Conductors, Capacitors & Dielectrics

Conductors in Electrostatic Equilibrium

• $\vec{E} = 0$ inside the conductor.
• All excess charge is on the surface.
• The surface is an equipotential.
• $E$ just outside the surface: $E = \sigma/\epsilon_0$.

Electric Potential

$V = -\int \vec{E} \cdot d\vec{l}$ $\vec{E} = -\nabla V = -\frac{dV}{dr}\hat{r}$ (for spherical symmetry)

Point charge: $V = kQ/r$.

Capacitance

$C = \frac{Q}{V}$

Parallel-plate: $C = \epsilon_0 A/d$.

Cylindrical: $C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}$ (radii $a$ and $b$).

Spherical: $C = 4\pi\epsilon_0 \frac{ab}{b-a}$.

Energy Stored

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

Energy density in an electric field: $u = \frac{1}{2}\epsilon_0 E^2$

Dielectrics

• Dielectric constant $\kappa$ increases capacitance: $C = \kappa C_0$.
• The field inside the dielectric is reduced: $E = E_0/\kappa$.
• $\epsilon = \kappa\epsilon_0$.

Capacitor Combinations

• Parallel: $C_{\text{eq}} = \sum C_i$ (same voltage).
• Series: $1/C_{\text{eq}} = \sum 1/C_i$ (same charge).

Problem: Derive the capacitance of a cylindrical capacitor with inner radius $a$, outer radius $b$, and length $L$.

Solution:

Using Gauss's law, the field between the cylinders (charge $Q$ on inner, $-Q$ on outer): $E = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{Q}{2\pi\epsilon_0 L r}$

Voltage: $V = -\int_b^a E\,dr = \int_a^b \frac{Q}{2\pi\epsilon_0 L r}\,dr = \frac{Q}{2\pi\epsilon_0 L}\ln\frac{b}{a}$

$C = \frac{Q}{V} = \frac{2\pi\epsilon_0 L}{\ln(b/a)}$

• Conductors: $E = 0$ inside, charge on surface, surface is equipotential.
• Capacitance derived from $C = Q/V$; use Gauss's law to find $E$, integrate for $V$.
• Energy density: $u = \frac{1}{2}\epsilon_0 E^2$.
• Dielectrics: increase $C$ by factor $\kappa$, reduce $E$ by factor $\kappa$.