---
title: "Displacement current density at 1 kHz"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:displacement-current-metal-1khz"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/displacement-current-metal-1khz
site: BCT Engineering
---

# Displacement current density at 1 kHz

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2070 Chaitra, 4 marks
- Video: [KTU Electromagnetics EE302, Module-4 Conduction current and displacement current densities](https://www.youtube.com/watch?v=pR0xqJGyUVs) — 100 Million Thoughts
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/displacement-current-metal-1khz
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Find the amplitude of the displacement current density inside a typical metallic conductor where f = 1 kHz, σ = 5×10⁷ mho/m, ε_r = 1 and the conduction current is J = 10⁷ sin(6283t − 444z) a_y A/m².

## Short answer

From E = J_c/σ = 0.2 sin(ωt-βz) V/m, displacement current density J_d = ε₀ ∂E/∂t = ε₀×0.2×ω cos(...) has amplitude 1.11×10⁻⁸ A/m².

## Answer

In the metal, the conduction current J_c = σE, and the displacement current density J_d = ∂D/∂t = ε ∂E/∂t.

**Given:** f = 1 kHz, ω = 2πf = 6283 rad/s, σ = 5×10⁷ S/m, ε = ε_r ε₀ = 8.854×10⁻¹² F/m, J_c = 10⁷ sin(6283t − 444z) a_y A/m².

**Step 1: E from J_c.**

```text
E = J_c / σ = (10⁷ / 5×10⁷) sin(6283t − 444z) a_y = 0.2 sin(6283t − 444z) a_y  V/m
```

**Step 2: D and J_d.**

```text
D = ε₀ E = 8.854×10⁻¹² × 0.2 sin(...) a_y
J_d = ∂D/∂t = 8.854×10⁻¹² × 0.2 × 6283 cos(6283t − 444z) a_y
    = 1.1126×10⁻⁸ cos(6283t − 444z) a_y   A/m²
```

**Step 3: ratio.**

```text
J_d / J_c = ωε/σ = 1.1126×10⁻⁸ / 10⁷ ≈ 1.1×10⁻¹⁵
```

So displacement current is negligible compared with conduction current in a metal at 1 kHz.

**Answer: amplitude of J_d = 1.11×10⁻⁸ A/m²** (leading J_c by 90°, since it varies as cosine while J_c varies as sine).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How do you get E from the conduction current density?
   A. E = sigma J_c
   B. E = J_c / sigma = 0.2 sin(6283t - 444z) V/m
   C. E = J_c / (w epsilon)
   D. E = J_c x 5 x 10^7
   Answer: B. J_c = sigma E, so E = 10^7/(5 x 10^7) sin(...) = 0.2 sin(...) V/m. (Key point: E from J_c)

2. What is the angular frequency used for f = 1 kHz?
   A. 1000 rad/s
   B. 3141 rad/s
   C. 6283 rad/s
   D. 62.83 rad/s
   Answer: C. w = 2 pi f = 6283 rad/s, matching the argument of the sine. (Key point: Angular frequency)

3. What is the amplitude of J_d = epsilon dE/dt?
   A. 1.11 x 10^-8 A/m^2
   B. 1.11 x 10^-15 A/m^2
   C. 10^7 A/m^2
   D. 5.5 x 10^-12 A/m^2
   Answer: A. J_d = 8.854 x 10^-12 x 0.2 x 6283 = 1.11 x 10^-8 A/m^2. (Key point: J_d amplitude)

4. What does J_d/J_c = w epsilon/sigma ~ 10^-15 tell us?
   A. Displacement current is dominant
   B. Displacement current is negligible in a metal at this frequency
   C. The metal is a perfect dielectric
   D. The currents cancel
   Answer: B. The displacement current is far smaller than the conduction current, so it can be neglected in metals. (Key point: Significance)

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