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Electromagnetic Waves: Electric Flux (Topic)

Electromagnetic Waves, Antennas, and RF: Electric Flux

EM05 introduced the Electric Field E(r) as a vector field in space. EM06 asks a new question: how much of that field passes through a surface?

The answer is described by electric flux. Flux combines two pieces of geometry:

  1. the strength and direction of the electric field, and
  2. the size and orientation of the surface.

The central mathematical operation is the dot product. For a small oriented surface element dA,

|--------------|
d-ΦE-=--E-⋅ dA.-
(1)

For a finite surface,

|------∫---------|
|                |
|ΦE =    E  ⋅ dA.|
--------S---------
(2)

This geometric language prepares the way for Gauss’s Law. EM06 develops flux itself first so that the later law can be understood as physics rather than as an unfamiliar integral formula [1235].

1 Why a surface needs an orientation

A scalar area tells us only how large a surface is. Flux also depends on which way the surface faces.

For a flat surface of area A, choose a unit normal vector n perpendicular to the surface. The corresponding area vector is

|---------|
A--=-A-ˆn.-|
(3)

For an infinitesimal surface patch,

|------------|
-dA--=-ˆn-dA.-|
(4)

The direction of the area vector is not tangent to the surface. It is perpendicular to the surface.

For an open surface, either of the two opposite normals may be chosen. Reversing the chosen normal reverses the sign of the flux.

PIC

Figure. A flat surface is assigned a normal direction n. The angle 𝜃 used in electric flux is the angle between E and the surface normal, not the angle between E and the surface itself.

2 Flux through a uniform flat surface

Suppose the electric field is uniform over a flat surface. Then the flux is

ΦE =  E ⋅ A.
(5)

Using the dot-product definition,

E ⋅ A = EA  cos 𝜃,
(6)

where 𝜃 is the angle between E and the chosen surface normal. Therefore,

|----------------|
|ΦE  = EA  cos𝜃. |
-----------------
(7)

This formula is the simplest electric-flux relation.

2.1 Maximum positive flux

If the field points in the same direction as the surface normal,

𝜃 = 0,
(8)

so

|----------|
-ΦE-=--EA.--
(9)

The field passes through the surface as directly as possible.

2.2 Zero flux

If the field lies parallel to the surface, then it is perpendicular to the normal:

𝜃 = 90∘.
(10)

Therefore,

|--------|
-ΦE-=--0.-
(11)

A strong electric field can therefore produce zero flux through a particular surface if the field runs along that surface rather than through it.

2.3 Negative flux

If the field points mostly opposite the chosen normal,

90∘ < 𝜃 ≤ 180 ∘,
(12)

then

cos𝜃 < 0,
(13)

and the flux is negative.

The sign is therefore an orientation statement, not a statement that the electric-field magnitude is negative.

3 Projected area interpretation

The factor

A cos 𝜃
(14)

has a simple geometric interpretation. It is the area of the surface projected onto a plane perpendicular to the electric field.

Thus

|-----------|
ΦE--=-EA--⊥--
(15)

with

|--------------|
|A ⊥ = A cos𝜃. |
---------------
(16)

A tilted surface presents a smaller effective area to the field.

PIC

Figure. A tilted surface of actual area A presents the projected area A cos 𝜃 to a uniform field. This provides a geometric interpretation of the dot product in ΦE = EA cos 𝜃.

4 Units of electric flux

From

ΦE =  EA,
(17)

we have

[E ] = N/C
(18)

and

[A] = m2.
(19)

Therefore,

|----------------|
|[ΦE ] = N m2 ∕C.|
-----------------
(20)

Electric flux is not measured in coulombs. It is a field-times-area quantity.

5 From one flat patch to a general surface

The simple formula

ΦE  = EA  cos 𝜃
(21)

Works only when the field is effectively uniform over the surface and the surface has one well-defined normal direction.

For a curved surface, or for a field that changes from point to point, divide the surface into many small patches.

For patch i,

ΔΦE,i ≈  Ei ⋅ ΔAi.
(22)

Adding all patches gives

      ∑
ΦE  ≈     Ei ⋅ ΔAi.
        i
(23)

In the limit of infinitesimal patches,

|----------------|
|      ∫         |
|ΦE =    E  ⋅ dA.|
--------S---------
(24)

This is a surface integral.

PIC

Figure. A curved surface is approximated by many small patches. Each patch has its own local normal and area vector dA. The total flux is the sum, in the continuum limit, of E dA over the surface.

6 What the dot product does locally

At every small patch,

dΦE  = E ⋅ dA =  E dA cos𝜃.
(25)

Only the component of E normal to the surface contributes. Define

En =  E ⋅ ˆn.
(26)

Then

|--------------|
|dΦE  = En dA. |
----------------
(27)

The tangential component of the electric field does not contribute to flux through that patch.

