Law

Dielectric constant

(E4.5.1) E4.5 · Capacitors with Dielectrics →

C = κ C_{0}

Statement

The capacitance rises by the factor κ ≥ 1 — paper about 3.7, water about 80.

Symbols and units

$κ$	dielectric constant of the filling	—
$C_{0}$	capacitance with vacuum in the gap	F
$C$	capacitance with the dielectric	F

When it applies

Linear dielectric, operated below its breakdown field.

Don't use when

Only for a linear dielectric operated below its breakdown field — exceeding the breakdown field causes the dielectric to conduct or fail.

Derivation

Keep the charge Q fixed and slide the slab in. Measurement shows the voltage drops by a fixed factor:

Δ V = \frac{Δ V _{0}}{κ}

The reason: the material polarizes, and its induced surface charge sets up a field opposing the original one:

E = \frac{E _{0}}{κ}

With Q unchanged and ΔV smaller, the capacitance must rise:

C = \frac{Q}{Δ V} = \frac{Q κ}{Δ V _{0}} = κ C_{0}

So for parallel plates:

C = \frac{κ ε _{0} A}{d}

Watch out for

Two different answers depending on the wiring. Battery disconnected (Q fixed): ΔV falls, stored energy falls, and the slab is pulled in. Battery connected (ΔV fixed): Q rises and stored energy rises.

In practice

A dielectric also raises the breakdown voltage and gives the plates mechanical support. Atomically: polar molecules rotate into line with the field, non-polar ones stretch into small dipoles; either way, induced charge layers appear on the slab faces and set up the opposing field.

Connects to

Parallel plates Torque and energy Stored energy