12
GENERAL PHYSICS 2
QUARTER 3
LEARNING ACTIVITY SHEET
Republic of the Philippines
Department of Education
COPYRIGHT PAGE
Learning Activity Sheet in EARTH SCIENCE
(Grade 12)
Copyright © 2020
DEPARTMENT OF EDUCATION
Regional Office No. 02 (Cagayan Valley)
Regional Government Center, Carig Sur, Tuguegarao City, 3500
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Writers
Content Editor
Focal Persons
: CHRISTOPHER MASIRAG, MILMAR EDRADA, LEONOR NATIVIDAD,
JOLLY-MAR CASTANEDA, JOHN DAVID MEDRANO, CHARLES
DAQUIOAG, KIMBERLY ANNE PAGDANGANAN, ALDRIN GREGADA,
MARJOHN ADDURU, ANGELIKA TORRES, JACKSON CASIBANG,
GRACE ANN CALIBOSO, TECHIE VERA CRUZ, FE CAGUMBAY,
SILVERIO MACARILAY, ARNOLD TEODORO, NASHRENE FORONDA
: MARIA LORESA TUMANGUIL- SDO TUGUEGARAO CITY, JOVY
DESEMRADA-SDO TUGUEGARAO CITY , RONNIE BIBAS- SDO NUEVA
VIZCAYA, NHORVIEN JAY P. LIBAO, SDO QUIRINO
: GERRY C. GOZE, PhD., Division Learning Area Supervisor
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ESTER T. GRAMAJE, Regional Learning Area Supervisor
RIZALINO G. CARONAN. Regional LR Supervisor
Printed by: DepEd Regional Office No. 02
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Address: Regional Government Center, Carig Sur, Tuguegarao City, 3500
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Email Address: region2@deped.gov.ph
Sample Table of Contents
Compentency
Describe using a diagram charging by rubbing and
charging by induction
Explain the role of electron transfer in electrostatic
charging by rubbing
Describe experiments to show electrostatic
charging by induction
Calculate the net electric force on a point charge
exerted by a system of point charges
Describe an electric field as a region in which an
electric charge experience a force
Calculate the electric field due to a system of point
charges usimng Coulumb’s law and the
superposition principle
Calculate electric influx
Use Gauss’s law to infer electric field due to
unniformly distributed charges on long wires,
spheres, and large plates
Solve problems involving electric charges, dipoles,
forces, fields and flux in context such as , but not
limited to, system of point charges, electrical
breakdown of air, charged pendelums, electrostatic
ink-jet printers
Relate the electric potential with work, potential
energy and electric field
Determine the electric potential fucntion at any
point due to highly symmetric continuous-charge
distributions
Infer the direction and strength of electriv field
vector, nature of electric field sources, and
electrostatic potential surfaces given the
equipotential lines
Calculate the electric field in the region given a
mathematical function describing its potential in a
region of space
Solve function involving electric potential energy
and electric potential in context such as, but not
limited to, electron guns in CRT TV picture tubes
and Van de Graaff generators
Deduce the effect of simple capacitors (e.g.
parallel-plate, spherical, cylindrical) on the
capacitance , charge and potential difference when
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the size, potential difference, or charged is
changed.
Calculate the equivalent capacitance of a network
of capacitors connected in series/parallel
Determine the total charge, the charge on, and the
potential difference across each capacitor in the
network given the capacitors connected in series/
parallel
Determine the potential energy stored inside the
capacitor given the geometry and potential
difference across capacitor
Describe the effects of inserting dielectric materials
on the capacitance, charge, and electric field of a
capacitor
Solve problem involving capacitors and dielectrics
in context such as, but not limited to , charged
plates, batteries, and camera flashlamps
Distinguish between conventional current and
electron flow
Apply the relationship charge=current x time to
new situations or to solve related problems
Describe the effect of temperature increase on the
resistance of a metallic conductor
Describe the ability of a material to conduct current
in terms of resistivity and conductivity
Apply the relationship of the proportionality
between resistance and the length and crosssectional area of a wire to solve problem
Differentiate ohmic and non-ohmic materials in
terms of their I-V curves
Differentiate emf of a source and potential
difference (PD) across a circuit
Given an emf source connected to a resistor,
determine the power supplied or dissipated by
each element in a circuit
Solve problem involving current, resistivity,
resistance, and Ohm’s law in context such as but
not limited to , batteries and bulbs, household
wiring and selection of fuses
Operate devices for measuring currents and
voltages
Draw circuit diagram with power sources (cell or
battery), switches, lamps, resistors, (fixed and
variable) fuses, ammeters and voltmeters
Evaluate the equivalent resistance, current and
voltage in a given network of resistors connected in
series and/or parallel
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STEM_GP12EM-IIId-26
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Calculate the current and voltage through and
across circuit elements using Kirchhoffs’s loop and
junction rules (at most 2 loops only)
Solve problems involving the calculation of
currents and potential difference in circuits
consisting of batteries, resistors and capacitors
Differentiate electric interactions from magnetic
interactions
Evaluate the total magnetic flux through an open
surface
Describe the motion of a charged particle in a
magnetic field in terms of its speed, acceleration,
cyclotron radius, cyclotron frequency and kinetic
energy
Evaluate the magnetic force on an arbitrary wire
segment placed in a uniform magnetic field
Evaluate the magnetic field vector at a given point
in space due to a moving point charge, an
infinitesimal current element, or a straight currentcarrying conductor
Calculate the magnetic field due to one or more
straight wire conductors using the superposition
principle
Calculate the force per unit length on a current
carrying wire due to the magnetic field produced by
other current-carrying wire
Evaluate the magnetic field vector at any point
along the axis of a circular current loop
Solve problems involving magnetic fields, forces
due to magnetic fields and the motion of charges
and current carrying-wires in context such as, but
not limited to , determining the strength of Earth’s
magnetic field, mass spectrometers, and solenoids
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GENERAL PHYSICS 2
Name: _______________________________________Grade Level: _________
Date: ________________________________________Score:______________
LEARNING ACTIVITY SHEET
ELECTRIC CHARGE
Background Information for the Learners (BIL)
The Origin of Electricity
The electrical nature of matter is inherent in atomic structure. An atom consists
of a small relatively massive nucleus that contains particles called protons and neutrons.
A proton has a mass of 1.673 x 10-27 kg, and a neutron has a slightly greater mass of
1.675 x 10-27 kg. Surrounding the nucleus is a diffuse cloud of orbiting particles called
electrons. An electron has a mass of 9.11 x 10 -31 kg. Like mass, electric charge is an
intrinsic property of protons and electrons, and only two types of charge have been
discovered, positive and negative. A proton has a positive charge, and an electron has
a negative charge. A neutron has no net charge.
Experiment reveals that the magnitude of the charge on the proton exactly equals
the magnitude of the charge of the electron; the proton carries a charge + e, and the
electron carries a charge of – e. The SI unit for measuring the magnitude of an electric
charge is the coulomb (C), and has been determined experimentally to have a value e
= 1.60 x 10-19 C.
The symbol e represents only the magnitude of the charge on a proton or an
electron and does not include the algebraic sign that indicates whether the charge is
positive or negative. In nature, atoms are normally found with equal numbers of protons
and electrons. Usually, then, an atom carries no net charge because the algebraic sum
of the positive charge of the nucleus and the negative charge of the electrons is zero.
When an atom, or any object, carries no net charge, the object is said to be electrically
neutral. The neutrons in the nucleus are electrically neutral particles.
The charge on the electrons or a proton is the smallest amount of free charge that
has been discovered. Charges of larger magnitude are built up on an object by adding
or removing electrons. Thus, any charge of magnitude q is an integer multiple on e; that
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is, q = Ne, where N is an integer. Because any electric charge q occurs in integers
multiples of elementary, invisible charges of magnitude e, electric charge is said to be
quantized.
To show the quantized nature of electric charge, let us consider this example.
How many electrons are there in one coulomb of negative charge?
The negative charge is due to the presence of excess electrons, since they carry negative
charge. Because an electron has a charge whose magnitude is e = 1.60 x 10 -19 C, the
number of electrons is equal to the charge e on each electron.
Thus, the number N of electrons is
𝑁=
𝑞
𝑒
=
1.00𝐶
1.60𝑥10−19 𝐶
= 6.25 x 1018
Triboelectric Charging
The presence of different atoms in an object provides different objects with
different electrical properties. One property is known as electron affinity. The property
of electron affinity refers to the relative amount of love that a material has for electrons.
If atoms of a material have a high electron affinity, then that material will have a relatively
high love for electrons. This property of electron affinity will be of utmost importance as
we explore one of the most common methods of charging - triboelectic charging, also
known as charging by friction or rubbing.
Charging by Induction
Induction charging is a method used to charge an object without touching the
object to any other charged object. An understanding of charging by induction requires
an understanding of the nature of a conductor and the polarization process.
Charging Two-Sphere System Using A Negatively Charged Object
One common demonstration performed to show how induction charging occur is
by using two metal spheres. The metal spheres are supported by insulating stands so
that any charge acquired by the spheres cannot travel to the ground. The spheres are
placed side by side (see diagram i. below) so as to form a two-sphere system. Being
made of metal (a conductor), electrons are free to move between the spheres - from
sphere A to sphere B and vice versa.
If a rubber balloon is charged negatively (perhaps by rubbing it with animal fur)
and brought near the spheres, electrons within the two-sphere system will be induced to
move away from the balloon. This is simply the principle that like charges repel. Being
charged negatively, the electrons are repelled by the negatively charged balloon. And
being present in a conductor, they are free to move about the surface of the conductor.
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Subsequently, there is a mass migration of electrons from sphere A to sphere B.
This electron migration causes the two-sphere system to be polarized (see diagram ii.
below). Overall, the two-sphere system is electrically neutral. Yet the movement of
electrons out of sphere A and into sphere B separates the negative charge from the
positive charge. Looking at the spheres individually, it would be accurate to say that
sphere A has an overall positive charge and sphere B has an overall negative charge.
Once the two-sphere system is polarized, sphere B is physically separated from sphere
A using the insulating stand. Having been pulled further from the balloon, the negative
charge likely redistributes itself uniformly about sphere B (see diagram iii. below).
Meanwhile, the excess positive charge on sphere A remains located near the
negatively charged balloon, consistent with the principle that opposite charges attract.
As the balloon is pulled away, there is a uniform distribution of charge about the surface
of both spheres (see diagram iv. below). This distribution occurs as the remaining
electrons in sphere A move across the surface of the sphere until the excess positive
charge is uniformly distributed.
https://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Induction
The Law of Conservation of Charge
The law of conservation of charge is easily observed in the induction charging
process. Considering the example above, one can look at the two spheres as a system.
Prior to the charging process, the overall charge of the system was zero. There were
equal numbers of protons and electrons within the two spheres. In diagram ii. above,
electrons were induced into moving from sphere A to sphere B. At this point, the
individual spheres become charged. The quantity of positive charge on sphere A equals
the quantity of negative charge on sphere B. If sphere A has 1000 units of positive
charge, then sphere B has 1000 units of negative charge. Determining the overall charge
of the system is easy arithmetic; it is simply the sum of the charges on the individual
spheres.
Overall Charge of Two Spheres = +1000 units + (-1000 units) = 0 units
The overall charge on the system of two objects is the same after the charging
process as it was before the charging process. Charge is neither created nor destroyed
during this charging process; it is simply transferred from one object to the other object
in the form of electrons.
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CHARGING TWO-SPHERE SYSTEM USING A POSITIVELY CHARGED OBJECT
What do you think will happen if there are two positively charged spheres? How
would the movement of electron be changed?
Study this figure:
https://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Induction
The positively charged balloon is brought near sphere A. Consider the graphic
below in which a positively charged balloon is brought near Sphere A. The presence of
the positive charge induces a mass migration of electrons from sphere B towards (and
into) sphere A. This movement is induced by the simple principle that opposites attract.
Negatively charged electrons throughout the two-sphere system are attracted to the
positively charged balloon. This movement of electrons from sphere B to sphere A leaves
sphere B with an overall positive charge and sphere A with an overall negative charge.
The two-sphere system has been polarized. With the positively charged balloon still held
nearby, sphere B is physically separated from sphere A. The excess positive charge is
uniformly distributed across the surface of sphere B. The excess negative charge on
sphere A remains crowded towards the left side of the sphere, positioning itself close to
the balloon. Once the balloon is removed, electrons redistribute themselves about sphere
A until the excess negative charge is evenly distributed across the surface. In the end,
sphere A becomes charged negatively and sphere B becomes charged positively.
The Importance of a Ground in Induction Charging
In the charging by induction cases discussed above, the ultimate charge on the
object is never the result of electron movement from the charged object to the originally
neutral objects. The balloon never transfers electrons to or receive electrons from the
spheres; nor does the glass rod transfer electrons to or receive electrons from the
spheres. The neutral object nearest the charged object (sphere A in these discussions)
acquires its charge from the object to which it is touched. In the above cases, the second
sphere is used to supply the electrons to sphere A or to receive electrons from sphere A.
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The role of sphere B in the above examples is to serve as a supplier or receiver of
electrons in response to the object that is brought near sphere A. In this sense, sphere
B acts like a ground.
Learning Competency:
Describe using a diagram charging by rubbing and charging by induction
(STEM_GP12EM-IIIa-1)
Activity 1: Charge it!
Directions: Follow the procedures as stated in the activity.
Objective: To demonstrate the transfer of electric charge form one object to another.
Materials:
2 small rubber balloons
Small piece of wool cloth
Small pieces of paper
Procedure:
1. Blow one balloon and tie it.
2. Rub one side of the balloon with the scrap of wool.
3. Move a finger toward the balloon in the charged spot. What do you observe?
_________________________________________________________________
_________________________________________________________________
__________________________________
4. Recharge your balloons, and try holding the charged parts near each other.
What do you observe? Explain your observations.
_________________________________________________________________
_________________________________________________________________
__________________________________
5. Rub one of the balloon and put on a table top (or the floor) and try gently rolling
it? What do you observe? Explain you observations.
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
___________________________________________________
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6. Prepare the small bits of paper. Place it on top of a table. Recharge your balloon
and hold it slightly above the small bits of paper. What do you observe? Explain
your observation.
_________________________________________________________________
_________________________________________________________________
Guide questions:
1. What happened on the part of the balloon that you rubbed with the scrap of
wool?
_________________________________________________________________
_________________________________________________________________
2. What is the role of the rubbing process in the activity?
______________________________________________________________
______________________________________________________________
_
Activity 2: Charging
by Induction
Objective: Describe how the presence of negatively charge object induces movement of
electrons.
Materials:
2 pcs Styrofoam cups
1 pcs rubber ballon
2 pcs softdrinks empty cans
scotch tape/double-sided tape
Procedure:
1. Label the softdrink cans as can A and can B.
2. Mount the softdrink can on top of the styro cup using a scotch tape or doublesided tape.
3. Place the can side by side.
4. Charge the ballon by rubbing it with animal fur or hair (this will make the rubber
ballon negatively charged)
5. Place the negatively charged balloon near to one of the cans.
6. Follow the figure below:
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7. Observe what happens. Write your observation in your notebook.
Guide questions:
1. What happens to the can when you brought the negatively charged rubber
ballon near it?
