Studying a plate capacitor with dielectric between the plates.

• The surface charge density of the plate is

  $$ \sigma_\text{plate}=\frac{Q}{A} $$

• If we now apply Gauss’ law

  $$ \int\vec D\cdot\text d\vec S=D_z A=Q $$

  with $D_z=(1+\chi_E)\epsilon_0 E_z$

• We can then obtain the components of the electric and polarization vectors

  $$ E_z=\frac{qQ}{(1+\chi_E)\epsilon_0 A} \qquad P_z=\frac{\chi_E Q}{(1+\chi_E) A} $$

  The potential difference is

  $$ \nu= E_z d=\frac{Qd}{(1+\chi_E) \epsilon_0 A} $$

• and Hence the capacitance

  $$ C=\frac{Q}{\mathcal{V}}=(1+\chi_E)\frac{\epsilon_0 A}{d}=(1+\chi_E) C_0 $$

  where $C_0$ is the capacitance without the dielectric.

  🗒️ Notes:

  • At constant $Q$ (battery not connected) we have calculated that the electric field is lowered by $(1+\chi_E)$
  • The capacitance increases
  • When the dielectric is not present $\chi_E=0$ the polarization vanishes

• We define the relative permittivity as

  $$ \epsilon_r \equiv 1+\chi_E $$

• the electric displacement vector becomes

  $$ \vec D= \epsilon_r \epsilon_0 \vec E $$

  🗒️ Note:

  $$ \begin{aligned} \sigma_\text{top}&=\vec P\cdot \hat n_\text{top} =-\frac{\epsilon_r-1}{\epsilon_r}\sigma_\text{plate} \\ \sigma_\text{bottom}&=\vec P \cdot \hat n_\text{bottom}=+\frac{\epsilon_r -1}{\epsilon_r} \sigma_\text{plate}

  \end{aligned} $$

• The energy of the capacitor is

  $$ U=\frac 12 C \mathcal V^2 =\frac 12 \frac{A \epsilon_r \epsilon_0}{d} (E_z d)^2=\frac 12 Ad D_z E_z =\frac 12 \int \vec D \cdot \vec E \,\text dV $$

  🗒️ Note: the expression for energy of an electric field is modified in the presence of a dielectric

  

  When the current is switched on the electric field between the plates causes the dipoles to align. At non zero temperature this alignment may only be partial with perhaps only a slight preponderance of dipoles aligned with the field

Interference between dielectrics

Consider the boundary between two regions $1$ and $2$ with relative permittivity’s $\epsilon_r^{(1)}$ and $\epsilon^{(2)}_r$ respectively, with no free charge on the boundary.

Let $S$ be the surface of thickness $d$ and area $\delta S$ encompassing the boundary an interface between dielectrics

• $As$ $\text d\to 0$ because there is no charge on the boundary

  $$ 0=\int\vec D\cdot \text d\vec S=-D^{(1)}\perp \delta S+D^{(2)}\perp \delta S $$

  Therefore $D_\perp$ is continuous

• Now let $L$ be a loop of length $\delta \ell$ and with $d$ which encompasses the boundary enclosing an interface between dielectrics. as $d\to0$

  $$ 0=\oint \vec E\cdot \text d\ell=-E_\parallel^{(1)}\delta S+D_\perp^{(2)}\delta S $$

  Which implies that $E_\parallel$ is continuous across the boundary

• We can apply these continuity arguments to the boundary. We may write the electric field in the two regions as

  $$ \vec E^{(1)}=E_1 \begin{pmatrix} \sin \theta_1 \\ \cos \theta_1 \end{pmatrix} \qquad \vec E^{(2)}=E_2\begin{pmatrix} \sin\theta_2 \\ \cos\theta_2 \end{pmatrix} $$

  If $E_\parallel$ is continuous then

  $$ E_1\sin\theta_1=E_2\sin\theta_2 $$

  if $D_\perp$ is continuous then

  $$ D_1\cos\theta_1=D_2\cos\theta_2 $$

• Hence we get

  $$ \frac{D_1}{E_1}\cot\theta_1=\frac{D_2}{E_2}\cot\theta_2 $$

• Therefore

  $$ \epsilon_r^{(1)} \cot\theta_1=\epsilon_r^{(2)} \cot \theta_2 $$

  

Inductance and relative permeability

Consider a solenoid of length $\ell$ carrying current $I$ and having $N$ turns per unit length. We choose an Amperian surface at a distance $r$ from the centre of the solenoid

💼 Case 1: