Module No. 2
COULOMB’S LAW
ELECTRICITY AND MAGNETISM
SCI 112
Course Description
This course is designed to discuss knowledge of basic relationship between electricity
and magnetism. It includes topics in electrostatics and magnetism, electric and magnetic fields
in matter, electrodynamics and electromagnetic waves. It provides the students the
mathematical relationship between current, voltage and resistance in an electric circuit.
Students may gain skills in solving problems needing high mathematical analysis apart from
the principles comprising this area of physics. Upon knowing the relationship between
electricity and magnetism students must be able to apply the concepts and principles to real
life situations for life-long learning.
Laboratory work is an integral part of this course. This involves dry (exercises,
problem sets) and wet laboratory activities that are done to confirm the correctness of
principles learned.
Total Learning Time: 6 hours per week (3 hours lecture and 3 hours laboratory)
The interaction between charged objects is a non-contact force that acts over some
distance of separation. Charge, charge and distance. Every electrical interaction involves a
force that highlights the importance of these three variables. Whether it is a plastic golf tube
attracting paper bits, two like-charged balloons repelling or a charged Styrofoam plate
interacting with electrons in a piece of aluminum, there is always two charges and a distance
between them as the three critical variables that influence the strength of the interaction.
Learning Outcomes
At the end of this module, the students are expected to:
1. state and explain Coulomb’s law
2. applied Coulomb’s law in problem solving
3. cite examples where coulomb’s law is applied
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Indicated content
1.
2.
3.
4.
5.
6.
Force as a vector quantity
Charles- Agustin de Coulomb
Coulomb’s Law
Coulomb’s Law Equation
Calculations Using Coulomb’s Law
Application of Coulombs Law
Discussion
 Force as a Vector Quantity
The electrical force, like all forces, is typically expressed using the unit Newton.
Being a force, the strength of the electrical interaction is a vector quantity that has both
magnitude and direction. The direction of the electrical force is dependent upon whether the
charged objects are charged with like charge or opposite charge and upon their spatial
orientation. By knowing the type of charge on the two objects, the direction of the force on
either one of them can be determined with a little reasoning. In the diagram below, objects A
and B have like charge causing them to repel each other. Thus, the force on object A is
directed leftward (away from B) and the force on object B is directed rightward (away from
A). On the other hand, objects C and D have opposite charge causing them to attract each
other. Thus, the force on object C is directed rightward (toward object D) and the force on
object D is directed leftward (toward object C). When it comes to the electrical force vector,
perhaps the best way to determine the direction of it is to apply the fundamental rules of
charge interaction (opposites attract and likes repel) using a little reasoning.
 Charles-Augustin de Coulomb (14 June 1736 – 23 August 1806)
was a French military engineer and physicist. He is best known as
the eponymous discoverer of what is now called Coulomb's law, the
description of the electrostatic force of attraction and repulsion,
though he also did important work on friction.
 The SI unit of electric charge, the coulomb, was named in his honor
in 1908.
He investigated the electric forces in the 1780’s using a torsion
balance much like used by Cavendish for his studies of gravitational force. Precise
instruments for the measurement of electric charge were not available in Coulomb’s time.
Nonetheless was able to prepare small spheres with different magnitude of charge in which
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the ratio of the charges was known. Although he had some difficulty with induced charges,
Coulomb was able to argue that the electric force of one tiny charged object exerts on a
second tiny charged object is directly proportional to the charge on each of them. That is, if
the charge on either one of the object is doubled, the force is doubled; and if the charge on
both of the object is doubled, the force increase four times the original value. This was the
case when the distance between the two charges remained the same. If the distance between
them was allowed to increase, he found that the force decreased with the square of the
distance between them. That is, if the distance was doubled, the force fell to one-fourth of its
original value. Thus Coulomb concluded, the magnitude of the force F that one small object
exert on a second one is proportional to the product of the magnitude of the charge on one, q1,
times the magnitude of the other, q2, and inversely proportional to the square of the distance r
between them. As an equation, we write Coulomb’s law as
𝐹=𝑘
𝑞1 𝑞2
𝑟2
where k is a proportionality constant.