_____________________________________________________
_____________________________________________________
_____________________________________________________
___
2. Describe the movement of the electrons in the experimental set-up.
_________________________________________________________________
_____________________________________________________________
3. Would you expect that can A would be attracted by the negatively charged
balloon? Explain why or why not?
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
____
4. What is the role of the balloon in the activity?
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
____
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Activity 3 : 10 minute
Video-Tutorial
Charging by Induction Video Tutorial
Objective: Explain how charging by induction happens.
Using
your
smartphones
or
laptops
with
internet
go
to
this
link:
https://www.youtube.com/watch?v=763tiBXvTGw&feature=youtu.be
The Charging by Induction Video Tutorial describes what charging by induction is and
explains how and why it occurs. Numerous examples, animations, and illustrations are
provided.
After watching the video lesson you will be able to answer the following questions:
1. What is charging by induction and how does it occur?
2. How can the results of charging by induction be predicted and explained?
Reflection:
Write your answer on the following:
1. I learned that
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
2. I enjoyed the lesson most on
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
3. I want to learn more on
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
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References:
Cutnell, J.D. and K. W. Johnsons. (2016). Physics, 9th Edition.
https://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Induction
https://www.youtube.com/watch?v=763tiBXvTGw&feature=youtu.be
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Answer key
Activity 1
Possible observation:
3. There is a crackling sound produced. This is due to the balloon’s negative
charge being released.
4. There is a production of crackling sound, and feel some resistance. This is
because of the two negative charges repelling each other.
5. The charged portion sticks to the floor or table, this is because the negatively
charged balloon is attracted to the neutrally charged floor.
6. The small bits of paper is attracted to the charged spot of the balloon. This is
because the positive ends of the small bits of paper are attracted to the
negatively charged balloon.
Answers to guide questions:
3. Rubbing the balloon with a wool cloth gives the spot on the balloon a
negative charge.
4. Ans. The rubbing process serves only to separate electrons and protons
already present in the materials.
Activity 2
Answers to guide questions:
5. The cans separated when the rubber ballon was brought near the end of one
of the cans. The balloon attracted can A.
6. The presence of negatively charge near the can induces electron movement
from can A to can B
7. The type of charge on the cans can be tested by seeing if they attract the
negatively charged balloon or repel the negatively charged balloon. Of course,
we would expect that Can A (being positively charged) would attract the
negatively charged balloon and Can B (being negatively charged) should repel
the negatively charged balloon.
8. During the process of induction charging, the role of the balloon is to simply
induce a movement of electrons from one can to the other can. It is used to
polarize the two-can system. The balloon never does supply electrons to can
A (unless your hear a spark, indicating a lightning discharge from the balloon
to the can).
Prepared by:
CHRISTOPHER A. MASIRAG
VICENTE D. TINIDAD NATIONAL HIGH SCHOOL
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GENERAL PHYSICS 2
Name: ________________________________________Grade Level: _________
Date: __________________________________________Score:______________
LEARNING ACTIVITY SHEET
ELECTROSTATIC CHARGING
Background Information for the Learners (BIL)
Charging by rubbing phenomenon in which friction transfers charged
particles from one body to another.
If two objects are rubbed together, especially if the objects are insulators and
surrounding air is dry, the objects acquire equal and opposite charges and an
attractive force develops between them.
• The object that loses electrons becomes positively charged.
• The other that gains electrons becomes negatively charged.
• The force is simply the attraction between charges of opposite sign.
Types of Electric Charges
Each type of charge attracts the opposite type but repels the same type. This
leads to the basic law of electrostatics: Unlike charges attract, like charges
repel.
• The SI unit of electric charge is the coulomb (C). It is a scalar quantity.
• Every electron has a charge of -1.6 x 10-19 C, and every proton has a
charge of +1.6 x 10-19 C.
Positively charged Particles
In this type of particles, numbers of positive ions are larger than the numbers of
negative ions. In other words, numbers of protons are larger than the number of
electrons.
p+>eTo neutralize positively charged particles, electrons from the surroundings
come to this particle until the number of protons and electrons become equal. Do not
forget protons cannot move!
Negatively Charged Particles
In this type of particles, numbers of negative ions are larger than the numbers
of positive ions. In other words, numbers of electrons are larger than the number of
protons.
e+>p11
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To neutralize negatively charged particles, since protons cannot move and
cannot come to negatively charged particles, electrons moves to the ground or any
other particle around itself.
Neutral Particles
These types of particles include equal numbers of protons and electrons. Be
careful, they have both protons, neutrons and electrons however, numbers of “+” ions
are equal to the numbers of “-” ions.
e+=pConductors
Some of the matters have lots of free electrons to move. It is easy for electrons
to flow from these materials. Metals are good conductors. Gold, copper, human
bodies, acid, base and salt solutions are example of conductors.
Insulators
These types of materials do not let electrons flow. Bonds of the electrons in the
insulators are tighter than the conductors thus, they cannot move easily. Glass, ebonite,
plastic, wood, air is some of the examples of insulators.
Atoms having same charge repel each other and atoms having opposite charges
attract each other.
Example: Charged spheres A, B and C behave like this under the effect of charged
rod D and E. If C is positively charged, find the signs of the other spheres and rods.
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We learned that opposite charges attract each other and same charges repel each
other. Using this explanation we can say that, if the sign of the C is “+” than rod E
must be “-” since it attracts C. B must be “+” since E also attract B. Rod D repels the
B so, we say that D must have same sign with B “+” , and finally D also repels A, thus
A is also “+”.
A(+), D(+), B(+), E(-), C(+)
LEARNING COMPETENCY
Explain the role of electron transfer in electrostatic charging by rubbing.
(STEM_GP12EM-IIIb-2)
Activity 1: Types
of Charging
Objective: Give examples of charging by friction and charging by contact.
Charging by Friction
When you rub one material to another, they are charged by friction. Material losing
electron is positively charged and material gaining electron is negatively charged.
Amount of gained and lost electron is equal to each other.
Charging by Contact
Charging by conduction occurs when two objects with different amounts of electric
charge come in contact and electrons move from one object to the other. There are
equal number of electrons and protons in a neutral matter. If something changes this
balance, we can say it is charged.
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Directions: Give examples of objects that demonstrate charging by friction and
charging by contact. (One example each is already given.)
Charging by Friction
1. Rubbing of comb and hair
2.
3.
4.
5.
Charging by Contact
1. Metal rod/bar touching a neutral
sphere
2.
3.
4.
5.
Guide questions:
7. Consider one of your answer in the first column (charging by friction), explain
how the electric charges are transferred from one object to the other.
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
____________
8. Choose one of your answer in the second column (charging by conduction) and
explain how the two objects with different amounts of electric charge come in
contact and electrons move from one object to the other.
________________________________________________________
________________________________________________________
________________________________________________________
________________________________________________________
Activity 2: Charging
by Friction
Objective: Perform activity that demonstrate charging by friction.
Materials do not always need to be rubbed together to create a charge imbalance. In
this activity, you will explore charging by friction through simple contact.
Materials:
Roll of clear plastic adhesive tape (scotch tape/clear packing tape)
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Procedure:
1. Pull two 8-10 cm of tape from the roll.
2. Hold one piece of tape in each hand and bring the two shiny (non-sticky) sides
of the tape close together without letting them touch. Record your observations.
3. Exhale onto both sides of each piece of tape several times (over its entire
length). Bring the two shiny sides of the tape close together again without
letting them touch. Observe what happens.
4. Using the same two pieces of tape, allow each piece of tape to stick to the top
of a clean desk without rubbing. Then quickly pull the pieces off the desk. Bring
the shiny, non-sticky side of one of the pieces close to the edge of the desk
without letting it touch the desk. Observe what happens.
5. Quickly bring the shiny, non-sticky sides of both pieces of tape close to each
other without letting them touch. Observe what happens.
Guide questions:
9. What do your observations in step 2 indicate about the electric charge on
the pieces of tape when they were first pulled off the roll? Explain.
___________________________________________________
______________________________________________________
______________________________________________________
10. What do your observations in step 3 indicate about the electric charge on the
pieces of tape? Explain.
______________________________________________________________
______________________________________________________________
__
11. Why is there a difference in the electric charge on the pieces of tape
between steps 2 and 3?
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
____
12. Write a question you have about the observations you made in this activity.
Exchange questions with a classmate through your social media account or
messenger and decide how you may find answers to your questions. Then design
and carry out simple experiments to answer your questions.
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
____
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NOTE: Practice personal hygiene protocols at all times
Activity 3 : Research
this!
Objective: Analyze how technology works to hinder the effect of charging by friction.
Many people use fabric softener dryer sheets to control static charge buildup on clothes.
As clothes made of different materials tumble inside a clothes dryer, they rub together
and become charged by friction. Fabric softener sheets prevent the buildup of static
charges.
1. Research how fabric softener sheets prevent the buildup of static charges.
2. Research what chemicals are used in fabric softener sheets and their effects on
people and the environment.
Guide questions:
1. Analyze how this technology works to hinder the effect of charging by friction.
Draw a diagram to support your analysis.
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
2. List some benefits and drawbacks of using fabric softener sheets. Suggest
alternatives to using fabric softener sheets.
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
3. Based on your research, decide whether fabric softener sheets are necessary.
Support your opinion.
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
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ADDITIONAL ACTIVITIES: Check Your Learning
1. Consider the following pairs of materials. Using the electrostatic series,
determine the charge that each material will gain when the two are rubbed
together. (a) glass and silk (b) ebonite and fur (c) human hair and a rubber
balloon (d) amber and cotton
2. Why do objects made from different materials develop an electric charge when
rubbed together? What is this method of charging called? Use a diagram to
illustrate your answer.
3. In your own words, explain charging by conduction. Include diagrams showing
how a positively charged object can be used to charge a neutral object.
4. Use a graphic organizer to compare charging by conduction to charging by
friction.
Reflection:
Write your answer on the following:
1.I learned that
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________
2.I enjoyed the lesson most on
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________
3.I want to learn more on
_______________________________________________________________________
_______________________________________________________________________
_____________________________________________________
17
NOTE: Practice personal hygiene protocols at all times
References:
Cutnell, J.D. and K. W. Johnsons. (2016). Physics, 9th Edition.
https://www.miniphysics.com/charging-by-rubbing.html
https://www.physicstutorials.org/home/electrostatics
http://www.mrcaslick.altervista.org/SNC1D/Textbook/11.2.pdf
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Answer key:
Activity 1
1. When they are rubbed together, the atoms in the comb gain electrons and the
atoms in the hair lose electrons.
2. Possible explanation: Two charged objects may come in contact, and electrons
may move from one object to the other. Electrons always move from the object
with a larger negative charge (less positive) to the object with the smaller negative
charge (more positive). This produces a more even distribution of electric charge
between the two objects.
Activity 2
Observations:
3. Repel each other.
4. The sticky tape is repelled by the edge of the table.
5. Repel each other.
Answers to guide questions:
1. The two tapes neither attract nor repel, this is because of the equal numbers
of positive and negative charges.
2. The repulsion between the 2 sticky tapes indicate that they have the same
electric charges.
3. The charges of two sticky tapes before exhaling on it has a neutral charge,
while the charge after exhalation makes the sticky tape acquire same charges
making them repel.
4. Answers may vary
Activity 3
Answers to guide questions:
1. The softener binds with the hydrogen- bonding network hence blocking the effect
of charging by friction.
19
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(https://www.semanticscholar.org/paper/Elucidation-of-the-softening-mechanism-of-fabric-Igarashi
Nakamura/cce20caa28a4d4f521d0c7d4615c973036444b91/figure)
3. Fabric softener dryer sheets coat clothes in a waxy substance that also makes the
clothes feel softer. Fabric softener softens clothes and adds a fragrance to them.
Fabric softeners can also reduce static cling. If you don't want to use cationic
fabric softeners, there are alternatives that work quite well. Add a half-cup cup of
white vinegar to the rinse cycle of your washer to soften clothes, naturally remove
static, and cut soap residue which can dull colored items. Or add a half-cup of
baking soda to the wash. You can also check with your local health food store or
look online for an all-natural softener. Last, because synthetic fibers are notorious
for static cling, wash and dry these items separately from cottons and remove
them from the dryer while slightly damp.
4. Answers may vary
Prepared by:
CHRISTOPHER A. MASIRAG
VICENTE D. TRINIDAD NATIONAL HIGH SCHOOL
20
NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 2
Name: ____________________________
Grade Level: ________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
ELECTRIC CHARGE, COULOMB’S LAW, ELECTRIC FIELDS, AND ELECTRIC
FLUX
Background Information for the Learners (BIL)
ELECTROSTATIC INDUCTION
Electrostatic induction is a process to produce static electricity in an object by
drawing near to an electrically charged material. The former will cause the electrical
charges to be reallocated in the material that will result in one side having an excess of
either positive (+) or negative (-) charges.
Electrostatic induction or Induction charging is a method used to charge an object
without touching. This method will cause the redistribution of electrical charges on a
material.
ELECTROSTATIC INDUCTION IN CONDUCTORS
Electrostatic induction is most effective when materials are conductors just like
metals. Metals are good conductors. In electrostatic induction, once you remove the
electrically charged object, the conductor loses its charge. Temporarily grounding the
conductor must be done to solve this phenomenon.
Electrical conductors in neutral state has an equal number of (+) and negative (-)
electrical charges. Equal number of positive ions and negative ions and electrons
interacts within the conducting material. When an static electrically charged is brought
near to an electrical conductor, the electrical charges on or near the surface of the
electrically charged object attracts the opposite charges in the conductor and repel the
like charges.
The law of attraction and repulsion is observed in this phenomenon. Unlike
charges attract, therefore a positive charge(+) will attract a negative charge(-). Like
charges repel, therefore a negative charge (-) will repel a negative charge (-) and vice
versa.
21
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Example
______ plastic rod
_
_
_ _ _
+ + + + + + + +
Figure 1
_ _ _ _ _ _ _ _
-
In figure 1, electrical charges in the conductor or in the metal are redistributed as
the rod is draw near in the metal plate. Nevertheless, by the time the electrically charged
object (plastic rod) is removed, the charges in the conductor interact or intermingle again.
So, the electrical charging is temporary.
ELECTROSCOPE
An electroscope is an instrument used to detect the presence of electric charge
on a body. It detects charge by the movement of a test object due to the Coulomb
electrostatic force on it.
Electrostatic induction is also applicable in electroscope. If you draw near a
charged object such as the plastic rod near an electroscope, the opposite charges will
move toward the metal end of an electroscope as shown in the illustration below. In the
same illustration, the negative charge(-) in the plastic rod attract the positive charge (+)
in the metal shaft of the electroscope. The electrical charges in the metal shaft are
redistributed while the negative charges are on the leaves at the lower shaft.
The leaves of the electroscope push apart because to the electrical force where the
charges are the same (negative charges repel negative charges).
______Plastic Rod
Metal shaft ______
+
+
++ +
+ +
_______ Electroscope
_________leave electroscope
22
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Electroscopes leaves separate because of electrical charges
Metal shaft ___
+
+
+
_______ Electroscope
___ leave of electroscope
Electroscopes leaves go back to its original position
When the charged plastic rod is removed, the leaves of the electroscope go to its
original position and the electrical charges in the latter will interact again. The leaves will
not repel anymore because the charges in the leaves are opposite.
