Electromagnetic Waves, Antennas, and RF: Electromagnetic Energy, Poynting Vector, and Intensity - Exercises and Complete Worked Solutions

EM17 introduced electromagnetic energy density, the Poynting vector, Poynting’s theorem, plane-wave intensity, vacuum impedance, and spherical spreading. This companion article turns those ideas into a sequence of worked problems. The emphasis is on connecting the field quantities E and B to measurable energy, power, and intensity while keeping the local conservation law visible throughout the calculations.

The central relations are

(1)

(2)

(3)

and

(4)

For a sinusoidal plane wave in vacuum,

(5)

where

(6)

These formulas are most useful when treated as consequences of electromagnetic energy conservation rather than as isolated expressions [1, 2, 3, 4, 5].

How to use this problem set

Attempt all exercises in Part I before consulting Part II. For each problem, first identify whether the requested quantity is an energy density, an energy flux, a total power, or a time average. This avoids one of the most common mistakes in this topic: mixing quantities with units of J/m³, W/m², and W.

Part I: Exercises

Exercise 1: electric-field energy density from a capacitor

A vacuum parallel-plate capacitor has

(7)

Neglect fringing. Find:

1. the capacitance;
2. the stored electrostatic energy from U_E = CV ²∕2;
3. the Electric Field magnitude;
4. the field energy density u_E;
5. the total energy obtained by multiplying u_E by the field-filled volume Ad.

Use the result to verify explicitly that the lumped-element and field-energy descriptions agree.

Exercise 2: magnetic-field energy density from a solenoid

An ideal vacuum solenoid has

(8)

Find:

1. the inductance L;
2. the magnetic field magnitude inside the solenoid;
3. the magnetic energy density u_B;
4. the total field energy u_BAℓ;
5. the inductive energy LI²∕2.

Verify that the two energy calculations agree.

Exercise 3: equal electric and magnetic energy in a vacuum plane wave

A vacuum plane wave has instantaneous electric-field magnitude E = 75 V/m at some point and time.

Using

(9)

calculate:

1. B;
2. u_E;
3. u_B;
4. the total instantaneous electromagnetic energy density u.

Show numerically that u_E = u_B.

Exercise 4: direction and magnitude of the Poynting vector

At a point in vacuum,

(10)

and

(11)

Find the instantaneous Poynting vector

(12)

State both its magnitude and direction. Then compare B with the plane-wave value E∕c.

Exercise 5: average intensity of a sinusoidal plane wave

A sinusoidal vacuum plane wave has peak electric-field amplitude

(13)

Find:

1. the peak magnetic-field amplitude B₀;
2. E_rms and B_rms;
3. the time-averaged intensity using
   
4. the same intensity using
   

Exercise 6: infer field amplitudes from measured intensity

A receiver measures an average electromagnetic intensity

(14)

for a nearly plane vacuum wave.

Find:

1. E_rms;
2. E₀;
3. B_rms;
4. B₀.

Exercise 7: power through a tilted receiving surface

A uniform time-averaged Poynting vector has magnitude

(15)

It crosses a flat receiver of area

(16)

The receiver’s outward area normal makes an angle

(17)

with the direction of S.

Find the magnitude of the electromagnetic power crossing the surface. Explain why the cosine uses the angle to the surface normal, not the angle to the surface itself.

Exercise 8: integral Poynting theorem and energy bookkeeping

Inside a fixed control volume, the electromagnetic field energy is increasing at the rate

(18)

At the same time, the fields transfer energy to matter inside the volume at the rate

(19)

Using the integral form of Poynting’s theorem,

(20)

find the net outward electromagnetic power through the boundary. Interpret the sign physically.

Exercise 9: local Joule heating from J ⋅ E

Within a small conducting region, the fields and current density are approximately uniform:

(21)

(22)

The Conductor volume is

(23)

Find:

1. the local electromagnetic power-transfer density J ⋅ E;
2. the total power transferred from the fields to the material in this volume;
3. the result if the current density were reversed while E remained unchanged.

Interpret the sign of J ⋅ E.

Exercise 10: derive Poynting’s theorem from Maxwell’s equations

Starting with

(24)

and

(25)

perform the following steps:

1. dot the first equation with E∕μ₀;
2. dot the second equation with B∕μ₀;
3. combine the results using
   
4. identify the time derivatives of u_E and u_B;
5. derive
   

State the physical meaning of each term.

