SPECIALIZED SUBJECT - STEM
GENERAL PHYSICS 2
___ SEMESTER, SY _______
QUARTER 1, MODULE 2
GAUSS’S LAW AND ELECTRIC POTENTIAL
i
General Physics 2
Self-Learning Module (SLM)
___Semester Quarter 1– Module 2: Gauss’s Law and Electric Potential
First Edition, 2021
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1
Lesson
1
GAUSS’S LAW
INTRODUCTION
Hello Senior High Learners! In this lesson, you will learn to:
1. Use Gauss’s law to infer electric field due to uniformly distributed charges on long
wires, spheres, and large plates STEM_GP12EM-IIIb-13.
LESSON AND PRACTICES
Gauss’s Law
Gauss’s law, also known as Gauss’s flux theorem, is a law relating the distribution of
electric charge to the resulting electric field. This law states that: The net outward normal
electric flux through any closed surface is proportional to the total electric charge
enclosed within that closed surface.
The law was formulated by a German Mathematician Carl Friedrich Gauss in 1835,
but was not published until 1867. It is one of the four Maxwell’s equations which form the basis
of classical electrodynamics, the other three being Gauss’s law for magnetism, Faraday’s law
of induction, and Ampère’s law with Maxwell’s correction. This law provides a means in
determining the electric field or electric field intensity as produced by charge (point or
distributed).
Since Coulomb’s Law is already popular in point charges, Gauss’s Law is more useful
in distributed charges specially if it is symmetrically distributed in a closed system, such as
changes in long wire or in a relatively infinite plane.
Electric field E for uniformly distributed system
Point of
Charge distribution
Electric Field
interest
1 𝑞
outside sphere:
𝐸=
4𝜋𝜀0 𝑟 2
Charge q distributed on the
𝑟>𝑅
surface of a
inside sphere:
𝐸= 0
conducting with radius R
𝑟<𝑅
2
Figure
Long wire, with uniform
distribution charge per unit
length 𝜆
at a distance r
from the wire
Infinite plane sheet, with a
charge per unit area 𝜎
at any distance
from the plate
Two conducting plates
oppositely charged, with
densities 𝜎 𝑎𝑛𝑑 − 𝜎
at any point
between the
conducting plate
𝐸=
𝐸=
1 𝑞
2𝜋𝜀0 𝑟
𝜎
2𝜋𝜀0
𝐸=
𝜎
𝜀0
Gauss's Laws
The total of the electric flux out of a closed surface is equal to the charge enclosed
divided by the permittivity.
𝑄
Φ𝐸 =
𝜀0
where:
Q = is total charge within the given surface.
ε0 = permittivity constant (8.85 x10-12)
Sample Problem
There are three charges q1, q2, and q3 having charge 6 pC, 5 pC and 3 pC enclosed
in a surface. Find the total flux enclosed by the surface.
Given:
𝑄
𝜀0
Φ𝐸
= q1 + q2 + q3 = 6 pC + 5 pC + 3 pC = 14 pC
= 8.85 x10-12
=?
Solution:
Φ𝐸 =
𝑄
14𝑝𝐶
=
= 𝟏. 𝟓𝟖𝟒 𝑵/𝑪 − 𝒎𝟐
𝜀0
8.85 𝑥 10−12
PRACTICE EXERCISES
CHECKPOINT 1:
Directions: Illustrate the following charge distribution using the given parameters. Use the
rubrics as your guide for your description/explanation.
3
Points
5
Criteria
Content
3
Grammar, usage
and mechanics
Neatness
2
•
•
•
Description
There is a clear and well-focused content. main idea
stands out and is supported by detailed info.
There has no error in grammar or spelling that
distract the reader from the content.
Work is illegible.
Given
Example: Charge q= 2 C distributed
on the surface of a conducting
with radius R= 1 cm
Illustration
1. From the previous given, in long
wire, with uniform distribution
charge per unit length 𝜆= 2 C/m
2. Infinite plane sheet, with a charge
per unit area
3. Two conducting plates oppositely
charged, with densities σ=2C/m2
and - σ=2C/m2
a. According to your illustration, what is the difference between the first and second
illustration of electric field?
b. According to your illustration, what can you say about the electric field of the third
and the fourth parameters?
