Field of Engineering

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Electric Flux and Gauss’s Law

Electric flux

Electric flux is a measure of the “flow” of electric field through a surface. (number of electric

field lines passing through a perpendicular area)

∅ = 𝑬⃗⃗ . 𝒏̂ 𝑨 = 𝑬𝒏 𝑨 = 𝑬𝑨𝒄𝒐𝒔𝜽

Where 𝐸𝑛 = 𝐸⃗ . 𝑛̂ is the component of 𝐸⃗ normal to the surface.

SI units of Electric flux is (N . m2/C )

In figure : the electric flux through the surface (1) is ∅ = 𝐸𝐴1

the electric flux through the surface (2 ) is ∅ = 𝐸𝐴2𝑐𝑜𝑠𝜃

the electric flux through the surface (3) is ∅ = 0 (because 𝐸⃗ ⊥ 𝑛̂ )

For a curved surface over which may vary 𝐸⃗ , the surface divided into a large number of small

elements, , area of each element is ∆𝐴𝑖 it can be modeled as a plane and the variation of the

electric field across the element can be neglected.

The flux of the electric field through this element is

∆∅𝑖 = 𝐸⃗ 𝑖. 𝑛̂𝑖 ∆𝐴𝑖

where: 𝑛̂𝑖 is the unit vector perpendicular to the surface element

and 𝐸⃗ 𝑖 is the electric field on the surface element.

the total flux through the surface ∅𝑛𝑒𝑡 = ∑ 𝐸⃗ 𝑖. 𝑛̂𝑖 ∆𝐴𝑖

If the area of each element approaches zero, the number of elements approaches infinity and the

sum is replaced by an integral. Therefore, the general definition of electric flux is

∅𝑛𝑒𝑡 = lim∆𝐴𝑖→0 ∑ 𝐸⃗ 𝑖. 𝑛̂𝑖 ∆𝐴𝑖 = ∮

𝐸⃗ . 𝑛̂ 𝑑𝐴
𝑠

The net flux ∅𝑛𝑒𝑡 through the closed surface is positive or negative, depending on whether 𝐸⃗ is

predominantly outward or inward at the surface.

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Gauss’s Law

Consider a point charge q surrounded by a spherical surface of radius (r) centered on the charge

as in Figure.

The electric field due to charge q is

𝐸⃗ =

𝑘𝑞

𝑟2 𝑟̂

Note that the electric field is perpendicular to the spherical surface at all

points on the surface. The electric flux through the surface is ∅𝑛𝑒𝑡 = ∮

𝐸⃗ . 𝑛̂ 𝑑𝐴
𝑠

∅𝑛𝑒𝑡 = ∮

𝑘𝑞

𝑟2 𝑟̂. 𝑟̂ 𝑑𝐴
𝑠

∅𝑛𝑒𝑡 =

𝑘𝑞

𝑟2 ∮ 𝑑𝐴
𝑠
=

𝑘𝑞

𝑟2 (4𝜋𝑟2) =

𝑞inside

𝜀0

This leads to the following general result, known as Gauss’ law

∅𝒏𝒆𝒕 = ∮

𝑬⃗⃗ . 𝒏̂ 𝒅𝑨
𝒔
=

𝒒𝐢𝐧𝐬𝐢𝐝𝐞

𝜺𝟎

where : 𝒒𝐢𝐧𝐬𝐢𝐝𝐞 is the algebraic sum of all charges enclosed by surface

Gauss’s law is mathematical formula to descript electric field lines.

The net number of lines out of any surface enclosing the charges is proportional to the net

charge enclosed by the surface.

The electric flux ∅𝐧𝐞𝐭 through any closed surface is equal to the net charge inside the surface,

qinside, divided by ε0.

In figure (a) dipole is enclosed by closed surface

Number of electric lines entering the surface = Number of electric lines leaving the surface

Thus, the net number of lines through surface = zero

𝒒𝐢𝐧𝐬𝐢𝐝𝐞 = 𝟎→∅𝒏𝒆𝒕 = 0
in figure (b) the net number of lines through surface is same as that for a single charge of +q

charge enclosed by the surface.

Figure (a)

Figure (b)

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Notes:

1.Whether there is a net outward or inward electric flux through a closed surface depends on

the sign of the enclosed charge.

2.Charges outside the surface do not give a net electric flux through the surface.

3.The net electric flux is directly proportional to the algebraic sum of all charges enclosed by

surface.

4.The net electric flux is independent of the size and shape of the closed surface.

5.The net number of electric field lines passing through an imaginary closed surface is

proportional to the amount of net charge enclosed within that surface.

6.Gauss’s law is valid for any distribution of charges and for any closed surface.

7.Gauss’ law can be used to calculate the electric field of a system of charges or a continuous

distribution of charge. It is useful in cases in which there is a high degree of symmetry, such

as spheres, cylinders, or planes.

