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Electromagnetic Waves: Electromagnetic Energy, the Poynting Vector, and Intensity (Topic)

Electromagnetic Waves, Antennas, and RF: Electromagnetic Energy, the Poynting Vector, and Intensity

EM16 showed that Maxwell’s equations support electromagnetic waves in vacuum. A wave equation tells us how a disturbance propagates, but it does not yet answer another fundamental physical question:

|--------------------------------------------------------------------------|
|How  much  energy is carried by the fields, and how does that energy move? |
---------------------------------------------------------------------------
(1)

The answer is encoded directly in Maxwell’s equations. Electric and magnetic fields possess energy density, and the flow of electromagnetic energy is described by the Poynting vector. Combining those ideas gives a local conservation law for electromagnetic energy called Poynting’s theorem.

For a plane electromagnetic wave in vacuum, the result becomes especially simple:

|-----1--------|
|S = ---E × B, |
-----μ0---------
(2)

|------------------|
|    1-   2   -B2- |
|u = 2 𝜖0E   + 2μ  ,|
-----------------0-
(3)

and, because E = cB for a vacuum plane wave,

|--------|
|S|-=-uc.-
(4)

These relations connect the field picture of Maxwell’s equations to measurable power, intensity, antenna radiation, and eventually RF link budgets [12345].

1 Energy stored in an electric field

The electrostatic energy density can be motivated from a parallel-plate capacitor. The energy stored in a capacitor is

      1
UE  = --CV 2.
      2
(5)

For a vacuum parallel-plate capacitor with plate area A and spacing d,

       A
C =  𝜖0--.
       d
(6)

Neglecting fringing, the Electric Field between the plates is approximately uniform:

E =  V-.
     d
(7)

Therefore

UE = 1
--
2(  A )
 𝜖0--
    dV 2 (8)
= 1-
2𝜖0A-
d(Ed)2 (9)
= 1
--
2𝜖0E2Ad. (10)

The volume occupied by the approximately uniform field is

V    =  Ad.
 field
(11)

Hence the energy per unit volume is

|-----U-----1-------|
uE  = --E = --𝜖0E2. |
------Ad----2--------
(12)

This result suggests a useful physical viewpoint: the capacitor stores energy in the electric field distributed through the space between the plates.

2 Energy stored in a magnetic field

A similar argument can be made using an ideal long solenoid. The magnetic energy stored in an inductor is

      1-  2
UB =  2LI  .
(13)

For a long vacuum solenoid of length , cross-sectional area A, and N turns,

         2
L  = μ0N--A-.
         ℓ
(14)

The magnetic field inside is approximately

       N--
B =  μ0 ℓ I.
(15)

Solve for the current:

I = -B-ℓ-.
    μ0N
(16)

Substitute into the inductive energy:

UB = 1-
2(        )
    N-2A-
  μ0  ℓ(      )
  -B-ℓ-
  μ0N2 (17)
=   2
B---
2μ0Aℓ. (18)

Since Aℓ is the field-filled volume,

|----------|
|      B2  |
|uB = ----.|
------2-μ0--
(19)

Thus a magnetic field also stores energy locally.

PIC

Figure. The familiar lumped-element energies UE = CV 22 and U B = LI22 can be rewritten as energy distributed through the electric and magnetic fields. In vacuum the corresponding field-energy densities are uE = 𝜖0E22 and u B = B2(2μ 0).

3 Total electromagnetic energy density in vacuum

When both fields are present, the total electromagnetic energy density is

|-----------------------------|
|               1    2   B2   |
u =  uE + uB =  -𝜖0E  +  ---. |
----------------2--------2μ0--
(20)

Here u has units of energy per volume:

          3
[u] = J/m  .
(21)

For fields that vary with position and time,

u =  u(r,t),
(22)

and the total field energy inside a volume V is

|---------∫------------|
UEM  (t) =    u (r,t) dV.|
------------V-----------
(23)

The next question is not how much energy is present, but how that energy moves from one place to another.

4 A conservation law must contain an energy flux

For any conserved quantity, a local conservation equation has the general form

rate of local accumulation + net outward flux = sources or sinks.
(24)

For electromagnetic energy, the accumulation term will be

∂u-.
∂t
(25)

The energy-flux term must have units of power per area,

W/m2,
(26)

because its surface integral must give power. Maxwell’s equations tell us exactly what this flux vector is.