Exercise 11: spherical spreading and the difference between power and amplitude

An isotropic radiator emits total power

(26)

Assume lossless far-field propagation.

Find the intensity at

(27)

and

(28)

Then determine:

1. the intensity ratio I₂∕I₁;
2. the electric-field amplitude ratio E₂∕E₁;
3. the magnetic-field amplitude ratio B₂∕B₁.

Explain why power density follows 1∕r² while field amplitude follows 1∕r.

Exercise 12: transmitter gain, power density, and received field strength

A transmitter radiates

(29)

through an antenna with gain

(30)

in the direction of a receiver 100 m away. Treat G as a linear power ratio and assume far-field free-space propagation.

Find:

1. the equivalent isotropically radiated power P_tG;
2. the intensity at the receiver;
3. E_rms;
4. the peak electric-field amplitude E₀.

Exercise 13: why power uses 10 log ₁₀ but field amplitude uses 20 log ₁₀

For a plane wave in a fixed medium,

(31)

Show that the intensity ratio in decibels can be written either as

(32)

or as

(33)

Then evaluate both expressions when the distance from an ideal spherical radiator doubles. Show that they give the same decibel change.

Exercise 14: standing waves can store energy without carrying net average power

Two equal-amplitude plane waves travel in opposite directions along z with the same polarization. Their superposition is

(34)

(35)

Find the instantaneous Poynting vector. Then show that its time average over one period is zero everywhere. Explain how this is consistent with the fields possessing nonzero electromagnetic energy.

Exercise 15: electromagnetic energy carried by a finite pulse

A plane electromagnetic pulse in vacuum has uniform cross-sectional area

(36)

and duration

(37)

During the pulse, approximate its average intensity as constant at

(38)

Find:

1. the average power transported through the cross section;
2. the total electromagnetic energy carried by the pulse;
3. the spatial pulse length L = cΔt;
4. the pulse volume AL;
5. the average electromagnetic energy density obtained from U∕(AL);
6. verify that this equals I∕c.

Exercise 16: Julia check of inverse-square and inverse-distance scaling

Write a short Julia program that evaluates, for 1 ≤ r ≤ 100 m,

(39)

and

(40)

for an isotropic radiator with P = 1 W.

Use logarithms to estimate the slopes of I(r) and E_rms(r) versus r on log-log axes. The expected slopes are −2 and −1, respectively.

Part II: Complete Worked Solutions

Solution 1: electric-field energy density from a capacitor

The capacitance is

C	= 𝜖0	(41)
	= (8.854 × 10−12)	(42)
	= 1.18 × 10−10 F.	(43)

Thus,

(44)

The stored energy is

UE	= CV 2	(45)
	= (1.18 × 10−10)(600)2	(46)
	= 2.13 × 10−5 J.	(47)

Therefore,

(48)

The electric field is

E	=	(49)
	=	(50)
	= 4.00 × 105 V/m.	(51)

The electric energy density is

uE	= 𝜖0E2	(52)
	= (8.854 × 10−12)(4.00 × 105)2	(53)
	= 0.708 J/m3.	(54)

The field-filled volume is

(55)

Hence

uEAd	= (0.708)(3.0 × 10−5)	(56)
	= 2.13 × 10−5 J.	(57)

This agrees with CV ²∕2, confirming that the capacitor energy may be regarded as energy stored in the electric field.

Solution 2: magnetic-field energy density from a solenoid

For an ideal long solenoid,

(58)

Thus,

L	= (1.2566 × 10−6)	(59)
	= 6.79 × 10−4 H.	(60)

Therefore,

(61)

The magnetic field is

B	= μ0I	(62)
	= (1.2566 × 10−6)(2.0)	(63)
	= 3.77 × 10−3 T.	(64)

The magnetic energy density is

uB	=	(65)
	= 5.65 J/m3.	(66)

The field volume is

(67)

Hence

uBAℓ	= (5.65)(2.40 × 10−4)	(68)
	= 1.36 × 10−3 J.	(69)

The inductive energy is

LI2	= (6.79 × 10−4)(2.0)2	(70)
	= 1.36 × 10−3 J.	(71)

Thus the field and lumped-element pictures agree.

