Lecture on Electric Potential and Conductivity

Electric Scalar Potential

In electric circuits, a convenient node is selected as ground and assigned zero reference voltage.
In free space and material media, infinity is chosen as the reference point with $V = 0$ .
At a point P, the electric potential is given by:
- $\hat{E} = \frac{q}{4 \pi \epsilon_0 R^2} \hat{R}$ (V/m)
- ${\rho} = \frac{q}{2 \pi r l \epsilon_0}$ (V/m)

Path Independence of Electrostatic Field

The line integral of the electrostatic field E along any closed path is zero.
${V{21}} + V{12} = 0$
$V{21} = V2 - V1 = -\int{P1}^{P2} E \cdot dl$
$\oint E \cdot dl = 0$ (Electrostatics)
Kirchhoff's voltage law is applicable.
A vector field with a zero line integral along any closed path is conservative or irrotational. Thus, the electrostatic field E is conservative.

Electric Potential Due to Charges

For a point charge, the electric potential V at range R is:
- $V = \frac{q}{4 \pi \epsilon_0 R}$ (V)
For continuous charge distributions:
- Volume charge distribution: $V = \frac{1}{4 \pi \epsilon0} \int{V'} \frac{\rho_v}{R'} dV'$
- Surface charge distribution: $V = \frac{1}{4 \pi \epsilon0} \int{S'} \frac{\rho_s}{R'} dS'$
- Line charge distribution: $V = \frac{1}{4 \pi \epsilon0} \int{l'} \frac{\rho_l}{R'} dl'$

Relating E to V

Differential relationship: $dV = -E \cdot dl$
For a scalar function V: $dV = \nabla V \cdot dl$ , where ${nabla}V$ is the gradient of V.
Relationship between E and V: $E = -\nabla V$
- $V = -\intP^{P2} E \cdot dl$ (V)
This allows determining E by first calculating V and then taking the negative gradient of V.

2013 Test 2 Q3

A uniform line charge density ${\rhol} = \rho1$ C/m exists on a circular ring of radius $r = a$ .
Find expressions for the potential, V(R) and electric field intensity, E(R) at the point $R = (0,0,z)$ .
$|R - R'| = \sqrt{a^2 + z^2}$
$dl' = a d\phi' \hat{\phi}$
$V(R) = \frac{1}{4 \pi \epsilon0} \int0^{2 \pi} \frac{\rhol dl'}{|R - R'|} = \frac{1}{4 \pi \epsilon0} \int0^{2 \pi} \frac{\rhol a d\phi'}{\sqrt{a^2 + z^2}} = \frac{\rhol a}{2 \epsilon0 \sqrt{a^2 + z^2}}$
$E(R) = \frac{1}{4 \pi \epsilon0} \int0^{2\pi} \frac{\rho_l dl' (R - R')}{|R - R'|^3}$
$E = -\nabla V = -\hat{z} \frac{\partial}{\partial z} \frac{\rhol a}{2 \epsilon0 \sqrt{a^2 + z^2}} = \hat{z} \frac{\rhol a z}{2 \epsilon0 (a^2 + z^2)^{3/2}}$ V/m

Poisson’s & Laplace’s Equations

In the absence of charges:
Differential form of Gauss's law:
- ${nabla} \cdot D = \rho$
- ${nabla} \cdot E = \frac{\rho}{\epsilon}$
Only if ${\epsilon}$ is homogeneous medium.

Laplace’s Equations

${\nabla}^2 V = \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} + \frac{\partial^2 V}{\partial z^2} = 0$
Cylindrical coordinates:
- ${\nabla}^2 V = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial V}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2} = 0$
Spherical coordinates:
- ${\nabla}^2 V = \frac{1}{R^2} \frac{\partial}{\partial R} \left(R^2 \frac{\partial V}{\partial R}\right) + \frac{1}{R^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial V}{\partial \theta}\right) + \frac{1}{R^2 \sin^2 \theta} \frac{\partial^2 V}{\partial \phi^2} = 0$

Application of Laplace's Equation

Find the potential function in the region between two cylindrical conductors.
Use cylindrical coordinates.
Assume ${pm}Q$ .
Find E and V.
Potential $V_{ab}$ .
${\nabla}^2 V = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial V}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2} = 0$
$E \neq 0, V_{ab} \neq 0$