Electrical force also has a magnitude or strength. Like most types of forces, there are
a variety of factors that influence the magnitude of the electrical force. Two like-charged
balloons will repel each other and the strength of their repulsive force can be altered by
changing three variables. First, the quantity of charge on one of the balloons will affect the
strength of the repulsive force. The more charged a balloon is, the greater the repulsive force.
Second, the quantity of charge on the second balloon will affect the strength of the repulsive
force. Gently rub two balloons with animal fur and they repel a little. Rub the two balloons
vigorously to impart more charge to both of them, and they repel a lot. Finally, the distance
between the two balloons will have a significant and noticeable effect upon the repulsive
force. The electrical force is strongest when the balloons are closest together. Decreasing the
separation distance increases the force. The magnitude of the force and the distance between
the two balloons is said to be inversely related.
 Coulomb's Law Equation
The quantitative expression for the effect of these three variables on electric force is
known as Coulomb's law. Coulomb's law states that the electrical force between two charged
objects is directly proportional to the product of the quantity of charge on the objects and
inversely proportional to the square of the separation distance between the two objects. In
equation n form, Coulomb's law can be stated as
𝑞1 𝑞2
𝐹=𝑘 2
𝑟
where q1 represents the quantity of charge on object 1 (in Coulombs), q2 represents the
quantity of charge on object 2 (in Coulombs), and r represents the distance of separation
between the two objects (in meters). The symbol k is a proportionality constant known as the
Coulomb's law constant. The value of this constant is dependent upon the medium that the
charged objects are immersed in. In the case of air, the value is approximately 9.0 x 10 9 N •
m2 / C2. If the charged objects are present in water, the value of k can be reduced by as much
as a factor of 80. It is worthwhile to point out that the units on k are such that when
substituted into the equation the units on charge (Coulombs) and the units on distance
(meters) will be cancelled, leaving a Newton as the unit of force.
The Coulomb's law equation provides an accurate description of the force between two
objects whenever the objects act as point charges. A charged conducting sphere interacts
with other charged objects as though all of its charge were located at its center. While the
charge is uniformly spread across the surface of the sphere, the center of charge can be
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considered to be the center of the sphere. The sphere acts as a point charge with its excess
charge located at its center. Since Coulomb's law applies to point charges, the distance d in
the equation is the distance between the centers of charge for both objects (not the distance
between their nearest surfaces).
The symbols q1 and q2 in the Coulomb's law equation represent the quantities of charge on the
two interacting objects. Since an object can be charged positively or negatively, these
quantities are often expressed as "+" or "-" values. The sign on the charge is simply
representative of whether the object has an excess of electrons (a negatively charged object)
or a shortage of electrons (a positively charged object). It might be tempting to utilize the "+"
and "-" signs in the calculations of force. While the practice is not recommended, there is
certainly no harm in doing so. When using the "+" and "-" signs in the calculation of force, the
result will be that a "-" value for force is a sign of an attractive force and a "+" value for force
signifies a repulsive force. Mathematically, the force value would be found to be positive
when q1 and q2 are of like charge - either both "+" or both "-". And the force value would be
found to be negative when q1 and q2 are of opposite charge - one is "+" and the other is "-".
This is consistent with the concept that oppositely charged objects have an attractive
interaction and like charged objects have a repulsive interaction. In the end, if you're thinking
conceptually (and not merely mathematically), you would be very able to determine the
nature of the force - attractive or repulsive - without the use of "+" and "-" signs in the
equation.
 Calculations Using Coulomb's Law
In physics courses, Coulomb's law is often used as a type of algebraic recipe to solve physics
word problems. Three such examples are shown here.
Example 1
Suppose that two point charges, each with a charge of +1.00 Coulomb are separated by a
distance of 1.00 meter. Determine the magnitude of the electrical force of repulsion between
them.
The first step of the strategy is the identification and listing of known information in
variable form. Here we know the charges of the two objects (q1 and q2) and the separation
distance between them (r). The next step of the strategy involves the listing of the unknown
(or desired) information in variable form. In this case, the problem requests information about
the force. So Felect is the unknown quantity. The results of the first two steps are shown in the
table below.