5 Deriving Poynting’s theorem from Maxwell’s equations

Begin with the Ampere–Maxwell law in vacuum with a possibly nonzero conduction current density J:

                    ∂E
∇ × B  = μ0J + μ0 𝜖0∂t-.
(27)

Take the dot product with E and divide by μ0:

1--                           ∂E-
μ0E  ⋅ (∇ × B ) = J ⋅ E + 𝜖0E ⋅ ∂t .
(28)

Now take Faraday’s law,

           ∂B
∇ ×  E = − ----,
            ∂t
(29)

and dot it with B∕μ0:

-1-B ⋅ (∇ × E) = − -1-B ⋅ ∂B-.
μ0                 μ0     ∂t
(30)

The Vector Identity that connects the two curl terms is

|------------------------------------------|
|∇ ⋅ (E  × B ) = B ⋅ (∇ × E) − E ⋅ (∇ × B ).|
-------------------------------------------
(31)

Substituting the two Maxwell Equations into this identity gives

-1-
μ0∇⋅ (E × B) = 1--
μ0B ∂B--
∂t J E 𝜖0E ∂E-
∂t. (32)

Now use

               2
E  ⋅ ∂E = 1-∂E--,
    ∂t    2  ∂t
(33)

and

    ∂B--   1∂B2--
B  ⋅∂t  =  2 ∂t .
(34)

Therefore

-1-
μ
 0∇⋅ (E × B) = ∂--
∂t(             )
  1-  2   -B2-
  2𝜖0E  + 2μ
             0J E. (35)

Move every term to the left:

|---(-------------)-------(----------)-------------|
|∂    1   2    B2           1                      |
|---  -𝜖0E  + ----  + ∇  ⋅  --E  × B   + J ⋅ E = 0.|
-∂t---2-------2-μ0----------μ0----------------------
(36)

This is the differential form of Poynting’s theorem.

6 Definition of the Poynting vector

The energy-flux vector appearing naturally in the derivation is

|--------------|
|S = -1-E × B. |
-----μ0---------
(37)

The Poynting vector points in the direction of electromagnetic energy flow. Its SI units are

[S] = W/m2.
(38)

With this definition and the energy density u, Poynting’s theorem becomes

|------------------------|
|∂u                      |
|--- + ∇ ⋅ S + J ⋅ E = 0.|
--∂t---------------------
(39)

Each term has units of power per volume:

     3
W/m   .
(40)

7 What does J E mean?

The electromagnetic force density on charges includes the electric contribution

fE = ρE.
(41)

For a small collection of moving charge, power is force dotted with velocity. In continuum form, the electrical power delivered to matter per unit volume is

p------=-J-⋅ E.|
--matter---------
(42)

If

J ⋅ E > 0,
(43)

the electromagnetic field is doing positive work on matter. Field energy is being converted into mechanical energy, thermal energy, chemical energy, or some other material form.

For an ordinary resistor with Ohm’s law

J = σE,
(44)

we obtain

J ⋅ E = σE2  > 0,
(45)

which is the local form of Joule heating.

If J E < 0, matter is transferring energy back into the electromagnetic field, as occurs in generators or radiating sources during appropriate parts of their operation.

8 Integral form of Poynting’s theorem

Integrate the local conservation equation over a fixed volume V :

∫  ∂u       ∫             ∫
   ---dV +     ∇ ⋅ S dV +    J ⋅ E dV = 0.
 V ∂t        V             V
(46)

For a fixed integration volume,

∫              ∫
   ∂u-dV  =  d-   u dV.
 V  ∂t       dt  V
(47)

Use the divergence theorem on the flux term:

∫             ∮

 V ∇ ⋅ SdV  =   ∂V S ⋅ dA,
(48)

where dA points outward. Therefore

--------------------------------------------
|d ∫          ∮            ∫               |
|--   u dV  +     S ⋅ dA +    J ⋅ E dV = 0.|
-dt--V---------∂V-----------V---------------
(49)

This equation has a direct bookkeeping interpretation:

|------------------------------------------------------------------------------------|
|field-energy accumulation  + net outward EM   power + power  delivered to matter = 0. |
-------------------------------------------------------------------------------------
(50)

PIC

Figure. Poynting’s theorem is an energy balance for a fixed control volume. Electromagnetic energy can accumulate inside the volume, flow through the boundary through S, or be transferred to matter through J E.

