Capacitance & Dielectrics

Reading: Chapter 24, CAPACITANCE AND DIELECTRICS

Study guide: Chapter 25

  1. Capacitance
  2. Dielectrics
  3. Energy, Energy density

Capacitance

 

Find a relationship between $\Delta V$ and $Q$...

Gausses law says... $Ea=q/\epsilon_0$, then

$$\Delta V=E\cdot d = \frac{\sigma}{\epsilon_0} d = \frac{Qd}{A\epsilon_0}$$

(That is... $Q=EA\epsilon_0$ **)

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

Getting a little sloppy.... the convention is to write the potential difference across the capacitor $\Delta V\equiv$ "$V$", then,$$Q=CV$$

Capacitance $C$ ($=\epsilon_0\frac{A}{d}$ for a flat plate capacitor):

[Capacitance of a 1 cm X 1 cm plate with a gap of 2 mm? What was $\epsilon_0$]

[Area of a square 1 farad capacitor?] .

Electric double-layer capacitors (EDLC or 'supercapacitors') can have capacitances up to several thousand F.

Energy stored in a capacitor

Potential energy of a charge in an electric potential: $$U=qV$$

$$dW=V\,dq = \frac{q}{C}dq$$

$$\Rightarrow U = \frac{Q^2}{2C} = \frac{1}{2}CV^2$$

In terms of $E$-field: $$U=\frac{Q^2}{2C}=\frac{(E\epsilon_0A)^2}{2\epsilon_0 A/d}$$

$\Rightarrow$ Energy density in terms of the electric field is $$u=\frac{1}{2}\epsilon_0 |E|^2.$$

[Check this for a cylindrical capacitor this week].

Dielectrics

(In a second kind of dielectric, the material does not consist of dipoles, but the field induces an average dipole moment.)

The field is reduced inside of the dielectric: $$E_D = \frac{E_0}{\kappa}$$where the dielectric constant, $\kappa$, of a material is a number greater than 1.

Permittivity: $\epsilon = \kappa \epsilon_0$

[What's the effective dielectric constant of a conductor??]

[How does capacitance change with dielectric material between plates?]

[How does energy stored change?]

If we run through our energy calculations as before, but with a dielectric material, we find that $$u=\frac{1}{2}\kappa |E^2| = \frac{1}{2}\epsilon |E^2|$$

Dielectric breakdown occurs in air when $|E| \gt 3\times 10^6 V/m$

[Estimate the potential difference, in volts, of our VdG with respect to ground?]

 

Formulary

$$\myv F=k\frac{qQ}{r^2}\uv r$$       $$k=\frac{1}{4\pi \epsilon_0}=9\times 10^9\frac{N\cdot m^2}{C^2}$$
$$\oint E_\perp \,dA=\frac{q}{\epsilon_0}$$ $$\Phi=\int_{\cal S} E_\perp dA$$       $$U=qV$$
$$\myv E=k\frac{q}{r^2}\uv r$$ $$\myv E=\frac{k\lambda}{2}\frac{1}{s}\uv s$$ $$\myv E=2\pi k\sigma\frac{\myv z}{|z|}$$
$$\Delta V = -\int \myv E \cdot d \myv l$$ $$V=k\frac{Q}{r}$$ $$Q=CV$$
$$u=\frac{1}{2}\epsilon_0 |E^2|$$ $$e=1.6\times 10^{-19}\,\text{C}$$ $$m_e=9.109\times 10^{-31}\,\text{kg}$$