Given:
Find:
q1 = 1.00 C
Felect = ???
q2 = 1.00 C
d = 1.00 m
The next and final step of the strategy involves substituting known values into the Coulomb's
law equation and using proper algebraic steps to solve for the unknown information. This step
is shown below.
Felect = k • q1 • q2 / r2
Felect = (9.0 x 109 N•m2/C2) • (1.00 C) • (1.00 C) / (1.00 m)2
b Felect = 9.0 x 109 N
The force of repulsion of two +1.00 Coulomb charges held 1.00 meter apart is 9 billion
Newton. This is an incredibly large force that compares in magnitude to the weight of more
than 2000 jetliners.
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This problem was chosen primarily for its conceptual message. Objects simply do not acquire
charges on the order of 1.00 Coulomb. In fact, more likely q values are on the order of 10-9 or
possibly 10-6 Coulombs. For this reason, a Greek prefix is often used in front of the Coulomb
as a unit of charge. Charge is often expressed in units of microCoulomb (µC) and
nanoCoulomb (nC). If a problem states the charge in these units, it is advisable to first convert
to Coulombs prior to substitution into the Coulomb's law equation. The following unit
equivalencies will assist in such conversions.
1 Coulomb = 106 microCoulomb
1 Coulomb = 109 nanoCoulomb
The problem-solving strategy used in Example 1 included three steps:
1. Identify and list known information in variable form.
2. List the unknown (or desired) information in variable form.
3. Substitute known values into the Coulomb's law equation and using proper algebraic steps
to solve for the unknown information. (In some cases and for some students, it might be
easier to first do the algebra using the variables and then perform the substitution as the
last step.)
This same problem-solving strategy is demonstrated in Example 2 below.
Two balloons are charged with an identical quantity and type of charge: -6.25 nC. They are
held apart at a separation distance of 61.7 cm. Determine the magnitude of the electrical force
of repulsion between them.
The problem states the value of q1 and q2. Since these values are expressed in units of
nanoCoulombs (nC), the conversion to Coulombs must be made. The problem also states the
separation distance (r). Since distance is given in units of centimeters (cm), the conversion to
meters must also be made. These conversions are required since the units of charge and
distance in the Coulomb's constant are Coulombs and meters. The unknown quantity is the
electrical force (F). The results of the first two steps are shown in the table below.
Given:
Find:
q1 = -6.25 nC = -6.25 x 10-9 C
Felect = ???
q2 = -6.25 nC = -6.25 x 10-9 C
r = 61.7 cm = 0.617 m
The final step of the strategy involves substituting known values into the Coulomb's law
equation and using proper algebraic steps to solve for the unknown information. This
substitution and algebra is shown below.
Felect = k • q1 • q2 / r2
Felect = (9.0 x 109 N•m2/C2) • (6.25 x 10-9 C) • (6.25 x 10-9 C) / (0.617 m)2
Felect = 9.23 x 10-7 N
Note that the "-" sign was dropped from the q1 and q2 values prior to substitution into the
Coulomb's law equation. As mentioned above, the use of "+" and "-" signs in the equation
would result in a positive force value if q1 and q2 are like charged and a negative force value if
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q1 and q2 are oppositely charged. The resulting "+" and "-" signs on F signifies whether the
force is attractive (a "-" F value) or repulsive (a "+" F value).
Example 3
Two balloons with charges of +3.37 µC and -8.21 µC attract each other with a force of 0.0626
Newton. Determine the separation distance between the two balloons.
The problem states the value of q1 and q2. Since these values are in units of microCoulombs
(µC), the conversion to Coulombs will be made. The problem also states the electrical force
(F). The unknown quantity is the separation distance (r). The results of the first two steps are
shown in the table below.
Given:
Find:
q1 = +3.37 µC = +3.37 x 10-6 C
r = ???