Boundary Conditions

Neglect fringe effects. $V(r,\phi,z) = V(r)$
${\nabla}^2 V = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial V}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2} = 0$
{\frac{d}{dr} \left( r \frac{dV}{dr} \right)=0
Boundary conditions:
- $V(r = a) = V_a$
- $V(r = b) = V_b$
Solution:
- $V(r) = K1 \ln(r) + K2$
- $V(r=a) = Va = K1 \ln(a) + K_2$
- $V(r=b) = Vb = K1 \ln(b) + K_2$
${V(r)} = V{ab} \frac{\ln(\frac{r}{a})}{\ln(\frac{b}{a})} + Va$
${K1} = \frac{V{ab}}{\ln(\frac{b}{a})}$
${K2} = Va - \frac{V_{ab}}{\ln(\frac{b}{a})} \ln(a)$
Boundary conditions:
- ${V(r=b)} = V_b$
- ${\frac{dV}{dr}} = \frac{K_1}{r}$

Conduction Current

Conductivity measures how easily electrons travel through a material under an applied electric field.
Conduction (volume) current density:
- $J = \sigma E$ (A/m²) (Ohm's law)
- $I = \int_S J \cdot ds$ (A), where S is the surface area.
- $As = n As$

Conductivity

A perfect dielectric is a material with ${\sigma} = 0$ .
A perfect conductor is a material with ${\sigma} = \infty$ .
Some materials (superconductors) exhibit such behavior.

Material	Conductivity, ${\sigma}$ (S/m)
Conductors
Silver	$6.2 × 10^7$
Copper	$5.8 × 10^7$
Gold	$4.1 × 10^7$
Aluminum	$3.5 × 10^7$
Iron	$10^7$
Mercury	$10^6$
Carbon	$3 × 10^4$
Semiconductors
Pure germanium	$2.2$
Pure silicon	$4.4 × 10^{-4}$
Insulators
Glass	$10^{-12}$
Paraffin	$10^{-15}$
Mica	$10^{-15}$
Fused quartz	$10^{-17}$

$J = \sigma E$ (A/m²)
Perfect dielectric: ${J = 0}$ , Perfect conductor: ${E = 0}$ .

Resistance

Homogeneous conductor
Constant cross section, A
Uniform E field
- $R = \frac{V}{I} = \frac{l}{\sigma A}$

Arbitrary Conductor Resistance

$R = \frac{V}{I} = \frac{\int E \cdot dl}{\int \sigma E \cdot ds}$

Lossy Coaxial Cable

Find the resistance between two cylindrical conductors.
Assume a leakage current I between the inner and outer conductor.
- $E = \frac{I}{2 \pi r l \sigma} \hat{r}$
- $J = \sigma E = \frac{I}{2 \pi r l} \hat{r}$
- $V{ab} = - \inta^b E \cdot dl = \frac{I}{2 \pi \sigma l} \ln \frac{a}{b}$

Resistance Calculation

\begin{aligned}
V{ab} &= -\intb^a E \cdot dr = \int_a^b \frac{I}{2 \pi r \sigma l} dr \
&= \frac{I}{2 \pi \sigma l} \ln\left(\frac{b}{a}\right)
\end{aligned}

$R{ab} = \frac{V{ab}}{I} = \frac{1}{2 \pi \sigma} \frac{\ln(\frac{b}{a})}{l}$

2003 Test 2 Q1

Use spherical coordinates.
Assume ${I_o}$ is known.
Find ${J, E, V_{ab}}$ .

Current Density

${overrightarrow{J}} = \frac{Io}{S(R)} \overrightarrow{aR} = \frac{Io}{2 \pi R^2 (1 - \cos \thetao)} \overrightarrow{a_R} A/m^2$

Electric Field Intensity and Potential Difference

$E = \frac{J}{\sigma} = \frac{Io}{2 \pi R^2 (1 - \cos \thetao) \sigma} \hat{R}$

$V{ab} = - \inta^b E \cdot dl = - \int{Ra}^{Rb} \frac{Io}{2 \pi R^2 (1 - \cos \thetao) \sigma} dR = \frac{Io}{2 \pi (1 - \cos \thetao) \sigma} \left[ \frac{1}{R} \right]{Ra}^{Rb}$

$V{ab} = \frac{Io}{2 \pi (1 - \cos \thetao) \sigma} \left[ \frac{1}{Ra} - \frac{1}{R_b} \right]$