GAUSS LAW

1. GAUSS'S LAW
GROUP 1

2. GAUSS'S LAW
Gauss Law states that the total electric flux out of a closed surface is equal to
the charge enclosed divided by the permittivity. The electric flux in an area is
defined as the electric field multiplied by the area of the surface projected in a
plane and perpendicular to the field.

3. CARL FRIEDRICH GAUSS
Carl Friedrich Gauss
worked in a wide variety
of fields in both
mathematics and
physics. He is a genius.
He had contribution in
the field of magnetism.

4. CARL FRIEDRICH GAUSS
• At the age of seven, Carl Friedrich Gauss started elementary school, and his potential was
noticed almost immediately
• In 1788 Gauss began his education at the Gymnasium with the help of Büttner and Bartels,
where he learnt High German and Latin. After receiving a stipend from the Duke of
Brunswick- Wolfenbüttel, Gauss entered Brunswick Collegium Carolinum in 1792.
• In 1795 Gauss left Brunswick to study at Göttingen University. Gauss's teacher there was
Kästner, whom Gauss often ridiculed. His only known friend amongst the students was
Farkas Bolyai. They met in 1799 and corresponded with each other for many years.
• Gauss left Göttingen in 1798 without a diploma, but by this time he had made one of his
most important discoveries - the construction of a regular 17-gon by ruler and compasses

5. CARL FRIEDRICH GAUSS
• Gauss married Johanna Ostoff on 9 October, 1805. Despite having a happy personal life for
the first time, his benefactor, the Duke of Brunswick, was killed fighting for the Prussian army.
• Gauss's work never seemed to suffer from his personal tragedy. He published his second
book, Theoria motus corporum coelestium in sectionibus conicis Solem ambientium Ⓣ, in
1809, a major two volume treatise on the motion of celestial bodies.
• Much of Gauss's time was spent on a new observatory, completed in 1816, but he still found
the time to work on other subjects.
• The latter work was inspired by geodesic problems and was principally concerned with
potential theory
• For an extensive survey of terrestrial magnetism, he invented an early type of magnetometer,
a device that measures the direction and strength of a magnetic field.

6. WHAT IS GAUSS'S LAW?
GROUP 1

7. According to the Gauss law, the total flux linked with a
closed surface is 1/ε0 times the charge enclosed by the
closed surface.
∮E→.d→s=1∈0q

8. FOR EXAMPLE, A POINT CHARGE Q IS PLACED INSIDE A
CUBE OF EDGE ‘A’. NOW, AS PER GAUSS LAW, THE FLUX
THROUGH EACH FACE OF THE CUBE IS Q/6Ε0.
The electric field is the basic concept of knowing about electricity.
Generally, the electric field of the surface is calculated by applying
Coulomb’s law, but to calculate the electric field distribution in a
closed surface, we need to understand the concept of Gauss law.
It explains the electric charge enclosed in a closed or the electric
charge present in the enclosed closed surface.

9. GAUSS'S LAW
FORMULA

10. Gauss’s law for electricity states that the electric flux Φ across any closed
surface is proportional to the net electric charge q enclosed by the surface.
The Gauss law formula is expressed by;
𝝓 =
𝑸
Ɛ𝟎
Where:
Φ = electric flux (Nm2/C)
Q = enclosed charge (C)
Ɛ𝟎 = permittivity of free space
(8.85 x 10-12 C2/Nm2 )

11. THE GAUSS
LAW THEOREM

12. The net flux through a closed surface is directly proportional
to the net charge in the volume enclosed by the closed surface.
Φ = → E.d → A = qnet/ε0
In simple words, the Gauss theorem relates the ‘flow’ of
electric field lines (flux) to the charges within the enclosed
surface. If no charges are enclosed by a surface, then the net
electric flux remains zero.
This means that the number of electric field lines entering
the surface equals the field lines leaving the surface.

13. The Gauss theorem statement also gives an important corollary:
The electric flux from any closed surface is only due to the sources (positive
charges) and sinks (negative charges) of the electric fields enclosed by the surface. Any
charges outside the surface do not contribute to the electric flux. Also, only electric
charges can act as sources or sinks of electric fields. Changing magnetic fields, for
example, cannot act as sources or sinks of electric fields.

14. GAUSS'S LAW IN
MAGNETISM

15. The net flux for the surface on the left is non-zero as it
encloses a net charge. The net flux for the surface on
the right is zero since it does not enclose any charge.

16. GAUSS'S LAW
EQUATION

17. Gauss law equation can be understood using an integral
equation. Gauss’s law in integral form is mentioned below:
∮ 𝐸 • 𝑑𝐴 =
𝑄
Ɛ0
⇢ (1)
Where,
 E is the electric field
 Q is the electric charge enclosed
 ε0 is the electric permittivity of free space
 A is the outward pointing normal area vector
Flux is a measure of the strength of a field passing through
a surface. Electric flux is given as:
Φ = ∮ 𝐸 • 𝑑𝐴 ⇢ (2)

18. APPLICATIONS
OF GAUSS’S
LAW

20. 1.Point Charge using Spherical Symmetry
Suppose a point charge +q rests in
space. Consider a sphere of radius r that
encloses the charge such that it lies at the
center of the sphere. From the symmetry
of the situation, it is evident that the
electric field will be constant on the
surface and directed radially outward.

