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GP2 Q3 MELC 6 MOD 2 - General Physics II
Introductory Physics (University of the Philippines System)
Studocu is not sponsored or endorsed by any college or university
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SENIOR HIGH SCHOOL
GENERAL PHYSICS 2
Quarter 3 – Module 2:
ELECTRIC FORCE
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General Physics 2 – Senior High School
Alternative Delivery Mode
Quarter 3 – Module 2: Electric Force
First Edition, 2020
Republic Act 8293, section 176 states that: No copyright shall subsist in any
work of the Government of the Philippines. However, prior approval of the
government agency or office wherein the work is created shall be necessary for
exploitation of such work for profit. Such agency or office may, among other things,
impose as a condition the payment of royalties.
Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand
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use these materials from their respective copyright owners. The publisher and
authors do not represent nor claim ownership over them.
Published by the Department of Education
Secretary: Leonor Magtolis Briones
Undersecretary: Diosdado M. San Antonio
DEVELOPMENT TEAM OF THE MODULE
WRITERS:
MARIA CHARLENE D. DIPAD
EDITORS:
WELIMEN C. OSEO
REVIEWERS:
MICHELLE H. GUADAMOR
WELIMEN C. OSEO
ILLUSTRATOR:
JERIEL G. MARTIREZ
LAYOUT ARTIST:
SEVERINO R. CANTUBA JR.
ROMAN B. JEBULAN
KEVIN H. OJOS
JERIEL G. MARTIREZ
DAVE B. FORTES
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SENIOR HIGH SCHOOL
GENERAL PHYSICS 2
Quarter 3 – Module 2:
Electric Force
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Most Essential Learning Competency
Calculate the net electric force on a point charge exerted
by a system of point charges.
(GP12EMIII a-6)
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Electric Force
Introduction
“In a macro world, gravitational attraction
between two objects is strong. In contrast, at the
subatomic level, the electrostatic attraction between
two charges such as an electron and a proton is far
greater than the gravitational attraction.”
As you go on this module, you will learn more
about this electrostatic force that governs the micro
world.
Most Essential Learning
Competency & Objectives
In this module, you will learn to calculate the net electric force on a point
charge exerted by a system of point charges (GP12EMIIIa-6).
Specifically, you should be able to:
1. Explain the relationship between electric force and magnitude of the
charges.
2. Explain the relationship between electric force and distance of
separation between charges.
3. Calculate the net electric force on a point charge due to two or morepoint charges.
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Pre-Test
Before we start our lesson, try to answer these
ten (10) questions on electric force.
low.
Directions: Read the questions carefully and choose the letter of the
correct answer.
1. Coulomb’s law states that electric force is directly proportional to
the magnitude of the forces and inversely proportional to the square of the
distance between them. Which equation correctly represents the statement
above?
a. F = kq1/r2
c. F = kq1q2/r
b. F = k q1q2/r2
d. F = kq2/r
2. Two equal positive isolated electric charges of value q are separated by a
distance, r, and exert a force F on each other. One of the charges is
increased to 2q; the other is increased to 3q. The force will change to
_______________.
a. (3/2) F
c. 3F
b. 2F
d. 6F
3. What will happen to the electrical force if the magnitude of one of the charges
is doubled? The force will be________.
a. doubled
c. tripled
b. quadrupled
d. not changed
4. What will happen to the electrical force if both charges are doubled? The
force will be________.
a. doubled
c. tripled
b. quadrupled
d. not change
5. What will happen to the electrical force if the distance of separation between
charges is doubled?
a. doubled
c. reduced to 1/2
b. quadrupled
d. reduced to 1/4
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6. Two protons, each with a charge of 1.6 x 10 -19 C are separated by a distance
of 5.3 x 10 -11 m. What is the magnitude of electrical force of repulsion
between them?
a. 8.2 x 10 -6 N
c. 8.2 x 10 -8 N
b. 8.2 x 10 -7 N
d. 8.2 x 10 -9 N
For questions 7 and 8, refer to the problem below.
Two point charges are located on the positive x –axis of a coordinate system.
Charge q1 = 1.0 nC is 2.0 cm from the origin and charge q2 = -3.0 nC is 4.0 cm from
the origin.
