INTERACTIVE EXAMPLE · SECTION 4.2 / Fields in biological media

Conduction or displacement?

One electric field can produce two current mechanisms. Which one matters most—and what does frequency have to do with it?

Graduate EE · Reading + exploration + practice · About 20 min

Illustrative lesson · Academic review pending

Start with a question

Same field. Faster oscillation. Different current?

Imagine a small, uniform region of material under a sinusoidal electric field. Keep the field amplitude the same, then increase its frequency. Does the material’s current response stay the same?

Pause for a moment: which material properties would you need to answer?

Build the idea

Two mechanisms, one total current.

In a simple linear medium, conduction follows the electric field, while displacement current follows the rate of change of electric displacement. Both appear in the Ampère–Maxwell law.

Conduction current

J_{\mathrm c}=\sigma E

(1)

Mobile charge produces a current proportional to conductivity. With real, constant conductivity, it is in phase with the field.

Displacement current

J_{\mathrm d}=\frac{\partial D}{\partial t}=\varepsilon\frac{\partial E}{\partial t}

(2)

A changing electric displacement contributes another current density. With real, constant permittivity, it leads the field by 90°.

Current densities in A/m². Here ε = ε₀εᵣ. Conductivity σ and permittivity ε are held constant.

Make a prediction

Predict before you explore.

Explore the model

Turn the frequency. Watch the balance change.

Start at 10 MHz. Try 100 MHz, then change conductivity. Compare the amplitudes and notice the quarter-cycle offset between the two currents.

INTERACTIVE EXPLORATIONσ + jωε

Current density over one cycle

Conduction currentDisplacement currentTotal current

time / period

One model cycle plays every 4 seconds for visibility. This is slowed playback, not real-time MHz motion.

Frequency10 MHz

MHz

Conductivity0.20 S/m

Fixed: εᵣ = 80 · E₀ = 1 V/m

PEAK AMPLITUDES

Conduction current

Displacement current

Displacement / conduction

Total-current phase lead Equal amplitudes near

Try these three observations

At σ = 0.20 S/m, compare 10 MHz with 100 MHz. Which amplitude changes?
Keep 100 MHz and increase σ. What happens to the amplitude ratio?
Use the equal-amplitudes frequency. A ratio of 1 means equal amplitudes, not equal instantaneous currents.

Know what this model assumes

A prescribed uniform field in a homogeneous, isotropic, linear medium; real σ and ε do not vary with frequency. These values are illustrative, not measured tissue data. This activity compares constitutive currents; it does not solve field propagation, interfaces, heating, or biological effects. Real tissues can be dispersive and anisotropic.

Explain the result

Frequency changes the balance, not just the speed.

With E(t) = E₀ cos(ωt), the conduction-current amplitude is σE₀. Differentiation gives a displacement-current amplitude of ωεE₀. Increasing frequency therefore changes their relative size, even when the applied field amplitude is unchanged.

\frac{|J_{\mathrm d}|}{|J_{\mathrm c}|}=\frac{\omega\varepsilon}{\sigma}

(3)

f_{\mathrm c}=\frac{\sigma}{2\pi\varepsilon}

(4)

A worked comparison

For σ = 0.20 S/m and εᵣ = 80, the amplitude ratio is about 0.223 at 10 MHz and 2.23 at 100 MHz. Conduction is larger in the first case; displacement is larger in the second. Equal amplitudes occur at about 44.94 MHz.

Go deeper: phasors and complex permittivity

Use the time convention e⁺ʲωᵗ and a positive frequency. The total-current phasor is (σ + jωε)E. Its phase lead is arctan(ωε/σ). Writing the same constitutive response as jωε*E gives ε* = ε − jσ/ω under this convention.

J=(\sigma+j\omega\varepsilon)E=j\omega\varepsilon^{*}E

(5)

The minus sign follows from the chosen time convention. Real tissue dispersion requires frequency-dependent material data; this example holds the properties fixed.

Put it into practice

Can you transfer the idea?

Take it with you

Three ideas to keep.

Conduction follows E; displacement follows ∂D/∂t.
The ratio ωε/σ compares amplitudes, not instantaneous values.
A model’s assumptions travel with every result you draw from it.

Before you leave

Without looking back, explain why increasing frequency changes the ratio. Then name one assumption you would revisit for a real tissue.

Completion is saved only in this browser. It is a study aid, not a grade.

Read, reproduce, revisit.

Revisit Maxwell’s correctionOpenStax · University Physics, volume 2, §16.1 · English Reproduce the current comparisonDownload a short Python activity · No extra packages

Optional computational activity

Reproduce the four frequency settings with Python. Change σ or εᵣ and predict the new crossover before running the code. Execution happens on your computer.

from math import pi

sigma = 0.20       # S/m, illustrative
epsilon = 80 * 8.8541878128e-12

for f_mhz in [1, 10, 100, 1000]:
    ratio = 2 * pi * f_mhz * 1e6 * epsilon / sigma
    print(f_mhz, round(ratio, 3))