-6
q2 = -8.21 µC = -8.21 x 10 C
Felect = -0.0626 N (use a - force value since it is
attractive)
As mentioned above, the use of the "+" and "-" signs is optional. However, if they are used,
then they have to be used consistently for the q values and the F values. Their use in the
equation is illustrated in this problem.
The final step of the strategy involves substituting known values into the Coulomb's law
equation and using proper algebraic steps to solve for the unknown information. In this case,
the algebra is done first and the substitution is performed last. This algebra and substitution is
shown below.
Felect = k • q1 • q2 / r2
r2 • Felect = k • q1 • q2
r2 = k • q1 • q2 / Felect
r = SQRT(k • q1 • q2) / Felect
r = SQRT [(9.0 x 109 N•m2/C2) • (-8.21 x 10-6 C) • (+3.37 x 10-6 C) / (-0.0626 N)]
r = Sqrt [ +3.98 m2 ]
r = +1.99 m
 Application of Coulomb’s Law
Haloid Becomes Xerox
In 1938 and a patent attorney named Chester Carlson who was working in the U.S.
Patent Office in New York City. Carlson was required to make copies by the hand of a large
number of papers, but his hands were arthritic. Tinkering in his kitchen at night, he made the
first photocopy machine. Between 1939 and 1944, Carlson tried to market his invention to
over 20 companies, including IBM and General Electric, and they all turned him down citing
no need for a copy machine. Eventually, in 1947, the Haloid Corporation, a small New York
manufacturer of photographic paper, obtained a license to develop a copy machine based on
Carlson's invention.
They consulted a professor of classical languages who came up with the name "xerography"
from the Greek words xeros for dry, and graphos for writing, and the Haloid Corporation
became the Xerox Corporation.
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The electrostatic process is what makes copies. It uses a selenium-coated aluminum
drum because selenium has an interesting property — it is an insulator when in the dark, and a
conductor when exposed to light. In the first stage of the xerography process, a negative
charge is induced under a thin layer of positively charged selenium. The surface of the drum
is then exposed to the image to be copied, and where the image is light, the positive charge is
neutralized, and where the image is dark, the positive charge remains. The image has now
been transferred to the drum. Then, a dry black powder, called toner, is sprayed with a
negative charge, which will be attracted to the positive areas of the drum. A blank piece of
paper is given a greater positive charge than the drum, so that it will pull the toner from the
drum, and finally, the paper and toner are passed through heated rollers that melt and
permanently adhere the toner to the paper.
Laser printer
In laser printers, a laser beam is scanned across a photo conducting drum, which leaves a
positively charged image, then the next steps are the same as in xerography. Because laser
light can be very precisely controlled, laser printers can produce very high-quality images.
The electrostatic process is also used in ink jet printers where a nozzle finely sprays tiny ink
droplets, which are then given an electrostatic charge. The droplets are directed using pairs of
charged plates, and they form letters and images on paper. Color inkjet printers use black,
cyan, magenta, and yellow jets.
Powder Coating, It's Not Just For Motorcycles
Another use of the electrostatic process is electrostatic painting or coating, also known as
"powder coating." The process uses a high voltage electrostatic charge which is applied to
both the object to be coated and the sprayer mechanism.
A coating of either powdered particles or an atomized liquid is accelerated toward the work
piece by a powerful electrostatic charge. The ionic bond of the coating to the object creates a
uniform coating that adheres extremely well.
Several powder colors can be applied before curing them all together, and this allows color
blending and bleeding that produce special effects. For that reason, powder coating is beloved
by motorcycle enthusiasts worldwide
Forces in Atoms
An atom is, in one respect, nothing other than a collection of electrical charges, positively
charged protons, and negatively charged electrons. Coulombic forces exist among these
particles. For example, a fundamental problem involved in a study of the atomic nucleus is
explaining how the enormous electrostatic force of repulsion among protons is overcome in
such a way as to produce a stable body.
Coulombic forces must be invoked also in explaining molecular and crystalline architecture.
The four bonds formed by a carbon atom, for example, have a particular geometric
arrangement because of the mutual force of repulsion among the four electron pairs that make
up those bonds. In crystalline structures, one arrangement of ions is preferred over another
because of the forces of repulsion and attraction among like-charged and oppositely-charged
particles respectively.