Solution 3: equal electric and magnetic energy in a vacuum plane wave

The magnetic field is

B	=	(72)
	=	(73)
	= 2.50 × 10−7 T.	(74)

The electric energy density is

uE	= 𝜖0E2	(75)
	= (8.854 × 10−12)(75)2	(76)
	= 2.49 × 10−8 J/m3.	(77)

For the magnetic energy density,

uB	=	(78)
	= 2.49 × 10−8 J/m3.	(79)

Thus

(80)

The total energy density is

(81)

This equality follows generally for a vacuum plane wave because B = E∕c and c⁻² = μ ₀𝜖₀.

Solution 4: direction and magnitude of the Poynting vector

The cross product is

(82)

Since

(83)

we obtain

S	= (4.80 × 10−5)z	(84)
	= 38.2 z W/m2.	(85)

Therefore,

(86)

The plane-wave prediction is

	=	(87)
	= 4.00 × 10−7 T,	(88)

which matches the specified B to the shown precision. The fields are therefore consistent with an instantaneous sample of a plane wave propagating in the +z direction.

Solution 5: average intensity of a sinusoidal plane wave

The peak magnetic amplitude is

B0	=	(89)
	=	(90)
	= 4.00 × 10−7 T.	(91)

The RMS amplitudes are

(92)

and

(93)

Using the peak-field expression,

I	= 𝜖0cE02	(94)
	= (8.854 × 10−12)(2.998 × 108)(120)2	(95)
	= 19.1 W/m2.	(96)

Using the RMS form,

I	=	(97)
	=	(98)
	= 19.1 W/m2.	(99)

Both methods agree.

Solution 6: infer field amplitudes from measured intensity

From

(100)

we obtain

Erms	=	(101)
	=	(102)
	= 30.7 V/m.	(103)

Therefore,

(104)

The magnetic fields follow from B = E∕c:

(105)

(106)

Solution 7: power through a tilted receiving surface

The power crossing a surface is

(107)

For uniform S over a flat area,

(108)

Hence

P	= (50)(0.80) cos 35∘	(109)
	= 32.8 W.	(110)

Therefore,

(111)

The dot product uses the area vector A = An, so the relevant angle is the angle between S and the surface normal. If an angle to the plane itself is given, it must first be converted to its complementary angle relative to the normal.

Solution 8: integral Poynting theorem and energy bookkeeping

Write

(112)

Poynting’s theorem gives

(113)

Therefore,

(114)

A negative outward flux means that the net electromagnetic flux is actually inward. Twenty watts of electromagnetic power enters the volume. Of that incoming power, 12 W increases stored field energy and 8 W is transferred to matter:

(115)

This is precisely the energy balance encoded by the sign convention in Poynting’s theorem.

Solution 9: local Joule heating from J ⋅ E

Because J and E are parallel,

J ⋅ E	= (3.0 × 106)(0.050)	(116)
	= 1.50 × 105 W/m3.	(117)

Thus,

(118)

The total material power is

Pmatter	= (J ⋅ E)V	(119)
	= (1.50 × 105)(2.0 × 10−5)	(120)
	= 3.0 W.	(121)

If the current density is reversed,

(122)

The negative sign means that, locally, matter is transferring energy to the electromagnetic field instead of receiving energy from it. Positive J ⋅ E corresponds to field energy being delivered to charges, as in ordinary Joule heating.

Solution 10: derive Poynting’s theorem from Maxwell’s equations

Start from the Ampere–Maxwell law:

(123)

Dot with E∕μ₀:

(124)

Now begin with Faraday’s law,

(125)

and dot with B∕μ₀:

(126)

Use the Vector Identity

(127)

Substituting the two Maxwell relations gives

(128)

Recognize

(129)

and

(130)

Define

(131)

and

(132)

Then

(133)

The three terms mean, respectively: local accumulation of electromagnetic field energy, net outward electromagnetic energy flux, and local transfer of electromagnetic energy to matter.