8.In electrostatic Gauss’s law and Coulomb’s law are equivalent, but in electrodynamics only

Gauss’s law is valid.

Example (1) Flux through a Piecewise-Continuous Closed Surface

An electric field is given by 𝑬⃗⃗ = +(𝟐𝟎𝟎 𝑵 𝑪 ) 𝑲̂⁄

throughout the region 𝒁 > 𝟎 and by

𝑬⃗⃗ = −(𝟐𝟎𝟎 𝑵 𝑪 ) 𝑲̂⁄

throughout the region 𝒁 < 𝟎 , cylindrical surface that has a length equal to

20 cm and a radius equal to has its center at the origin and its axis along the axis, so that one

end is at and the other is at Figure.

(a)What is the net outward flux through the closed surface? (b) What is the net charge inside

the closed surface?

∅𝑟𝑖𝑔ℎ𝑡 = 𝐸⃗ 𝑟𝑖𝑔ℎ𝑡 . 𝑛̂𝑟𝑖𝑔ℎ𝑡 𝐴

= +200 𝐾̂ . 𝐾̂ (𝜋𝑅2)
= +200 (3.14 × 5 × 10−2) = 1.57 𝑁. 𝑚2/𝐶

∅𝑙𝑒𝑓𝑡 = 𝐸⃗ 𝑙𝑒𝑓𝑡 . 𝑛̂𝑙𝑒𝑓𝑡 𝐴

= −200 𝐾̂ . (−𝐾̂ )(𝜋𝑅2)
= 200 (3.14 × 5 × 10−2) = 1.57 𝑁. 𝑚2/𝐶

∅𝑐𝑢𝑟𝑣𝑒𝑑 = 𝐸⃗ 𝑐𝑢𝑟𝑣𝑒𝑑. 𝑛̂𝑐𝑢𝑟𝑣𝑒𝑑 𝐴

∅𝑐𝑢𝑟𝑣𝑒𝑑 = 0 (because 𝐸⃗ ⊥ 𝑛̂ everywhere on the curved piece) 𝐾̂. 𝑗̂ = 0

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∅𝒏𝒆𝒕 = ∅𝑟𝑖𝑔ℎ𝑡 + ∅𝑙𝑒𝑓𝑡 + ∅𝑐𝑢𝑟𝑣𝑒𝑑
∅𝒏𝒆𝒕 = 1.57 + 1.57 + 0 = 3.14 𝑁. 𝑚2/𝐶
Note: The net flux does not depend on the length of the cylinder. This result is expected for an

electric field that does not vary with distance from the plane.

(b)∅𝑛𝑒𝑡 =

𝑞inside

𝜀0

𝑞inside = ∅𝑛𝑒𝑡 × 𝜀0
𝑞inside = 3.14 × 8.85 × 10−12= 2.78 × 10−11 𝐶
= 27.8 𝑝𝐶

Using symmetry to calculate E with Gauss's law

Calculate 𝐸⃗ of highly symmetrical charge distribution is more easily by using Gauss’ law than

Coulomb’s law

Thera are three classes of symmetry

1.A charge configuration has cylindrical (or line) symmetry if the charge density depends only

on the distance from a line.

The Gaussian surface for this configuration is a cylinder coaxial with the symmetry line

2.plane symmetry if the charge density depends only on the distance from a plane,

the Gaussian surface for this configuration is a cylinder bisected by the symmetry plane and

with its symmetry axis normal to the symmetry plane.

3.spherical (or point) symmetry if the charge density depends only on the distance from a

point.the Gaussian surface for this configuration is a sphere centered on the symmetry point.

Example (2) 𝐄⃗ Due to a Uniformly Charged Slab

A very large (infinite), uniformly charged slab of plastic of thickness 2a occupies the region

between the z=-a plane and the z=a plane. Find the electric field everywhere due this charge

configuration. The charge per unit volume of the plastic is 𝝆.

Solution:

charge configuration has plane symmetry , so Gaussian surface is chosen to be a cylinder bisected

by the symmetry plane and with its symmetry axis normal to the symmetry plane.

