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.

• 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]