7 Open surfaces and closed surfaces

An open surface has an edge. Examples include a disk, a rectangle, or a hemisphere without its flat base. Its normal orientation must be specified.

A closed surface completely encloses a volume. Examples include a sphere, a cube, or a sealed irregular surface.

For a closed surface, the standard convention is

|--------------------|
-dA--points-outward.-|
(28)

The total flux through a closed surface is written

|------∮---------|
|                |
|ΦE =   S E ⋅ dA.|
------------------
(29)

With the outward-normal convention:

  • field leaving the enclosed volume contributes positive flux;
  • field entering the enclosed volume contributes negative flux.

8 A uniform field through a closed box

Consider a constant electric field E = E0x and a rectangular box.

The right face has outward normal +x, so its flux is positive. The left face has outward normal x, so its flux is negative with equal magnitude. The other faces have normals perpendicular to E, so their flux is zero.

Hence the total closed-surface flux is

|Φ--=--0.|
--E-------
(30)

This does not mean the electric field is zero. It means as much field passes into the box as passes out.

9 Field lines are a picture, not the definition

Electric-field lines are often used to visualize flux. A surface crossed by many field lines is drawn as having large flux, while a surface nearly parallel to the lines is drawn as having small flux.

This picture is useful, but flux is not literally a count of physical lines. Field lines are a visualization convention. The mathematical definition is

|------∫---------|
|                |
|ΦE =   S E ⋅ dA.|
------------------
(31)

The field itself is continuous even though a diagram contains only a finite number of drawn lines.

10 Flux of a point-charge field through a centered sphere

EM05 showed that a point charge q at the origin produces

        --1--q-
E (r) = 4π 𝜖0r2ˆr.
(32)

Now surround the charge by a sphere of radius r centered on the charge.

At every point on the sphere, the outward area vector is radial:

dA  = ˆr dA.
(33)

Therefore,

E ⋅ dA =  E dA.
(34)

The field magnitude is the same everywhere on the sphere, so

         ∮
Φ   = E    dA.
  E       S
(35)

The area of the sphere is

4πr2.
(36)

Thus

ΦE = (        )
  -1---q-
  4π𝜖0 r2 (4πr2) (37)
= q
𝜖-
 0. (38)

Therefore,

|---------|
|     -q  |
ΦE  = 𝜖0. |
-----------
(39)

The radius cancels. A larger sphere has weaker field magnitude, but it also has proportionally larger area.

PIC

Figure. For a sphere centered on a point charge, the electric field and outward area vector are parallel everywhere. The 1∕r2 decrease of field magnitude is exactly balanced by the 4πr2 growth of spherical area.

This result is a preview of Gauss’s law. At this stage it has been derived only for a sphere centered on a point charge, using the known Coulomb field. EM07 will state and analyze the much more general law for arbitrary closed surfaces.

11 Worked examples

Example 1: uniform field normal to a surface

A uniform electric field has magnitude

E =  200N/C.
(40)

It passes normally through a flat surface of area

           2
A  = 0.50m  .
(41)

With 𝜃 = 0,

ΦE = EA cos 0 (42)
= (200)(0.50) (43)
= 100 N m2C. (44)

Hence

|------------------|
-ΦE-=--100N-m2-∕C.--
(45)

Example 2: tilted flat surface

Let

                             2            ∘
E =  300N/C,      A  = 0.20m  ,     𝜃 = 60 ,
(46)

where 𝜃 is measured from E to the surface normal.

Then

ΦE = EA cos 𝜃 (47)
= (300)(0.20) cos 60 (48)
= 30 N m2C. (49)

Thus

|-----------------|
|            2    |
ΦE--=-30-N-m--∕C.--
(50)

Example 3: field parallel to the surface

If a field of any magnitude lies exactly parallel to a flat surface, then the angle to the normal is 90. Therefore,

|----------------------|
|ΦE  = EA  cos90 ∘ = 0.|
-----------------------
(51)

A nonzero field can therefore have zero flux through a particular surface.

Example 4: reversing the surface orientation

Suppose a flat surface has flux

              2
ΦE =  +12 N m  ∕C
(52)

for the chosen normal n.

If the surface orientation is reversed,

ˆn →  − ˆn,
(53)

then

dA →  − dA.
(54)

Hence

|--------------------|
|Φ   →  − 12 N m2 ∕C.|
---E-----------------
(55)

The physical field has not changed. Only the orientation convention changed.

Example 5: a nonuniform field through a plane

Let

E(x ) = αx ˆx
(56)

and consider a rectangular surface lying in the plane x = a, with outward normal +x and area A.

Every point on that surface has the same coordinate x = a, so

E =  αaˆx.
(57)

Thus

ΦE = SE dA (58)
= SαadA (59)
= αaA. (60)

Therefore,

|-----------|
ΦE--=-αaA.---
(61)

The field is nonuniform in space generally, but it is uniform over this particular constant-x surface.

