Electromagnetic Waves, Antennas, and RF: Electric Flux - Exercises and Complete Worked Solutions

This companion article provides self-study exercises for EM06, Electric flux. All exercises are stated first. Complete worked solutions follow in Part II.

The problems reinforce four ideas developed in EM06:

1. a surface must be given an orientation through its normal vector;
2. only the component of E normal to the surface contributes to flux;
3. a general surface requires a surface integral; and
4. a closed surface uses outward-pointing area vectors.

For a flat surface of area A with unit normal n,

(1)

For a small surface patch,

(2)

The differential flux is

(3)

and for a uniform field over a flat surface,

(4)

where 𝜃 is measured between the Electric Field and the chosen surface normal.

For a general open surface,

(5)

while for a closed surface the outward-normal convention gives

(6)

These are the same definitions and conventions used in EM06 [1, 2, 3, 4].

How to use this problem set

Attempt all exercises in Part I before reading Part II. For every flux problem, explicitly identify

1. the field vector E;
2. the surface normal n;
3. whether the surface is open or closed;
4. the angle between the field and the normal, not the field and the surface; and
5. the sign of E ⋅ dA.

In Exercises 12 and 13, use the centered point-charge spherical result derived directly from Coulomb’s field in EM06. Do not invoke the general form of Gauss’s Law; that theorem is reserved for EM07.

Part I: Exercises

Exercise 1: area vector and orientation

A flat surface has area

(7)

and chosen unit normal

(8)

Find the area vector A. Then write the area vector if the orientation is reversed.

Exercise 2: field normal to a surface

A uniform electric field has magnitude

(9)

It passes through a flat surface of area

(10)

in the same direction as the chosen surface normal. Find the electric flux.

Exercise 3: tilted flat surface

A uniform field of magnitude

(11)

crosses a flat surface of area

(12)

The angle between E and the chosen surface normal is

(13)

Find the flux.

Exercise 4: angle given relative to the surface

A field of magnitude

(14)

crosses a flat surface of area

(15)

The field makes an angle of 30^∘ with the surface itself. Find the electric flux for the normal that makes an acute angle with the field.

Exercise 5: reversing orientation

For one chosen normal, the electric flux through an open surface is

(16)

What is the flux if the surface orientation is reversed? Has the electric field changed?

Exercise 6: compute flux with a vector dot product

A uniform electric field is

(17)

A flat surface has area

(18)

and unit normal

(19)

Compute the electric flux using Φ_E = E ⋅ A.

Exercise 7: zero flux with a nonzero field

A uniform field is

(20)

A rectangular surface lies in the xy plane and has chosen normal +z.

1. Find the flux through the rectangle.
2. Explain why the result does not imply that the electric field is zero.

Exercise 8: a nonuniform field over a plane

The electric field is

(21)

A rectangular surface lies in the plane x = 2 m, with chosen normal +x. Its coordinate limits are

(22)

Evaluate

(23)

Exercise 9: uniform field through a closed box

A rectangular box is placed in a uniform field

(24)

The two faces perpendicular to the x axis each have area A. Determine the flux through each of those two faces, the flux through the other four faces, and the total flux through the closed box.

Exercise 10: a nonuniform field through a rectangular box

Let

(25)

with

(26)

A rectangular box spans

(27)

Compute the flux through all six faces directly and find the net closed-surface flux.

Exercise 11: flux through a hemisphere in a uniform field

A hemisphere of radius

(28)

occupies the +x side of a sphere. Its curved surface uses the outward normal. A uniform field

(29)

passes through it.

Using the projected-area interpretation, find the flux through the curved hemispherical surface.

Exercise 12: centered point charge and spherical flux

A point charge

(30)

is at the center of a spherical surface. Using the Coulomb-field derivation from EM06,

(31)

find the electric flux through the sphere.

Exercise 13: compare two spherical radii

The same positive point charge is surrounded by two centered spherical surfaces of radii

(32)

Find:

1. the ratio E(r₁)∕E(r₂);
2. the ratio A₂∕A₁ of spherical areas;
3. the ratio Φ₁∕Φ₂ of total fluxes.

Explain why the result is consistent with the inverse-square field.

Exercise 14: diagnose conceptual statements

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

1. “Electric flux is the number of physical electric-field lines crossing a surface.”
2. “A nonzero electric field always produces nonzero flux through every surface.”
3. “The angle in EA cos 𝜃 is measured from the field to the surface normal.”
4. “Reversing the normal of an open surface reverses the sign of its flux.”
5. “For a closed surface, outward normals are the standard orientation.”
6. “The result q∕𝜖₀ derived in EM06 for a centered point charge and sphere is already the full general statement of Gauss’s law.”

Part II: Complete Worked Solutions

Solution 1: area vector and orientation

The area vector is

(33)

Therefore,

A	= (0.40) m2	(34)
	= m2.	(35)

Thus

(36)

Reversing the orientation changes n to −n, so

(37)

The scalar area has not changed. Only the chosen orientation has changed.

Solution 2: field normal to a surface

Because the field points in the same direction as the chosen normal,

(38)

Therefore,

ΦE	= EA cos 0	(39)
	= (250)(0.32)	(40)
	= 80 N m2∕C.	(41)

Hence

(42)

Solution 3: tilted flat surface

Use

(43)

Then

ΦE	= (250)(0.32) cos 60∘	(44)
	= 80	(45)
	= 40 N m2∕C.	(46)

Thus

(47)

The tilt reduces the effective projected area by the factor cos 60^∘ = 1∕2.

