2024-02-15 Electrostatics In matter Cont.

Logistics:

• No class next Tuesday
• Midterm next Thursday
  • Covers classes 2-10
  • Exam format will be pen and paper
  • Resources provided will be “Very Useful Mathematical Formulas” and “Legendre Polynomials and Their Important Properties”
  • 4 problems
  • 60 Percent is easily earned, 20-30% intermediate, 10% difficult
  • No self provided crib sheets
  • Back exams will be posted 2024-02-12 Electrostatics in Matter review
• Dialectics and polarization

Describing linear dielectrics

• We can use electric displacement $\mathbf{D}$ and properties of electric susceptibility ${\chi }_e$ and electric permittivity $ϵ$

Previously:

$$\vec{p}={\alpha }_1\vec{E}⟹\vec{P}=\frac{⟨{\vec{p}}_i⟩}{V}$$

Alternatively:

$$\vec{P}={ϵ}_0{\chi }_e\vec{E}$$

An isotropic dielectric has a scalar value for electric susceptibility. This means that the susceptibility is independent from the crystal structure of the material and is not dependent on the polarization.

An anisotrpoic dielectric is dependent on polarization

A linear dielectric denotes a material in which ${\chi }_e$ is independent of the $\vec{E}$

A nonlinear dielectric has changing electric properties based on applied external electric fields .

Displacement is given by the following:

$$\vec{D}={ϵ}_0\vec{E}+\vec{P}={ϵ}_0\vec{E}+{ϵ}_0{\chi }_e\vec{E}$$

Factoring out ${ϵ}_0$ we get

$$\vec{D}={ϵ}_0\vec{E}(1+{\chi }_e)$$

Where $1+{\chi }_e$ is known as the relative dielectric constant and denoted by ${ϵ}_r$

$$\vec{D}={ϵ}_0{ϵ}_r\vec{E}$$ $${ϵ}_0\approx 8.85\times 10^{-12}\frac{As}{Vm}$$

Boundary Conditions for Dielectric Interfaces

If the dielectric has a surface charge, the perpendicular element of electric field is discontinuous

$$E_{\perp }^o-E_{\perp }^i=\frac{\sigma }{{ϵ}_0}$$

We can have electric field inside the material in a dielectric so the parallel component of the electric field is continuous

$$E_{\parallel }^o=E_{\parallel }^i$$

Super and Ultra Capacitors

• The energy storage of a capacitor is proportional to it’s capacitance, which is based on the materials and geometry of the capacitor.
• Capacitance:

$$C=\frac{{ϵ}_r{ϵ}_{0}A}{d}$$

• Energy storage:

$$U_e=\frac{1}{2}C{V}^2$$

a. Find the electric displacement in each slab

$$\vec{D}={ϵ}_0\vec{E}(1+{\chi }_e)$$ $$\vec{D}={ϵ}_0{ϵ}_r\vec{E}$$ $${\vec{E}}_1=\frac{\sigma }{{ϵ}_1{ϵ}_0}$$ $${\vec{D}}_1={ϵ}_0{ϵ}_1\frac{\sigma }{{ϵ}_1{ϵ}_0}$$ $${\vec{D}}_1\text{ and }{\vec{D}}_2=\sigma$$

b. Find the electric field:

$${\vec{E}}_1=\frac{\sigma }{{ϵ}_1{ϵ}_0}$$ $${\vec{E}}_2=\frac{\sigma }{{ϵ}_2{ϵ}_0}$$

c. Find the polarization in each slab

$$\vec{P}={ϵ}_0{\chi }_e\vec{E}$$ $$\vec{P}=\vec{D}-{ϵ}_0\vec{E}$$ $${\vec{P}}_1=\sigma -\frac{{ϵ}_0\sigma }{{ϵ}_0{ϵ}_1}$$ $${\vec{P}}_2=\sigma -\frac{{ϵ}_0\sigma }{{ϵ}_0{ϵ}_2}$$

d.

$$V=E_{1}a+E_{2}a$$ $$\frac{\sigma }{{ϵ}_1{ϵ}_0}a+\frac{\sigma }{{ϵ}_2{ϵ}_0}$$ $$V=\frac{7\sigma a}{6{ϵ}_0}$$

e.