Year 1 Sem 2/LC Electromagnetism I/Lecture 10

Lecture 10: Electrostatics X: Dielectrics
Thu 19 Feb 2026 11:00

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Energy Density

Consider a parallel plate capactitor We have previously determined that:

$$U = \frac{1}{2}CV^2$$

And that:

$$C = \frac{Q}{V} = \frac{\epsilon_0 A}{d}$$

$$V = Ed$$

Hence:

$$U = \frac{1}{2} \frac{\epsilon_0 A}{d} E^2 d^2 = \frac{1}{2} A d \epsilon_0 E^2$$

Noting that $Ad$ is just the volume of the gap, we have that the energy per unit volume is given by:

$$\boxed{u_E = \frac{1}{2} \epsilon_0 E^2}$$

this gives the energy density of an electric field (for any E-field in a vacuum).

Force Per Capacitor Plates

Again for a parallel plate capacitor, we want to find the attractive force between the plates. Integrating coulombs law directly would be overly messy, so we use the definition:

$$U = - \int \pmb{F} \cdot d \pmb{x}$$

Where $x$ is the axis between the plates. Hence $F = -du/dx$:

$$U(x) = Ax \epsilon_0 E^2$$

$$F = - \frac{1}{2} A \epsilon_0 E^2$$

Using the definition:

$$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A \epsilon_0}$$

$$\implies F = - \frac{1}{2} A \epsilon_0 \left(\frac{Q}{A \epsilon_0}\right)^2$$

$$F = - \frac{1}{2} QE$$

And as the force is negatively signed, that confirm that it is attractive.

Dielectrics

What is a dielectric?

In a perfect conductor, there’s a large quantity of free charges. Under an applied E-field, these charges move rapidly along the direction of the field.

A dielectric is a poor conductor where free charge is negligible. When an E-field is applied, the charges in the atom are displaced, while remaining bound to the atom. This is called polarisation.

They are placed between the plates of a capacitor to keep the charged surfaces physically separated and to raise the capacitance. For EMI (and EMII) we only consider well behaved dielectrics which follow the acronym “HILS”

Homogeneous.
Isotropic, same in all directions hence $P \parallel E$.
Linear, $P \propto E$.
Stationary, no moving parts.

Polarisation

Polarisation is the effect of an E-field on a dielectric to pull the positively charged nucleus in one direction and the negatively charged electron cloud in the other. The electron cloud and the nucleus remain bound.

The E-field slightly pulls the nucleus in the direction of the E-field and the electron cloud against it. This creates a dipole moment in the direction of the E-field. This polarisation creates an E-field which acts in the opposite direction to the E-field, cancelling a small part of it.

All of these dipoles align and act together oppositely to the E-field. This reduces the field in the electric.

If the dielectric has $N$ dipoles, we define the total polarisation as:

$$\pmb{P} = N \pmb{p} = Nq \pmb{a}$$

The field inside the dielectric is less than the external applied field.

Consider a large dielectric block placed into a field. While the block remains neutral, the polarisation causes a small depletion region, approximately the width of the dipoles at each end of the block (width $a$). One end of the block becomes depleted of negative charges and one end positive charges.

The charge on the surface (surface area $S$) is given by:

$$\text{surface charge } = \text{ volume of depleted region } \times \text{ molecular density } \times \text{charge per molecule} = (aS)Nq$$

However we’ve already defined $P = Nqa$, so the surface charge on the RHS (for a right to left E-field) is:

$$q_r = PS$$

And the LHS:

$$q_l = - PS$$

Hence the surface charge density is given by $\sigma_i \pm P$.

If we add a dielectric to a capacitor between the plates, the dielectric surface with negative surface charge is in contact with the capacitor plate of positive surface charge. The two E-fields partially cancel and, while the net E-field still points in the same direction (positive plate to negative plate) it has a smaller magnitude.

Considering the positive plate, the net surface charge is $\sigma - \sigma_i$. With no dielectric, the E-field is:

$$E_0 = \frac{\sigma}{\epsilon_0}$$

With the dielectric:

$$E = \frac{\sigma - \sigma_i}{\epsilon_0} = \frac{\sigma - P}{\epsilon_0}$$

For a well-behaved dielectric, the polarisation is proportional to the electric field $P \propto E$, hence the polarisation must be some constant times the electric field:

$$\pmb{P} = \chi_E \epsilon_0 E$$

Where $\chi_E \epsilon_0$ is the constant of proportionality and $\epsilon_0$ is included to make sure that $\chi_E$ is dimensionless. $\chi_E$ is defined as the susceptibility. We therefore have:

$$\epsilon_0 E = \sigma - P$$

$$\epsilon_0 E = (\epsilon_0 E_0) - (\chi_E \epsilon_0 E)$$

$$\epsilon_0 E = \epsilon_0 E_0 - \chi_E \epsilon_0 E$$

$$\boxed{(1 + \chi_E) \epsilon_0 E = \epsilon_0 E_0}$$

We define another dimensionless constant, the relative permittivity (AKA the dielectric constant):

$$\epsilon_r = (1 + \chi_E)$$

$$\implies \epsilon_r \epsilon_0 E = \epsilon_0 E_0$$

The E-field in the dielectric is:

$$E = \frac{E_0}{\epsilon_r}$$

From Gauss’ law we have:

$$E = \frac{\sigma}{\epsilon_r \epsilon_0} = \frac{Q}{\epsilon_r \epsilon_0 A}$$

$$V = V_+ - V_- = Ed = \frac{Qd}{\epsilon_0 \epsilon_r}$$

$$\implies C = \frac{Q}{V} = \frac{\epsilon_0 \epsilon_r A}{d}$$

Since $\epsilon_r > 1$, this increases the capacitance. We can modify Gauss’ law for a well behaved dielectric:

$$\int_S \pmb{E} \cdot d \pmb{S} = \frac{Q_\text{enc}}{\epsilon_r \epsilon_0}$$

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