ELECTROSTATIC INDUCTION IN NON-CONDUCTORS
Non-conductors or insulators can undergo also the process of electrostatic
induction. These non-conductors can be given static electric charge nevertheless
electrostatic induction in non-conducting materials is least effective because the
movement of charge is constraint.
Although electrostatic induction is possible to nonconducting or dielectric
materials, the movement of electrical charges is much more constrained in
nonconductors than in conducting materials. In, conductors’ electrons are allowed to
move freely that cause electricity.
In a nonconductor, separation of charged particles does not work because
electrons are constrained. Nevertheless, if the nonconductor consists of polar molecules
the electrostatic induction may be possible. The latter will cause the molecules to be with
the positive charges (+) and the other side with negative charges (-). Polar molecules are
molecules that has one side more positive that the other side.
For example: water is polar molecule so water can be slightly attracted to a static electric
charge that is why if you draw near a charged object to a water the stream of water will
eventually bend.
A tissue and small pieces of Styrofoam which are nonconductors can be also attracted
by a charge object through electrostatic induction.
`
Learning Competency
Describe experiments to show electrostatic charging by induction. (STEM_GP12EMIIIa-3)
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NOTE: Practice personal hygiene protocols at all times
Activity 1: FACT OR BLUFF
Directions: Label the following statements as Fact if the statement is true or Bluff if the
statement is false. If the statement is false, underline the word/s that make it false and
change it to make it true.
_____ 1. Electrostatic induction can exist in nonconductors.
_____ 2. Negative charge will attract negative charge.
_____ 3. Positive charge will attract negative charge.
_____ 4. Electrostatic induction can produce static electricity when you draw near an
electrically charged object to a material.
_____ 5. Induction is not possible when the objects are not in contact.
____ 6. Electrostatic induction is more effective in conductors than in nonconductors.
____ 7. When a static electrically charged is brought near to an electrical conductor, the
electrical charges on or near the surface of the electrically charged object attract the
opposite charges in the conductor.
_____ 8. The law of attraction and repulsion is observed in electrostatic induction.
_____ 9. Electrostatic induction is NOT possible to nonconducting or dielectric materials.
_____ 10. The movement of electrical charges is much more constrained in
nonconductors than in conducting materials
ACTIVITY 2.
Draw the Charges
DIRECTIONS: Draw the correct orientation of charges in electrostatic induction in the
illustration given below. Draw your answer in the empty box at the right. (10 points)
ro
r
___
____++++
__
NOTE: Assume that the upper illustration is charged object and the
lower part is metal plate.
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Describe what happen on the charges on the metal plate.
Answer:
_____________________________________________________________
_____________________________________________________________
________________________
ACTIVITY 3. Electrostatic Induction in Electroscope
DIRECTIONS: Draw the correct orientation of charges in electrostatic induction in
electroscope. Draw your answer in the empty box. (10 points)
1. BRINGING NEGATIVELY CHARGE OBJECT TO AN ELECTROSCOPE
Describe what happen on the the electroscope during electrostatic induction.
Answer:
_____________________________________________________________
_____________________________________________________________
_________________________
2. WHEN CHARGED OBJECT IS REMOVED FROM ELECTROSCOPE
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Describe what happen on the electroscope during electrostatic induction.
Answer:
_____________________________________________________________
_____________________________________________________________
________________________________________
ACTIVITY 4. ESSAY
DIRECTIONS: Write your idea briefly but substantially in the following situations.
1. Electromagnetic induction in a plastic rod and metal plate.
2. Electromagnetic induction in electroscope and charge object.
3. Electromagnetic induction in charged object and nonconductors.
ACTIVITY 5. INDUCTION ON A NONCONDUCTOR
Experiment
Materials:
comb and tissue paper
Procedures:
1.
2.
3.
4.
5.
6.
Tear off several bits of tissue paper.
Bring the comb near to the bits of tissue paper.
Describe what happens.
Then rub your hair with the comb.
Bring near the comb to the tissue paper.
Describe what happens
Guide Questions:
1. Why does the comb attract the pieces of tissue paper when you rub the comd
into your hair?
2. Explain the electrification that takes place in the comb.
26
NOTE: Practice personal hygiene protocols at all times
Reflection:
Write your answer on the following:
1.I learned that
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________
2.I enjoyed the lesson most on
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________
3.I want to learn more on
_______________________________________________________________________
_______________________________________________________________________
_____________________________________________________
27
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REFERENCES:
“Electric Field: Concept of a Field Revisited.” Accessed January 19, 2021.
https://courses.lumenlearning.com/physics/chapter/18-4-electric-field-concept-of-a-fieldrevisited/.
“Electroscope
-
Google
Search.”
Accessed
January
19,
2021.
https://www.google.com/search?q=electroscope&oq=electroscope&aqs=chrome..69i57j
0l7.8818j0j7&sourceid=chrome&ie=UTF
Contributors to Wikimedia projects. “Electrostatic Induction.” Wikipedia, January 16,
2021. https://en.wikipedia.org/wiki/Electrostatic_induction.
Keil, Dennis. “Electrostatic Induction.” College-Physics - Lernportal. Accessed January
19, 2021. http://www.college-physics.com/book/electric-field/electrostatic-induction/.
IOPSpark. “Charging by Electrostatic Induction.” Accessed January 19, 2021.
https://spark.iop.org/charging-electrostatic-induction
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ANSWER KEY:
Activity 1. FACT OR BLUFF
1. Fact
2. Bluff
3. Fact
4. Fact
5. Bluff
6. Fact
7. Fact
8. Fact
9. Bluff
10. Fact
ACTIVITY 2. Draw the Charges
Answer:
_
_
_ _ _
+ + + +
_ _ _ _
-
When a static electrically charged is brought near to an electrical conductor, the
electrical charges on or near the surface of the electrically charged object attract the
opposite charges in the conductor and repel the like charges.
ACTIVITY 3. Electrostatic Induction in Electroscope
1. BRINGING A NEGATIVELY CHARGED OBJECT TO AN ELECTROSCOPE
+
+
++ +
+ +
The negative charge (-) object attract the positive charge (+) in the metal shaft of the
electroscope. The electrical charges in the metal shaft are redistributed while the
negative charges are on the leaves at the lower shaft.
29
NOTE: Practice personal hygiene protocols at all times
The leaves of the electroscope push apart because to the electrical force where the
charges are the same (negative charges repel negative charges).
2. WHEN CHARGED OBJECT IS REMOVED FROM ELECTROSCOPE
+
+
+
________
_leave of electroscope
When the charged object is removed, the leaves of the electroscope go to its original
position and the electrical charges in the latter will interact again. The leaves will not
repel anymore because the charges in the leaves are opposite.
ACTIVITY 4. ESSAY
1. Answers may vary.
2. Answers may vary.
3. Answers may vary.
ACTIVITY 5. INDUCTION ON A NONCONDUCTOR
1. When you rub the comb into your hair the comb became negatively charged.
This will create an electric field into the comb that will polarise and attract the
tissue paper so that the part closer to the comb will be the positive(+) and the
other will be(-).
2. When electrification occurs, electrons are not created but they are
transferred. In the case of the comd attracting the tiny bits of tissue paper
when you rubbed it into your hair electrons from your hair got transferred and
now the comb induces a dipole in the bits of paper and so the paper get
attracted.
Prepared by:
MILMAR T. EDRADA
Dassun National High School
30
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: ________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
COULOMB’S LAW
Background Information for the Learners (BIL)
COULOMB’S LAW
Coulomb’s law also known as Coulomb’s inverse-square law measures the
amount of force between stationary charged particles. The law was discovered by French
physicist Charles-Augustin de Coulomb in 1785. Coulomb’s law is very significant in the
progress of electricity and magnetism since it quantifies electrical charges. Hence, the
unit for charge is in unit Coulomb (C) in honor to the proponent of the law.
Coulomb force is extremely basic, since most charges are due to point-like
particles. It is responsible for all electrostatic effects and are due to point-like particles. It
is responsible for all electrostatic effects and underlies most macroscopic forces.
The Coulomb force is strong compared with the gravitational force, another
fundamental force. Nevertheless, compared to gravitational force it can cancel, since it
can be attractive or repulsive depending on the charges.
The electrostatic force between two subatomic particles is far greater than the
gravitational force between the same two particles.
When dealing with charged objects or charges, we also talk of forces between
them. These forces can be repulsive for unlike charges and can be a repulsive force for
like charges. To find for the force for charges we can employ the Coulomb’s Law is
F=
k q 1 q2
d2
In this equation, F is the electrical force exist between charges, q 1 is the charge of the
first object or a particle, q2 is the charge of the second object or particle and d is the
distance between the object or charges.
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k q 1 q2
d2
F α q1 ( If the charge on an object is doubled, the electrical force also doubled)
In the equation
F=
F α q1q2 ( If the charge of the two object or particles doubled, the electrical force
is quadrupled)
F α 1 ( If the distance of the two charges are doubled, the electrical force are quartered
d2
).
Coulomb’s law calculates the magnitude of the force F between two point charges,
q1 and q2 separated by a distance d. In SI units, the constant k is equal to 9 x 10 9 N.m2/
C2. The electrostatic force a vector quantity and is expressed in units of newtons (N).
In the equation
F=
k q 1 q2
d2
, it is apparent that the electrostatic force between any two points is directly
proportional to the product of the magnitude of the charges and inversely proportional to
the square of the distance between them.
Sample Problem 1:
A point charge has a magnitude of 3 x 10 -7 C. A second charge has a magnitude
of – 1.5 x 10 -7 C and is 0.11 meter away from the first charge. Determine the electrostatic
force that the charges exerted to each other.
Given: q1 = 3 x 10 -7 C
q2 = – 1.5 x 10 -7 C
k = 9 x 109 N.m2/ C2
d = 0.11m
F=?
Solution:
F=
k q 1 q2
d2
F= (9 x 109 N.m2/ C2) (3 x 10 -7 C) (– 1.5 x 10 -7 C)
(0.11m)2
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F= 0.000405N
0.0121
F=0.0334 N
Sample Problem 2.
Determine the magnitude of electric force between a nucleus ( 6 protons having
a charge of 1.6 x 10 -19 C and its internal electron having a charge of – 1.6 x 10 19
C if the distance that separate them is 1.0 x 10
-7 m.
Given: q1 = 1.6 x 10 -19 C
q2 = – 1.6 x 10 -19
k = 9 x 109 N.m2/ C2
d = 1.0 x 10 -7 m.
F=?
Solution:
F=
k q 1 q2
d2
F= (9 x 109 N.m2/ C2){ (1.6 x10-19 C)(6)} (–1.6x10-19C)
(1x10-7m)2
F= (9 x 109 N.m2/ C2)(9.6x10-19C)(–1.6x10-19C)
1x10-14m2
F= -1.3824 x 10-27 N
1x10-14
F= 1.3824-41 N
Sample Problem 3. Determine the distance of two electrons having a charge of
-3.0 C if the force between them is 19.2N?
Given: q1 = 3.0C
q2 = 3.0 C
F = 19.2 N
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NOTE: Practice personal hygiene protocols at all times
k = 9 x 109 N.m2/ C2
d= ?
Solution:
F=
k q 1 q2
d2
d2= kq1q2
F
√d2= √ kq1q2
F
d = √(9 x 109 N.m2/ C2)( 3C )( 3C )/ 9.2 N
d = √8.1x1010m/9.2
d = √4218750000
d = 64951.9m
Sample Problem 4.
A negative charge of -4.0 x 10 -3 C exerts an attractive force of 12 N on a second
charge that is 0.050 m away. Determine the magnitude of the second charge?
Given: q1 = -4.0 x 10 -3 C
d= - 0.050 m
F= 12 N
q2 = ?
Solution:
k q 1 q2
d2
F=
Fd2 = k q1 q2
k q1
k q1
q2 = Fd2
k q1
q2=
(12N)(0.050m)2
(9 x 109 N.m2/ C2)( -4.0 x 10 -3 C)
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q2=
0.03
-3.6x107C
q2= -83333.33 C
Learning Competency
Calculate the net electric force on a point charge exerted by a system of point charges
(STEM_GP12EM-IIIa-6)
Activity 1: RELATIONSHIP OF ELECTRIC FORCE, ELECTRICAL CHARGE AND
DISTANCE OF CHARGES.
DIRECTIONS: Choose the correct term in the parenthesis. Underline the correct
term. (2 points each)
1. The magnitude of distance of charges is ( inversely proportional , directly
proportional) to the electrical force.
2. If the charge of both particles is doubled, the force ( unchanged, halved, doubled,
quadrupled)
3. If the charged of one of the particles id doubled, the force is ( unchanged, halved,
doubled, quadrupled)
4. If the distance between the particles is doubled, the force becomes ( one fourth, half,
double, 4 times)
5. The product of charges is ( inversely proportional , directly proportional) to the
electrical force.
ACTIVITY 2. SOLVING FOR THE ELECTRIC FORCE
DIRECTIONS: Solve for the following problems. Show your solutions for each
problem (5 points each)
1. Calculate the electric force between two charges -3.79 x 10 19 C and 5.67 x 10 –
18 C
placed 3.5 x 10 -6C away from each other.
2. Find the force between charges of + 8.0µC and -40.0 µC located 10 cm apart.
Note: 1 µC = 1x 10 -6 C.
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ACTIVITY 3. SOLVING FOR A NET ELECTRIC FORCE ON A POINT CHARGE BY A
SYSTEM OF POINT CHARGES
DIRECTIONS: Solve for the following problems. Show your solutions for each
problem (5 points each)
1. Calculate the magnitude of electric force between an iron nucleus ( 26 protons
having a charge of 1.6 x 10 -19 C and its internal electron having a charge of of
– 1.6 x 10 -19 C if the distance that separate them is 1.0 x 10
-12 m.
2. Determine the electrostatic force between a 5 protons to an electron if the
distance that separate them is 1 x 10 -10m.
ACTIVITY 4. SOLVING FOR THE DISTANCE OF POINT CHARGES
DIRECTIONS: Solve for the following problems. Show your solutions for each
problem (5 points each)
1. What is the distance of two electrons having a charge of -1.6 x 10 -19 C if the
force between them is 1.0 x 10 -12 N?
2. What is the distance of two spheres, each with a charge of 2. 5 x 10
-5 C,
when
the force between them is 0.60 N
ACTIVITY 5. SOLVING FOR THE MAGNITUDE OF CHARGE
DIRECTIONS: Solve for the following problems. Show your solutions for each
problem (5 points each)
1. An electric force 3.24 x 10 -10 N existing between a 18 proton charges and a
mystery particle separated by a distance of 2. 53 x 10 -10 N. Calculate the charge
of the mystery particle?
2. A negative charge of -5.0 x 10 -4 C exerts an attractive force of 12 N on a second
charge that is 0.040 m away. What is the magnitude of the second charge?
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REFLECTION
1. I learned that ____________________________________________________
_______________________________________________________________
_______________________________________________________
2. I enjoyed most on _________________________________________________
_______________________________________________________________
_______________________________________________
3. I want to learn more on _____________________________________________
_______________________________________________________________
_______________________________________________
37
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References:
“Coulomb’s Law – Problems and Solutions | Solved Problems in Basic Physics.” 2018.