INSIGHTS
Directions: Please write what you have learned from the discussion by completing the openended question below. Write your learning/s neatly and legibly in a clean yellow pad paper.
Upon reading the lessons above, I learned that
and I realized that
4
Lesson
2
ELECTRIC FIELD (E) AND
FIELD FORCES (F)
INTRODUCTION
Hello Senior High Learners, I am glad you come this far. In this lesson, you will also learn
to:
1. Solve problems involving electric charges, dipoles, forces, fields, and flux in contexts
such as, but not limited to, systems of point charges, electrical breakdown of air,
charged pendulums, electrostatic ink-jet printers STEM_GP12EM-IIIb-14.
LESSON AND PRACTICES
Electric Field (E) and Field Forces (F)
The region of space around an electrically charged body is called electric field. This
electric field exerts a force on another charged body and is usually illustrated by imaginary
lines called electric field lines of force or simply lines. The strength of an electric field E at
any point may be defined as the electric force F exerted per unit positive electric charge q at
that point, or simply
𝑘
𝑞𝑆
𝐸=𝑘 2
𝑟
𝐹
𝐸 = (𝑁/𝐶 𝑜𝑟 𝑉/𝑚)
𝑞
𝐸=
𝑞 𝑞𝑆
𝑟2
𝑞
To detect the presence of electric field at a particular region, another charged body,
usually test charge (q) is placed and if the test charge experiences a force of electrical origin,
then an electric field is present at that region.
Quantities
𝐹𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 electric force
Units
𝑁 newtons
𝑞
charge
𝐶
𝑒
𝑘𝐶
Coulomb constant
𝑁∙
𝐸
electric field strength
Conversions
coulomb (SI unit of charge)
fundamental unit of charge
𝑚2
𝐶2
N/C
𝑚𝑒𝑡𝑒𝑟 2
𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑥 𝐶𝑜𝑢𝑙𝑢𝑚𝑏
newtons/coulomb
1𝐶 = 1 𝑥 10−6𝜇𝐶
5
2
= 6.3 𝑥 1018𝑒
= 1.60 𝑥 10−19𝐶
= 8.99 𝑥 109 𝑁 ∙
𝑚2
Sample Problem
A charge 𝑞1 = +7.00𝜇𝐶 is at the origin, and a charge 𝑞2 = −5.00𝜇𝐶 is on the x-axis 0.300
𝑚 from the origin, is at the right. Find the electric field strength at point P, which is on the yaxis 0.400 𝑚 from the origin.
Given: 𝑞1 = +7.00𝜇𝐶 = 7.00 𝑥 10−6 𝐶
𝑞 2 = −5.00𝜇𝐶 = −5.00 𝑥 10−6 𝐶
𝑟1 = 0.400 𝑚
𝑟2 = 0.500 𝑚
Unknown: E at P (𝑦 = 0.400 𝑚)
Calculate the electric field strength produced by each charge.
Because we are finding the magnitude of the electric field, we can neglect the sign of
each charge.
PRACTICE EXERCISES
CHECKPOINT 2:
Directions: Solve the following problems and write your answers on a separate piece of
paper.
1. A proton and an electron in a hydrogen atom are separated on the average by about
5.3 𝑥 10−11 𝑚. What is the magnitude and direction of the electric field set up by the
proton at position of electron?
2. An electric field of 2.00 𝑥 104𝑁/ 𝐶 is directed along the positive x-axis.
a.
What is the electric force on an electron in this field?
b.
What is the electric force on a proton in this field?
INSIGHTS
Directions: Please write what you have learned from the discussion by completing the openended question below. Write your learning/s neatly and legibly in a clean yellow pad paper.
