Capacitors with Dielectrics

Slide an insulator between the plates of a capacitor and its capacitance goes up by a factor K. This lesson shows why, and what it does to Gauss’s law and to stored energy.

Filling the gap#

A parallel-plate capacitor (area A, gap d) carries ±Q and its gap is filled with a dielectric of constant K. The plate charges alone would make E₀ = Q/ε₀A. The dielectric cuts this to E = E₀/K = Q/ε₀AK, so the potential difference is V = Ed = Qd/ε₀AK and

This is often used as the definition of K: the factor by which a material multiplies the capacitance when it fills the gap.

How much bound charge?#

The bound charges ±Q_p on the slab faces make a field Q_p/ε₀A opposing E₀, so E = (Q − Q_p)/ε₀A. Setting this equal to Q/ε₀AK:

Gauss’s law with dielectrics#

Draw a Gaussian pillbox around the + plate’s inner face in a filled capacitor. It encloses the free charge Q and bound charge −Q_p, so ∮E⃗·dS⃗ = (Q − Q_p)/ε₀ = Q/ε₀K. Moving K to the left:

This holds for any homogeneous, isotropic dielectric. A version valid for every material uses the polarization. The bound charges make E⃗_p = −P⃗/ε₀, so ε₀E⃗ + P⃗ = ε₀E⃗₀, and since E⃗₀ comes from the free charge alone:

A point charge inside a dielectric#

Surround a charge q in a large dielectric with a sphere of radius r. Using (31.14), KE · 4πr² = q/ε₀:

The field is K times weaker than in vacuum. Physically, the medium’s negative ends crowd round a positive q, putting bound charge −q(1 − 1/K) on the tiny cavity around it, so the field outside sees only q/K.

Energy in a dielectric#

For a filled parallel-plate capacitor, U = ½CV² = ½(Kε₀A/d)V², spread through volume Ad:

For the same field, a dielectric stores K times more energy per volume than vacuum: the interacting dipoles hold the extra.