Physics.gurumuda.net. March 2, 2018. https://physics.gurumuda.net/coulombs-lawproblems-and-solutions.htm.
“Coulombs Law Worksheets & Teaching Resources | Teachers Pay Teachers.” n.d.
Www.teacherspayteachers.com.
Accessed
February
1,
2021.
https://www.teacherspayteachers.com/Browse/Search:coulombs+law/Page:2.
“NGSS Physics: Static Electricity - Coulombs Law.” n.d. Www.physicsclassroom.com.
Accessed February 1, 2021. https://www.physicsclassroom.com/NGSS-Corner/ActivityDescriptions/Coulombs-Law.
“Physics
Tutorial:
Coulomb’s
Law.”
n.d.
Www.physicsclassroom.com.
https://www.physicsclassroom.com/class/estatics/Lesson-3/Coulomb-sLaw#:~:text=Coulomb.
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ANSWER KEY
Activity 1: RELATIONSHIP OF ELECTRIC FORCE, ELECTRICAL CHARGE AND
DISTANCE OF CHARGES.
1. INVERSELY PROPORTIONAL
2. QUADRUPLED
3. DOUBLED
4. ONE FOURTH
5. DIRECTLY PROPORTIONAL
ACTIVITY 2. SOLVING FOR THE ELECTRIC FORCE
1. 58 x 10 -15 N
2. -28.8 N
ACTIVITY 3. SOLVING FOR A NET ELECTRIC FORCE ON A POINT CHARGE BY A
SYSTEM OF POINT CHARGES
1. -5. 99 x 10 -3 N
2. 1.2 x 10 -47 N
ACTIVITY 4. SOLVING FOR THE DISTANCE OF POINT CHARGES
1. 1.52 x 10 -8 m
2. 3.06 m
ACTIVITY 5. SOLVING FOR THE MAGNITUDE OF A POINT CHARGE
1. + 50 C
2. -4.27 x 10-9C
Prepared by:
MILMAR T. EDRADA
Dassun National High School
39
NOTE: Practice personal hygiene protocols at all times
GENERAL PHYISCS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Electric Field
Background Information for the Learners (BIL)
Electric field
A region of space in which an electric charge will experience a force. The direction
of the field at a point in space is the direction in which a positive test charge would move
if placed at that point.
Representing electric fields
We can represent the strength and direction of an electric field at a point
using electric field lines. This is similar to representing magnetic fields around magnets
using magnetic field lines.
Positive charge acting on a test charge
The magnitude of the force that a test charge experiences due to another charge
is governed by Coulomb's law. In the diagram below, at each point around the positive
charge, +Q, we calculate the force a positive test charge, +q, would experience, and
represent this force (a vector) with an arrow. The force vectors for some points
around +Q are shown in the diagram along with the positive test charge +q (in red)
located at one of the points.
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NOTE: Practice personal hygiene protocols at all times
At every point around the charge +Q, the positive test charge, +q, will experience
a force pushing it away. This is because both charges are positive and so they repel each
other. We cannot draw an arrow at every point, but we include enough arrows to illustrate
what the field would look like. The arrows represent the force the test charge would
experience at each point. Coulomb's law is an inverse-square law which means that the
force gets weaker the greater the distance between the two charges. This is why the
arrows get shorter further away from +Q.
Negative charge acting on a test charge
For a negative charge, −Q, and a positive test charge, +q, the force vectors
would look like:
Notice that it is almost identical to the positive charge case. The arrows are the
same lengths as in the previous diagram because the absolute magnitude of the charge
is the same and so is the magnitude of the test charge. Thus, the magnitude of the force
is the same at the same points in space. However, the arrows point in the opposite
direction because the charges now have opposite signs and attract each other.
Electric fields around isolated charges - summary
Now, to make things simpler, we draw continuous lines that are tangential to the
force that a test charge would experience at each point. The field lines are closer together
where the field is stronger. Look at the diagram below: close to the central charges, the
field lines are close together. This is where the electric field is strongest. Further away
from the central charges where the electric field is weaker, the field lines are more spread
out from each other.
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NOTE: Practice personal hygiene protocols at all times
We use the following conventions when drawing electric field lines:
•
•
•
•
•
•
Arrows on the field lines indicate the direction of the field, i.e. the direction in which
a positive test charge would move if placed in the field.
Electric field lines point away from positive charges (like charges repel) and
towards negative charges (unlike charges attract).
Field lines are drawn closer together where the field is stronger.
Field lines do not touch or cross each other.
Field lines are drawn perpendicular to a charge or charged surface.
The greater the magnitude of the charge, the stronger its electric field. We
represent this by drawing more field lines around the greater charge than for
charges with smaller magnitudes.
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NOTE: Practice personal hygiene protocols at all times
Some important points to remember about electric fields:
•
•
•
•
There is an electric field at every point in space surrounding a charge.
Field lines are merely a representation – they are not real. When we draw them,
we just pick convenient places to indicate the field in space.
Field lines exist in three dimensions, not only in two dimensions as we've drawn
them.
The number of field lines passing through a surface is proportional to the charge
contained inside the surface.
Electric fields around different charge configurations
We have seen what the electric fields look like around isolated positive and
negative charges. Now we will study what the electric fields look like around
combinations of charges placed close together.
Electric field around two unlike charges
We will start by looking at the electric field around a positive and negative charge
placed next to each other. Using the rules for drawing electric field lines, we will sketch
the electric field one step at a time. The net resulting field is the sum of the fields from
each of the charges. To start off let us sketch the electric fields for each of the charges
separately.
A positive test charge (red dots) placed at different positions directly between the
two charges would be pushed away (orange force arrows) from the positive charge and
pulled towards (blue force arrows) the negative charge in a straight line. The orange
and blue force arrows have been drawn slightly offset from the dots for clarity. In reality
they would lie on top of each other. Notice that the further from the positive charge, the
smaller the repulsive force, F+ (shorter orange arrows) and the closer to the negative
charge the greater the attractive force, F− (longer blue arrows). The resultant forces are
shown by the red arrows. The electric field line is the black line which is tangential to
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NOTE: Practice personal hygiene protocols at all times
the resultant forces and is a straight line between the charges pointing from the positive
to the negative charge.
Now let's consider a positive test charge placed slightly higher than the line
joining the two charges. The test charge will experience a repulsive force (F+ in orange)
from the positive charge and an attractive force (F− in blue) due to the negative charge.
As before, the magnitude of these forces will depend on the distance of the test charge
from each of the charges according to Coulomb's law. Starting at a position closer to
the positive charge, the test charge will experience a larger repulsive force due to the
positive charge and a weaker attractive force from the negative charge. At a position
half-way between the positive and negative charges, the magnitudes of the repulsive
and attractive forces are the same. If the test charge is placed closer to the negative
charge, then the attractive force will be greater and the repulsive force it experiences
due to the more distant positive charge will be weaker. At each point we add the forces
due to the positive and negative charges to find the resultant force on the test charge
(shown by the red arrows). The resulting electric field line, which is tangential to the
resultant force vectors, will be a curve.
Now we can fill in the other field lines quite easily using the same ideas. The electric
field lines look like:
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Electric field around two like charges (both positive)
For the case of two positive charges Q1 and Q2 of the same magnitude, things
look a little different. We can't just turn the arrows around the way we did before. In this
case the positive test charge is repelled by both charges. The electric fields around
each of the charges in isolation looks like.
Now we can look at the resulting electric field when the charges are placed next
to each other. Let us start by placing a positive test charge directly between the two
charges. We can draw the forces exerted on the test charge due to Q1Q1 and Q2Q2 and
determine the resultant force.
The force F1 (in orange) on the test charge (red dot) due to the charge Q1 is equal
in magnitude but opposite in direction to F2 (in blue) which is the force exerted on the
test charge due to Q2. Therefore, they cancel each other out and there is no resultant
force. This means that the electric field directly between the charges cancels out in the
middle. A test charge placed at this point would not experience a force.
Now let's consider a positive test charge placed close to Q1 and above the
imaginary line joining the centers of the charges. Again, we can draw the forces exerted
on the test charge due to Q1 and Q2 and sum them to find the resultant force (shown in
red). This tells us the direction of the electric field line at each point. The electric field line
(black line) is tangential to the resultant forces.
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NOTE: Practice personal hygiene protocols at all times
If we place a test charge in the same relative positions but below the imaginary
line joining the centers of the charges, we can see in the diagram below that the resultant
forces are reflections of the forces above. Therefore, the electric field line is just a
reflection of the field line above.
Since Q2 has the same charge as Q1, the forces at the same relative points close
to Q2 will have the same magnitudes but opposite directions i.e. they are also reflections.
We can therefore easily draw the next two field lines as follows:
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Working through a number of possible starting points for the test charge we can
show the electric field can be represented by:
Electric field around two like charges (both negative)
We can use the fact that the direction of the force is reversed for a test charge if
you change the sign of the charge that is influencing it. If we change to the case where
both charges are negative, we get the following result:
Charges of different magnitudes
When the magnitudes are not equal the larger charge will influence the direction
of the field lines more than if they were equal. For example, here is a configuration where
the positive charge is much larger than the negative charge. You can see that the field
lines look more similar to that of an isolated charge at greater distances than in the earlier
example. This is because the larger charge gives rise to a stronger field and therefore
makes a larger relative contribution to the force on a test charge than the smaller charge.
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NOTE: Practice personal hygiene protocols at all times
A small test charge q placed near a charge Q will experience a force due to the
electric field surrounding Q. The magnitude of the force is described by Coulomb's law
and depends on the magnitude of the charge Q and the distance of the test charge
from Q. The closer the test charge q is to the charge Q, the greater the force it will
experience. Also, at points closer to the charge Q, the stronger is its electric field. We
define the electric field at a point as the force per unit charge.
According to Coulomb's law, the magnitude of the electric force exerted on the test
charge q is
𝐹=
𝑘𝑄𝑞
(Eq. 1)
𝑟2
Inserting this expression into our relation for the electric field (Eq. 1) gives
𝐹=
𝑘𝑄𝑞
𝑟2
= 𝑞𝐸
which leads to
𝐸=
𝑘𝑄
𝑟2
where E = electric field; N/C
F= electric force; F
Q = charge; C
r = distance between charges
Solved Problems:
1. What is the magnitude of a point charge that would create an electric field of 5N/C at
points
150 cm away?
Given:
E = 5N/C
1𝑚
𝑟 = 150𝑐𝑚 100𝑐𝑚 = 1.5𝑚
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K = 9 x 109
𝑁−𝑚2
𝐶1
Required, Q
Solution: 𝐸 =
𝑘𝑄
𝑟2
𝑄=
𝐸𝑟 2 = 5𝑁∕𝐶(1.5𝑚)2
=
𝑁−𝑚2
𝑘
9×109
1.25 x 109 C
𝐶2
2. A charge 1.5 𝜇C present in an electric field produces a force of 0.06N. What is the
intensity of the electric field?
𝑞 = 1.5 𝜇C
Given:
F = 0.06N
Required, E
Solution: 𝐹⃗ = 𝑞𝐸⃗⃗
𝐹
0.06𝑁
𝐸 = 𝑞 = 1.5𝑢𝐶 = 4.0 x 104N/C
3. Calculate the electric field strength 30 cm from a 55 nC charge.
+ 5 nC
30 cm
.
Given:
X
.
.
.
.
.
Q= 55𝑛𝐶
r = 30 cm = 0.3m
Required, E
Solution: 𝐸 =
𝑘𝑄
𝑟2
𝐸=
𝑞𝑥,09
𝑁−𝑚2
5𝑥|0−9 𝐶
𝐶2
0.3𝑚2
= 4.99 x 102N/C
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Two charges of Q1=+3nC and Q2=−4nC are separated by a distance of 50cm. What is
the electric field strength at a point that is 20 cm from Q1Q1 and 50cm from Q2? The
point lies between Q1 and Q2.
Q1= 3×10-9C
Given:
r1 = 20 cm = 0.2m
Q2 = 4×10−9 C
r2 = 30 cm = 0.3m
Required, E for Q1
E for Q2
Solution for Q1
𝐸=
=
𝑘𝑄
𝑟2
(qx,09
N−m2 )
(3x|0−9 C)
C2
0.2m2
= 6.74 x 102 N/C
Solution for Q2
𝐸=
=
𝑘𝑄
𝑟2
(qx,09
N−m2 )
(4x|0−9 C)
C2
0.3m2
= 3.99 x 102 N/C
We need to add the two electric fields because both are in the same direction.
The field is away from Q1 and towards Q2. Therefore,
Etotal = 6.74×102N/C + 3.99×102N/C = 1.08×102 N/C
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Learning Competency:
Describe an electric field as a region in which an electric charge experiences a
force (STEM_GP12EM-IIIa-7).
Learning Activity 1-
Check your Understanding
Directions: Answer the following questions briefly but substantially.
1. Does the electric field depend on a test charge?
_____________________________________________________
_____________________________________________________
_____________________________________________________
__________________________________________________.
2. In exploring the electric field with a test charge, we have often
assumed for convenience that test charge was positive. Does this
really make any difference in determining electric field?
_____________________________________________________
_____________________________________________________
_____________________________________________________
__________________________________________________.
3. Why do we need the electric field?
_____________________________________________________
_____________________________________________________
_____________________________________________________
______________________________________.
Learning Activity 2- Solved for me
Directions: Read carefully and analyze the following questions before solving.
Show your complete solutions.
1. What is the magnitude and direction of the force exerted on a 3.50 μC charge by
a 250 N/C electric field that points due east?
2. (a) What magnitude point charge creates a 10,000 N/C electric field at a distance
of 0.250 m? (b) How large is the field at 10.0 m?
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3. a) Find the direction and magnitude of an electric field that exerts (a) 4.80 × 10−17 N
westward force on an electron. (b) What magnitude and direction force does this
field exert on a proton?
4. Calculate the electric field strength 20 m from a 7 nC charge.
X
+ 7 nC
.
.
20 cm
.
.
.
.
20 cm
5. Two charges of Q1=−6 pC and Q2=−8 pC are separated by a distance of 3km. What
is the electric field strength at a point that is 2 km from Q1 and 1km from Q2? The point
lies between Q1 and Q2.
Learning Activity 3- Mine Right or Mine wrong
Directions: Write MR if the statement/question is true and MW is the
statement/question is false.
1. The field was created by a positive charge and here acts on a negative
charge.
2. The number of field lines passing through a surface is proportional to
the charge contained inside the surface.
3. The magnitude of the force that a test charge experiences due to
another charge is governed by Newton's law.
4. The field lines are closer together where the field is stronger.
5. The greater the magnitude of the charge, the weaker its electric field.
We represent this by drawing more field lines around the greater charge
than for charges with smaller magnitudes.