 Using Coulomb’s Law to Clean the Air
Coal-burning power plants produce large amounts of potential pollution in the form of small
particles (soot). Modern smokestacks use devices called scrubbers to remove these particles
from the smoke they emit. Scrubbers use a two-step process: electrons are first added to the
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soot particle, and an electric force then pulls the particle out of the smoke stream. In this
example, we analyze the force on a soot particle after electrons are added.
Laboratory Activities
1. Static Electricity
2. Problem Set
Exercises/Drill
1. What is the magnitude of the repulsive force of two protons in a nucleus if the charge on
each proton is 1.6 x 10 -19 coulomb and the protons are 3 x 10 -15 meter apart?
2. Determine the magnitude and direction of the electric force on the electron of a hydrogen
atom exerted by the single proton (q2 = + e) that is the atom’s nucleus. Assume the average
distance between the revolving electron and the proton is r= 0.53 x 10 -10 m.
3. Two objects are both negatively charged with 0.02 C each and are 70 cm apart. What kind
of force exists between them and how much?
4. Find the magnitude of the force between two charges of 1.0 C each that are 1.0 m apart.
5. The force between two identical charges separated by 1 cm is equal to 90 N. What is the
magnitude of the two charges?
Part I. Direction: Encircle the best answer.
𝑞 𝑞
1. In the formula F = k 𝑟1 2 2, the value of the constant k
a. is the same under all circumstances
b. depend on the medium the charges are located in
c. is different for positive and negative charges
d. has the numerical value 1.6 x 10 -19
2. Two charges of +q are 1 cm apart. If one of the charges is replaced by a charge of –q, the
magnitude of the force between them is
a. zero
b. smaller
c. the same
d. larger
3. A charge of +q is placed 2 cm, from a charge of –q. A second of +q is the placed next to the
first. The force on the charge of -q
a. decrease to half its former magnitude
b. remains the same
c. increase to twice its former magnitude
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d. increase to four times its former magnitude
4. Two charges repel each other with a force of 10 -6 N when they are 10 cm apart. When they
are brought close together until they are 2 cm apart, the force between them becomes
a. 4 x 10 -8 N
b.5 x10 -6 N c. 8 x 10 -6 N
d. 2.5 x 10 -5 N
5. Two charges, one positive and the other negative, are initially 2 cm apart and are then pulled
away from each other until they are 6cm apart. The force between them is now smaller by a
factor of
a. √3
b. 3
c. 9
d.27
Part II. Direction: Solve the following problem.
1. Two neutrally charged bodies are separated by 1 cm. Electrons are removed from
one body and placed on the second body until a force of 1×10-6 N is generated
between them. How many electrons were transferred between the bodies?
2. Calculate the force of attraction between the lone electron and proton in a
hydrogen atom. The average distance between them is 5. 0 x 10 -11 m.
3. Two unlike charges attract each other with a force of 400 N. What will be the force
of attraction if the distance between them is increased to twice its original value?
4. A metal sphere is given a charge of -3.0 x 10 -6 C and a second sphere a charge of
+2.0 x 10 -6 C. The spheres are separated by a distance of 0.010 m, (a) What is the
force between them? Is this an attractive or repulsive force? (b) The sphere are
brought into contact? The spheres are again separated by 0.010 m. What I now the
force between them? Is this an attractive or repulsive force?
5. Three charges, each 5. 0 x 10 -6 C are situated at the corners of an equilateral
triangle of side 1.0 m. Find the force that charge q3 experienced due to the other
two charges.
Part III. Direction:Write TRUE if the statement is true. If false, change the underlined
word to make the statement true on the space provided before each number.
_______________1. Electrostatic force is a conservative force.
_______________2. As charged particle gains kinetic energy, it loses an equal amount of
mmmmmmmmmmmpotential energy.
_______________3. A positive charge gains potential energy when it is moved in a direction
mmmmmmmmmm opposite the electric field.
_______________4. Electric potential energy is characteristic of the field only independent of
mmmmmmmmmm test charge that may be placed within the field.