Solution 4: angle given relative to the surface

The given 30^∘ is measured from the field to the surface, not to the normal. The normal is perpendicular to the surface, so the angle to the normal is

(48)

Hence

ΦE	= EA cos 60∘	(49)
	= (180)(0.50)	(50)
	= 45 N m2∕C.	(51)

Therefore,

(52)

This exercise illustrates why the angle convention must be checked before applying the cosine formula.

Solution 5: reversing orientation

Reversing the normal sends

(53)

Therefore,

(54)

Thus

(55)

The electric field has not changed. Only the bookkeeping orientation of the open surface changed.

Solution 6: compute flux with a vector dot product

First construct the area vector:

A	= An	(56)
	= 0.40 m2	(57)
	= m2.	(58)

Now take the dot product:

ΦE	= E ⋅ A	(59)
	= (120)(0.24) + (−50)(0.32) + (30)(0)	(60)
	= 28.8 − 16.0	(61)
	= 12.8 N m2∕C.	(62)

Therefore,

(63)

The z component of the field contributes nothing because the chosen surface normal has no z component.

Solution 7: zero flux with a nonzero field

The field points along +x while the surface normal points along +z. These directions are perpendicular, so

(64)

Thus

(65)

The field is still

(66)

which is nonzero. The flux vanishes because the field runs parallel to the surface instead of through it.

Solution 8: a nonuniform field over a plane

The rectangle lies in the plane x = 2 m and has normal +x, so

(67)

Therefore,

(68)

The total flux is

ΦE	= ∫ 00.50 ∫ 01(10 + 4y) dy dz	(69)
	= ∫ 00.50 01dz	(70)
	= ∫ 00.5012 dz	(71)
	= 6.0 N m2∕C.	(72)

Hence

(73)

The simple product EA is not sufficient here because the field magnitude changes with y across the surface.

Solution 9: uniform field through a closed box

For the face whose outward normal is +x,

(74)

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

(75)

On the remaining four faces, the normals are perpendicular to E, so

(76)

for each of those faces.

Adding all six contributions,

(77)

Therefore,

(78)

The field is nonzero, but equal flux enters and leaves the box.

Solution 10: a nonuniform field through a rectangular box

The field is

(79)

Only the two faces perpendicular to the x axis contribute. Their area is

(80)

At the right face, x = 3 m and the outward normal is +x. The field magnitude there is

(81)

Therefore,

(82)

At the left face, x = 1 m and the outward normal is −x. The field magnitude there is

(83)

Thus,

(84)

The other four faces have normals perpendicular to the field, so their flux is zero. Hence

Φclosed	= 30 − 10	(85)
	= 20 N m2∕C.	(86)

Therefore,

(87)

Unlike the uniform-field case, the field is stronger on the right side than on the left, so the outward and inward contributions do not cancel.

Solution 11: flux through a hemisphere in a uniform field

For a uniform field, the flux through the curved hemisphere equals the field magnitude times the area projected onto a plane perpendicular to the field.

The projection of the hemisphere onto the yz plane is a disk of area

(88)

Therefore,

ΦE	= EπR2	(89)
	= (300)π(0.20)2	(90)
	= 12π N m2∕C	(91)
	≈ 37.7 N m2∕C.	(92)

Hence

(93)

The flux is positive because the outward normals on the +x hemisphere have positive x components.

Solution 12: centered point charge and spherical flux

EM06 derived, directly from Coulomb’s field for a centered point charge,

(94)

Using

(95)

and

(96)

we obtain

ΦE	=	(97)
	≈ 3.39 × 102 N m2∕C.	(98)

Thus

(99)

This result is independent of the sphere radius for the centered point-charge geometry.

Solution 13: compare two spherical radii

For a point charge,

(100)

Therefore,

	= 2	(101)
	= 2	(102)
	= 16.	(103)

Hence

(104)

Spherical area scales as r², so

	= 2	(105)
	= 16.	(106)

Thus

(107)

Flux is the product of field magnitude and spherical area in this centered geometry. The factor of 16 decrease in field magnitude from r₁ to r₂ is exactly canceled by the factor of 16 increase in area. Therefore,

(108)

Solution 14: diagnose conceptual statements

1. Incorrect. Electric-field lines are a visualization convention. Electric flux is defined mathematically by

   (109)

2. Incorrect. A nonzero field can give zero flux through a surface if the field is tangent to the surface everywhere, so that E ⋅ dA = 0.
3. Correct. The angle in

   (110)

   is measured between E and the chosen surface normal.

4. Correct. Reversing the normal changes dA to −dA, so the flux changes sign.
5. Correct. Closed surfaces conventionally use outward-pointing area vectors.
6. Incorrect. EM06 derived q∕𝜖₀ only for a sphere centered on a point charge, using Coulomb’s field. EM07 will state and analyze the general law for arbitrary closed surfaces and enclosed charge distributions.

Common mistakes

• Using the angle to the surface rather than the normal. Convert to the complementary angle before using EA cos 𝜃.
• Treating area as a scalar when orientation matters. Flux uses dA = ndA.
• Assuming a strong field guarantees large flux. Tangential field contributes zero normal flux.
• Forgetting the outward-normal convention on a closed surface.
• Using EA for a field that varies over the surface. In that case, evaluate the surface integral.
• Calling field lines physical objects. They are only a visualization of the continuous vector field.
• Invoking general Gauss’s law too early. EM06 and EM06E use only flux definitions plus the special centered-sphere result derived from Coulomb’s field.

What EM06E reinforces

The fundamental local statement is

(111)

For a uniform field over a flat surface,

(112)

For a general surface,

(113)

and for a closed surface,

(114)

The centered point-charge sphere provides the important preview

(115)

but the general relationship between closed-surface flux and enclosed charge belongs to EM07, Gauss’s Law.

References