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REFLECTION
1. I learned that _________________________________________________
____________________________________________________________
_______________________________________________________
2. I enjoyed most on _________________________________________________
_______________________________________________________________
_______________________________________________
3. I want to learn more on _____________________________________________
_______________________________________________________________
_______________________________________________
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References
©2013, 2010Brooks/Cole, Cengage Learning
College Physics: Reasoning and Relationship
2nd Edition
General Physics 2 K to 12 Philippine Edition
Printed by Rex Book Sore
General Physics 2
COPYRIGHT 2020
Christopher G. Reyes
https://courses.lumenlearning.com/physics/chapter/18-4-electric-field-concept-of-a-fieldrevisited/\
https://www.quora.com/Does-the-electric-field-depend-on-a-test-charge
https://intl.siyavula.com/read/science/grade-11/electrostatics/09-electrostatics-03
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Answer key
Learning Activity 1- Check your Understanding
1. No. A test charge is conceptually so small that it affects nothing. In
principle, a test charge is used to measure the electric field. A static electric
field is determined by the source of the field and the displacement from it.
2. Answer may vary
3. Answer may vary
Learning Activity 2- Solved for me
1.
2.
3.
4.
8.75 × 10−4 N
(a) 6.94 × 10−8 C; (b) 6.25 N/C
(a)300 N/C (east); (b) 4.80 × 10−17 N (east)
0.15 N/C
5. −5,8×10−8 N/C: direction of the -8-pC charge
Learning Activity 3- Mine Right or Mine wrong
1.
2.
3.
4.
5.
MR
MR
MW- Coulomb’s
MR
MW-stronger
Prepared by:
LEONOR C. NATIVIDAD
Baggao National High School
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GENERAL PHYSCIS 2
Name: _________________________________
Grade Level: _________
Date: __________________________________
Score: ______________
LEARNING ACTIVITY SHEET
Coulomb’s Law: Superposition principle
Background Information for the Learners (BIL)
Coulomb’s Law
The magnitude F of the electrostatic force exerted by one-point charge on
another point charge is directly proportional to the magnitudes Q1 and Q2 of the
charges and inversely proportional to the square of the distance r between them.
Q1
Q2
r
𝐹=𝑘
|𝑄1 ||𝑄2 |
𝑟2
1
where, k = 4𝜋𝜀 = 9 × 109 𝑁𝑚2 ∕ 𝐶
0
Electric charge is a property that accompanies fundamental particles, whenever
they exist.
When an electric charge qo is held in the vicinity of another charge Q, qo either
experiences a force of attraction or repulsion. We say that this force is set up due to the
electric field around the charge Q.
The electric field due to a given electric charge Q is defined as the space
around the charge in which electrostatic force of attraction or repulsion due to the
charge Q can be experienced by another charge q.
Electric Field Due to a Point Charge Formula
The magnitude of the electric field (E) produced by a point charge with a charge of
magnitude Q, at a point a distance r away from the point charge, is given by the
equation,
𝐸=
𝑘𝑄
𝑟2
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NOTE: Practice personal hygiene protocols at all times
where k is constant = 9 x 109 N m2/C2
r= distance from Q
Q = charge
The direction of the electric field produced by a point charge is away from the charge
if the charge is positive, and toward the charge if the charge is negative.
Electric field is a vector, so when there are multiple point charges present, the net
electric field at any point is the vector sum of the electric fields due to the individual
charges.
The concept of the field was firstly introduced by Faraday. The electric field intensity
at any point is the strength of the electric field at that point. It is defined as the force
experienced by a unit positive charge at a particular point.
Here, if force acting on this unit positive charge +qo at a point r the electric
field intensity is given by:
𝐸⃗⃗ (𝑟⃗) =
𝐹⃗ (𝑟⃗)
𝑞0
Here, E is a vector quantity and is in the direction of the force and along
the direction in which the test charge +q tends to move.
The electric field for +qo is directed radially outwards from the charge while for qo, it will be radially directed inwards.
Superposition principle
Coulomb’s law explains the interaction between two-point charges. If there are
more than two charges, the force on one charge due to all the other charges needs to be
calculated. Coulomb’s law alone does not give the answer. The superposition principle
explains the interaction between multiple charges.
According to this superposition principle, the total force acting on a given
charge is equal to the vector sum of forces exerted on it by all the other charges.
Consider a system of n charges, namely q1, q2, q3 …. qn. The force on q1 exerted by
the charge q2
𝑭𝟏𝟐 = 𝒌
𝒒𝟏 𝒒𝟐
𝒓̂𝟐𝟏
𝒓𝟐𝟐𝟏
Here 𝑟̂21 is the unit vector from q2 to q1 along the line joining the two charges
and r21 is the distance between the charges q1 and q2. The electrostatic force between
two charges is not affected by the presence of other charges in the neighborhood.
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The force on q1 exerted by the charge q3 is
𝑭𝟏𝟑 = 𝒌
𝒒𝟏 𝒒𝟑
𝒓̂𝟑𝟏
𝒓𝟐𝟑𝟏
By continuing this, the total force acting on the charge q1 due to all other
charges is given by
⃗⃗⃗⃗
𝐹1 ᵗᶱˢᵗ = 𝐹12 + 𝐹13 + 𝐹14 + ⋯ 𝐹1𝑛
𝑞 𝑞
𝑞 𝑞
𝑞 𝑞
𝑞 𝑞
⃗⃗⃗⃗
𝐹1 ᵗᶱˢᵗ = 𝑘{ 𝑟12 2 𝑟̂21 + 𝑟12 3 𝑟̂31 + 𝑟12 4 𝑟̂41 + ⋯ 𝑟12 𝑛 𝑟̂𝑛1 }
21
31
41
(1.3)
𝑛1
EXAMPLE
Consider four equal charges q1,q2, q3 and q4 = q = +1μC located at four different
points on a circle of radius 1m, as shown in the figure. Calculate the total force acting
on the charge q1 due to all the other charges.
y
q2
xq
q3
1
q4
Solution
According to the superposition principle, the total electrostatic force on charge q1 is the
vector sum of the forces due to the other charges,
⃗⃗⃗⃗
𝐹1 ᵗᵒᵗ = ⃗⃗⃗⃗
𝐹1 2 + ⃗⃗⃗⃗⃗⃗
𝐹13 + ⃗⃗⃗⃗⃗⃗
𝐹14
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The following diagram shows the direction of each force on the charge q1.
y
q2
𝐹⃗14
r21
ϴ
q3
q1
𝐹⃗13
x
ϴ
Type equation here.
r31
r41
𝐹⃗12
q4
𝐹⃗14
𝐹14 𝑠𝑖𝑛𝛳
𝐹⃗13 𝐹⃗14 𝑐𝑜𝑠𝛳
ϴ
q1
ϴ
𝐹12 𝑠𝑖𝑛𝛳
𝐹⃗12 𝑐𝑜𝑠𝛳
𝐹⃗12
The charges q2 and q4 are equi-distant from q1. As a result, the strengths (magnitude)
of the forces 𝐹⃗12 and 𝐹⃗14 are the same even though their directions are different.
Therefore, the vectors representing these two forces are drawn with equal lengths. But
the charge q3 is located farther compared to q2 and q4. Since the strength of the
electrostatic force decreases as distance increases, the strength of the force 𝐹⃗13 is lesser
than that of forces 𝐹⃗12 and 𝐹⃗14 . Hence the vector representing the force 𝐹⃗13 is drawn with
smaller length compared to that for forces 𝐹⃗12 and 𝐹⃗14 .
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NOTE: Practice personal hygiene protocols at all times
From the figure, r21 = √2𝑚 = r41 and r31 = 2m
The magnitude of the forces is given by
𝑘𝑞 2 9 × 10−12
𝐹13 = 2 =
4
𝑟31
𝐹13 = 2.25 x 10-3 N
𝐹12 =
𝑘𝑞 2
9 × 10−12
=
𝐹
=
14
2
2
𝑟21
𝐹12 = 4.5 x 10-3 N
From the figure, the angle θ = 45º. In terms of the components, we have
𝐹⃗12 = 𝐹12 𝑐𝑜𝑠 𝜃𝑖̂ − 𝐹12 𝑠𝑖𝑛 𝜃 𝑗̂
=4.5 × 10−3 ×
1
√2
𝑖̂ − 4.5 × 10−3 ×
1
√2
𝑗̂
𝐹⃗13 = 𝐹13 𝑖̂ = 2.25 𝑥 10-3 N𝑖̂
𝐹⃗14 = 𝐹14 𝑐𝑜𝑠 𝜃𝑖̂ + 𝐹14 𝑠𝑖𝑛 𝜃 𝑗̂
= 4.5 × 10−3 ×
1
√2
𝑖̂ + 4.5 × 10−3 ×
1
√2
𝑗̂
Then the total force on q1 is,
𝐹⃗1 ᵗᵒᵗ = (𝐹12 𝑐𝑜𝑠 𝜃𝑖̂ − 𝐹12 𝑠𝑖𝑛 𝜃 𝑗̂ + 𝐹13 𝑖̂ + 𝐹14 𝑐𝑜𝑠 𝜃 𝑖̂ + 𝐹14 𝑠𝑖𝑛 𝜃 𝑗̂
𝐹⃗1 ᵗᵒᵗ = (𝐹12 𝑐𝑜𝑠 𝜃 + 𝐹13 +𝐹14 𝑐𝑜𝑠 𝜃)𝑖̂ + (−𝐹12 𝑠𝑖𝑛 𝜃+𝐹14 𝑠𝑖𝑛 𝜃)𝑗̂
Since F12 = F14, the jth component is zero.
Hence, we have
𝐹⃗1 ᵗᵒᵗ = (𝐹12 𝑐𝑜𝑠 𝜃 + 𝐹13 +𝐹14 𝑐𝑜𝑠 𝜃)𝑖̂
Substituting the values in the above equation,
4⋅5
4.5
𝐹⃗1 ᵗᵒᵗ = ( + 2.25 + ) 𝑖̂
√2
√2
= (4.5√2 + 2.25)𝑖̂
𝐹⃗1 ᵗᵒᵗ = 8.61 x 10-3 N 𝑖̂
The resultant force is along the positive x axis
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Learning Competency:
Calculate the electric field due to a system of point charges using coulombs law
and the superposition principle (STEM_GP12EM-IIIa-10).
Learning Activity 1- Check your Understanding
Directions: Choose the letter that correspond the correct answer. Write your answer in
another sheet of paper.
1. Electric field lines of force
A. exist everywhere
B. exist only in the immediate vicinity of electric charges
C. exist only when both positive and negative charges are near one
another
D. is imaginary
2. Identify which of the following diagrams best represents the electric field lines around
a positive charge?
A.
B.
C.
D.
3. The constant k in Coulomb's law depends upon
A. nature of medium
C. system of units
C. intensity of charge
D. both a and b
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4.The point-like charges carrying charges
of +3×10−9C and −5×10−9C are 2 m apart.
Determine the magnitude of the force between them and state whether it is
attractive or repulsive.
A. F=3.37×10−8 N
C. F=3.57×10−8 N
B. F=3.47×10−8 N
D. F=3.67×10−8 N
5. The electric field due to a charge at a distance of 3m from it is 500N/C. The
magnitude of the charge is[4πε01=9×109Nm2/C2]
A. 2.5𝜇C
C. 1.0 𝜇C
B. 2.0 𝜇C
D. 0.5 𝜇C
Learning Activity 2- Apply your Skills
Directions: Read and carefully analyze the given problems. Be sure to identify all the
given and unknown quantities. Use separate sheet of paper for your answer. Scoring is
provided for your reference.
Given: 1 point
Formula with solutions: 2 points
Answer with correct unit: 2 points
1. Two charges +3μC and +12μC are fixed 1 m apart, with the second one to the
right. Find the magnitude and direction of the net force on a −2-nC charge when
placed at the following locations:
(a) halfway between the two
(b) half a meter to the left of the +3μC charge
(c) half a meter above the +12μC charge in a direction perpendicular to the
line joining the two fixed charges
2. Point charges q1=50μC and q2=−25μC are placed 1.0 m apart. What is the
force on a third charge q3=20μCplaced midway between q1 and q2?
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3. A particle of charge 2.0×10−8C2.0×10−8C experiences an upward force of
magnitude .0×10−6N when it is placed in a particular point in an electric field.
(a) What is the electric field at that point?
(b) If a charge q=−1.0×10−8C is placed there, what is the force on it?
4. If the electric field is 100N/C at a distance of 50 cm from a point charge q, what
is the value of q?
5. Point charges q1=50μC and q2=−25μC are placed 1.0 m apart.
(a) What is the electric field at a point midway between them?
(b) What is the force on a charge q3=20μC situated there?
Learning Activity 3 -Field Map Representation
Directions: In this exercise, you will practice drawing electric field lines. Make sure you
represent both the magnitude and direction of the electric field adequately. Note that the
number of lines into or out of charges is proportional to the charges.
(a) Draw the electric field lines map for two charges +20μC and −20μC situated 5
cm from each other.
(b) Draw the electric field lines map for two charges +20μC and +20μC situated 5
cm from each other.
(c) Draw the electric field lines map for two charges +20μC and −30μC situated 5
cm from each other.
63
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REFLECTION
1. I learned that _________________________________________________
____________________________________________________________
_______________________________________________________
2. I enjoyed most on _____________________________________________
_______________________________________________________________
_______________________________________________
3. I want to learn more on _________________________________________
_______________________________________________________________
_______________________________________________
64
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References
©2013, 2010Brooks/Cole, Cengage Learning
College Physics: Reasoning and Relationship
2nd Edition
General Physics 2 K to 12 Philippine Edition
Printed by Rex Book Sore
General Physics 2
COPYRIGHT 2020
Christopher G. Reyes
https://www.vedantu.com/physics/electric-field-due-to-point-charge
http://www.brainkart.com/article/Coulomb---s-Law--Superposition-principle
phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(Op
enStax)/Map%3A_University_Physics_II_
_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/05%3A_Electric_Charges_
and_Fields/5.0E%3A_5.E%3A_Electric_Charges_and_Fields_(Exercises)
https://www.google.com/search?q=The+electric+field+due+to+a+point+charge+at+a+di
stance+r+is&sa=X&ved=2ahUKEwiC54H5ptXuAhUPPnAKHSiHDkwQ1QIwGnoECCoQ
AQ&biw=1366&bih=657
https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Supplemental_Modu
les_(Electricity_and_Magnetism)
65
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Answer key
Learning Activity 1-
1. D
Check your Understanding
2. B
3. A
4. A
5. D
Learning Activity 2- Apply your Skills
1.
q1 = 3μC
a.
q2 = 12μC
F31=2.16×10−4N
F31=2.16×10−4N to the left,
F32=8.63×10−4N
F32=8.63×10−4N to the right,
Fnet=6.47×10−4N
Fnet=6.47×10−4N to the right;
b. F31=2.16×10−4N
F31=2.16×10−4N to the right,
F32=9.59×10−5N
F32=9.59×10−5N to the right,
Fnet=3.12×10−4N
Fnet=3.12×10−4N to the right,
c. F⃗ 31x=−2.76×10−5Ni^,
F⃗ 31y=−1.38×10−5Nj^
F⃗ 32y=−8.63×10−4Nj^
F⃗ net=−2.76×10−5Ni^−8.77×10−4Nj^
2. F=53.94N
3. a. E=2.0×10−2 NC
b. F=2.0×10−19 N
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4. q=2.78×10−9C
5. If the q2 is to the right of q1, the electric field vector from both charges point to
the right.
a. E=2.70×106N/C
b. F=54.0N
Learning Activity 3- Field Demonstration
Prepared by:
LEONOR C. NATIVIDAD
Baggao National High School
67
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
ELECTRIC FLUX
Background Information for the Learners (BIL)
What is Electric flux?