Upon reading the lessons above, I learned that
and I realized that
6
Lesson
3
ELECTRIC POTENTIAL AND
ELECTRIC POTENTIAL
ENERGY
INTRODUCTION
You made it again Senior High Learners, I am so proud of you and this time, you will learn to:
1. Relate the electric potential with work, potential energy, and electric field
STEM_GP12EM-IIIb-15; and
2. Determine the electric potential function at any point due to highly symmetric
continuous-charge distributions STEM_GP12EM-IIIc-17
LESSON AND PRACTICES
We just scratched the surface (or at least rubbed it) of electrical phenomena. In
previous quarter, we introduced the concepts of work and energy in the context of mechanics,
this time we’ll combine this concept with what we have learned about electric charge, electric
force and electric field. Two terms commonly used to describe electricity are energy and
voltage. The energy and voltage are not the same thing.
Potential energy is the energy stored by an object because of its position relative to
other objects, its electric charge, or other factors. A common type of potential energy includes
the electric potential energy.
In this lesson, you will be able to relate the electric potential with work, potential
energy, and electric field. You will also solve problems involving electric potential for point
charge and continuous charge distribution.
In an electric field, a charge has potential energy relative to its position. When a
positive charge q is accelerated in an electric field, the charge has electric potential energy
(see Figure 1). It is like an object being accelerated in a gravitational field, as if the charge
were going down an electrical hill, although the sources of the forces are very different.
7
Figure 1. A charge accelerated by an electric field is
analogous to a mass going down a hill. In both cases,
potential energy decreases as kinetic energy
increases, –ΔU = ΔK. Work is done by a force, but
since this force is conservative, we can write W=–ΔU.
Source: https://openstax.org/books/university-physics-volume-2/pages/7-1-electric-potential-energy
If the electric force is conservative, then the work done by this force on a charged
particle moving from point A to point B can be expressed in terms of the electric potential
energy 𝑼,
𝑎
𝑊𝑎𝑏 = 𝑊𝑎→𝑏
𝑏
= ∫ 𝐹⃑ ∙ 𝑑𝑙⃗ = ∫ 𝑞𝑜 𝐸⃑⃑ ∙ 𝑑𝑙⃗ = −∆𝑈 = 𝑈𝑎 − 𝑈𝑏
𝑏
𝑎
The work done is equal to the negative of the change in electric potential energy.
1.
2.
Whether the test charge is positive or negative, the following general rules apply:
U increases if a test charge moves in the direction opposite to the electric force acting
on it.
U decreases if a test charge moves in the same direction as the electric force acting
on it.
Electric Potential Energy of Point Charges
The electric potential energy for two-point charges, q1 and q0, separated by a distance
r (see Figure 2) is given as
1 𝑞1 𝑞0
𝑈=
4𝜋𝜀0 𝑟
Figure 2 Two Point Charges
8
The electric potential energy for a test charge qo in the electric field produced by a collection
of charges (see Figure 3) is given by
𝑛
𝑞0
𝑞𝑖
𝑞0 𝑞1 𝑞2 𝑞3
𝑞𝑛
𝑈=
∑ =
( + + +. . . + )
4𝜋𝜀0
𝑟𝑖 4𝜋𝜀0 𝑟1 𝑟2 𝑟3
𝑟𝑛
𝑖
Figure 3 A Test Charge in a Collection of Charges
A more general equation to determine the total electric potential energy of the system:
𝑛
𝑞𝑖 𝑞𝑗
1
𝑈=
∑
4𝜋𝜀0
𝑟𝑖𝑗
𝑖<𝑗
Electric Potential
Electric Potential, V, is defined as the potential energy per unit charge. The electric potential
energy is shown by two elements, the charge possessed by the object itself and the position
relative to an object with respect to electrically charged objects. The magnitude of electric
potential is dependent on the amount of work done in moving the object from one point to
another against the electric field.
When an object is moved against the electric field it gains amount of energy which is
the electric potential energy. In a given charge the electric potential energy is the total work
done by an external agent in bringing the system of charges from continuity to the present
without any acceleration. It is obtained by dividing potential energy to the quantity of charge.
Electric potential energy is a scalar quantity. It is measured in terms of Joules.