21. SPHERICAL SYMMETRY
∮ 𝐸 • 𝑑𝐴 =
𝑄𝑒𝑛𝑐𝑙
Ɛ0
∮ 𝐸 • 𝑑𝐴 = 𝐸 (4𝜋𝑅2
)
𝑄𝑒𝑛𝑐𝑙
Ɛ0
=
𝑞
Ɛ0
𝐸 (4𝜋𝑅2
) =
𝑞
Ɛ0
1
4𝜋𝑅2
[𝐸 (4𝜋𝑅2
) =
𝑞
Ɛ0
]
1
4𝜋𝑅2
𝑬 =
𝒒
𝟒𝝅Ɛ𝟎 𝑹𝟐

22. 2.) Infinite Line Charge using Cylindrical
Symmetry
Let us consider an infinitely long wire
with linear charge density λ and length L.
To calculate electric field, we assume a
cylindrical Gaussian surface. As the
electric field E is radial in direction, the
flux through the end of the cylindrical
surface will be zero.

23. CYLINDRICAL SYMMETRY
∮ 𝐸 • 𝑑𝐴 =
𝑄𝑒𝑛𝑐𝑙
Ɛ0
∮ 𝐸 • 𝑑𝐴 = 𝐸 (2𝜋𝑟𝐿)
𝑄𝑒𝑛𝑐𝑙
Ɛ0
=
𝜆𝐿
Ɛ0
𝐸 2𝜋𝑟𝐿 =
𝜆 𝐿
Ɛ0
𝐸 2𝜋𝑟𝐿 =
𝜆 𝐿
Ɛ0
𝐸 2𝜋𝑟 =
𝜆
Ɛ0
1
2𝜋𝑟
[𝐸 2𝜋𝑟 =
𝜆
Ɛ0
]
1
2𝜋𝑟
𝑬 =
𝝀
𝟐𝝅Ɛ𝟎𝒓
Charge density
𝜆 =
𝑄
𝐿
𝑄 = 𝜆 𝐿

24. 3. Infinite Plane Charge using Planar
Symmetry
Let us consider an infinite plane sheet,
with surface charge density σ and cross-
sectional area A. The position of the
infinite plane sheet is as below:
The direction of the electric field due to
an infinite charge sheet is perpendicular
to the plane of the sheet. Let us
consider a cylindrical Gaussian surface,
whose axis is normal to the plane of the
sheet. We can evaluate the electric field
E from Gauss’s Law.

25. PLANAR SYMMETRY
∮ 𝐸 • 𝑑𝐴 =
𝑄𝑒𝑛𝑐𝑙
Ɛ0
𝐸 • 𝑑𝐴 = (𝐸)(2 • 𝜋𝑅2
)
𝑄𝑒𝑛𝑐𝑙
Ɛ0
=
(𝜎)(𝐴)
Ɛ0
(𝐸)(2 • 𝜋𝑅2
) =
(𝜎)(𝜋𝑅2
)
Ɛ0
(𝐸)(2 • 𝜋𝑅2
) =
(𝜎)(𝜋𝑅2
)
Ɛ0
(𝐸)(2) =
𝜎
Ɛ0
1
2
[(𝐸)(2 [) =
𝜎
Ɛ0
]
1
2
𝑬 =
𝝈
𝟐Ɛ𝟎
A = 𝜋𝑅2
Charge density
𝜎 =
𝑄
𝐴
𝑄 = 𝜎𝐴

26. APPLICATIONS OF
GAUSS LAW IN DAILY
Reporter: Julleana May O. Mestedio

27. Electrostatic Painting
The electrostatic painting is the
process of using magnetic field to
apply paint to metals and various
types of plastics, it uses positively
charged paint particles from the
specialized gun to coat grounded
metal surfaces.

28. Electrostatic Smoke Precipitators
The electrostatic smoke precipitators is a filter-less
device that is used to control air pollution by eliminating
pollutants such as soot and ashes and other high
particulate matter.
It works by creating an electrostatic force and
ionizing the particles. It also works in the presence of
two electrodes. One is a positive electrode and the other
is a negative electrode. The negative electrode is in the
form of mesh wire and the positive electrode is in the
form of plates. Both are placed vertically and
alternatively to each other in the precipitator.

29. Photocopy Machines and Printers
The Laser Jet Printer is an electrostatic
digital printing process. Designed for high
print quality, horizontal and vertical
printing, printing graphics, text, letters,
memos and spreadsheets.
Laser Jet Printer

30. The following steps detail how a laser printer works:
• A photo, graphic or text image is sent to the printer, which
begins the process of transferring that image to paper
using a combination of positive and negative static electric
charges.
• The revolving drum gets a positive charge.
• The system's electronics convert the image into a laser
beam.
• The laser beam bounces off a mirror onto the drum,
drawing the image on the drum by burning a negative
charge in the shape of the image.
• Then the drum picks up the positively charged toner from
the toner cartridge. The toner sticks to the negatively
charged image on the drum.
• Paper entering the printer receives a negative charge.