The concept of flux describes how much of something goes through a given area.
More formally, it is the dot product of a vector field with an area. You may conceptualize
the flux of an electric field as a measure of the number of electric field lines passing
through an area (Figure 1). Thus, Electric flux is the rate of flow of the electric field
through a given surface. The larger the area, the more field lines go through it and, hence,
the greater the flux; similarly, the stronger the electric field is (represented by a greater
density of lines), the greater the flux. On the other hand, if the area rotated so that the
plane is aligned with the field lines, none will pass through and there will be no flux.
Photo source: https://phys.libretexts.org/@api/deki/files/8028/CNX_UPhysics_23_01_Flux.jpg?revision= 1
Figure 1: The flux of an electric field through the shaded area captures information about the
“number” of electric field lines passing through the area. The numerical value of the electric flux
depends on the magnitudes of the electric field and the area, as well as the relative orientation of
the area with respect to the direction of the electric field.
A macroscopic analogy that might help you imagine this is to put a hula hoop in a flowing
river. As you change the angle of the hoop relative to the direction of the current, more
or less of the flow will go through the hoop. Similarly, the amount of flow through the hoop
68
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depends on the strength of the current and the size of the hoop. Again, flux is a general
concept; we can also use it to describe the amount of sunlight hitting a solar panel or the
amount of energy a telescope receives from a distant star, for example.
Electric Flux through Open Surfaces
https://howtomechatronics.com/wp-content/uploads/2018/09/1.Electric-Flux-and-Gausss-Law-Electric-Flux-through-an-OpenSurface.png
Figure 2. The red lines represent a uniform electric field. We will bring in that field a
rectangle, which is an open area, and we will divide it into very small elements, each with
size 𝑑𝐴 (differential of area).
𝑑Φ = 𝐸. 𝑑𝐴⃗
𝑑Φ = 𝐸𝑑𝐴𝑐𝑜𝑠𝜃
The electric flux that passes through this small area 𝒅𝝋 (also called a differential of flux),
is defined as a dot product of the magnitude of the electric field E and the magnitude of
⃗⃗⃗, times the angle between these two vectors 𝜽.
the vector area 𝒅𝑨
⃗⃗ a vector, with a magnitude 𝒅𝑨
⃗⃗. The vector
Now we are going to make the area 𝒅𝑨
⃗⃗.
direction is always perpendicular to the small element 𝒅𝑨
Φ = ∫ 𝑑Φ
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Φ = ∮ 𝐸. 𝑑𝐴
The total flux is going to be the integral of 𝑑𝜑, or the integral over that entire area of
𝑬. 𝒅𝑨.
It is a scalar quantity and the end result can be positive or negative. If the flux is going
from the inside to the outside, we call that a positive flux, if it is going from the outside
to the inside, that is a negative flux.
𝑁𝑚2
Φ=
𝐶
The unit of electric flux is Newton meters squared per Coulomb (𝑵𝒎𝟐 ⁄𝑪).
).
https://howtomechatronics.com/wp-content/uploads/2018/09/5.Electric-Flux-and-Gausss-Law-3-Rectangles-with-DifferentOrientation-into-an-Electric-Field-768x280.png
Figure 3. Electric flux through open surfaces in different orientations
To get a better understanding of what electric flux is, let us consider electric fields
passing through these three rectangles with different orientations.
In the first case, the area is perpendicular to the electric field, and the angle between
their vectors 𝜃 is 0°. 𝑐𝑜𝑠 0° is 1, so the electric flux is going to be 𝑬𝒅𝑨. Here we have
maximum flux.
𝒅𝚽 = 𝑬𝒅𝑨𝒄𝒐𝒔𝜽
𝒅𝚽 = 𝑬𝒅𝑨𝒄𝒐𝒔𝟎°
𝒅𝚽 = 𝑬𝒅𝑨
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In the second case, the angle between 𝑬 and 𝒅𝑨 𝜽 is 60°, and 𝑐𝑜𝑠 60°cos is 0.5, so the
electric flux will be half 𝑬𝒅𝑨.
𝒅𝚽 = 𝑬𝒅𝑨𝒄𝒐𝒔𝜽
𝒅𝚽 = 𝑬𝒅𝑨𝒄𝒐𝒔𝟔𝟎°
𝑬𝒅𝑨
𝒅𝚽 =
𝟐
In the third case, the area is parallel to the electric field, which means that their vectors
are perpendicular to each other, and the angle 𝜽 between them is 90°. cos90° is 0, so
the electric flux here will be 0. This means that nothing goes through that rectangle, so
here we have zero flux.
𝒅𝚽 = 𝑬𝒅𝑨𝒄𝒐𝒔𝜽
𝒅𝚽 = 𝑬𝒅𝑨𝒄𝒐𝒔𝟗𝟎°
𝒅𝚽 = 𝟎
Problem #1
A uniform electric field 𝐸 = 8000 𝑁⁄𝐶 is passing
through a flat square area 𝐴 = 10𝑚2 . Determine the
electric flux.
https://physics.gurumuda.net/wp-content/uploads/2018/03/Electric-fluxthrough-area-and-closed-surface-%E2%80%93-problems-and-solutions-1.png
Given:
•
The magnitude of the electric field (𝐸) = 8000 𝑁⁄𝐶
•
Area (𝐴) = 10𝑚2
•
𝜃 = 0° (the angle between the electric field direction and a line drawn a
perpendicular to the area)
Unknown: Electric flux (Φ)
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Solution:
The formula of electric flux :
𝑑Φ = 𝐸𝑑𝐴𝑐𝑜𝑠𝜃
Φ = 𝐸𝐴𝑐𝑜𝑠𝜃
Φ = electric flux (𝑁𝑚2/ 𝐶), E = electric field (N/C), A = area (𝑚2 ), 𝜃 = angle between
electric field line with the normal line.
Electric flux :
Φ = 𝐸𝐴𝑐𝑜𝑠𝜃
𝑁
Φ = (8000 ) (10𝑚2 )(𝑐𝑜𝑠0°)
𝐶
𝑵𝒎𝟐
𝑵𝒎𝟐
𝟒
𝚽 = 𝟖𝟎𝟎𝟎𝟎
𝒐𝒓 𝟖𝒙𝟏𝟎
𝑪
𝑪
Problem #2
A square surface with side length of 3.7 mm is in a
uniform electric field with magnitude E=2400 N/C, as
shown in the figure. The angle between the normal to
the surface and the electric field is 30°. What is the
electric flux through the surface, assuming that the
normal is directed outward?
https://ds055uzetaobb.cloudfront.net/brioche/solvable/bdc8
e6790a.4cd16ed770.RxZunD.jpg?width=500
Given:
𝐸 = 2400 𝑁⁄𝐶
𝐴 = 𝑆 2 = (3.7𝑚𝑚)2 = 13.69𝑚𝑚2 = 3.7𝑥10−6 𝑚2
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The vector area 𝐴⃗ and electric field 𝐸⃗⃗ are shown on the diagram below. The angle θ
between them is 180° − 30° = 150°
Solution:
⃗⃗⃗⃗. 𝐸⃗⃗ = 𝐸𝐴𝑐𝑜𝑠𝜃
Φ=𝐴
= (2400 𝑁⁄𝐶 )(3.7𝑥10−6 𝑚2 )(𝑐𝑜𝑠150°)
= −𝟐. 𝟖𝟓𝒙𝟏𝟎−𝟐
𝑵𝒎𝟐
𝑪
Note:
𝛷𝐸 is positive if flux lines are flowing outward that is if 𝜃 < 90°
𝛷𝐸 is negative if flux lines are flowing inwards that is if 𝜃 > 90°
Quiz Time!
Multiple choices: Choose the letter of the correct answer that suits to the given question.
1. The electric flux Φ through a surface
a. is the amount of electric field piercing the surface.
b. does not depend on the area involved.
c. is the electric field multiplied by the area.
d. is the line integral of the electric field around the edge of the surface
2. The area vector for a flat surface
a. is parallel to the surface and has a magnitude equal to the length of a side of
the surface.
b. is parallel to the surface and has a magnitude equal to the area of the surface.
c. is perpendicular to the surface and has a magnitude equal to the length of a
side of the surface.
d. is perpendicular to the surface and has a magnitude equal to the area of the
surface.
3. Which quantity and unit are correctly paired?
a. electric field strength and N/C
b. electrostatic force and electrons
c. electricity and Coulombs
d. electric field strength and E
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4. When is the flux on a surface zero?
a. When it is perpendicular to an electric field
b. When it is parallel to an electric field
c. When it is at an angle to an electric field
d. It is never zero in an electric field
5. Which among of the choices is/are not FALSE about electric flux?
a. Electric flux can be defined on how much of something is passing through
some surface.
b. Electric flux is the amount of charge in a certain surface.
c. Electric flux is the direction of the electric field in a certain charge.
d. All of the above
6. Which among the statement is TRUE?
a. A flux flowing from inside to outside is negative.
b. A flux flowing from inside to outside is positive.
c. A flux flowing from outside to inside is positive.
d. A flux flowing from inside to outside can be negative or positive.
7. The formula to solve the electric flux is:
a. Electric flux is equal to Electric Field over Area of the surface subtracted to
cosine theta.
b. Electric flux is equal to cosine theta over Electric field multiplied by Area of the
surface.
c. Electric flux is equal to Area of the surface multiplied by cosine theta over
Electric field.
d. Electric flux is equal to Electric field multiplied by Area of the surface times
cosine theta.
8. Flux will be minimum when the electric field lines to the vector area are
a. parallel
c. at angle
b. perpendicular
d. at distance
9. System International unit of electric flux is
a. 𝑁𝑚2 𝐶 −1
c. 𝑁𝑚1 𝐶 −1
b. 𝑁𝑚2 𝐶
d. 𝑚2 𝐶 −1
10. If the electric field lines are parallel to the vector area, the electric flux will be
a. minimum
d. zero
b. maximum
e. constant
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Learning Competency
Calculate electric flux (STEM_GP12EM-IIIb-12)
Activity 1: DRAW MY ANGLE!
Directions: Draw the corresponding orientation of each shapes with the given angle.
Illustrate vector lines and electric field using arrow heads.
1. Rectangle; 𝑐𝑜𝑠𝜃 = 35°
2. Circle; 𝑐𝑜𝑠𝜃 = 0°
3. Square; 𝑐𝑜𝑠𝜃 = 45°
4. Triangle; 𝑐𝑜𝑠𝜃 = 90°
Activity 2: SOLVE THE SHAPE OF YOU!
Directions: Solve for the value of Area and Electric flux.
Rectangle
Dimensions
Square
Circle
s=89cm
r=89cm
1.8𝑥10−24
8.99𝑥106
2.89𝑥1024
75°
90°
L=24cm
W= 15cm
Triangle
b=25cm
h=30cm
Area (𝒎𝟐 )
Electric Field (𝑵⁄𝑪)
Angle
45°
9.0𝑥109
50°
Electric Flux
75
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Activity 3: PROBLEM SOLVING
Directions: Solve the following problem with complete solutions
Problem no. 1
An anemoscope is a device invented to
show the direction of the wind, or to
foretell a change of wind direction or
weather. Consider a situation where an
anemoscope is in a uniform electric
field of magnitude 𝐸 = 4.5 𝑚𝑁⁄𝐶 E=4.5
mN/C, as depicted in the above figure. If the rim of the anemoscope is a circle with
radius r=13 cm, and is perpendicular to the electric field, what is the magnitude of the
electric flux through the fabric of the anemoscope?
Problem no. 2
Calculate the electric flux through the rectangle
of sides 5 cm and 10 cm kept in the region of a
uniform electric field 100 N/C The angle θ is 60°.
Suppose θ becomes zero, what is the electric
flux?
REFLECTION
1. I learned that__________________________________________________
_______________________________________________________________
_________________________________________________________
2. I enjoyed most on ______________________________________________
_______________________________________________________________
_______________________________________________
3. I want to learn more on __________________________________________
_______________________________________________________________
________________________________________________
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REFERENCES
Ali, F. (2020). Electric Flux, Gauss Law: Solved Example Problems. Retrieved February
13, 2021, from https://www.brainkart.com/article/Electric-Flux,-Gauss-Law--SolvedExample-Problems_38381/
D. (2018, October 1). Electric Flux and Gauss’s Law. Retrieved January 20, 2021, from
https://howtomechatronics.com/learn/electricity/electric-flux-gausss-law/
Electric Flux, Gauss Law: Solved Example Problems. (2019, March 13). Retrieved
February 13, 2021, from https://www.brainkart.com/article/Electric-Flux,-Gauss-Law-Solved-Example-Problems_38381/
Libretexts. (2020, November 5). 6.2: Electric Flux. Retrieved January 20, 2021, from
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physi
cs_(OpenStax)/Map%3A_University_Physics_II__Thermodynamics_Electricity_and_Magnetism_(OpenStax)/06%3A_Gauss’s_Law/6.0
2%3A_Electric_Flux
Paul,
J.
(2019,
February
26).
Quiz.
Retrieved
February
13,
2021,
from
https://www.scribd.com/document/400547728/Quiz
Practice Electricity and Magnetism | Brilliant. (2014, November 26). Retrieved February
13, 2021, from https://brilliant.org/electricity-and-magnetism/?subtopic=electrodynamics
77
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ANSWER KEY:
Quiz Time
1. A
2. D
3. A
4. B
5. A
6. B
7. D
8. B
9. A
10. B
Activity 1 Draw my angle! (Possible illustrations)
1.
2.
3.
4.
Activity 2 SOLVE THE SHAPE OF YOU!
Rectangle
Dimensions
Square
Circle
s=89cm
r=89cm
𝟎. 𝟑𝟔
𝟎. 𝟕𝟗
𝟐. 𝟕𝟗
1.8𝑥10−24
8.99𝑥106
75°
90°
𝟏. 𝟕𝒙𝟏𝟎−𝟐𝟓
0
L=24cm
W= 15cm
Area (𝒎𝟐 )
Electric Field (𝑵⁄𝑪)
Angle
Electric Flux𝑵𝒎𝟐 ⁄𝑪
2.89𝑥1024
45°
𝟓. 𝟕𝒙𝟏𝟎𝟐𝟒
Triangle
b=25cm
h=30cm
𝟎. 𝟎𝟑𝟕
9.0𝑥109
50°
𝟐. 𝟏𝒙𝟏𝟎𝟖
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Activity 3 PROBLEM SOLVING
Problem no. 1
The flux through the flat surface encircled by the rim is given Φ = 𝜋𝑟 2 𝐸. Thus, the flux
through netting is
𝚽 ′ = −𝚽 = −𝝅𝒓𝟐 𝑬
= −𝝅(𝟎. 𝟏𝟑𝒎)𝟐 (𝟒. 𝟓𝒙𝟏𝟎−𝟑 𝑵⁄𝑪)
= −𝟐. 𝟑𝟗𝑿𝟏𝟎−𝟒 𝑵𝒎𝟐 ⁄𝑪
Problem no. 2
Given:
𝐴 = (𝐿)(𝑊) = (0.005𝑚)(0.1𝑚) = 0.5𝑚2
𝐸 = 100 𝑁⁄𝐶
𝜃 = 60°, 0°
𝚽 =?