𝑉=
𝑈
𝑞𝑜
Unit of Electric Potential: I Volt, V = 1 J/C
Given the sizes of each charge and the distance between them, the electric potential
energy they have relative to each other can be calculated. This is assuming the two charges
can be treated as point charges, which are where all the charge is concentrated at an exact
point in space.
9
1. The potential of infinity is defined to be zero.
2. If a point charge is positive, the electric potential of the charge is positive. When moving a
charge from infinity to this point, the electric potential increases above a zero level.
3. If a point charge is negative, the electric potential of the charge negative. When moving a
charge from infinity to this point, the electric energy decreases below a zero level.
Moving along the direction of the electric field, 𝐸⃑⃑, in both positive and negative point
charges, the electric potential V decreases. Otherwise, the potential 𝑉 increases (see Figure
4).
Figure 4 Electric Field and Electric Potential of a (A) Positive Charge (B) Negative Charge
Equations for Calculating Electric Potential
Working Equation
Description
for single point charge
q
1 q
V =k =
r 4 o r
V = k
i
V = k
for several point charges
qi
qi
1
=

ri 4 o i ri
for continuous charge distribution
dq
1
dq
=

r
4 o r
dq =  dl → linear charge distribution
dq =  dA → surface charge distribution
dq =  dV → volume charge distribution
The potential difference between two points can be expressed as a line integral given by
Electron Volt
One electron volt (1 eV) is the kinetic energy gained by an electron moving through a
potential difference of one volt (1 V).
1 eV = 1.602 x 10 -19 J
10
PRACTICE EXERCISES
CHECKPOINT 3: Problem Solving
Directions. Answer the following questions. Write your solution and answer on a separate
sheet of paper. Use the rubrics as your guide in presenting your solution.
1. How much work is required to move a charge of 4 nC from a point 2m away to a point 0.5 m
away from a point charge of 60 nC? What is the potential difference between these points?
2. A solid conducting sphere of radius 30 cm has a charge of 4 µC. If the potential is zero at
infinity. Find the value of the potential at the following distances from the center of the
sphere:
a) 45 cm
b) 30 cm
c) 15 cm
INSIGHTS
Summary
•
•
•
•
•
•
A charged object has electric potential energy by virtue of its location in an electric field.
Electric potential energy is a form of mechanical energy. It is expressed in electron volts
(eV).
The electric potential, or voltage, at any point in an electric field is the electric potential
energy per unit charge for a charged object at that point.
The work done by the electric force on a charged particle moving in an electric field is
equal to the negative of the change in electric potential energy.
Electric Potential can be calculated by summing or integrating over the charges.
Electron volt (1 eV) is the kinetic energy gained by an electron moving through a
potential difference of one volt (1 V),
Study Log
Based on the lesson you have just been through, please answer accordingly.
What questions do I
What I have
What did I find most
Keywords
learned?
interesting?
still have?
Electric Potential
Energy
Electric Potential
11
WRITTEN WORKS
Multiple Choice. Answer the question that follows. Choose the letter of the best answer from
among the given choices. Write your answer on your answer sheet.
1. Which of the following quantities is not a scalar in character?
A. Charge
C. Energy
B. Electric field
D. Potential difference
2. A system of two charges has a negative potential energy. What does it signify?
A. Both charges are positive
B. Both charges are negative
C. Both charges are positive or both are negative
D. One charge is positive & the other is negative
3. An electron & a proton are accelerated through the same potential difference, which of
the following has the correct status?
A. The electron has lower KE
C. The electron has lower speed
B. The proton has lower KE
D. The proton has lower speed.
4. The electric potential energy due to several point charges is given by the equation:
U = k Σ qi / r2i. What does r indicate?
A. The distance from the point charge to the point at which the potential energy is
evaluated.