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∅𝒏𝒆𝒕 = ∮

𝐸⃗ . 𝑛̂ 𝑑𝐴
𝑠
=

𝑞inside

𝜀0

∅𝑛𝑒𝑡 = ∅𝑟𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒 + ∅𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 + ∅𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑖𝑑𝑒
∅𝑟𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒 = ∅𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 = ∫ 𝐸𝑛 𝑑𝐴 = 𝐸𝑛𝐴

∅𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑖𝑑𝑒 = 0 (because 𝐸⃗ ⊥ 𝑛̂ everywhere on the curved piece)

∅𝑛𝑒𝑡 = 𝐸𝑛𝐴 + 𝐸𝑛𝐴 + 0
2𝐸𝑛𝐴 =

𝑞inside

𝜀0

𝐸𝑛 =

𝑞inside
2𝐴𝜀0

𝑞inside = 𝜌 𝑉 = { 𝜌𝐴2𝑧 𝑓𝑜𝑟 𝑧 ≤ 𝑎

𝜌𝐴2𝑎 𝑓𝑜𝑟 𝑧 ≥ 𝑎 }

𝐸𝑛 =

𝑞inside
2𝐴𝜀0=

{

𝜌𝐴2𝑧

2𝐴𝜀0=

𝜌𝑧

𝜀0 𝑓𝑜𝑟 𝑧 ≤ 𝑎

𝜌𝐴2𝑎

2𝐴𝜀0=

𝜌𝑎

𝜀0 𝑓𝑜𝑟 𝑧 ≥ 𝑎

𝐸𝑛 is magnitude of 𝐸⃗ in z direction

𝐸⃗ =

{

−

𝜌𝑎

𝜀0𝑘̂ 𝑓𝑜𝑟 𝑧 ≤ 𝑎

𝜌𝑧

𝜀0𝑘̂ 𝑓𝑜𝑟 − 𝑎 ≤ 𝑧 ≤ 𝑎

𝜌𝑎

𝜀0 𝑘̂ 𝑓𝑜𝑟 𝑧 ≥ 𝑎

Example (3) 𝐄⃗ Due to a Thin Spherical Shell of Charge

Find the electric field due to a uniformly charged thin spherical shell of radius R and total

charge Q.

Solution:

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charge configuration has spherical symmetry , so Gaussian surface is chosen to be sphere of

radius (r)

∅𝒏𝒆𝒕 = ∮

𝐸⃗ . 𝑛̂ 𝑑𝐴
𝑠
=

𝑞inside

𝜀0

∮𝐸𝑟 𝑟̂. 𝑟̂ 𝑑𝐴
𝑠
=

𝑞inside

𝜀0

𝐸𝑟 (4𝜋𝑟2) =

𝑞inside

𝜀0
→𝐸𝑟 =

𝑞inside
4𝜋𝜀0𝑟2

𝑞inside = {0 𝑓𝑜𝑟 𝑟 < 𝑅

𝑄 𝑓𝑜𝑟 𝑟 > 𝑅

𝐸⃗ = 𝐸𝑟 𝑟̂ = {

0 𝑓𝑜𝑟 𝑟 < 𝑅 (𝑖𝑛𝑠𝑖𝑑𝑒 𝑠ℎ𝑒𝑙𝑙)
𝑄

4𝜋𝜀0𝑟2 𝑓𝑜𝑟 𝑟 > 𝑅 (𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑠ℎ𝑒𝑙𝑙)

Outside the charged shell, the electric field is the same as that of a point charge Q at the shell’s

center. the electric field is discontinuous at r=R

Example (4) Electric Field Due to a Point Charge and a Charged Spherical Shell

A spherical shell of radius R=3 m has its center at the origin and has a surface charge density

σ=3nC/m2 of A point charge q=250nC is on the y axis at y=2m. Findthe electric field on the axis

at (a) x=2m and (b) x=4m

Solution:

Point x=2m is located inside the shell so 𝐸⃗ at this point due to point charge only

𝐸⃗ 1 = 𝑘

𝑞

𝑟12 𝑟1̂

𝐸⃗ 1 = 9 × 109×

250×10−9

(2√2)2 𝑟̂1 = (218𝑁/𝐶 ) 𝑟̂1

𝐸⃗ 1 = (218 cos 𝜃 𝑖̂ − 218 sin 𝜃 𝑗̂ ) 𝑁/𝐶

= 218 (

2

2√2) 𝑖̂ − 218 (

2

2√2) 𝑗̂ 𝑁/𝐶

𝐸⃗ 1 = (199 𝑖̂ − 199 𝑗̂ ) 𝑁/𝐶

Point (x=4m) is located outside the shell so 𝐸⃗ at this point due to point charge (𝐸⃗ 𝑝) and charged

shell (𝐸⃗ 𝑠ℎ𝑒𝑙𝑙 )

𝐸⃗ 2 = 𝐸⃗ 𝑝 + 𝐸⃗ 𝑠ℎ𝑒𝑙𝑙

𝐸⃗ 𝑃 = 𝑘

𝑞

𝑟22 𝑟̂2

𝐸⃗ 𝑃 = 9 × 109×

250×10−9

20
𝑟̂2 = (112 𝑁/𝐶 ) 𝑟̂2

𝐸⃗ 𝑃 = (112 cos 𝜃 𝑖̂ − 112 sin 𝜃 𝑗̂ ) 𝑁/𝐶

= 112 (

4

√20) 𝑖̂ − 112 (

2

√20) 𝑗̂ 𝑁/𝐶

q

r1

E1

2m

2m

q

r2

Epoint charge

2m

4m →←

Eshell

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𝐸⃗ 𝑃 = (100 𝑖̂ − 50 𝑗̂ ) 𝑁/𝐶

