Dielectrics

Slide an insulator into the gap and the capacitance jumps. Almost every real capacitor uses one.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §26.5.

Before you start

What you need first

Topic 12 — the parallel-plate formula $C = \varepsilon_0 A/d$.
Topic 1 — insulators polarize in a field.

What you'll be able to do

Say what a dielectric does to capacitance.
Use $C = \kappa\,\varepsilon_0 A/d$.
Read a dielectric-constant ($\kappa$) table.
Explain why a dielectric boosts $C$.

The idea

A dielectric boosts capacitance

Slide an insulator — plastic, paper, glass — into the gap between the plates. This is a dielectric, and it makes the capacitance bigger by a factor $\kappa$ (kappa), the material's dielectric constant.

$$ C = \kappa\,\frac{\varepsilon_0 A}{d} $$

The slab polarizes: induced $-$ faces the $+$ plate, induced $+$ faces the $-$ plate.

where:
Symbol	Meaning	SI unit
κ	dielectric constant of the material (a pure number $\ge 1$)	—
A, d	plate area and gap (as before)	m², m
ε₀	permittivity of free space $=8.85\times10^{-12}$	C²/(N·m²)

Some dielectric constants

Material	$\kappa$ (about)
Vacuum / air	1 (no boost)
Paper	3.7
Glass	5
Water	80

$\kappa$ has no units. Paper more than triples the capacitance; water multiplies it by about 80. Air is so close to 1 that an "air-gap" capacitor is treated as vacuum.

Why a dielectric raises C

The plates' field polarizes the dielectric (Topic 1): its molecules line up, leaving a layer of induced $-$ charge facing the $+$ plate and induced $+$ facing the $-$ plate.

Those induced charges set up a field opposing the plates' field, so the net field — and hence the voltage $V = Ed$ — drops for the same stored charge $Q$. Since $C = Q/V$, a smaller $V$ at the same $Q$ means a bigger $C$.

📐 Worked example 1

Capacitor with a dielectric

Plates of area $A = 0.010$ m² sit $d = 0.0020$ m apart, filled with plastic of $\kappa = 2.0$. Find the capacitance.

1Use $C = \kappa\,\varepsilon_0 A/d$:

$$ C = (2.0)\frac{(8.85\times10^{-12})(0.010)}{0.0020} $$

2Work it out:

$$ C \approx 8.9\times10^{-11}\ \text{F} = 89\ \text{pF} $$

The $\kappa = 2$ doubled what an air gap would give.

📐 Worked example 2

Filling an air capacitor

An air-gap capacitor has $C_0 = 50$ pF. Glass ($\kappa = 5$) is slid into the gap, filling it. Find the new capacitance.

1The dielectric just multiplies by $\kappa$:

$$ C = \kappa\,C_0 = (5)(50\ \text{pF}) = 250\ \text{pF} $$

✏️ Try it yourself

A capacitor has plates of area $A = 0.020$ m² with a gap $d = 0.0010$ m, filled with paper ($\kappa = 3.7$).

(a) Find its capacitance.
(b) How many times bigger is this than the same capacitor with just air?

(a) $C = \kappa\dfrac{\varepsilon_0 A}{d} = (3.7)\dfrac{(8.85\times10^{-12})(0.020)}{0.0010} \approx 6.5\times10^{-10}\ \text{F} = 650\ \text{pF}.$ (b) Exactly $\kappa = 3.7$ times bigger — the air-filled value would be about $180$ pF.

Common mistakes

Mistake	Fix
Giving $\kappa$ units.	It is a pure number (a ratio), always $\ge 1$.
Thinking a dielectric lowers $C$.	It raises $C$ by the factor $\kappa$.
Using $\kappa$ for an air gap.	Air is $\kappa \approx 1$ — no boost.
Forgetting why: it's polarization.	Induced charges oppose the field, dropping $V$ and so raising $C = Q/V$.

Recap — the whole topic on one screen

Idea	What you own now
Effect	A dielectric multiplies capacitance by $\kappa$.
Formula	$C = \kappa\,\dfrac{\varepsilon_0 A}{d}$; $\kappa$ dimensionless $\ge 1$.
Why	Polarization sets up an opposing field → lower $V$ → higher $C$.

Next topic · new theme

Theme 3 — Current & DC circuits

So far charge has sat still. Next it flows: we meet electric current, resistance, and the circuits that put capacitors, batteries, and resistors to work — the heart of electronics, and the last stop before the midterm.

→ Topic 16 · Electric current & drift speed

Mistake	Fix
Giving \(\kappa\) units.	It is a pure number (a ratio), always \(\ge 1\).
Thinking a dielectric lowers \(C\).	It raises \(C\) by the factor \(\kappa\).
Using \(\kappa\) for an air gap.	Air is \(\kappa \approx 1\) — no boost.
Forgetting why: it's polarization.	Induced charges oppose the field, dropping \(V\) and so raising \(C = Q/V\).

Idea	What you own now
Effect	A dielectric multiplies capacitance by \(\kappa\).
Formula	\(C = \kappa\,\dfrac{\varepsilon_0 A}{d}\); \(\kappa\) dimensionless \(\ge 1\).
Why	Polarization sets up an opposing field → lower \(V\) → higher \(C\).