Wave velocity in a dielectric-filled cable A uniform cable has relative permittivity ε_r = 4 (μ_r ≈ 1). What is the phase velocity of electromagnetic waves in this cable?

Difficulty: Easy

Correct Answer: 1.5 × 10^8 m/s

Explanation:


Introduction / Context:
Phase velocity in a medium directly affects signal delay, line length for a given electrical length, and dispersion analysis. Designers regularly convert between free-space parameters and substrate/cable parameters using ε_r and μ_r.


Given Data / Assumptions:

  • Relative permittivity ε_r = 4.
  • Relative permeability μ_r ≈ 1 (nonmagnetic dielectric).
  • Free-space speed of light c = 3 × 10^8 m/s.


Concept / Approach:

The phase velocity in a dielectric is v = c / √(ε_r μ_r). With μ_r ≈ 1, v = c / √ε_r. Substituting ε_r = 4 gives √ε_r = 2, hence v = c/2 = 1.5 × 10^8 m/s. This result is fundamental to transmission-line timing and length calculations in RF and high-speed digital designs.


Step-by-Step Solution:

1) Formula: v = c / √(ε_r μ_r) → with μ_r = 1, v = c / √ε_r.2) Compute √ε_r = √4 = 2.3) v = 3 × 10^8 / 2 = 1.5 × 10^8 m/s.


Verification / Alternative check:

Velocity factor VF = v/c = 1/√ε_r = 0.5. Many cables with ε_r ≈ 4 exhibit VF ≈ 0.5, confirming the calculation.


Why Other Options Are Wrong:

3 × 10^8 m/s corresponds to air (ε_r = 1). 0.75 × 10^8 and 1 × 10^8 m/s do not satisfy v = c/√4. 2.25 × 10^8 m/s equates to ε_r ≈ 1.78, not 4.


Common Pitfalls:

Mixing phase and group velocities or forgetting to take the square root of ε_r.


Final Answer:

1.5 × 10^8 m/s.

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