𝐸⃗ 𝑠ℎ𝑒𝑙𝑙 =

𝑄

4𝜋𝜀0𝑟2 𝑖̂

=

𝜎 (4𝜋𝑅2)

4𝜋𝜀0𝑟2 𝑖̂ =

3×10−9 × 32

8.85×10−12 ×42 𝑖̂

𝐸⃗ 𝑠ℎ𝑒𝑙𝑙 = (190𝑁/𝐶 )𝑖̂

𝐸⃗ 2 = (100 𝑖̂ − 50 𝑗̂ ) + (190 )𝑖̂

𝐸⃗ 2 = (290 𝑖̂ − 50 𝑗̂ ) 𝑁/𝐶

Example (5) 𝐄⃗ Due to a Uniformly Charged Solid Sphere

Find the electric field everywhere for a uniformly charged solid sphere that has a radius R

and a total charge Q that is uniformly distributed throughout the volume of the sphere

Solution:

The charge configuration has spherical symmetry. We choose a spherical Gaussian surface of

radius (r)

∅𝒏𝒆𝒕 = ∮

𝐸⃗ . 𝑛̂ 𝑑𝐴
𝑠
=

𝑞inside

𝜀0

∮𝐸𝑟 𝑟̂. 𝑟̂ 𝑑𝐴
𝑠
=

𝑞inside

𝜀0

𝐸𝑟 (4𝜋𝑟2) =

𝑞inside

𝜀0
→ 𝐸𝑟 =

𝑞inside
4𝜋𝜀0𝑟2

We find 𝑞inside Gaussian surface in case 𝑟 ≤ 𝑅 as following

𝜌 =

𝑄

𝑉𝑠𝑝ℎ𝑎𝑟𝑒=

𝑞inside

𝑉𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛

𝑞inside = 𝑄

𝑉𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛
𝑉𝑠𝑝ℎ𝑎𝑟𝑒→𝑞inside = 𝑄

4
3𝜋𝑟3

4
3𝜋𝑅3 = 𝑄

𝑟3

𝑅3

𝑞inside = {𝑄

𝑟3

𝑅3 𝑓𝑜𝑟 𝑟 ≤ 𝑅

𝑄 𝑓𝑜𝑟 𝑟 ≥ 𝑅

𝐸⃗ = 𝐸𝑟 𝑟̂ = {

𝑄𝑟3

𝑅3

4𝜋𝜀0𝑟2 =

𝑄 𝑟

4𝜋𝜀0𝑅3 𝑓𝑜𝑟 𝑟 ≤ 𝑅 (𝑖𝑛𝑠𝑖𝑑𝑒 𝑠ℎ𝑒𝑙𝑙)

𝑄

4𝜋𝜀0𝑟2 𝑓𝑜𝑟 𝑟 ≥ 𝑅 (𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑠ℎ𝑒𝑙𝑙)

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Example (6 ) Electric Field Due to Infinite Line Charge

Use Gauss’s law to find the electric field everywhere due to an infinitely long line charge of

uniform charge density λ.

Solution:

we know the magnitude of the field depends only on the radial distance from the line charge. We

therefore choose a cylindrical Gaussian surface coaxial with the line charge has a length (L) and a

radius (r).

∅𝒏𝒆𝒕 = ∮

𝐸⃗ . 𝑛̂ 𝑑𝐴
𝑠
=

𝑞inside

𝜀0

∅𝑛𝑒𝑡 = ∅𝑟𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒 + ∅𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 + ∅𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑖𝑑𝑒

∅𝑟𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒 = ∅𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 = 0 (because 𝐸⃗ ⊥ 𝑛̂ )

∅𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑖𝑑𝑒 = ∮

𝐸𝑟 𝑟̂. 𝑟̂ 𝑑𝐴
𝑠
= 𝐸𝑟(2𝜋𝑟𝐿)

𝐸𝑟(2𝜋𝑟𝐿) =

𝑞inside

𝜀0

𝑞inside = 𝜆 𝐿

𝐸𝑟(2𝜋𝑟𝐿) =

𝜆 𝐿

𝜀0→𝐸𝑟 =

𝜆

2𝜋𝜀0𝑟

Conductors in Electrostatic Equilibrium

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1-The electric field is zero inside a charged conductor.

2-Any excess charge on an isolated conductor resides entirely on the surface of the conductor.

3-The electric field at the surface of a charged conductor is perpendicular to the surface and

has a magnitude σ/ε0.

4-Excess charge tends to accumulate at sharp points, or locations of highest curvature, on

charged conductors. As a result, the electric field is greatest at such locations.

these properties can be prove by using Gauss's law