9 Example 1: steady power absorbed by a resistor

Suppose a resistor is enclosed by a fixed surface. In steady operation the electromagnetic energy stored inside the chosen volume is not changing appreciably, so

   ∫
-d    u dV = 0.
dt  V
(51)

If the resistor absorbs 5.0 W,

∫

   J ⋅ E dV = +5.0 W.
 V
(52)

Poynting’s theorem therefore requires

∮
    S ⋅ dA = − 5.0W.
 ∂V
(53)

The negative outward flux means that 5.0 W of electromagnetic power flows into the control volume and is delivered to the resistor.

This is a useful conceptual correction to the informal idea that electrical energy simply “travels inside the wire.” In the field description, electromagnetic energy flows through the surrounding fields and is transferred into matter where J E is positive.

10 Poynting vector for a vacuum plane wave

EM16 established that a plane wave propagating in direction k satisfies

     1-^
B  = c k × E.
(54)

Therefore

E = cB,
(55)

and the electric field, magnetic field, and propagation direction are mutually perpendicular.

The Poynting vector is

    -1-
S = μ  E × B.
      0
(56)

Because E B,

S =  EB--.
     μ0
(57)

Use B = E∕c:

     E2--
S =  μ0c.
(58)

Since

-1--= 𝜖 c,
μ0c    0
(59)

we may also write

|----------|
|        2 |
S-=--𝜖0cE-.-
(60)

Equivalently, using E = cB,

|----------|
|    -c- 2 |
S =  μ  B .|
------0-----
(61)

PIC

Figure. For a vacuum plane wave, E, B, and the energy-flow direction are mutually perpendicular. The Poynting vector points in the same direction as wave propagation.

11 Electric and magnetic energies are equal in a vacuum plane wave

The electric energy density is

u   = 1-𝜖E2.
  E   2  0
(62)

The magnetic energy density is

      B2
uB = ----.
     2 μ0
(63)

For a plane wave,

B =  E-.
     c
(64)

Thus

uB = -1--
2μ0E2-
 c2 (65)
= 1-
2E2(     )
  -1---
  μ0c2. (66)

But

 1
-----= 𝜖0.
μ0c2
(67)

Therefore

|------------------|
|           1      |
|uB = uE  = -𝜖0E2. |
------------2-------
(68)

The total energy density is consequently

|----------------|
|        2   B2  |
|u = 𝜖0E  =  μ--.|
--------------0--
(69)

Comparing this with the Poynting magnitude,

S =  𝜖0cE2,
(70)

gives

|--------|
-S-=-uc.-|
(71)

This has an intuitive interpretation: a plane wave with energy density u transports that energy at speed c.

12 Sinusoidal waves: instantaneous energy flow

Consider a linearly polarized harmonic wave propagating in +z:

E (z,t) = E0 cos(kz − ωt)^x,
(72)

B (z,t) = B0 cos(kz − ωt)^y,
(73)

with

B0 =  E0-.
       c
(74)

The instantaneous Poynting vector is

S = -1-
μ0E × B (75)
= E0B0
--μ---
   0 cos 2(kz ωt)z. (76)

Using B0 = E0∕c,

|------------------------------|
|S(z,t) = 𝜖0cE20 cos2(kz −  ωt)^z.|
--------------------------------
(77)

Because cos 2 is never negative, the energy flux remains in the +z direction even while the field components themselves reverse sign every half cycle.

13 Intensity is the time-averaged Poynting flux

For rapidly oscillating electromagnetic waves, instruments often respond to energy averaged over many periods rather than to the instantaneous carrier oscillation. The time average of

   2
cos (kz − ωt)
(78)

over one period is

⟨   2          ⟩   1
 cos (kz − ωt ) =  2.
(79)

Therefore the average Poynting vector is

       1
⟨S⟩ =  -𝜖0cE20^z.
       2
(80)

The intensity of the plane wave is the magnitude of this time-averaged energy flux:

|--------------------|
|I = |⟨S⟩| = 1𝜖 cE2 .|
-------------2-0---0-|
(81)

Using the magnetic-field amplitude,

|------------|
|      c   2 |
|I = 2μ--B 0.|
--------0----
(82)

The distinction is important:

  • S(r,t) is the instantaneous electromagnetic power-flow density;
  • Sis its time average;
  • intensity I commonly denotes the magnitude of the time-averaged Poynting vector for a periodic traveling wave.

