---
title: "Displacement current density, dE/dt, induced B"
subject: "Engineering Physics"
chapter: 9
key: "semester_1/engineering_physics:displacement-current-density-dedt-plates"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/displacement-current-density-dedt-plates
site: BCT Engineering
---

# Displacement current density, dE/dt, induced B

Engineering Physics · Chapter 9: Electromagnetic Waves

- Asked in: 2068 Chaitra, 5 marks
- Video: [Displacement current density | Complete derivation in less than 7 minutes](https://www.youtube.com/watch?v=edJqpjADIkA) — Solely Science
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/displacement-current-density-dedt-plates
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-9

## Question

A parallel plate capacitor with circular plates of 10cm radius is charged producing uniform displacement current of magnitude 20A/m². Calculate (i) dE/dt in the region (ii) Displacement current density and (iii) Induced magnetic field.

## Short answer

dE/dt = J_d/ε₀ = 2.26×10¹² V m⁻¹ s⁻¹. The displacement current density is 20 A/m². Induced magnetic field at the plate edge is B = μ₀ J_d R/2 = 1.26×10⁻⁶ T.

## Answer

**Given:** plate radius R = 10 cm = 0.10 m, displacement current density J_d = 20 A/m² (uniform). ε₀ = 8.85×10⁻¹² F/m, μ₀ = 4π×10⁻⁷ H/m.

#### (ii) Displacement current density
This is given: **J_d = 20 A/m²**. The total displacement current is I_d = J_d πR² = 20 × π × 0.01 = 0.628 A.

#### (i) dE/dt
Displacement current density is J_d = ε₀ dE/dt.

```text
dE/dt = J_d / ε₀ = 20 / 8.85×10⁻¹²
      = 2.26×10¹² V/(m·s)
```

#### (iii) Induced magnetic field
Ampere-Maxwell law for a circle of radius r: B(2πr) = μ₀ J_d (πr²), so B = μ₀ J_d r/2. At the edge of the plates, r = R:

```text
B = μ₀ J_d R / 2
  = (4π×10⁻⁷ × 20 × 0.10) / 2
  = 1.26×10⁻⁶ T
```

**Answer: (i) dE/dt ≈ 2.26×10¹² V m⁻¹s⁻¹; (ii) J_d = 20 A/m²; (iii) B ≈ 1.26×10⁻⁶ T at the plate edge.**

## Check yourself

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

1. How is displacement current density related to the changing field?
   A. J_d = dE/dt / ε₀
   B. J_d = μ₀ dE/dt
   C. J_d = ε₀ dE/dt
   D. J_d = ε₀ E
   Answer: C. Dividing I_d = ε₀ A dE/dt by A gives J_d = ε₀ dE/dt. (Key point: Relation of J_d and dE/dt)

2. With J_d = 20 A/m², dE/dt equals:
   A. J_d ε₀ ≈ 1.8×10⁻¹⁰
   B. μ₀ J_d ≈ 2.5×10⁻⁵
   C. J_d/μ₀ ≈ 1.6×10⁷
   D. J_d/ε₀ ≈ 2.26×10¹² V/(m·s)
   Answer: D. Dividing 20 by 8.85×10⁻¹² gives about 2.26×10¹². (Key point: Value of dE/dt)

3. For uniform J_d, the field at radius r inside the plates is:
   A. B = μ₀ J_d r/2
   B. B = μ₀ J_d r
   C. B = μ₀ J_d/(2π r)
   D. B = μ₀ J_d r²
   Answer: A. From B(2πr) = μ₀ J_d π r². (Key point: Induced B formula)

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