductor. A conductor is the limiting "infinitely polarizable" dielectric.

If the external field is zero, what is the polarization of a polar dielectric like water?

Zero on average — the permanent dipoles point in random directions and cancel, so

P = 0

despite each molecule carrying a dipole. It takes a field's torque to produce net alignment.

A dielectric only half-fills the gap (a slab of thickness

t < d

). Is

E = E_{0} / κ

everywhere in the gap?

No — the reduced field

E_{0} / κ

holds only inside the slab; in the remaining vacuum the field is still

E_{0}

. The two regions are in series and must be handled separately.

What is

σ_{b}

in the trivial case

κ = 1

(vacuum or air)?

σ_{b} = σ_{f} (1 - 1/ κ) = σ_{f} (1 - 1) = 0

— no polarizable matter means no bound charge and no field reduction, consistent with

E = E_{0}

Battery disconnected: you pull the dielectric back out. Does the stored energy go up or down?

Up — removing the slab drops

C

, and with

Q

fixed

U = Q^{2} /2 C

rises. You had to do work against the field's inward pull to extract it, and that work became stored energy.

Two identical capacitors, one filled with

κ = 2

, both charged to the same

Q

. Which has the larger field inside?

The empty one — with

Q

(hence

σ_{f}

) fixed, the filled capacitor's field is halved to

E_{0} /2

by its bound charge. Same charge, weaker field, because the dielectric opposes it.

If a dielectric's

χ_{e}

were negative (a hypothetical "diamagnetic-like" dielectric), what would happen to

E

κ = 1 + χ_{e} < 1

would give

E > E_{0}

— the field would increase. Ordinary dielectrics never do this because inducing dipoles always costs energy in a way that opposes, not reinforces, the field.

Recall One-line self-test before you leave

The single question that untangles most traps ::: "What is being held fixed — the charge $Q$ (battery off) or the voltage $V$ (battery on)?" Everything else follows from that.

Connections

Hinglish parent
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