Example 6: net flux through a box in a uniform field

Take

E =  E0ˆx.
(62)

Let the two faces normal to the x axis each have area A.

The right face contributes

ΦR  = +E0A.
(63)

The left face contributes

ΦL  = − E0A.
(64)

The other four faces contribute zero. Thus

|-----------|
Φclosed = 0. |
------------
(65)

Example 7: centered point charge and spherical surface

Let

            −9
q = 2.0 × 10   C.
(66)

For a centered sphere, the flux found above is

      -q
ΦE  = 𝜖0.
(67)

Using

               −12  2      2
𝜖0 ≈ 8.854 × 10   C  ∕(N m  ),
(68)

we obtain approximately

|------------------------|
ΦE--≈-2.26-×-102-N-m2-∕C.-
(69)

The answer does not depend on the radius of the centered sphere.

12 Common mistakes

12.1 Using the angle to the surface instead of the normal

In

ΦE  = EA  cos𝜃,
(70)

𝜃 is the angle between E and n.

If a problem gives the angle between the field and the surface itself, convert it to the complementary angle before using the cosine form.

12.2 Treating area as a scalar when orientation matters

The flux integrand uses

dA  = ˆn dA,
(71)

not merely dA.

12.3 Assuming zero flux means zero field

A field parallel to a surface produces zero flux through that surface even though the field can be large.

12.4 Assuming flux is always positive

Flux can be positive, negative, or zero depending on the chosen orientation and local field direction.

12.5 Counting drawn field lines literally

Field-line diagrams illustrate direction and relative density. Flux is defined by an integral, not by counting artistic lines in a sketch.

12.6 Applying EA cos 𝜃 to any surface without checking assumptions

The simple formula requires a uniform field over a flat surface, or at least conditions under which the field and normal are effectively constant. The general formula is

      ∫
Φ  =    E  ⋅ dA.
 E     S
(72)

13 Why electric flux matters later

Electric flux is much more than a geometric exercise. The same mathematical structure appears repeatedly in Electromagnetism.

Gauss’s law will relate closed-surface electric flux to enclosed charge:

closed electric flux ← →  charge inside.
(73)

Later, magnetic flux will use the analogous quantity

B ⋅ dA,
(74)

and Faraday’s law will relate changing magnetic flux to circulation of the electric field.

Still later, electromagnetic power flow through a surface will use the Poynting vector:

     ∫

P  =    S ⋅ dA.
      S
(75)

Thus the same geometric idea developed here becomes essential for fields, induction, waves, antennas, apertures, and RF power flow.

Summary

The essential results of EM06 are:

  • A surface becomes oriented by assigning a unit normal n.
  • The infinitesimal area vector is
    dA  = ˆn dA.

  • For a uniform field through a flat surface,
    ΦE  = EA  cos𝜃.

  • The angle is measured between the field and the surface normal.
  • Only the normal component of the field contributes to flux.
  • The general surface-integral definition is
          ∫

ΦE =   S E ⋅ dA.

  • Closed surfaces use outward-pointing area vectors by convention.
  • For a sphere centered on a point charge,
    ΦE  = q∕𝜖0,

    directly from Coulomb’s field and spherical geometry.

The next lesson, EM07, introduces Gauss’s law and explains when symmetry allows electric fields to be obtained from closed-surface flux with remarkable efficiency.

References

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

[2]   Samuel J. Ling, Jeff Sanny, and William Moebs, University Physics, Volume 2, OpenStax, 2016, sections on electric flux and Gauss’s law.

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

[4]   Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, Volume II, Addison-Wesley, 1964, chapters on electrostatics and Gauss’s law.

[5]   Massachusetts Institute of Technology, 8.02 Physics II: Electricity and Magnetism, MIT OpenCourseWare, materials on electric flux and Gauss’s law.


"Electromagnetic Waves: Electric Flux" is owned by bloftin.
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Keywords:  electric flux, electric field, area vector, surface normal, dot product, projected area, differential area, surface integral, closed surface, electric field lines, Coulomb field, Gauss law preparation, electrostatics

Attachments:
Electromagnetic Waves: Electric Flux - Exercises and Complete Worked Solutions (Example) by bloftin

Cross-references: waves, power, Electromagnetism, charge, diagram, volume, works, magnitude, relation, vector, scalar, formula, Gauss's Law, dot product, operation, flux, field, vector field, Electric Field, EM05
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This is version 1 of Electromagnetic Waves: Electric Flux, born on 2026-09-16.
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Physics Classification41.20.Cv (Electrostatics; Poisson and Laplace equations, boundary-value)
 41.20.-q (Applied classical electromagnetism)
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