7. What is the magnitude of electrostatic force exerted by q1 on q3? q2 on q3?
F13
1.12 x 10 -4 N
1.12 x 10 -5 N
1.12 x 10 -6 N
1.12 x10 -7 N
a.
b.
c.
d.
F23
8.4 x 10 -5 N
8.4 x 10-6 N
8.4 x 10 -7 N
8.4 x 10 -8 N
8. What is the net electric force on q3?
a. 2.8 x 10 -4 N
b. 2.8 x 10 -5 N
c. 2.8 x 10 -6 N
d. 2.8 x 10 -7 N
For numbers 9-10, refer to the problem below.
A system of three charges is shown in the figure below. Calculate the net
electrical force of charges q2 (-1.0 x 10-5C) and q3 ( 3.0x 10-5C) on q1 ( 2.0 x 10-5).
2.0 m
+q1
+q3
1.0 m
-q2
9. What is the magnitude of F2 and F3?
F2
F3
a.
1.5 N
1.0 N
b.
1.8 N
1.4 N
c.
2.0 N
1.8 N
d.
2.8 N
2.0 n
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10. What is the magnitude of the resultant force?
a. 1.8 N, 450 below the negative x-axis
b. 2.3 N, 450 above the negative x-axis
c. 1.8 N, 520 below the negative x-axis
d. 2.3 N, 520 below the negative x -axis
How did you find the test? If you got a
score of 10/10, you can proceed to the next
module. It is alright if you don’t get a perfect
score. As you go on with this module, you will
learn more about electric forces. So get ready
as you go on with another journey ahead of
you.
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Learning Activities
E licit/ Engage
In the previous module, you have learned
about charges and the methods of charging. Let
us recall the key ideas by answering the task
given below.
What do you think will happen to the leaves of an electroscope? Will they move
towards each other or will they move apart? Why do you say so?
______________________________________________________
______________________________________________________
______________________________________________________
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Were you able to use the concept on
Law of Charges in answering the question? We
need this concept to better understand the
lesson in this module.
ENGAGE
Charges are contained inside an atom. Protons
carry the positive charge while electrons carry the negative
charge. When two materials are rubbed together just like
in the examples given in the previous module, they
become electrically charged. When objects are charged, it
has an ability to affect the behavior of objects surrounding
it. But, how do charged objects interact with each other?
Write down your initial ideas on how charged
objects interact with each other.
How do charged objects interact?
You will learn more about electric forces as
you go on the succeeding activities.
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E xplore
Activity 1: Coulomb’s Law for the Electric Force
Picture Analysis 1: Electric Force and Magnitude of Point Charges
Study the three (3) diagrams below. As you analyze the figures, take
note of the magnitude/size of the point charge and the force. The length of the arrow
represents the magnitude of the force. Then, answer the following guide questions.
F
F
+
+
+
+
F
F F
F
-
-
-
F
F
Guide Questions:
F
F
-
+
+
-
F
F
1. How would you describe the forces acting between point charges?
___________________________________________________________
___________________________________________________________
2. If the magnitude of the point charges becomes bigger, what do you think will
happen to force between the charges?
___________________________________________________________
___________________________________________________________
3. How would you relate electric force and the magnitude of the point charges?
____________________________________________________________
____________________________________________________________
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Picture Analysis 2: Electric Force and Distance
In the next task, you will analyze how electric force is affected by the distance.
Again, take note the magnitude of the force represented by the length of the arrow.
After studying the diagram, answer the guide questions below.
+
+
+
+
Figure 1
Guide Questions
1. What happens to the electric force as the distance of separation between
point charges decreases? How about when distance increases?
_____________________________________________________________
_____________________________________________________________
2. How would you relate electric force and the distance of separation between
point charges?
_____________________________________________________________
_____________________________________________________________
3. Considering the diagrams in picture analysis 1 and 2, how would you relate
electric force to the magnitude of point charges and distance of separation
between them?
_____________________________________________________________
_____________________________________________________________
Were you able to answer the guide
questions? All of these is describe as Coulomb’s
law. In the next activity, you will find out more about
this law.