31. • As the paper passes the drum, the paper's
negative charge attracts toner from the
positively charged drum; the toner literally
sits on top of the paper.
• The paper's charge is removed and a fuser
permanently bonds the toner onto the
paper.
• The printed paper is released from the
printer.
• The electrical charge is removed from the
drum, and the excess toner is collected.

32. Xerography
• Image-forming process that relies on
a photoconductive substance whose
electrical resistance decreases when
light falls on it.
• The process was invented in the
1930s by U.S. physicist Chester F.
Carlson (1906–1968) and developed
in the 1940s and ’50s by Xerox Corp.
(then called Haloid).

33. Xerography
• Light passing through or reflected from a document
reaches a selenium-coated drum surface onto which
negatively charged particles of ink (toner) are
sprayed, forming an image of the document on the
drum.
• As a sheet of paper is passed close to the drum, a
positive electric charge under the sheet attracts the
negatively charged ink particles, transferring the
image to the copy paper. Heat briefly applied fuses
the ink particles to the paper.

34. Ink Jet Printer
• commonly used to print computer-
generated text and graphics, also
employs electrostatics.
• Ink jet printers can produce color
images by using a black jet and three
other jets with primary colors, usually
cyan, magenta, and yellow, much as a
color television produces color (this is
more difficult with xerography,
requiring multiple drums and toners).

35. DNA Structure
• DNA-DNA interactions are dominated
by long range electrostatic forces
due to the highly charged nature of
DNA.
• Each DNA is free to rotate along its
cylindrical axis.
• Overlapping of the respective
hydration shell of opposing DNA
helices leads dehydration forces.

36. Van De Graff
What is Van De Graff Generator?
• is a device that can be used to separate
charges and create potential differences
in the range of megavolts.
• It was invented by American physicist
Robert J. Van de Graaff in 1931 for use in
nuclear physics research.

37. • Fundamentally, a Van de Graaff generator
consists of a flexible dielectric belt (silk is
commonly used) running over two metal
pulleys. One of these pulleys is
surrounded by a hollow metal sphere.
Two electrodes are positioned near the
bottom of the lower pulley and inside the
sphere, over the upper pulley. One comb
is connected to the sphere, and another
is connected to the ground. In this figure,
a high, positive DC potential is applied to
the upper roller.

38. The schematic of the Van De Graff
generator
• A very large excess charge can be
deposited on the sphere, because it
moves quickly to the outer surface.
Practical limits arise because the large
electric fields polarize and eventually
ionize surrounding materials, creating
free charges that neutralize excess charge
or allow it to escape. Nevertheless,
voltages of 15 million volts are well within
practical limits.

39. Problem solving
Examples
on Gauss Law

40. Example 1: In the x-direction, there is a homogeneous electric field of size E = 50
N/C. Calculate the flux of this field across a plane square area with an edge of 5 cm
in the y-z plane using the Gauss theorem. Assume that the normal is positive along
the positive x-axis.
Solution:
Given:
Electric field, E = 50 N/C
Edge length of square, a = 5 cm = 0.05 m
The flux of the field across a plane square, ϕ = ∫ E cosθ ds
As the normal to the area points along the electric field, θ = 0.
Also, E is uniform so, Φ = E ΔS = (50 N/C) (0.05 m)2 = 0.125 N m2 C-1.
Hence, the flux of the given field is 0.125 N m2 C-1.

41. Example 2: There are three charges, q1, q2, and q3, having charges 4 C, 7 C, and 2
C enclosed in a surface. Find the total flux enclosed by the surface.
Total charge Q,
Q= q1 + q2 + q3
= 4C + 7C + 2C
= 13 C
𝜙 =
𝑄
Ɛ0
𝜙 =
13𝐶
8.854 𝑥 10−12 𝐶2/𝑁𝑚2
𝜙 = 1.584 𝑁𝑚2
𝐶−1

42. Example 3: Calculate the electric field of the hollow spherical conductor of radius
3m with a charge of +20 uC at the center of the sphere. (b) What is the surface
charge density? (3) what is the electric field 5m away from the center?
Solution:

43. Solution:

44. Solution:
c. What is the electric field of the 1.5 meters away from the center of the sphere?
𝐸𝐴 =
𝑄
𝜀0
𝐸 4𝜋𝑟2
=
𝜎(4𝜋𝑅2
)
𝜀0
𝐸 =
𝜎𝑅2
𝜀0𝑟2

45. Solution:
c. What is the electric field of the 1.5 meters away from the center of the sphere?
𝐸 =
20 × 10−6
(0.40m)2
8.5 × 10−12(1.5m)2
𝐸 = 1.61 × 10 5
N/c

46. THANK
YOU!