Block 14 - The steady Conduction Field

Learning objectives

Preparation at Home

Well, again

• read through the present chapter and write down anything you did not understand.
• Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting).

For checking your understanding please do the following exercises:

• …

90-minute plan

1. Warm-up (10 min):
   1. Quick recap of Block 11 field pictures (parallel plates, coax) → link to resistance by replacing $\varepsilon$ with $\sigma$.
   2. Mini check: which vector integrates over length/area? ($\vec{E}$ along paths, $\vec{J}$ across areas)
2. Core concepts (20 min):
   1. Definitions: steady conduction, $\vec{j}=\sigma\vec{E}$, current $I$.
   2. From potential drop to Ohm’s law in fields.
3. Guided derivations (25 min):
   1. Parallel-plate bar
   2. Coaxial conductor
4. Practice (30 min):
   1. Short exercises: compute $R$ for a busbar, and for a coax segment; compare materials (copper vs. aluminum).
   2. “What-if” variations: halve $l$, double $A$, change $\sigma$; predict $R$ qualitatively before computing.
5. Wrap-up (5 min):
   1. Summary box (key formulas, units); Common pitfalls checklist and Q&A.

Conceptual overview

Core content

In the discussion of the electrostatic field in principle, no charges in motion were considered.
This lead to multiple fomulas, which are aggregated in the following diagram:

One outcome was, that the capacitance is defined as:

\begin{align*} C &= {{Q}\over{U}} \\ &= {{ \rlap{\Large \rlap{\int_A} \int} \, \LARGE \circ} \;\; \vec{D} \, {\rm d} \vec{A}\over{\int \vec{E} \,{\rm d} \vec{s} }} \end{align*}

Now the motion of charges shall be considered explicitly.
With the knowledge of th electrostatic field, we want to see, whether we can calulate the resistance of more complicated geometries.

For this we want to introduce the current density $J$: The current density here describes how charge carriers move together (collectively).
The stationary current density describes the charge carrier movement if a direct voltage is the cause of the movement.
Then, a constant direct current flows in the stationary electric flow field. Thus, there is no time dependency on the current:

$\large{{{\rm d}I}\over{{\rm d}t}}=0$

Important: Up to now it was considered, that charges had moved through a field in the past or could be moved in the future. Now, the exact moment of moving the charge is considered.

By comparison, we see now, that the resistance can be defined as:

\begin{align*} {{1}\over{R}} &= {{I}\over{U}} \\ &= {{ \rlap{\Large \rlap{\int_A} \int} \, \LARGE \circ} \;\; \vec{J} \, {\rm d} \vec{A}\over{\int \vec{E} \,{\rm d} \vec{s} }} \end{align*}

Given the results from block 11 we can derive:

• for a current between parallel plates
  • The current density is given as: \begin{align*} J = {{I}\over{A}} = \sigma \cdot E = {\rm const.} \end{align*}
  • This leads to the electric field: \begin{align*} E = {{J}\over{\sigma}} \end{align*}
  • The resistance value is given as: \begin{align*} {{1}\over{R}}&= {{ \rlap{\Large \rlap{\int_A} \int} \, \LARGE \circ} \;\; \vec{J} \, {\rm d} \vec{A}\over{\int \vec{E} \,{\rm d} \vec{s} }} = {{J} \cdot \rlap{\int_A}\int \; {\rm d} {A}\over{{E} \cdot \int \,{\rm d} {s} }} \end{align*}\begin{align*} \boxed{ {{1}\over{R}}= {{\sigma A}\over{l}} }_\text{between parallel plates} \end{align*}

• for a current between coaxial plates
  • The current density is given as: \begin{align*} J = {{I}\over{2\pi \cdot l \cdot r}} \end{align*}
  • The resistance value is given as: \begin{align*} \boxed{ {{1}\over{R}}=\dfrac{2\pi\sigma l}{\ln(r_a/r_i)} }_\text{between coaxial plates}\end{align*}

Common pitfalls

• Mixing $\vec{D}$ (electrostatics) with $\vec{j}$ (conduction). Use $\vec{D}=\varepsilon\vec{E}$ for capacitors, $\vec{j}=\sigma\vec{E}$ for resistive flow.
• Forgetting surface orientation in $I=\iint_A \vec{j}\cdot{\rm d}\vec{A}$ (normal must align with the chosen current reference arrow).
• Confusing material parameters: $\sigma$ vs. $\rho$ with $\rho=\dfrac{1}{\sigma}$. Writing both in the same formula yields unit errors.
• Using the wrong area: for coax, the relevant area element is the *lateral* surface $2\pi r\,l$ (not $\pi r^2$).
• Dropping units or not checking dimensions; e.g., verify $G=\dfrac{\sigma A}{l}$ gives $\rm S$ and $R$ gives $\Omega$.

Exercises

Embedded resources