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Activity 2: Calculating Electric Force
Part I: Force on a Point Charge due to two-point charges
Study the table below and find out how the magnitude of electric force is
computed.
constant,
permittivity of free
space
(K)
Magnitude
of charge 1
Magnitude of
charge 2
(q1)
(q2)
Distance
between
two
charges
(r)
Square
of the
distance
between
two
charges
Magnitude
of electric
force
(F)
(r2)
9 x 109 N.m2/C2
25 x 10-9 C
-75 x 10-9C
0.030 m
9x10-4m2
0.019 N
1. How did you compute the value of the electric force? What equation did you
use?
______________________________________________________________
______________________________________________________________
2. In column no. 5, notice that the distance between charges is squared, what
does it imply about the relationship between electric force and distance
between point charges?
______________________________________________________________
______________________________________________________________
3. Using the equation you’ve formulated, solve the problem presented below.
In a hydrogen atom, the electron is separated from the proton by
average distance of about 5.3 x 10 – 11 m. Calculate the magnitude of the
electrostatic force of attraction exerted by the proton on the electron. ( The
charge of proton is 1.6 x 10-19 C while the charge of an electron is -1.6 x 10-19 C)
given:
find:
solution
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Part II: Force on a Point Charge due to multiple charges
In part I, you’ve calculated the electric force on a point
charge, q1, due to another point charge, q2. Suppose that
another point charge, q3, is also present, what would be the net
force on q1 due to q2 and q3?
It is easier to deal with such problem in parts. As you
continue the exploration, please be guided of the following
steps.
Problem Solving Tips
1. Find the magnitude and direction of the force exerted on q1 by q2
(ignoring q3).
2. Determine the force exerted on q1 by q3 (ignoring q2).
3. The net force on q1 is the vector sum of these forces.
Use these tips in calculating the force on a point charge due
to point charges 2 and 3.
Three Charges on a Line
Three point charges lie along the x-axis in a vacuum. Determine the
magnitude and direction of the net electrostatic force on q1.
q₂
q1
0.20 m
- 4.0 µC
0.15 m
3.0 µC
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q3
- 7.0 µC
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The Free Body Diagram (FBD) is as shown.
F1 2
q1
F1 3
Guide Questions:
1. Why do you think the force exerted on q1 by q2 is directed to the left?
____________________________________________________________
____________________________________________________________
2. Why do you think F13 is directed to the right?
___________________________________________________________
___________________________________________________________
3. Using the equation you’ve formulated in Activity 1, compute for the magnitude
of F12.
4. What is the magnitude of F13?
5. Since F12 points in the (-) x-direction and F13 points in the (+) x direction, the
net force is the vector sum of F12 and F13. Show your calculation on how to
compute for the net force in the space provided below.
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E xplain
Electric force or electrostatic force exists
between point charges. The force exerted by
one point charge to another acts along the line
between charges. It can be repulsive or an
attractive force. A repulsive force exists
between like charges and an attractive force for
unlike charges.
How do we calculate the magnitude of
this force? Coulomb’s Law is used to calculate
the magnitude of electric force. It states that
electric force is directly proportional to product of the charges and inversely
proportional to the square of the distance between them. The law shows that electric
force is dependent upon the size of the charges and their distance of separation. As
shown in picture analysis 1, electric force is dependent upon the magnitude of the
charge. As the size of the charges increases, the electric force also increases and
vice versa. In picture analysis 2, electric force is dependent on the distance of
separation between charges. As the distance decreases, the electric force increases
and vice versa.
Coulomb’s Law follows the inverse square law. It means that if the distance of
separation between two objects increases, the repulsive or attractive force between
the objects decreases. Decreasing the separation distance between objects,
increases the electric force between the two objects.
Applying the inverse square relationship in Coulomb’s law, you will notice that
electrostatic force between two point charges varies inversely with the square of the
distance of separation between the two charges. That is, the factor by which the
electrostatic force is changed is the inverse of the square of the factor by which the
separation distance is changed. So if the separation distance is doubled (increased
by a factor of 2), then the electrostatic force is decreased by a factor of four (2 raised
to the second power). And if the separation distance is tripled (increased by a factor
of 3), then the electrostatic force is decreased by a factor of nine (3 raised to the
second power).
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In a quantitative form, Coulomb’s Law is expressed as:
where:
Fe = force on each charge in Newton (N)
q1 & q2 = interacting point charges in
Coulombs (C)
r = distance of separation between point
charges in meters (m)
k = constant of proportionality.