𝜽 = 𝟔𝟎°
𝚽 = ⃗⃗⃗⃗
𝑨 . ⃗𝑬⃗ = 𝑬𝑨𝒄𝒐𝒔𝜽
𝚽 = (𝟏𝟎𝟎 𝑵⁄𝑪)(𝟎. 𝟎𝟎𝟓𝒎𝟐 )(𝒄𝒐𝒔𝟔𝟎°)
= 𝟎. 𝟐𝟓 𝑵𝒎𝟐 ⁄𝑪
𝜽 = 𝟎°
⃗⃗⃗⃗. 𝑬
⃗⃗ = 𝑬𝑨𝒄𝒐𝒔𝜽
𝚽=𝑨
𝚽 = (𝟏𝟎𝟎 𝑵⁄𝑪)(𝟎. 𝟎𝟎𝟓𝒎𝟐 )(𝒄𝒐𝒔𝟎°)
= 𝟎. 𝟓 𝑵𝒎𝟐 ⁄𝑪
Prepared by:
JOHN DAVID B. MEDRANO
APARRI EAST NATIONAL HIGH SCHOOL
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NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 1
Name: __________________________________ Grade Level: _______________
Date: ___________________________________ Score: ____________________
LEARNING ACTIVITY SHEET
THE USE OF GAUSS’S LAW
Background Information for the Learners (BIL)
Electric Flux and Gauss’s Law
In order to explain Gauss’s law, we must first define a quantity called the electric
flux. In words, we say that there is an electric flux through a surface whenever electric
field lines pass through the surface. In the simplest case, with the electric field directed
perpendicular to the surface, the magnitude of the flux is the product of the electric field
and the area of the surface. Electric flux is denoted by the symbol 𝚽𝑬 .
Figure 1.0 Finding the electric flux through a surface.
Some examples of electric flux calculations are given pictorially in Figure 1.0. For
simplicity, these examples assume the electric field is constant in both magnitude and
⃗⃗⃗ is perpendicular to a flat surface having a total area A, and
direction. In Figure 1.0-A, 𝑬
the flux is 𝚽𝑬 = 𝑬𝑨. In this case, the flux is just the magnitude of the electric field
multiplied by the area of the surface. Flux is a scalar quantity.
⃗⃗ is parallel to the surface (Figure 1.0-B), no electric field lines cross. The surface
If ⃗𝑬
and 𝚽𝑬 = 𝟎. In general, if the electric field makes an angle 𝜃 with the direction normal to
the surface (Figure 1.0-C), the magnitude of the flux is proportional to the component of
the field perpendicular to the surface, and
𝚽𝑬 = 𝑬𝑨 𝒄𝒐𝒔 𝜽
(Equation 1.0)
The three examples in figure 1.0A-C all involve simple, flat surfaces. We’ll usually
be interested in the flux through a closed surface such as a box or a sphere. The flux due
⃗⃗ through these closed surfaces is shown in parts D and E of Figure
to a constant field ⃗𝑬
1.0. By convection, the flux through a surface is positive if the field is directed out of the
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region contained by the surface; whereas the flux is negative if ⃗𝑬⃗ is directed into the
region. For the case with constant field, the total flux through the entire closed surface is
zero in each case. For the box in Figure 1.0-D, there is a negative flux through the face
of the box on the left and a positive flux through the face on the right. The total flux is the
sum of these two contributions and is zero. Likewise, there is a negative flux through the
left side of the spherical surface in Figure 1.0-E and a positive flux through the right side,
and the total flux again is zero.
Gauss’s Law
Gauss’s law asserts that the electric flux through any closed surface is proportional
to the total charge q inside the surface, with
𝚽𝑬 =
𝒒
𝜺𝟎
(Equation 1.1)
The constant of proportionality that relates flux and charge is 𝜺𝟎 , the same physical
constant that enters Coulomb’s law.
Since electric flux depends on the magnitude and direction of the electric field, the
⃗⃗ while the right-hand side depends on the
left-hand side of the equation depends on ⃗𝑬
⃗⃗, but it is not immediately obvious
charge. We would like to use Gauss’s law to calculate𝑬
how to do so. In particular, 𝚽𝑬 is the total flux through closed surface, so its value
depends on the magnitude and direction of the electric field at all points on the surface.
⃗⃗ for a Point Charge
Using Gauss’s Law to find 𝑬
Let’s first consider the familiar case of a single point charge. To apply Gauss’s
law, we must first choose the surface, called a Gaussian surface that will be used in the
flux calculation. Equation 1.0 holds for any closed surface, so our strategy is to choose a
surface that will make the calculation of 𝚽𝑬 as simple as possible. To this end, it is almost
always best choose a surface that matches the symmetry of the problem. For a point
charge, the electric field lines have a spherical symmetry (Fig. 1.1), meaning that the
magnitude of the electric field depends only on the distance r from the charge and that
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⃗𝑬
⃗⃗ is directed radially, either outward or inward with respect to the central point. A surface
that matches this symmetry is a sphere centered on the charge as sketched in Figure
⃗⃗ is the same at all points on the sphere,
1.0. Because of the symmetry, the magnitude of ⃗𝑬
and ⃗𝑬⃗ is everywhere perpendicular to the sphere.
Figure 1.1 To calculate the electric field near a point charge using Gauss’s law, we
choose a spherical Gaussian surface centered on the point charge.
⃗⃗ is perpendicular to a surface, the flux is equal to the magnitude of ⃗𝑬⃗
When ⃗𝑬
multiplied by the area of the surface. So, for the flux in Figure 1.1, we have
𝚽𝑬 = 𝑬𝑨𝒔𝒑𝒉𝒆𝒓𝒆
(Equation 1.2)
where 𝑨𝒔𝒑𝒉𝒆𝒓𝒆 is the area of our spherical Gaussian surface. If the radius of this sphere
is r, the 𝑨𝒔𝒑𝒉𝒆𝒓𝒆 = 𝟒𝝅𝒓𝟐 and the flux is
𝚽𝑬 = 𝟒𝝅𝒓𝟐 𝑬
(Equation 1.3)
According to Gauss’s law, this flux is proportional to the total charge contained
within the surface. Using Equation 1.1, we have
𝚽𝑬 = 𝟒𝝅𝒓𝟐 𝑬 =
𝒒
𝜺𝟎
(Equation 1.4)
We can now solve for E and find
𝑬=
𝒒
𝟒𝝅𝜺𝟎 𝒓𝟐
(Equation 1.5)
The key to this application of Gauss’s law was our choice of the Gaussian surface.
This choice made the calculation of 𝚽𝑬 straightforward because 𝑬 has the same value
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over the entire surface and the electric field’s direction is always perpendicular to the
surface.
More Applications of Gauss’s Law
A. Consider a very long, straight line of charged. The line of charge has total length
L (where L is very large) and the total charge Q. How can we find the electric field
produced by this charge distribution?
We can calculate the flux through our chosen Gaussian surface which is a
cylinder. For a cylinder of radius r and length h, the area of the curve part of the
cylinder is 𝐴 = 2𝜋𝑟ℎ. The flux through this curved surface is thus
𝚽𝑬 = 𝑬𝑨 = 𝟐𝝅𝑬𝒓𝒉
We now find the total charge inside the Gaussian surface and then apply Gauss’s
law. The total charge within the cylinder is equal to the charge per unit length Q/L
multiplied by the length h of the cylinder, so
𝒒=
𝑸
𝒉
𝑳
Applying Gauss’s law then gives
𝚽𝑬 = 𝟐𝝅𝑬𝒓𝒉 =
𝒒 𝑸𝒉/𝑳
=
𝜺𝟎
𝜺𝟎
wherein E:
𝑸
𝑬 = 𝟐𝝅𝜺
𝟎 𝑳𝒓
(Equation 1.6)
B. Consider a large, flat sheet of charge. If this sheet has positive charge per unit
area 𝜎, find the electric field produced by the sheet.
We choose a Gaussian surface. The cylinder’s axis is oriented perpendicular to
⃗⃗
the plane. The flux through the curved sidewall of the surface is zero because 𝑬
is parallel to this part of the cylinder. The electric field ⃗𝑬⃗ is perpendicular to the
ends of the cylinder, so if the ends each have an area A, the flux is 𝚽𝑬 = 𝑬𝑨
through each end. For a positively charged plane, the electric flux through each
end of the cylinder is positive because the electric field lines pass outward through
the cylinder. The total electric flux through the cylinder is thus
𝚽𝑬 = 𝟐𝑬𝑨
where the factor of two is needed to include the flux through both ends.
Applying Gauss’s law then gives
𝚽𝑬 =
𝒒
𝜺𝟎
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𝟐𝑬𝑨 =
𝝈𝑨
𝜺𝟎
Solving for the electric field, we get
𝑬=
𝝈
(Equation 1.7)
𝟐𝜺𝟎
C. Calculation of the electric field between two thin metal plates. This arrangement is
called a parallel-plate capacitor.
In this case, the charges on the two plates attract each other and all the charge
on each plate resides on the inner surface of the plate, with none on the outer
surfaces.
𝑬=
𝑸
(Equation 1.8)
𝜺𝟎 𝑨
Examples of Problems Solving;
1. A 4.0 cm2 in the x-y plane sits in a uniform electric field E = (2.0 i + 3.0 j + 5.0 k)
N/C. Find the electric flux through the square.
Solution:
𝑁
𝒎𝟐
Φ𝐸 = 𝐸𝐴 = (5.0 ) 𝑥 (16 𝑐𝑚2 ) = 𝟖. 𝟎 𝒙 𝟏𝟎−𝟑 𝑵.
𝐶
𝑪
2. A thundercloud produces a vertical electric field of magnitude 28.0 kN/C at ground
level. You hold a 22.0 cm x 28.0 cm sheet of paper horizontally below the cloud.
a. What is the electric flux through the sheet?
𝑁
𝒎𝟐
Φ𝐸 = 𝐸𝐴 = (28.0 𝑥 103 ) (0.0616 𝑚2 ) = 𝟏. 𝟕𝟐 𝒙 𝟏𝟎𝟑 𝑵.
𝐶
𝑪
0
b. What would be the flux be if you tilt the sheet of paper by 30 ?
Φ𝐸 = 𝐸𝐴 𝑐𝑜𝑠𝜃 =( 1.72 𝑥 103 𝑁.
𝑚2
𝐶
)(0.8660) = 𝟏. 𝟒𝟗 𝐱 𝟏𝟎𝟑 𝑵.
𝒎𝟐
𝑪
c. What would be the flux be if you hold the sheet of paper vertically?
Since the electric field would be parallel to the paper, the flux would be zero.
3. A long copper wire with radius of 1.0 mm carries a uniform surface charge density
of 5.0 x 10-6 C/m2.
a. Find the total charge in a 1.0-meter-long section of the wire.
2𝜋𝑟ℎ = 2𝜋(1.0 𝑥 10−3 𝑚)(1.0 𝑚) = 6.28 𝑥 10−3 𝑚2
Therefore, the charge is (5.0 𝑥
10−6 𝐶
𝑚2
) (6.28 𝑥 10−3 𝑚2 ) = 𝟑. 𝟏 𝒙 𝟏𝟎−𝟖 𝑪
b. Find the magnitude of the electric field at a distance of 15 cm from the wire.
𝐶
3.1 𝑥 10−8 𝑚
𝑄
𝑵
𝐸=
=
= 𝟑. 𝟖 𝒙 𝟏𝟎𝟑
2𝜋𝜀0 𝑟 2𝜋𝜀0 (0.15 𝑚)
𝑪
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Learning Competency
Use Gauss’s Law to infer electric field due to uniformly distributed charges on long wires,
spheres, and large plates. STEM_GP12EM-IIIb-13
Learning Activity #1: COMPLETE ME!
Directions: For 5 points, complete the concept below.
The key to this application of _________________was our choice of the
_________________.
This
choice
made
the
calculation
of
_______________________________straightforward because ________________________________has the
same value over the entire surface and the electric field’s direction is always
_________________ to the surface.
Learning Activity #2: INFERRING
Directions: Consider a hollow spherical conductive shell of radius (R) 0.2 m with a fixed
charge of +2.0 x 10-6 C uniformly distributed on its surface.
a. What is the electric field at all points inside the sphere? Express your answer as
a function of the distance r from the center of the sphere. (2 points)
b. What is the electric field outside the sphere? Express your answer as a function
of the distance r from the center of the sphere. (2 points)
c. What if the sphere is a solid conductive sphere? What is the electric field at all
points inside the sphere? Express your answer as function of the distance r from
the center of the sphere. (3 points)
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d. (Connected to c) What is the electric field at all points outside the sphere? Express
your answer as a function of the distance r from the center of the sphere. (3 points)
Learning Activity #3: PROBLEM SOLVING
Directions: Solve the following problems accurately with complete solutions. (5 points
each)
1. A cylindrical metal can has a height of 27 cm and a radius of 11 cm. the electric
field is directed outward along the entire surface of the can (including the top and
bottom), with a uniform magnitude of 4.0 x 105 N/C. How much charge does the
can contain?
2. An insulating sphere with a radius of 20 cm carries a uniform volume charge
density of 1.5 x 10-6 C/m3. Find the magnitude of the electric field at a point inside
the sphere that lies 8.0 cm from the center.
3. A square metal plate with a thickness of 1.5 cm has no net charge and is placed
in a region of uniform electric field 8.0 x 104 N/C directed perpendicularly to the
plate. Find the resulting surface charge density on each face of the plate.
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REFLECTION
1.I learned that ________________________________________________________
_______________________________________________________________
_______________________________________________________
2.I enjoyed most on ____________________________________________________
_______________________________________________________________
__________________________________________________
3.I want to learn more on ________________________________________________
_______________________________________________________________
_______________________________________________
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References:
2018. General Physics 2, Rex Book Store, Inc. pages 23-33.
Gauss’s
Law
review
paper
retrieved
from
https://www.pearsonhighered.com/content/dam/region-na/us/highered/en/products-services/course-products/young-freedman-14e-info/pdf/samplechapter--ch22.pdf
Worksheet
on
Gauss’s
law
retrieved
from
http://faculty.bard.edu/~belk/phys142s09/Chapter24Solutions.pdf
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Answer Key
Learning Activity #1: COMPLETE ME!
The key to this application of Gauss’s law was our choice of the Gaussian surface.
This choice made the calculation of 𝚽𝑬 straightforward because 𝑬 has the same value
over the entire surface and the electric field’s direction is always perpendicular to the
surface.
Learning Activity #2: INFERRING
a. E = 0
𝑞
b. 𝐸 = 4𝜋𝜀 𝑟 2
0
c. 𝐸 =
d. 𝐸 =
𝑞𝑟
3𝜀0
𝜎𝑅 3
3𝜀0 𝑟 2
Learning Activity #3: PROBLEM SOLVING
1. C = +9.3 x 10-7
2. E = 4.5 x 103 N/C
3. E = 0, 𝜎 = 7.1 x 10-7 C/m2
Prepared by:
JOLLY MAR D. CASTANEDA
Baggao National Agricultural School Sta. Margarita Annex
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GENERAL PHYSICS 1
Name: ________________________________ Grade Level: _________________
Date: _________________________________ Score: ______________________
LEARNING ACTIVITY SHEET
SOLVE PROBLEMS INVOLVING ELECTRIC CHARGES DIPOLES,
FORCES, FIELDS AND FLUX
Background Information for the Learners (BIL)
Review on the Concepts of Electric Forces and Fields
A. Electric Charges
There are only two types of charge, which we call positive and negative. Like
charges repel, unlike charges attract, and the force between charges decreases
with the square of the distance.