14 RMS fields and the vacuum impedance

For a sinusoidal electric field,

        E0
Erms = √---.
         2
(83)

Therefore

I = 1
--
2𝜖0cE02 (84)
= 𝜖0cErms2. (85)

Define the vacuum wave impedance

|-----∘------------|
|Z  =    μ0-= μ  c.|
| 0      𝜖0     0  |
-------------------
(86)

Numerically,

Z0  ≈ 376.73 Ω.
(87)

Since

-1-
Z0 =  𝜖0c,
(88)

we obtain the useful RF relation

-----------
|      2   |
|I = E-rms-.|
------Z0---|
(89)

In terms of peak electric-field amplitude,

|------2--|
I =  E-0. |
-----2Z0---
(90)

These are electromagnetic analogues of familiar power relations in circuit theory. The impedance Z0 links electric and magnetic field amplitudes in a traveling wave.

15 Example 2: intensity of a 1 V/m plane wave

Suppose a sinusoidal vacuum plane wave has peak electric-field amplitude

E0 =  1.00 V/m.
(91)

Its magnetic-field amplitude is

B0 = E0-
 c (92)
=       1.00
----------------8
2.99792458 × 10 (93)
= 3.34 × 109 T. (94)

Thus

|--------------|
-B0-≈--3.34-nT.-|
(95)

The average intensity is

I = 1-
2𝜖0cE02 (96)
1.33 × 103 W/m2. (97)

Therefore

|------------------|
|I ≈ 1.33 mW/m2.   |
-------------------
(98)

A field amplitude that sounds modest in volts per meter can therefore be translated directly into an energy-flow density.

16 Power crossing an arbitrary surface

If an electromagnetic field crosses a surface A, the instantaneous power through that surface is

|--------------|
|    ∫         |
P  =    S ⋅ dA.|
------A---------
(99)

For a uniform plane wave normally incident on a flat area A,

P  = SA.
(100)

For a sinusoidal wave, the average power is

|----------|
Pavg = IA. |
------------
(101)

If the wave arrives at an angle 𝜃 relative to the surface normal,

P    = IA cos 𝜃.
 avg
(102)

The factor A cos 𝜃 is the projected area presented to the power flow.

17 Spherical spreading and the inverse-square law

Now consider an ideal isotropic source radiating total time-averaged power Prad uniformly in all directions. Far enough from the source, imagine a sphere of radius r surrounding it.

Conservation of energy requires the same total radiated power to cross every such sphere:

       ∫
Prad =    ⟨S ⟩ ⋅ dA.
         A
(103)

For an isotropic source the intensity is uniform over the sphere, so

Prad = I(r)(4πr2).
(104)

Therefore

|------------|
|       Prad-|
I-(r)-=--4πr2.-
(105)

This is the electromagnetic inverse-square law for power density in lossless free-space spherical spreading.

If the radius doubles,

I(2r) = --Prad-- = 1-I(r).
        4 π(2r)2   4
(106)

PIC

Figure. In lossless spherical spreading the same radiated power crosses spheres whose area grows as 4πr2. Intensity therefore falls as 1∕r2. Because intensity is proportional to field amplitude squared, the field amplitudes fall as 1∕r.

18 Why field amplitude falls as 1∕r

The inverse-square law describes power density. It does not say that the electric-field amplitude falls as 1∕r2.