_______________5.The electric potential is varies everywhere on the surface of a
charged conductor in equilibrium.
_______________6. A surface on which all points are at the same potential is called
mmmmmmmmmm equipotential surface.
_______________7-8. The force of attraction or repulsion between two charged objects is
mmmmmmmmmm inversely proportional to the product of their charge magnitudes and
directly proportional to the square of the distance between them.
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_______________8.
_______________9. Charging an object by conduction requires no contact with the object
mmmmmmmmm inducing the charge
_______________10. When an object loses electron, it becomes negatively charged.
Additional Reading
Topic:
1. Capacitance
Guide Question:
1. Discuss capacitance. Give example problem and its application.
References
a. Books
1. Physics Principles with Applications (2016). Giancoli, Douglas. Pearson
Education South Asia Pte Ltd.
2. Physicss II (2006). Hirsch, A. et.al . Thomson Learning
3. College Physics (2015). Serway, Raymond and Vuille, Chris. CENGAGE
Learning Philippine Edition
4. Physics Fundamentals 2 (2012). Serway, Raymond and Vuille, Chris. CENGAGE
Learning Philippine Edition
b. Journals
c. Website
1.
2.
3.
4.
https://www.physicsclassroom.com/class/estatics/Lesson-3/Coulomb-s-Law
https://www.accessscience.com/content/coulomb-s-law/164800
http://sciencenotes.org/coulombs-law-example-problem/
Coulomb - Applications - Charged, Repulsion, Example, and Forces - JRank
Articles https://science.jrank.org/pages/1836/CoulombApplications.html#ixzz6SRAhhZQg
Name:________________________Course: ______________ Date:_________Rating:_____
Static Electricity
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I.
Objectives:
1. Define static electricity
2. Differentiate a positively charged body from a negatively charge body
3. Develop awareness on the importance of static electricity
II.
Materials:
plastic comb, silk cloth, water in a faucet, balloon, plastic cover, tiny bits of paper,
scissors
III.
Procedure:
Part A: Bending Water Trick
1. Rub a plastic comb to charge it with static electricity.
2. Turn on a faucet so that water runs in a stream.
3. Hold the charged comb close to the water and observe. What happened to the
water? Describe.
Part B: Sticky Balloons
1. Rub a balloon several times on a woolen jumper or sheet.
2. Hold it against a wall. What happened? Describe.
Part C: It’s Magic!
1. Cut paper into tiny bits.
2. Rub a plastic comb with silk cloth of plastic cover.
3. Bring it near the tiny bits of paper. Observe what happens.
IV.
Observation:
Part A:
Part B:
Part C:
Guide questions:
1. What is static electricity?
2. When can you say that a body is positively charged or negatively charged?
V.
VI.
Conclusion
Enrichment
1. Cite the importance of static electricity in your life.
Name:________________________Course: ______________ Date:_________Rating:_____
Problem Solving in Coulomb’s Law
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I.
Objectives:
1. State and explain Coulomb’s Law
2. Derive the equation for Coulomb’s Law
3. Calculate related problems accurately
II.
Materials:
Calculator, activity sheet
III.
Procedure:
1. Given the following problems below, calculate them accurately. Show your
process. You may use the back portion of the activity paper or extra sheet for your
solution.
IV.
Problem Set
1. Two charges, one of +5X10-7 C and the other of -2X10-7 C, attract each other with a
force of 100N. How far apart are they?
2. A charge of -1X10-6 C is placed halfway between a charge of +5X10-6 C that is 20 cm
apart. Find the magnitude of the force that exists between these charges. What kind of
force is present?
3. A force of 2.80X10-4 N exists between two charges. If one of the charges has a charge
of 1.5X10-7C, find the magnitude of another charge if they are 85 cm apart.
Guide Questions
1. State Coulomb’s Law.
2. Describe the mathematical equation for Coulomb’s law.
3. If the distance that separates the charges was doubled, what happens to the
magnitude of force?
V.
Conclusion
VI.
Enrichment
1. Cite some applications of the concepts of charge.
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