Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 10: Electrostatics X: Dielectrics</b><br>
Thu 19 Feb 2026 11:00
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[[Year 1 Sem 2/LC Electromagnetism I/Lecture 09|← Previous lecture]] · [[Year 1 Sem 2/LC Electromagnetism I|Module]] · [[Year 1 Sem 2/LC Electromagnetism I/Lecture 11|Next lecture →]]

== Energy Density ==

Consider a parallel plate capactitor We have previously determined that: <div class="pn-import-equation"><nowiki>$$U = \frac{1}{2}CV^2$$</nowiki></div>

And that: <div class="pn-import-equation"><nowiki>$$C = \frac{Q}{V} = \frac{\epsilon_0 A}{d}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$V = Ed$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$U = \frac{1}{2} \frac{\epsilon_0 A}{d} E^2 d^2 = \frac{1}{2} A d \epsilon_0 E^2$$</nowiki></div>

Noting that <nowiki>$Ad$</nowiki> is just the volume of the gap, we have that the energy per unit volume is given by: <div class="pn-import-equation"><nowiki>$$\boxed{u_E = \frac{1}{2} \epsilon_0 E^2}$$</nowiki></div> 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: <div class="pn-import-equation"><nowiki>$$U = - \int \pmb{F} \cdot d \pmb{x}$$</nowiki></div>

Where <nowiki>$x$</nowiki> is the axis between the plates. Hence <nowiki>$F = -du/dx$</nowiki>: <div class="pn-import-equation"><nowiki>$$U(x) = Ax \epsilon_0 E^2$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$F = - \frac{1}{2} A \epsilon_0 E^2$$</nowiki></div>

Using the definition: <div class="pn-import-equation"><nowiki>$$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A \epsilon_0}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies F = - \frac{1}{2} A \epsilon_0 \left(\frac{Q}{A \epsilon_0}\right)^2$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$F = - \frac{1}{2} QE$$</nowiki></div>

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

== Dielectrics ==

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<b>What is a dielectric?</b>

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.

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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”

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Homogeneous.

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Isotropic, same in all directions hence <nowiki>$P \parallel E$</nowiki>.

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Linear, <nowiki>$P \propto E$</nowiki>.

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Stationary, no moving parts.

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== 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 <nowiki>$N$</nowiki> dipoles, we define the total polarisation as: <div class="pn-import-equation"><nowiki>$$\pmb{P} = N \pmb{p} = Nq \pmb{a}$$</nowiki></div> 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 <nowiki>$a$</nowiki>). One end of the block becomes depleted of negative charges and one end positive charges.

The charge on the surface (surface area <nowiki>$S$</nowiki>) is given by: <div class="pn-import-equation"><nowiki>$$\text{surface charge } = \text{ volume of depleted region } \times \text{ molecular density } \times \text{charge per molecule} = (aS)Nq$$</nowiki></div>

However we’ve already defined <nowiki>$P = Nqa$</nowiki>, so the surface charge on the RHS (for a right to left E-field) is: <div class="pn-import-equation"><nowiki>$$q_r = PS$$</nowiki></div> And the LHS: <div class="pn-import-equation"><nowiki>$$q_l = - PS$$</nowiki></div>

Hence the surface charge density is given by <nowiki>$\sigma_i \pm P$</nowiki>.

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.

[[File:EM-Y1S2-lec10-02.png|thumb|center|600px]]

Considering the positive plate, the net surface charge is <nowiki>$\sigma - \sigma_i$</nowiki>. With no dielectric, the E-field is: <div class="pn-import-equation"><nowiki>$$E_0 = \frac{\sigma}{\epsilon_0}$$</nowiki></div> With the dielectric: <div class="pn-import-equation"><nowiki>$$E = \frac{\sigma - \sigma_i}{\epsilon_0} = \frac{\sigma - P}{\epsilon_0}$$</nowiki></div>

For a well-behaved dielectric, the polarisation is proportional to the electric field <nowiki>$P \propto E$</nowiki>, hence the polarisation must be some constant times the electric field: <div class="pn-import-equation"><nowiki>$$\pmb{P} = \chi_E \epsilon_0 E$$</nowiki></div> Where <nowiki>$\chi_E \epsilon_0$</nowiki> is the constant of proportionality and <nowiki>$\epsilon_0$</nowiki> is included to make sure that <nowiki>$\chi_E$</nowiki> is dimensionless. <nowiki>$\chi_E$</nowiki> is defined as the susceptibility.

We therefore have: <div class="pn-import-equation"><nowiki>$$\epsilon_0 E = \sigma - P$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\epsilon_0 E = (\epsilon_0 E_0) - (\chi_E \epsilon_0 E)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\epsilon_0 E = \epsilon_0 E_0 - \chi_E \epsilon_0 E$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\boxed{(1 + \chi_E) \epsilon_0 E = \epsilon_0 E_0}$$</nowiki></div>

We define another dimensionless constant, the relative permittivity (AKA the dielectric constant): <div class="pn-import-equation"><nowiki>$$\epsilon_r = (1 + \chi_E)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies \epsilon_r \epsilon_0 E = \epsilon_0 E_0$$</nowiki></div>

The E-field in the dielectric is: <div class="pn-import-equation"><nowiki>$$E = \frac{E_0}{\epsilon_r}$$</nowiki></div>

From Gauss’ law we have: <div class="pn-import-equation"><nowiki>$$E = \frac{\sigma}{\epsilon_r \epsilon_0} = \frac{Q}{\epsilon_r \epsilon_0 A}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$V = V_+ - V_- = Ed = \frac{Qd}{\epsilon_0 \epsilon_r}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies C = \frac{Q}{V} = \frac{\epsilon_0 \epsilon_r A}{d}$$</nowiki></div> Since <nowiki>$\epsilon_r > 1$</nowiki>, this increases the capacitance. We can modify Gauss’ law for a well behaved dielectric:

<div class="pn-import-equation"><nowiki>$$\int_S \pmb{E} \cdot d \pmb{S} = \frac{Q_\text{enc}}{\epsilon_r \epsilon_0}$$</nowiki></div>

[[File:EM-Y1S2-lec10-03.png|thumb|center|600px|Typical Values for <nowiki>$\epsilon_r$</nowiki>]]

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Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc