Electromagnetic Waves, Antennas, and RF: Gauss’s Law - Exercises and Complete Worked Solutions

This companion article provides self-study exercises for EM07, Gauss’s Law. All exercises are stated first. Complete worked solutions follow in Part II.

The central law is

(1)

The most important methodological distinction is

(2)

For symmetric charge distributions, useful results from EM07 are

(3)

for spherical symmetry,

(4)

for an ideal infinite line charge, and

(5)

for an ideal infinite uniformly charged sheet.

The differential form is

(6)

These are the same definitions and conventions used in EM07 [1, 2, 3, 5].

How to use this problem set

Attempt every exercise in Part I before reading Part II. For each Gaussian-surface problem, explicitly answer these questions before writing the flux integral:

1. What symmetry does the physical charge distribution possess?
2. What direction must E point because of that symmetry?
3. On what parts of the Gaussian surface is E constant?
4. On what parts is E ⋅ dA zero?
5. What charge is actually enclosed?

Choosing a convenient surface does not create symmetry that the charge distribution does not have.

Part I: Exercises

Exercise 1: net enclosed charge and total flux

A closed Gaussian surface encloses three point charges:

(7)

Find:

1. the net enclosed charge;
2. the total electric flux through the closed surface.

Use

(8)

Exercise 2: off-center charge inside an arbitrary closed surface

A point charge q = +4.0 nC lies inside an irregular closed surface but is not at its geometric center.

1. Find the total electric flux through the surface.
2. Can the field magnitude be assumed constant on the surface?
3. Can the flux integral be replaced by EA for this arbitrary surface? Explain.

Exercise 3: external charge and zero enclosed charge

A closed box contains no charge. A positive point charge lies outside the box and produces a nonzero Electric Field throughout much of the box.

What is the net electric flux through the entire closed box? Does your answer imply that the electric field is zero on the box?

Exercise 4: field outside a uniformly charged sphere

A spherically symmetric insulating sphere of radius

(9)

contains total charge

(10)

Find the electric-field magnitude at

(11)

State the field direction.

Exercise 5: field inside a uniformly charged solid sphere

The same sphere as in Exercise 4 has charge uniformly distributed throughout its volume. Find the electric-field magnitude at

(12)

Show explicitly how the enclosed charge is determined.

Exercise 6: derive the field of an infinite line charge

An ideal infinite line carries uniform linear charge density λ > 0 along the z axis.

Using a coaxial cylindrical Gaussian surface of radius s and length L:

1. identify which parts of the cylinder contribute flux;
2. determine the enclosed charge;
3. derive E(s).

Exercise 7: numerical infinite line-charge field

For

(13)

find the electric-field magnitude at

(14)

Use the result derived in Exercise 6.

Exercise 8: field of an infinite charged sheet

An ideal infinite sheet carries positive surface charge density σ.

Using a pillbox Gaussian surface that straddles the sheet, derive

(15)

Explain why the curved side of the pillbox contributes no flux.

Exercise 9: two oppositely charged infinite sheets

Two parallel infinite sheets carry surface charge densities

(16)

Using superposition and the single-sheet result, find the field magnitude:

1. between the sheets;
2. outside the pair.

State the direction of the field between the sheets.

Exercise 10: choose the useful Gaussian surface

For each idealized charge distribution below, identify the most useful Gaussian surface and explain why it matches the field symmetry:

1. a point charge at the origin;
2. an infinite straight line charge;
3. an infinite uniformly charged plane;
4. two unequal point charges separated by a finite distance.

For part (d), state whether Gauss’s law remains valid and whether it is likely to solve directly for E.

Exercise 11: uniform external field through a closed box

A rectangular box has two opposite faces of area

(17)

normal to a uniform field

(18)

Find the flux through:

1. the right face;
2. the left face;
3. the other four faces;
4. the complete closed box.

What enclosed charge is implied?

Exercise 12: infer charge density from divergence

In a region of space,

(19)

where a is a constant.

Find:

1. ∇⋅ E;
2. the corresponding volume charge density ρ using differential Gauss law.

Exercise 13: verify integral and differential Gauss law for a simple field

Consider

(20)

where a is constant. For a sphere of radius R centered at the origin:

1. use the differential form of Gauss’s law to find ρ;
2. find the total enclosed charge from ρ;
3. compute the flux directly from the spherical surface;
4. verify that the two sides of Gauss’s law agree.

Exercise 14: Gauss-law synthesis

For each statement, decide whether it is correct. If incorrect, rewrite it accurately.

1. “Gauss’s law is valid only for symmetric charge distributions.”
2. “If a closed surface has zero net flux, the electric field must be zero everywhere on it.”
3. “Any charge outside a Gaussian surface contributes nothing to the electric field on that surface.”
4. “A spherical Gaussian surface always allows the replacement ∮ E ⋅ dA = E(4πr²).”
5. “In a charge-free region, ∇⋅ E = 0.”
6. “The integral and differential forms of Gauss’s law describe the same physical law.”

Part II: Complete Worked Solutions

Solution 1: net enclosed charge and total flux

The enclosed charge is the algebraic sum:

Qenc	= q1 + q2 + q3	(21)
	= (2.0 − 5.0 + 1.0) nC	(22)
	= −2.0 nC.	(23)

Thus

(24)

Gauss’s law gives

(25)

Therefore,

ΦE	=	(26)
	≈−2.26 × 102 N m2∕C.	(27)

Hence

(28)

The negative sign means the net signed flux is inward.

Solution 2: off-center charge inside an arbitrary closed surface

Gauss’s law depends only on enclosed charge:

(29)

Thus

ΦE	=	(30)
	≈ 4.52 × 102 N m2∕C.	(31)

Therefore,

(32)

The field magnitude cannot generally be assumed constant on the irregular surface. The field also need not be everywhere normal to the surface. Therefore the total flux is known, but the integral cannot generally be replaced by EA.

Solution 3: external charge and zero enclosed charge

The box encloses no charge:

(33)

Therefore,

(34)

This does not imply E = 0. The external charge can produce a nonzero field on the surface. Field entering the box contributes negative flux and field leaving contributes positive flux; the signed contributions cancel over the complete closed surface.

Solution 4: field outside a uniformly charged sphere

Because the charge distribution is spherically symmetric, choose a spherical Gaussian surface of radius r = 0.25 m. Since r > R, the Gaussian surface encloses the full charge:

(35)

Therefore,

(36)

so

(37)

Using

(38)

we obtain

E	= (8.99 × 109)	(39)
	≈ 8.63 × 102 N/C.	(40)

Hence

(41)

Because Q > 0, the field points radially outward.

Solution 5: field inside a uniformly charged solid sphere

Uniform volume charge density means

(42)

The Gaussian sphere of radius r < R encloses only the fraction of charge proportional to its volume:

Qenc	= ρ	(43)
	= Q.	(44)

With r = 0.050 m and R = 0.10 m,

(45)

Thus

(46)

Gauss’s law gives

(47)

Therefore,

E	= (8.99 × 109)	(48)
	≈ 2.70 × 103 N/C.	(49)

Hence

(50)

Equivalently, the inside-sphere formula is

(51)

Solution 6: derive the field of an infinite line charge

Cylindrical symmetry forces the electric field to point radially away from the line and to depend only on perpendicular distance s:

(52)

Choose a coaxial cylinder of radius s and length L.

On the curved side, the field is parallel to the outward area normal, so

(53)

The curved area is

(54)

On the end caps, dA points along ±z while E points radially sideways, so

(55)

The enclosed charge is

(56)

Thus Gauss’s law becomes

(57)

Canceling L gives

(58)

For λ > 0, the field points radially outward.

Solution 7: numerical infinite line-charge field

Use

(59)

Substituting,

E	=	(60)
	≈ 6.74 × 102 N/C.	(61)

Therefore,

(62)

For positive λ, the field points radially away from the line.

Solution 8: field of an infinite charged sheet

Planar symmetry requires the field to be perpendicular to the sheet and to have the same magnitude on both sides.

Choose a pillbox with two flat caps of area A. On the curved side, the field is tangent to the surface, so

(63)

Each flat cap contributes

(64)

Therefore the total flux is

(65)

The enclosed charge is

(66)

Gauss’s law gives

(67)

Canceling A yields

(68)

For σ > 0, the field points away from the sheet on both sides.

Solution 9: two oppositely charged infinite sheets

A single sheet produces magnitude

(69)

Between the +σ and −σ sheets, both fields point from the positive sheet toward the negative sheet, so they add:

Ebetween	= +	(70)
	= .	(71)

Thus

(72)

Outside the pair, the two fields point in opposite directions with equal magnitudes, so

(73)

Solution 10: choose the useful Gaussian surface

1. A point charge at the origin has spherical symmetry. Use a sphere centered on the charge.
2. An infinite straight line charge has cylindrical symmetry. Use a coaxial cylinder.
3. An infinite uniformly charged plane has planar symmetry. Use a pillbox whose flat faces are parallel to the plane.
4. Two unequal separated point charges generally possess insufficient symmetry for a simple Gaussian surface to make E constant over the useful parts of the surface. Gauss’s law remains exactly valid, but it does not usually solve directly for E.

The key logic is

(74)

Solution 11: uniform external field through a closed box

The field is

(75)

For the right face, the outward normal is +x, so

(76)

Thus

(77)

For the left face, the outward normal is −x, so

(78)

The other four faces have normals perpendicular to E, so

(79)

for each of them.

The complete flux is

(80)

Therefore,

(81)

Gauss’s law then gives

(82)

Solution 12: infer charge density from divergence

The field is

(83)

Its divergence is

∇⋅ E	= + +	(84)
	= a + 2a − a	(85)
	= 2a.	(86)

Hence

(87)

Differential Gauss law gives

(88)

so

(89)

Solution 13: verify integral and differential Gauss law for a simple field

The field is

(90)

Its divergence is

(91)

Therefore differential Gauss law gives

(92)

For a sphere of radius R, the enclosed charge is

Qenc	= ρ	(93)
	= (3𝜖0a)	(94)
	= 4π𝜖0aR3.	(95)

Thus

(96)

On the spherical surface, E = aRr, so the field is radial and constant in magnitude. The flux is

ΦE	= E(4πR2)	(97)
	= (aR)(4πR2)	(98)
	= 4πaR3.	(99)

Therefore,

(100)

The right-hand side of Gauss’s law is

(101)

which matches the direct flux calculation exactly.

Solution 14: Gauss-law synthesis

1. Incorrect. Gauss’s law is valid for every closed surface and every charge distribution. Symmetry is needed only to simplify the integral enough to solve easily for the field.
2. Incorrect. Zero net flux does not imply zero field. Positive and negative flux contributions may cancel over the closed surface.
3. Incorrect. External charges can contribute strongly to the electric field on the surface; they contribute zero net flux through the complete closed surface.
4. Incorrect. The replacement ∮ E⋅dA = E(4πr²) requires spherical field symmetry with constant E and radial alignment over the sphere.
5. Correct. If ρ = 0, differential Gauss law gives

   (102)

6. Correct. The divergence theorem connects the integral and differential forms, so they represent the same physical law at global and local levels.

Common mistakes

• Using total charge in the problem instead of the charge enclosed by the chosen Gaussian surface.
• Assuming Gauss’s law itself requires symmetry.
• Pulling E outside a surface integral without proving that it is constant over that part of the surface.
• Forgetting that dA points outward on a closed surface.
• Assuming zero total flux implies zero electric field.
• Assuming external charges produce no field on a Gaussian surface.
• Choosing a Gaussian surface first and then assigning symmetry that the physical charge distribution does not possess.

What EM07E reinforces

The universal statement is

(103)

The useful-symmetry chain is

(104)

And the local form is

(105)

The next lesson, EM08, introduces electric current and current density, moving from static charge toward moving charge, magnetic fields, and eventually electromagnetic radiation.

References