Dielectrics and Polarisation

A dielectric is an insulator placed in an electric field. Unlike a conductor, its charges cannot flow freely — but they can shift slightly, producing an internal polarisation that opposes the external field. This is why dielectrics increase the capacitance of capacitors.

Concept

Non-polar molecules ($CO_2, H_2$) have coincident centres of positive and negative charge in their natural state. An external field stretches them slightly, creating an induced dipole moment $\vec p = \alpha \vec E$, where $\alpha$ is the molecular polarisability.

Polar molecules ($H_2O, HCl$) have permanent dipole moments due to asymmetric charge distribution. In zero external field, thermal motion randomly orients them, giving zero net polarisation. In a field, they tend to align with $\vec E$, but thermal jostling fights alignment — the equilibrium average gives a net polarisation along $\vec E$.

Polarisation $\vec P$ is the dipole moment per unit volume. For a uniform dielectric slab in a uniform external field $\vec E_0$, the polarisation creates bound surface charges $\pm\sigma_b$ on the faces, which set up an opposing field $\vec E_p$ inside. The net field inside is $\vec E = \vec E_0 - \vec E_p < \vec E_0.$

Dielectric constant $K$ (also written $\varepsilon_r$): $K = \frac{\vec E_0}{\vec E} = \frac{C_\text{with dielectric}}{C_\text{vacuum}}.$ $K \geq 1$ always; $K = 1$ for vacuum, $\approx 1.0006$ for air, $\approx 80$ for water, very high for ferroelectrics.

Derivation

Consider a parallel-plate capacitor with plate area $A$, separation $d$, and free surface charge density $\sigma_f$ on the plates. Without dielectric: $E_0 = \sigma_f/\varepsilon_0.$

Insert a dielectric. Bound charges $\pm\sigma_b$ appear on the dielectric surfaces. The net surface charge facing across the gap is $\sigma_f - \sigma_b$, so the field inside the dielectric is $E = \frac{\sigma_f - \sigma_b}{\varepsilon_0} = \frac{\sigma_f}{K\varepsilon_0} = \frac{E_0}{K}.$ Capacitance becomes $C = \frac{\sigma_f A}{V} = \frac{\sigma_f A}{E d} = \frac{K \varepsilon_0 A}{d} = K C_0.$

Worked Example

A parallel-plate capacitor has $C_0 = 5\,\mu F$ in air. Inserted a slab of mica ($K = 6$) filling the gap entirely. New capacitance: $C = K C_0 = 6\times 5 = 30\,\mu F.$

If the capacitor is connected to a $12\,V$ source, charge on plates rises from $5\times 12 = 60\,\mu C$ to $30\times 12 = 360\,\mu C$.

If instead the capacitor is disconnected before inserting the dielectric, charge is fixed; voltage drops from $12$ to $12/K = 2\,V$, and field drops by factor of $K$.

Common Confusions

• Dielectric reduces field, increases capacitance. Both are due to the same polarisation.
• Bound charges are not free to move. They appear as a result of microscopic dipole alignment.
• $K$ is dimensionless and $\geq 1$. Vacuum has $K = 1$ exactly.
• Charge on plates with battery connected (constant V) increases. Charge on isolated plates (constant Q) stays — voltage decreases.

Key Takeaways

• Non-polar molecules acquire induced dipoles in $\vec E$; polar molecules align permanent dipoles.
• Polarisation $\vec P$ = dipole moment per volume.
• Net field inside dielectric: $E = E_0/K$.
• Capacitance increases by factor $K$: $C = KC_0$.
• $K$ ranges from 1 (vacuum) to several thousand (ferroelectrics).