---
title: "State and prove Poynting's theorem"
subject: "Electromagnetism"
chapter: 4
key: "semester_3/electromagnetism:poynting-theorem-state-prove"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/poynting-theorem-state-prove
site: BCT Engineering
---

# State and prove Poynting's theorem

Electromagnetism · Chapter 4: Wave equation and Wave propagation

- Asked in: 2076 Asoj, 6 marks; 2069 Chaitra, 1 marks (asked 2 times)
- Video: [Poynting Theorem Explained: Basics, Derivation, Proof, and Power Calculation](https://www.youtube.com/watch?v=d7eZIiLR43E) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/poynting-theorem-state-prove
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4

## Question

State and prove Poynting's theorem.

## Short answer

Poynting's theorem states ∮(E×H)·dS = −d/dt ∫(½εE²+½μH²)dv − ∫σE² dv. Proof uses ∇·(E×H)=H·(∇×E)−E·(∇×H) with Maxwell's equations, then integrates over volume and applies divergence theorem. The Poynting vector P = E×H gives power flow per unit area.

## Answer

#### Statement
The net power flowing out of a closed surface equals the rate of decrease of the electromagnetic energy stored in the enclosed volume, minus the ohmic power dissipated inside it. The Poynting vector P = E × H (W/m²) gives the power flowing per unit area:

∮ (E × H)·dS = −d/dt ∫ (½εE² + ½μH²) dv − ∫ σE² dv

#### Proof
Start with the vector identity

```text
∇·(E × H) = H·(∇×E) − E·(∇×H)
```

Substitute Maxwell's equations ∇×E = −∂B/∂t = −μ ∂H/∂t and ∇×H = J + ∂D/∂t = σE + ε ∂E/∂t:

```text
∇·(E × H) = −μ H·∂H/∂t − σ E² − ε E·∂E/∂t
```

Since H·∂H/∂t = ½ ∂(H²)/∂t and E·∂E/∂t = ½ ∂(E²)/∂t:

```text
∇·(E × H) = −∂/∂t (½μH² + ½εE²) − σE²
```

Integrate over a volume v bounded by the closed surface S and use the divergence theorem, ∫∇·(E×H) dv = ∮(E×H)·dS:

```text
∮_S (E × H)·dS = −d/dt ∫_v (½εE² + ½μH²) dv − ∫_v σE² dv
```

#### Meaning of the terms
- **Left side:** total power leaving the volume through S.
- **½εE² + ½μH²:** electric and magnetic energy densities (J/m³). The first term on the right is the decrease of stored energy per second.
- **σE² = J·E:** power dissipated as heat per unit volume (Joule loss).

If a source inside the volume also supplies power, an extra term −∫ E·J_source dv is added on the right. For time-harmonic fields the average power density is P_avg = ½ Re(E_s × H_s*) W/m².

## Check yourself

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

1. Poynting's theorem says the net power flowing out of a closed surface equals:
   A. The increase of stored energy plus ohmic loss
   B. The decrease in stored EM energy per second minus the ohmic power dissipated
   C. The ohmic loss only
   D. The electric energy only
   Answer: B. Power out = -d/dt (stored energy) - integral of sigma E^2 dv. (Key point: Statement)

2. The proof starts from which identity?
   A. div(E x H) = H . curl E - E . curl H
   B. div(E x H) = E . curl H + H . curl E
   C. curl(E x H) = 0
   D. div(E . H) = E x H
   Answer: A. Substituting Maxwell's curl equations into div(E x H) = H.(curl E) - E.(curl H) gives the energy balance. (Key point: Starting identity)

3. What do the terms (1/2) epsilon E^2 and (1/2) mu H^2 represent?
   A. Power losses
   B. Electric and magnetic energy densities
   C. Poynting vectors
   D. Charge densities
   Answer: B. They are the stored electric and magnetic energy per unit volume (J/m^3). (Key point: Energy densities)

4. The term sigma E^2 = J . E is:
   A. Stored electric energy
   B. Radiated power
   C. Power dissipated as heat per unit volume
   D. Magnetic energy density
   Answer: C. It is the Joule heating loss in the volume. (Key point: Loss term)

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