For a plane-wave-like far field,

     E20-
I =  2Z0.
(107)

If spherical spreading gives

       1
I(r) ∝ -2,
       r
(108)

then

  2      1
E0 (r) ∝  -2.
         r
(109)

Taking the positive square root of the amplitude relation gives

|----------|
E  (r) ∝ 1.|
--0------r--
(110)

Because B0 = E0∕c,

|-----------|
|        1- |
B0 (r) ∝ r. |
------------
(111)

This is the key connection between power spreading and amplitude spreading:

|------------------------------------------------|
field amplitude  ∼ 1-   =⇒    power  density ∼  1-.|
------------------r---------------------------r2--
(112)

19 Deriving the far-field electric amplitude of an isotropic radiator

Combine

       Prad
I(r) = ---2-
       4πr
(113)

with

       2
I =  E-0.
     2Z0
(114)

Then

E20-   Prad-
2Z0 =  4πr2.
(115)

Solve for the peak electric-field amplitude:

 2   2Z0Prad
E0 = -----2--.
       4πr
(116)

Therefore

|----------∘---------|
|        1   Z0Prad  |
|E0(r) = --  -------.|
---------r-----2π----
(117)

For the RMS field,

|------------∘---------|
|          1-  Z0Prad- |
|Erms(r) = r     4π   .|
-----------------------
(118)

The 1∕r dependence is therefore not an independent empirical rule. It follows from conservation of radiated power plus the fact that electromagnetic intensity is proportional to the square of field amplitude.

20 Example 3: a 10 W isotropic radiator at 100 m

Let

Prad = 10.0W,      r = 100 m.
(119)

The intensity is

I = --10.0---
4π(100)2 (120)
= 7.96 × 105 W/m2. (121)

Thus

|-----------------|
I = 79.6 μW/m2.   |
------------------
(122)

The RMS electric field is

Erms = ∘ ----
  IZ0 (123)
= ∘ ----------−5---------
  (7.96 × 10  )(376.73) (124)
0.173 V/m. (125)

The peak amplitude is

      √ --
E0 =    2Erms ≈ 0.245 V/m.
(126)

Therefore

|----------------|
E0  ≈ 0.245 V/m. |
------------------
(127)

The corresponding magnetic amplitude is

      E0-            −10
B0  =  c  ≈ 8.17 × 10    T.
(128)

If the distance is doubled to 200 m, intensity falls by a factor of four while E0 and B0 each fall by a factor of two.

21 Directional radiation and gain

Real antennas are generally not isotropic. They redistribute radiated power with direction. In the far field, the power density in a particular direction is commonly written

|----------------------|
|I(r,𝜃,ϕ) = PtG-(𝜃,ϕ-),|
---------------4πr2----|
(129)

where Pt is transmitter power delivered to the radiating system under the convention being used and G(𝜃,ϕ) is antenna gain in the specified direction.

The product

PtG
(130)

is the quantity that leads to effective isotropic radiated power in a chosen direction. Combining the gain form with

    E20
I = ----
    2Z0
(131)

gives

|------------∘----------------|
E  (r,𝜃,ϕ) =   2Z0PtG--(𝜃,ϕ). |
--0-----------------4πr2------|
(132)

This relation is a direct bridge from Maxwell’s field quantities to the power-density language used in antenna and RF engineering.

22 The far-field qualification matters

The simple relations

                  E2           1
E =  cB,     S =  ---,    I ∝  -2,
                  Z0           r
(133)

are traveling-wave or radiation-zone relations. Close to an antenna, reactive electric and magnetic fields can store and return energy rather than carry it irreversibly outward.

In such a near-field region:

  • E and B need not have the simple plane-wave amplitude ratio;
  • the instantaneous Poynting vector can have complicated spatial structure;
  • some field energy can oscillate back and forth near the source;
  • a simple 1∕r2 intensity law need not describe every field component.

Far from a localized radiator, the radiative terms dominate and the field approaches the transverse-wave behavior developed in EM16 and this article.

23 A useful distinction: energy density, flux, intensity, and power

These quantities are related but should not be confused.

Energy density

    1    2    B2
u = --𝜖0E   + ----,
    2        2μ0
(134)

with units J/m3.

Instantaneous Poynting vector

     1
S = ---E × B,
    μ0
(135)

with units W/m2.

Intensity for a periodic traveling wave

I = |⟨S⟩|,
(136)

also with units W/m2.

Power through a surface

     ∫
P  =    S ⋅ dA,
      A
(137)

with units W.

The progression is therefore

|--------------------------------------------------------------------------|
|field amplitudes −→  energy  density and  flux −→  intensity − →  total power. |
---------------------------------------------------------------------------
(138)

24 Extension to simple material media

In a simple linear, nondispersive medium, the energy density is commonly written

u = 1-E ⋅ D + 1B  ⋅ H,
    2         2
(139)

and the Poynting vector is

|------------|
|S = E ×  H. |
-------------
(140)

In vacuum,

                    B
D  = 𝜖0E,     H  =  --,
                    μ0
(141)

so these reduce to the formulas derived above.