Solution 11: spherical spreading and the difference between power and amplitude

For an isotropic radiator,

(134)

At r₁ = 2.0 m,

I1	=	(135)
	= 1.99 × 10−1 W/m2.	(136)

At r₂ = 20 m,

I2	=	(137)
	= 1.99 × 10−3 W/m2.	(138)

Therefore,

(139)

For a plane-like far-field wave,

(140)

and similarly I ∝ B². Hence

(141)

and

(142)

The sphere area grows as 4πr², so the same total power is distributed over an area proportional to r². Since intensity is quadratic in field amplitude, the amplitudes fall only as 1∕r.

Solution 12: transmitter gain, power density, and received field strength

The equivalent isotropic power in the receiver direction is

(143)

The corresponding far-field intensity is

I	=	(144)
	=	(145)
	= 4.77 × 10−4 W/m2.	(146)

Using

(147)

we obtain

Erms	=	(148)
	= 0.424 V/m.	(149)

Thus

(150)

These values assume lossless free-space propagation in the antenna far field and use the specified directional gain.

Solution 13: why power uses 10 log ₁₀ but field amplitude uses 20 log ₁₀

Because

(151)

we have

10 log 10	= 10 log 10	(152)
	= 20 log 10.	(153)

If distance doubles, spherical spreading gives

(154)

and

(155)

Therefore,

10 log 10(1∕4)	= −6.02 dB,	(156)
20 log 10(1∕2)	= −6.02 dB.	(157)

The factor of 20 for amplitude is not a different physical loss law; it arises because power is proportional to amplitude squared.

Solution 14: standing waves can store energy without carrying net average power

The Poynting vector is

(158)

Since x ×y = z,

S(z,t)	= z	(159)
	= cos(kz) sin(kz) cos(ωt) sin(ωt)z.	(160)

Using

(161)

we obtain

(162)

The average of sin(2ωt) over a full period is zero, so

(163)

The electric and magnetic fields still possess nonzero energy density. In a Standing Wave, energy moves locally back and forth between regions and between electric and magnetic storage, but there is no net time-averaged transport in either propagation direction.

Solution 15: electromagnetic energy carried by a finite pulse

The average transported power is

P	= IA	(164)
	= (3.0 × 104)(0.010)	(165)
	= 300 W.	(166)

The pulse energy is

U	= PΔt	(167)
	= (300)(20 × 10−9)	(168)
	= 6.0 × 10−6 J.	(169)

Thus,

(170)

The spatial pulse length is

L	= cΔt	(171)
	= (2.998 × 108)(20 × 10−9)	(172)
	= 6.00 m.	(173)

Its volume is

(174)

Therefore the average energy density inside the pulse is

u	=	(175)
	=	(176)
	= 1.0 × 10−4 J/m3.	(177)

Finally,

	=	(178)
	= 1.00 × 10−4 J/m3,	(179)

which agrees with u = I∕c for the averaged plane-wave quantities.

Solution 16: Julia check of inverse-square and inverse-distance scaling

One direct implementation is:

using Printf

P = 1.0
Z0 = 376.730313668

r = 10 .^ range(0.0, 2.0, length=101)
I = P ./ (4*pi .* r.^2)
E = sqrt.(Z0 .* I)

# Endpoint log-log slopes
slope_I = (log(I[end]) - log(I[1])) / (log(r[end]) - log(r[1]))
slope_E = (log(E[end]) - log(E[1])) / (log(r[end]) - log(r[1]))

@printf("slope of I(r) = %.6f\n", slope_I)
@printf("slope of E(r) = %.6f\n", slope_E)

@printf("I(1 m) = %.6e W/m^2\n", I[1])
@printf("I(100 m) = %.6e W/m^2\n", I[end])
@printf("E(1 m) = %.6e V/m RMS\n", E[1])
@printf("E(100 m) = %.6e V/m RMS\n", E[end])

Because the formulas are exact powers of r, the logarithms are

(180)

and

(181)

Therefore the expected numerical slopes are

(182)

This computational check makes the distinction between power spreading and amplitude spreading especially clear.

What EM17E1 adds to the series

EM17 established the energy interpretation of the electromagnetic field. EM17E1 makes that interpretation operational. The worked problems connect stored field energy, local energy transfer, boundary flux, plane-wave intensity, antenna power density, and geometric spreading through one conservation framework.

The key chain is

(183)

The next article can build naturally from energy flow to electromagnetic momentum, radiation pressure, or from intensity to antenna reception and effective aperture.

References