B. The radius of the two pair of charges.
C. The radius from the center of the charge
D. None of the above
5. Which of the following statements are TRUE?
I. When a positive charge moves in the direction of an electric field, the potential energy
decreases and the field does positive work
II. When a positive charge moves in the direction of an electric field, the potential energy
increases and the field does negative work
III. When a negative charge moves in the direction of an electric field, the potential
energy increases and the field does negative work
IV. When a negative charge moves in the direction of an electric field, the potential
energy decreases and the field does positive work
A. I & IV
B. I &III
C. II&III
D. II&IV
6. In terms of energy unit, which does not belong?
A. eV
B. J
C. Nm
D. V
7. How is the direction of the electric field lines related to the equipotential surface?
A. Parallel
C. Depends on the sign of the charge
B. Perpendicular
D. Cannot be determine
8. When moving from infinity to towards a negative point charge, the electric potential.
A. Increases
C. Remains the same
B. Decreases
D. Cannot be determine
9. A point charge q = 2.00μC is located at the origin. Find the electric potential due to this
charge at point x = 4.0 m.
A. 1500 V
B. 2500 V
C. 3500 V
D. 4500 V
12
10. Two protons in the nucleus of a 238U atom are 6.0 fm (6.0x10-15m) apart. What is the
potential energy (in joules) associated with the electric force that acts between these two
particles?
A. 3.8x10-14
B. 4.5x10-15
C. 5.0x10-9
D. 5.0x10-8
11. Two point charges are arranged along the x-axis; q1=2.0 µC is at x=0.80m, and q2=2.0
µC at x=0.80m. The net electrical potential measured at the origin due to these point
charges is
A. zero
B. 4500V
C. 22500V
D. 45kV
12. Two like charges of charge +10 nC are placed at the two corners of an equilateral triangle
of side 3 cm. What is the magnitude of the total potential at the third corner?
A. 0
B. 6 V
C. 600 V
D. 6000 V
For numbers 13-15, refer to the figure below where Q1=+2.0µC, Q2 = -3.0µC.
13. What is the potential at point A?
A. -180,000 V
B. -225,000 V
14. What is the potential at point B?
A. -180,000 V
B. -225,000 V
15. What is the potential difference VA-VB?
A. -90, 000 V
B. -45,000 V
C. 225,000 V
D. 180,000 V
C. 225,000 V
D. 180,000 V
C. 45,000 V
D. 90, 000 V
PERFORMANCE TASK
Searching Time:
Using GOOGLE or preferably DEPED COMMON find the answers of the following questions
below.
Points
5
Criteria
Content
3
Grammar, usage
and mechanics
Neatness
2
•
•
•
Description
There is a clear and well-focused content. main idea
stands out and is supported by detailed info.
There has no error in grammar or spelling that
distract the reader from the content.
Work is illegible.
13
1. A Van de Graff Generator is a laboratory device with a large hollow metal sphere
supported by a cylindrical insulating stand. It is for building up high voltages. Why does
your hair stand out when you are charged by such device?
2. An electron gun is a device that is the heart of most TVs and computer monitors. How
does it work inside the device?
REFERENCES
Madeleine J. Ngitngit, General Physics 2, Quarter 3-Module 2 Electric Potential: Department
of Education-Division of Cagayan de Oro, 2020.
Suarez, V., et al, Workbook in Electromagnetism 8th Edition (Mindanao University of Science
and Technology, Cagayan de Oro City, 2015)
Young, H.D. and Freedman R. A., University Physics with Modern Physics 10th edition pp. 731761 (Pearson Education Asia Pte Ltd, 2002)
Villamor, R., 2003. Electronics Engineering. Sampaloc, Manila, Philippines: Ronnie Belarmino
Villamor & HR Publishing, pp.16 & 18
“Coulomb’s law – problems and solutions”, accessed August 6,
https://physics.gurumuda.net/coulombs-law-problems-and-solutions.htm
2021.
“Electric flux: Problems with Solutions for AP Physics”, By Physexams.Com, accessed
August 7, 2021. https://physexams.com/lesson/electric-flux-problems-solutions_18
“7.1
Electric
Potential
Energy”,
accessed
August
8,
2021.
https://openstax.org/books/university-physics-volume-2/pages/7-1-electric-potentialenergy
“Electric
field
worksheet”,
accessed
August
http://orbitsimulator.com/Physics6/ElectricFieldWorksheet0.htm
14
8,
2021.
KEY TO PRACTICE EXERCISES
15