Caution is required for strongly dispersive, lossy, nonlinear, or anisotropic media because the relation between stored field energy and the constitutive response can require additional terms or more careful definitions. The vacuum result remains the cleanest starting point.

25 Common mistakes

  • Confusing energy density u in J/m3 with intensity I in W/m2.
  • Forgetting that the Poynting vector is a vector: its direction matters as much as its magnitude.
  • Using S = EB∕μ0 without first checking that E and B are perpendicular.
  • Using E = cB for arbitrary static or near-field configurations. This is a plane-wave or radiation-zone relation in vacuum.
  • Forgetting the factor 12 when converting sinusoidal peak amplitudes to time-averaged intensity.
  • Mixing peak and RMS field amplitudes. For a sinusoid, Erms = E0√ --
  2.
  • Assuming that field amplitude obeys a 1∕r2 law. In spherical radiation, intensity scales as 1∕r2 while field amplitudes scale as 1∕r.
  • Applying I = P∕(4πr2) to a directional antenna without including its radiation pattern or gain.
  • Applying far-field inverse-square relations inside a reactive near field.
  • Misreading the sign of J E. Positive J E means electromagnetic energy is being delivered to matter.

26 What EM17 adds to the series

EM16 established that Maxwell’s equations support waves. EM17 adds the energy interpretation of those waves.

The vacuum electromagnetic energy density is

|------------------|
|    1    2    B2  |
|u = --𝜖0E   + ----.|
-----2--------2μ0--
(142)

The electromagnetic energy-flux vector is

|--------------|
|S = -1-E × B. |
|    μ0        |
----------------
(143)

Maxwell’s equations imply the conservation law

|∂u----------------------|
|--- + ∇ ⋅ S + J ⋅ E = 0.|
--∂t---------------------|
(144)

For a sinusoidal plane wave in vacuum,

|---------------------------|
|    1     2   E20    E2rms  |
I =  -𝜖0cE0 =  ----=  ----. |
-----2---------2Z0-----Z0---
(145)

Finally, conservation of radiated power over an expanding sphere gives

|--------|
|    -1  |
|I ∝ r2 ,|
---------
(146)

while the electromagnetic field amplitudes scale as

|--------------------|
E0  ∝ 1-,    B0 ∝  1.|
------r------------r--
(147)

These results form the energy bridge from Maxwell’s equations to antennas, RF propagation, received power, and link-budget calculations.

References

[1]   David J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017.

[2]   Edward M. Purcell and David J. Morin, Electricity and Magnetism, 3rd ed., Cambridge University Press, 2013.

[3]   Samuel J. Ling, Jeff Sanny, and William Moebs, University Physics, Volume 2, OpenStax, 2016, sections on electromagnetic energy, Poynting’s theorem, and electromagnetic waves.

[4]   Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, Volume II, Addison-Wesley, 1964, chapters on electromagnetic energy and energy flow.

[5]   Massachusetts Institute of Technology, 8.02 Physics II: Electricity and Magnetism, MIT OpenCourseWare, materials on Maxwell’s equations, electromagnetic waves, energy density, and the Poynting vector.


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Cross-references: static, formulas, system, square, impedance, speed, magnitude, boundary, divergence, operation, generators, Ohm's law, work, velocity, charges, force, identity, Maxwell Equations, curl, Vector Identity, dot product, conduction, flux, position, volume, Electric Field, radiation, power, field, relations, theorem, vector, energy, magnetic fields, wave equation, waves, Maxwell's equations, EM16
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This is version 1 of Electromagnetic Waves: Electromagnetic Energy, the Poynting Vector, and Intensity, born on 2026-09-19.
Object id is 1241, canonical name is ElectromagneticWavesElectromagneticEnergyThePoyntingVectorAndIntensity.
Accessed 3 times total.

Classification:
Physics Classification41.20.Jb (Electromagnetic wave propagation; radiowave propagation )
 41.20.-q (Applied classical electromagnetism)
 84.40.-x (Radiowave and microwave technology)
Pending Errata and Addenda
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