9 x 10 9 Nm2/C2
Traditionally, Coulomb’s Law is written as:
This equation is used in solving the problem presented during the explore
part.
In a hydrogen atom, the electron is separated from the proton by average
distance of about 5.3 x 10 – 11 m. Calculate the magnitude of the electrostatic force of
attraction exerted by the proton on the electron. ( The charge of proton is 1.6 x 10 -19 C
while the charge of an electron is -1.6 x 10-19 C)
given: q1 = 1.6 x 10-19 C
q2 -1.6 x 10-19 C
r = 5.3 x 10 – 11 m
find: F
solution: F = k
/𝑞1𝑞2/
𝑟2
= 9 x 10 9 n m2/C2 ( 1.6 x 10 -19 C)2 / (5.3 x 10-11m)2
= 8.20 x 10-8 N
To calculate the net force exerted on a point charge due to multiple charges,
you will apply the Principle of Superposition of Forces. By using this principle, you
can apply Coulomb’s Law to any collection of charges. The principle explains that
the total force acting on that charge is the vector sum of the forces that the two
charges would exert individually. This principle is demonstrated in Problem 2 in the
explore part.
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Three Charges on a Line
Three point charges lie along the x-axis in a vacuum. Determine the
magnitude and direction of the net electrostatic force on q1.
q₂
q1
0.20 m
- 4.0 µC
0.15 m
3.0 µC
given: q1 = 3.0 x 10 -6 C
q2 = -4.0 x 10 -6 C
q3 = -7.0 x 10 -6 C
q3
- 7.0 µC
r12 = 0.20 m
r13 = 0.15 m
find: net electric force on q1
Solution:
a. F12 = k q1q2/r122
= 9 x 109 Nm2/C2 ( 3.0 x 10-6 C) ( 4 x 10 -6 C)/ (0.20 m)2
= 2.7 N
b. F13= k q1q3/r132
= 9x109Nm2/C2 (3.0 x 10-6C) ( 7 x 10-6)/ (0.15m)2
= 8.4 N
Since F12 points in the ( -) x direction and F13 points in the (+) x direction, the net F is:
Fnet = F12 + F13
= (-2.7 N) + (8.4N)
= 5.7 N points to the right.
Have you noticed that Coulomb’s Law which
describes the electric force is comparable to Newton’s Laws
of Universal Gravitation? Both follow the inverse-square law.
The difference is that electric force can be attractive or
repulsive while gravitational force is always attractive.
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E laborate
One important clinical application concerning
electric forces is electrocardiography or ECG. Let’s
find out the physics behind it.
Electrocardiography or ECG is a technique for analyzing
the condition of the heart. As each muscle cell in the heart
contracts, positive and negative charges are separated. The
polarization of charges is shown in the diagram below.
The effect of the charges separated across all of the
contracting muscle cells at one instant time of time can be
represented by a single positive and a single negative charge.
The magnitude and relative location of these charges depends on the number and
orientations of muscle cells that are contracting at any one time. Since the heart
repeats a cycle of contraction about one each second, it produces charge
distribution that change magnitude and location in a repetitive fashion-once each
second.
At any instant of time, the electric charges on the heart exert forces on
ions in the body surface tissue that cause the ions to move.
++ -
electrode
electrode
For example, the charges on the heart force sodium ions in the tissue toward
the left and chlorine ions toward the right. Electrodes placed on the arms absorb
charges of different signs. The heart acts much like a battery pushing opposite
charges in opposite directions in the body tissue. An electrocardiogram is a
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recording of the charge separation on the body’s surface caused by the electric
charge of the heart. Abnormalities in the size or timing sequence of different phases
of the heartbeat cycle are easily detected by these electrodes.
Can you now explain to your family how ECG works?
Now, it’s time to write down what have you learned in this
module.
1. How would you relate electric
force to the magnitude of charges
and
distance
between charges?
of
2. What problem solving technique
would help you to calculate the
electric force on a point charge
due to multiple charges?
separation
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E xtend
Do you want to have a challenge? I know you can do
this! In the problem presented below, you will apply what
you’ve learned in Coulomb’s Law, Principle of Superposition
of Forces and the component method in resolving force
vectors.
Challenge Problem 1: Three Charges in a Plane
Three point charges lie in the x,y plane in a vacuum. Find the magnitude and
direction of the net electrostatic force on q1.
The force exerted on q1 by q2 is represented by F12 and is an attractive force
because the two charges have opposite signs. It points along the line between
charges. The force exerted on q1 by q3 is F13 and is also an attractive force. It points
along the line between q1 and q3.
1. Using Coulomb’s Law, compute for the magnitude of F12 and F13.
a. F12 =
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b. F13 =
2. The net force is the vector sum of F12 and F13. Use the component method
to find the magnitude and direction of the net force.
Force
x-component
y-component
F12
F13
Fx =
Fy =
The net force is:
The direction of the net force is:
Good job for solving the challenge problem. You are
now ready to take our post-test. But before taking the test,
kindly review the physics terms used in this module.
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Vocabulary List
You may refer to the following vocabulary words used in this module:
Coulomb’s Law – established by Charles Augustin de Coulomb. It states that the
magnitude of the electric force between two point charges is
directly proportional to the product of the charges and inversely
proportional to the square of the distance between them.
– typically made of gold foil leaves hung from a conducting metal
stem and is insulated from the room air in a glass-walled
container.
inverse-square law – principle in physics that the effect of certain forces on an
object varies by the inverse square of the distance between the
object and source of the force.
point charge
–charged bodies that are very small in comparison with the
distance in between them.
electroscope
principle of superposition of forces – principle that states that any number of
forces applied at a point on a body have the same effect as s
ingle force equal to the vector sum of the forces.
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Post-Test
E valuate
Directions: Read the questions carefully and choose the letter of the correct answer.
1. Coulomb’s law states that electric force is directly proportional to the
magnitude of the forces and inversely proportional to the square of the
distance between them. Which equation correctly represents the statement
above?
a. F = kq1/r2
c. F = kq1q2/r
b. F = k q1q2/r2
d. F = kq2/r
2. Two equal positive isolated electric charges of value q are separated by a
distance, r, and exert a force F on each other. One of the charges is
increased to 2q; the other is increased to 3q. The force will change to
_______________.
a. (3/2) F
c. 3F
b. 2F
d. 6F
3. What will happen to the electrical force if the magnitude of one of the charges
is doubled? The force will be________.
a.doubled
c. tripled
b. quadrupled
d. not changed
4. What will happen to the electrical force if both charges are doubled? The
force will be________.
a. doubled
c. tripled
b. quadrupled
d. not change
5. What will happen to the electrical force if the distance of separation between
charges is doubled?
a. doubled
c. reduced to 1/2
b. quadrupled
d. reduced to 1/4
6. Two protons, each with a charge of 1.6 x 10 -19 C are separated by a distance
of 5.3 x 10 -11 m. What is the magnitude of electrical force of repulsion
between them?
a. 8.2 x 10 -6 N
c. 8.2 x 10 -8 N
b. 8.2 x 10 -7 N
d. 8.2 x 10 -9 N
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For questions 7 and 8, refer to the problem below.
Two point charges are located on the positive x –axis of a coordinate system.
Charge q1 = 1.0 nC is 2.0 cm from the origin and charge q2 = -3.0 nC is 4.0 cm
from the origin.
7. What is the magnitude of electrostatic force exerted by q1 on q3? q2 on q3?
F13
1.12 x 10 -4 N
1.12 x 10 -5 N
1.12 x 10 -6 N
1.12 x10 -7 N
a.
b.
c.
d.
F23
8.4 x 10 -5 N
8.4 x 10-6 N
8.4 x 10 -7 N
8.4 x 10 -8 N
8. What is the net electric force on q3?
a. 2.8 x 10 -4 N
b. 2.8 x 10 -5 N
c. 2.8 x 10 -6 N
d. 2.8 x 10 -7 N
For numbers 9-10, refer to the problem below.
A system of three charges is shown in the figure below. Calculate the net
electrical force of charges q2 (-1.0 x 10-5C) and q3 ( 3.0x 10-5C) on q1 ( 2.0 x 10-5).
2.0 m
+q1
+q3
1.0 m
-q2
9. What is the magnitude of F2 and F3?
F2
F3
a.
1.5 N
1.0 N
b.
1.8 N
1.4 N
c.
2.0 N
1.8 N
d.
