---
title: "Retarded potentials of infinitesimal dipole"
subject: "Propagation and Antenna"
chapter: 1
key: "semester_6/propagation_and_antenna:retarded-potential-infinitesimal-dipole"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/retarded-potential-infinitesimal-dipole
site: BCT Engineering
---

# Retarded potentials of infinitesimal dipole

Propagation and Antenna · Chapter 1: Radiation and Antenna Fundamentals

- Asked in: 2072 Asoj, 2+4 marks
- Video: [Retarded Current,Vector and Scalar Potential of a short dipole.](https://www.youtube.com/watch?v=lG0cG8KTkyE) — Raji
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/retarded-potential-infinitesimal-dipole
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-1

## Question

What is retarded potential? Write the expression of retarded scalar potential and vector potential of infinitesimal dipole antenna in Fraunhofer far field region.

## Short answer

Retarded potential uses source values at the earlier time t − R/v; in phasor form the delay is e^(−jkR). For an infinitesimal dipole in the Fraunhofer region, the far-field vector potential is Az = μ I0 l e^(−jkr)/(4πr), and the scalar potential is V = (I0 l cosθ/4πε)(1/(c r) + 1/(jω r²)) e^(−jkr), which reduces to the 1/r term for r ≫ λ.

## Answer

**Retarded potential** is the potential (scalar V or vector A) at a point computed from the source value at the earlier time t − R/v, since the effect of a time-varying source takes R/v seconds to arrive. In phasor form the delay appears as a factor e^(−jkR).

```text
V = (1/4πε) ∫ ρ e^(−jkR)/R dv
A = (μ/4π)  ∫ J e^(−jkR)/R dv
```

#### Infinitesimal dipole
Length l ≪ λ along the z-axis, uniform current I0, end charges ±q with I0 = jωq.

**Retarded vector potential** (exact for small l):

```text
Az = μ I0 l e^(−jkr) / (4π r)
Ar = Az cosθ = μ I0 l cosθ e^(−jkr)/(4πr)
Aθ = −Az sinθ = −μ I0 l sinθ e^(−jkr)/(4πr)
```

**Retarded scalar potential** (from the two end charges, or from the Lorenz gauge):

```text
V = (q l cosθ / 4πε r²) (1 + jkr) e^(−jkr)
  = (I0 l cosθ / 4πε) [1/(c r) + 1/(jω r²)] e^(−jkr)
```

#### In the Fraunhofer (far-field) region
For r ≫ λ (kr ≫ 1, r > 2D²/λ) only the 1/r terms remain:

```text
Az ≈ μ I0 l e^(−jkr) / (4π r)
Aθ ≈ −μ I0 l sinθ e^(−jkr) / (4π r)
V  ≈ I0 l cosθ e^(−jkr) / (4π ε c r)
   = η I0 l cosθ e^(−jkr) / (4π r)
```

(using 1/(εc) = η = 120π Ω).

From these the far fields are

```text
Eθ ≈ −jω Aθ = jη k I0 l sinθ e^(−jkr)/(4πr)
Hφ ≈ Eθ / η
```

which give the sinθ (doughnut) pattern of the dipole.

## Check yourself

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

1. In phasor form, the retarded potential delay appears as:
   A. e^(+jkR)
   B. e^(-jkR)
   C. cos kR
   D. 1/kR
   Answer: B. The delay R/v appears as the phase factor e^(-jkR). (Key point: Retarded potential meaning)

2. What is the retarded vector potential of an infinitesimal dipole along z?
   A. Az = μ I0 l e^(-jkr)/(4π r)
   B. Az = I0 l/(4π r^2)
   C. Az = μ I0 l/(2π)
   D. Az = μ I0 l sinθ e^(-jkr)/r
   Answer: A. Az = μ I0 l e^(-jkr)/(4πr), with Ar = Az cosθ and Aθ = -Az sinθ. (Key point: Dipole vector potential)

3. What happens in the Fraunhofer region (kr much greater than 1)?
   A. Only 1/r^3 terms remain
   B. All terms stay equal
   C. The potentials vanish
   D. Only the 1/r terms remain
   Answer: D. For r much greater than λ the higher inverse powers are negligible, leaving 1/r terms. (Key point: Far-field simplification)

4. How is the far-field Eθ obtained from Aθ?
   A. Eθ ≈ Aθ/jω
   B. Eθ ≈ ∇ × Aθ
   C. Eθ ≈ -jω Aθ
   D. Eθ ≈ jω Aθ·r
   Answer: C. In the far field Eθ ≈ -jωAθ = jηkI0 l sinθ e^(-jkr)/(4πr), and Hφ = Eθ/η. (Key point: Far-field E from A)

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