Dielectrics and polarisation

So far we have treated the capacitor with vacuum (or air) between the plates. Real capacitors — from the ceramic ones in electronic circuits to the large electrolytic ones in power supplies — have a dielectric between the plates: an insulating material (paper, plastic, ceramic, aluminium oxide) that modifies their properties in a surprising way.

Polarisation of the material

The molecules of a dielectric do not let free charges flow — it is an insulator — but under the effect of the plates’ field $\vec E_0$ their electron clouds shift slightly relative to the nuclei: each molecule becomes a small electric dipole oriented along $\vec E_0$. Some molecules are already polar by nature (for example water, $\mathrm{H_2O}$): in that case the field merely orients them.

The dielectric’s molecular dipoles align along the plates’ field $\vec E$; the induced charges accumulate on the faces facing the electrodes.

The collective result is that the dipoles’ charges accumulate on the faces of the dielectric near the plates: on the upper face induced negative charges appear (facing the positive plate), on the lower face positive charges. These polarisation charges produce a field $\vec E_\text{pol}$ opposite to $\vec E_0$. The total field in the dielectric is therefore reduced:

$E = E_0 - E_\text{pol} = \frac{E_0}{\varepsilon_r}$

where $\varepsilon_r \geq 1$ is the material’s relative dielectric constant, a pure number.

Principle — Capacitance with a dielectric

Inserting a dielectric of constant $\varepsilon_r$ between the plates of a parallel-plate capacitor of vacuum capacitance $C_0$ increases the capacitance by a factor $\varepsilon_r$: $\ev{C = \varepsilon_r\,C_0 = \frac{\varepsilon_r\,\varepsilon_0\,A}{D}}$

The weaker field, for the same charge, means a smaller potential difference $\Delta V$, and hence — since $C = Q/\Delta V$ — a larger capacitance. The dielectric “helps” the capacitor hold more charge per volt.

Key formula

$\begin{aligned} C &= \varepsilon_r C_0 \\ \varepsilon &= \varepsilon_r \varepsilon_0 \quad \text{(absolute permittivity)} \\ U &= \tfrac12 C V^2 \ \text{scales as } C \end{aligned}$

The values of $\varepsilon_r$ vary enormously from material to material:

Material	$\varepsilon_r$
Vacuum / air	$1{,}00$
Paper	$\sim 3$
Glass	$\sim 5$–$10$
Mica	$\sim 7$
Water (at $20\,^\circ$C)	$\sim 80$
Barium titanate	$\sim 1200$

Water, with $\varepsilon_r \approx 80$, is an extraordinary solvent for exactly this reason: reducing the Coulomb force between ions by a factor of $80$ is enough to break apart a crystal such as $\mathrm{NaCl}$, dissolving the salt.

Links

Topics: Electric field and potential Concepts: Dielectrics and polarisation · Capacitance and capacitor

Related exercises: Ranking the energy of capacitors · Dielectric in a capacitor · Ranking capacitors with different geometries