2.8 N
2.0 n
10. What is the magnitude of the resultant force?
a. 1.8 N, 450 below the negative x-axis
b. 2.3 N, 450 above the negative x-axis
c. 1.8 N, 520 below the negative x-axis
d. 2.3 N, 520 below the negative x -axis
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Answer Keys
Pre-Test Answer Keys
1.
2.
3.
4.
5.
B
D
A
B
D
6. C
7. A
8. D
9. B
10. D
Learning Activities Answer Keys
Elicit
•
When a negatively charges object is brought near an electroscope, it
causes the free moving electrons in the electroscope to move down
into the leaves leaving the knob (top part) positive. Since like charges
repel, the leaves with both negative charge repeal each other and
move apart.
Engage
Activity 1: Coulomb’s Law for the Electric Force
Picture Analysis 1: Electric Force and magnitude of the point charges
1. Forces can be attractive or repulsive. Its magnitude depends on
the magnitude of the charges and its direction depends on the
signs of the point charges.
2. If the magnitude of the point charges becomes bigger, the force
between charges increases.
3. Electric force is directly proportional to the magnitude of the point
charges. Increasing the size of the charges increases the force.
Decreasing its size, decreases the force.
Picture Analysis 2: Electric Force and distance
1. As the distance of separation between point charges decreases,
the force increases. As the distance increases, the electric force
between charges decreases.
2. Electric force is inversely proportional to the distance of separation
between point charges.
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3. Electric force is directly proportional to the magnitude of the
charges and inversely proportional to the distance of separation
between them.
Activity 2: Calculating Electric Force
Part I: Force on a Point Charge due to two point charges
1. The equation for Coulomb’s Law is used to compute for the magnitude of
electric force.
2. Electric force is inversely proportional to the square of the distance of
separation between point charges.
3. F = 8.2 x 10 -8N
Part II: Force in a Point Charge due to multiple charges
1. q1 and q2 have opposite signs so they attract one another, thus, the
force exerted on q1 by q2 is F12 and it points to the left.
2. The force exerted by q3 on q1 is F13 is also an attractive force so it points
to the right.
3. F12 = 2.7 N
4. F13 = 8.4 N
5. Since F12 points in the (–) x direction and F13 points in the (+) x direction,
the net force is +5.7 N. The net force points to the right.
Extend
1. F12 = 9.6 N
2. F13 = 18 N
3. Fx = 21 N; Fy =9.2 N
The magnitude of F and the angle of the net force are 23 N, 24 0.
Post-Test Answer Keys
1.
2.
3.
4.
5.
B
D
A
B
D
6. C
7. A
8. D
9. B
10. D
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References
Printed Resources
Basic ECG Theory, Recordings, and Interpretation by Anthony Dupre, Sarah Vincent
and Paul A. Iaizzo. Retrieved from http://eknygos.lsmuni.lt/springer/675/191201.pdf on July 2, 2020
Cutnell, John D. and Johnson, Kenneth W. Physics 8th edition, ohn Wiley and Sons,
Inc. 2008 pp 539-542
Department of Education.(May 2016) K to 12 senior high school specialized subject
–General
Physics
2
(Curriculum
Guide)
Retrieved
from https://www.deped.gov.ph/wp-content/uploads/2019/01/General-Physics2.pdf
Young Hugh D., Freedman, Roger A. and Ford, A. Lewis. University Physics with
Modern Physics 12th Edition. Pearson Education South Asia Pte Ltd. 2009 pp
716-720
Heuvelen, Alan Van. Physics A General Introduction/Second Edition. Little, Brown
and Company 1986
Teaching Guide for Senior High School General Physics 2
Online Resources
http://scienceres-edcpeduc.sites.olt.ubc.ca/files/2012/07/sec_phys_electrostatics_coulombLaw940x705.jpg ( Coulomb’s Law equation)
https://www.chegg.com/homework-help/questions-and-answers/coulomb-s-lawproportionality-constant-permitivity-free-space-come-also-4-pi-come-thankq29607791 (permittivity constant)
https://www.researchgate.net/figure/Coulombs-law-is-applicable-to-theNACAP_fig8_24328478 figure uploaded by Melvin R. Carruth (Coulomb’s Law with
epsilon nought)
https://www.dictionary.com/browse/inverse-square--law
https://www.physicsclassroom.com/class/estatics/Lesson-3/Inverse-Square-Law
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