• The vast majority of positive charge in nature is carried by protons, whereas the
vast majority of negative charge is carried by electrons. The electric charge of one
electron is equal in magnitude and opposite in sign to the charge of one proton.
• An ion is an atom or molecule that has nonzero total charge due to having unequal
numbers of electrons and protons.
• The SI unit for charge is the coulomb (C), with protons and electrons having
charges of opposite sign but equal magnitude; the magnitude of this basic charge
is e- =1.602 × 10−19 C
• Both positive and negative charges exist in neutral objects and can be separated
by bringing the two objects into physical contact; rubbing the objects together
can remove electrons from the bonds in one object and place them on the other
object, increasing the charge separation.
• For macroscopic objects, negatively charged means an excess of electrons and
positively charged means a depletion of electrons.
• The law of conservation of charge states that the net charge of a closed system
is constant.
B. Conductors, Insulators, and Charging by Induction
•
•
•
•
A conductor is a substance that allows charge to flow freely through its atomic
structure.
An insulator holds charge fixed in place.
Polarization is the separation of positive and negative charges in a neutral object.
Polarized objects have their positive and negative charges concentrated in
different areas, giving them a charge distribution.
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C. Coulomb’s Law
Coulomb’s law gives the magnitude of the force between point charges. It is
𝑞1 𝑞2
𝑞1 𝑞2
𝐹=𝑘 2 =
𝑟
4𝜋𝜀0 𝑟 2
Where, q1 and q2 are two point charges separated by a distance r. This
Coulomb force is extremely basic, since most charges are due to point-like
particles. It is responsible for all electrostatic effects and underlies most
macroscopic forces.
D. Electric Field
Whenever you have a charge Q placed anywhere in space, it will be surrounded
by a region such that if you will put any other charge q at any point P in this region,
the charge q will be acted upon by an electric force ⃗⃗⃗⃗
𝐹𝑒 . We call this region around
Q the electric field 𝐸⃗⃗ of Q. The strength of this electric field is operationally defined
as the ratio of the electric force ⃗⃗⃗⃗
𝐹𝑒 to the charge q placed at that point in the field.
In symbols,
⃗⃗⃗⃗
𝐹𝑒
𝐸⃗⃗ =
𝑞
E. Electric Flux
There is an electric flux through a surface whenever electric field lines pass
through the surface. In the simplest case, with the electric field directed
perpendicular to the surface, the magnitude of the flux is the product of the electric
field and the area of the surface. Electric flux is denoted by the symbol 𝚽𝑬 . When
electric field (E) is perpendicular to a flat surface having a total area A, and the
flux is 𝚽𝑬 = 𝑬𝑨. In this case, the flux is just the magnitude of the electric field
multiplied by the area of the surface. Flux is a scalar quantity.
F. Gauss’s Law
Gauss’s law asserts that the electric flux through any closed surface is
proportional to the total charge q inside the surface, with
𝒒
𝚽𝑬 =
𝜺𝟎
The constant of proportionality that relates flux and charge is 𝜺𝟎 , the same physical
constant that enters Coulomb’s law.
Since electric flux depends on the magnitude and direction of the electric
⃗⃗ while the right-hand side
field, the left-hand side of the equation depends on 𝑬
⃗⃗, but it is
depends on the charge. We would like to use Gauss’s law to calculate𝑬
not immediately obvious how to do so. In particular, 𝚽𝑬 is the total flux through
closed surface, so its value depends on the magnitude and direction of the electric
field at all points on the surface.
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Learning Competency
Solve Problems Involving Electric Charges Dipoles, Forces, Fields and Flux in Context
such as but not limited to systems of point charges electrical breakdown of air charges,
electric breakdown of air charged pendulums, electrostatic ink set printers.
(STEM_GP12EM-IIIb-14)
Activity #1: KEY TERMS
Directions: Use each term/phrase in a sentence.
Key Terms
Sentence
1. Conductor
2. Electric Charge
3. Electric Field
4. Electrostatics
5. Induced charges
6. Induction
7. Insulator
8. Lines of force
Activity #2: CRITICAL THINKING
Directions: Answer the following questions comprehensively. (5 points each)
1.
When glass rod is rubbed with silk, the rod becomes positively charged, but when a
rubber rod is rubbed with fur, the rubber becomes negatively charged. Suppose you
have a charged object but don’t know whether its charge is positive or negative.
Explain how you could use a glass rod and piece of silk to determine the sign of the
unknown charge on the object.
2.
Two point charges are separated by a distance r. If the separation is reduced by a
factor of 1.5, by what factor does the electric force between them change?
3.
If there are more electric field lines leaving a Gaussian surface than entering, what can
you conclude about the net charge enclosed by that surface?
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4.
A point charge is placed at the center of an uncharged metallic spherical shell insulated
from ground. As the point charge is moved off center, describe what happens to
(a) the total induced charge on the shell and
(b) the distribution of charge on the interior and exterior surfaces of the shell.
5.
Two solid spheres, both of radius R, carry identical total charges, Q. One sphere is a
good conductor while the other is an insulator. If the charge on the insulating sphere
is uniformly distributed throughout its interior volume, how do the electric fields outside
these two spheres compare? Are the fields identical inside the two spheres?
Activity #3: BE AN EXPERT!
Directions: Solve the following problems accurately withy complete solutions. (5 points
each)
1. A positively charged particle with Q= 5 𝜇C is placed between two negatively
charged particles with q1= 1 𝜇C (left) and q2=9 𝜇C (right). The distance between
q1 and q is 5 cm and the distance from q to q2 is 9 cm. What is the total force
acting on the middle particle? Find the value and the direction.
2. A small particle with mass m=1 mg and positive charge Q=1 𝜇C is placed just near
the ground. What should be the surface charge density on the ground to keep the
particle above it in a stationary position? Assume that the ground is a nonconductor.
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3. What is the magnitude of a point charge that would create an electric field of 1.00
N/C at points 1.00 m away?
4. A 4.0 cm2 in the x-y plane sits in a uniform electric field E = (2.0 i + 3.0 j + 5.0 k)
N/C. Find the electric flux through the square.
5. A long copper wire with radius of 1.0 mm carries a uniform surface charge density
of 5.0 x 10-6 C/m2.
c. Find the total charge in a 1.0-meter-long section of the wire.
d. Find the magnitude of the electric field at a distance of 15 cm from the wire.
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PERFOMANCE TASK: Electric Fields in a Smokestack Scrubber
Problem: Smokestack scrubber removes undesirable particles by first adding some
excess electrons and then using electric forces to pull the particles out of the air. Consider
a soot particle of mass 𝑚𝑠𝑜𝑜𝑡 = 1.0 𝑥 10−12 𝑘𝑔 that travels upward in a smokestack and a
charge of 𝑞𝑠𝑜𝑜𝑡 = 1.1 𝑥 10−17 𝐶. Assume the electric field in the scrubber is produced by
two parallel, square plates of width 𝐿 = 1.0 𝑚 and separation 𝑑 = 0.010 𝑚, with charges
±𝑄.
(a) What must be the value of the electric field between the plates so that the force on
the soot particle is equal to the weight of the particle? (A real scrubber would use
a collection of many pairs of such plates in parallel)
(b) What charge Q on the scrubber’s plates is required to produce the electric field in
part (a)?
Recognize the Principle (How will you find the magnitude of E) (5 points)
Sketch the Problem (5 points)
Identify the Relationships in part (a) (5 points) –“The formula”
Solution for part (a) (5 points)
Identify the Relationships in part (b) (5 points) –“The formula”
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Solution for part (b) (5 points)
Conclusion: (5 points)
REFLECTION
1.I learned that ________________________________________________________
_______________________________________________________________
_______________________________________________________
2.I enjoyed most on ____________________________________________________
_______________________________________________________________
_______________________________________________
3.I want to learn more on ________________________________________________
_______________________________________________________________
_______________________________________________
96
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References:
Padua, Alicia L. et. al, 2003, States of Equilibrium, Practical and Explorational
Physics: Modular Approach, pp. 244-254.
2018. General Physics 2, Rex Book Store, Inc. pages 2-40.
97
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Answers key
Learning Activity #1: KEY TERMS
Answers may vary
Learning Activity #2: CRITICAL THINKING
1. When a glass rod is rubbed with silk, the rod becomes positively charged and as
2.
3.
4.
5.
the sum of the total charge must be zero this means that silk has become
negatively charged.
The force between two charges depends upon distance. According to
the Coulomb's law the forces between two charges is inversely proportional to the
distance between the charges.
More field lines leaving a surface than coming in means a net positive charge
within the surface. Remember, E-field lines originate on positive charges and
terminate on negative charges.
a. The total induced charge -- the sum of the charge on the interior and on the
exterior -- will be zero.
b. The distribution of the charges will change but there will be the same amount
on the exterior as on the interior -- but of the opposite sign.
Outside the radius R, the electric fields are identical. Inside the radius R, the
electric field inside the conductor is zero. Inside the radius R, the electric field
inside the conductor increases linearly with radius from zero at the center.
Learning Activity #3: BE AN EXPERT!
1. 3.2 x 107 N, right
2. 1.75 x 10-13 C/m2
3. 1.1 x 10-10 C
4. 8.0 𝑥 10−3 𝑁.
5. A. 3.1 𝑥 10
−8
𝑚2
𝐶
𝑁
𝐶, B. 3.8 𝑥 103 𝐶
Prepared by:
JOLLY MAR D. CASTANEDA
Baggao National Agricultural School Sta. Margarita Annex
98
NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
ELECTRIC POTENTIAL
Background Information for the Learners (BIL)
What is Electric Potential?
Electric field lines "flow" from positive charges to negative charges. A positive charge is
like an open faucet and a negative charge is like an open drain. Anyone with a working
sink can make a crude model of an electric dipole in their kitchen or bathroom with the
flick of the wrist. Similar analogies exist for wind, heat, and dissolved substances.
Think for a moment, of the other things that flow and think of what causes them to flow.
This will be the answer to our next conceptual problem. Let us set up a table that
compares similar phenomena. In all cases, there will be something that flows and
something that causes the flow.
the flow of…
A river
(a liquid water)
the wind
(atmospheric gases)
heat
(internal energy)
dissolved substances
(solutes)
is caused by a difference in…
altitude
atmospheric pressure
temperature
concentration
In each case, the thing that is flowing can be described by a vector field (a quantity that
has magnitude and direction at any location) and the thing that causes the flow can be
described by a difference in a scalar field (a quantity that has magnitude only at any
location).
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the flow of…
is caused by a difference in…
a vector field
a scalar field
The "flow" of the electric field is "caused" by a difference in electric potential.
the flow of…
is caused by a difference in…
electric field (test charges)
electric potential
The electric potential, or voltage, is the difference in potential energy per unit charge
between two locations in an electric field.
When a charged particle moves in an electric field, the field exerts a force that can do
work on the particle. This work can always be expressed in terms of electric potential
energy. Just as gravitational potential energy depends on the height of a mass above the
earth’s surface, electric potential energy depends on the position of the charged particle
in the electric field.
Work and Potential Energy
When a free positive charge q is accelerated by an electric field, it is given kinetic energy
(Figure 1). The process is analogous to an object being accelerated by a gravitational
field, as if the charge were going down an electrical hill where its electric potential energy
is converted into kinetic energy, although of
course the sources of the forces are very
different.
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, −∆𝑈 = ∆𝐾. Work is done by a
force, but since this force is conservative, we can
write 𝑊 = −∆𝑈.
https://phys.libretexts.org/@api/deki/files/8098/CNX_UPhysics
_24_01_PotEnWork.jpg?revision=1
A difference in electric potential gives rise to an
electric field. The electric field is the force per charge acting on an imaginary test charge
at any location in space. The work done placing an actual charge in an electric field gives
the charge electric potential energy. (This concept is called the work-energy theorem.)
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By the transitive property, electric potential gives rise to electric potential energy; and by
the reflexive property, the electric potential is the energy per charge that an imaginary
test charge has at any location in space.
Derivation of the Equation
Start from the work-energy theorem. When work is done (𝑊), energy changes (∆𝐸).
𝑾 = ∆𝑬
(Equation 1)
More specifically, when work is done against the electric force (𝐹̅𝐸 ), electric potential
energy changes (∆𝑈𝐸 ). Recall that work is force times displacement (𝑑). There's a bar
over the force symbol to indicate that we will be using the average value. This is one of
the limitations of derivations done without calculus.
̅ 𝑬 𝒅 = ∆𝑼𝑬
𝑭
(Equation 2)
̅ 𝑬 𝒅 = 𝟏 ∆𝑼𝑬
𝑭
(Equation 3)
Divide both sides by charge (𝑞).
𝟏
𝒒
𝒒
Rearrange things a bit.
̅𝑬
𝑭
𝒒
𝒅=
∆𝑼𝑬
(Equation 4)
𝒒
The ratio of force to charge on the left is called electric field (𝐸). The only thing that is
changed is we are dealing with average values right now. The ratio of energy to charge
on the right is called electric potential (𝑉).
̅=
𝑬
̅𝑬
𝑭
(Equation 5)
𝒒
∆𝑽 =
∆𝑼𝑬
𝒒
(Equation 6)
The electric field is the force on a test charge divided by its charge for every location in
space. Because it is derived from a force, it is a vector field. The electric potential is the
101
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electric potential energy of a test charge divided by its charge for every location in space.
Because it is derived from an energy, it is a scalar field. These two fields are related.
The electric field and electric potential are related by displacement. Field times
displacement is potential…
̅ 𝒅 = ∆𝐕
𝑬
(Equation 7)
or field is potential over displacement if you prefer.
̅ = ∆𝑽
𝑬
𝒅
(Equation 8)
Quiz Time!
Multiple choices: Choose the letter of the correct answer that suits to the given question.
1. It is the potential energy per unit charge.
a. Gravity
c. Electric potential
b. Electric field
d. Electric force
2. A change in potential energy of a charge moved from one point to another,
divided by the charge; units are joules per coulomb.
a. Work
c. Potential energy
b. Electric field
d. Potential difference
3. In what direction will an object accelerate when released with initial velocity
upward?
a. Downward only if the ratio of the g to initial velocity is large enough.
b. Upward or downward depending on its mass.
c. Downward
d. Upward
4. As a proton moves in the direction the electric field lines...
a. it is moving from low potential to high potential and gaining electric
potential energy.
b. it is moving from high potential to low potential and gaining electric
potential energy.
c. it is moving from low potential to high potential and losing electric
potential energy.
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d. it is moving from high potential to low potential and losing electric
potential energy.
5. Electric potential ____________ as distance increases.
a. increases
c. decreases
b. remain constant
d. zero
6. The direction of electric field lines shows the
a. direction of the force on a test positive charge.
b. strength of the field.
c. size of the field.
d. all of the above
7. The difference in electrical potential energy between two places is called_____
a. Voltage
c. Electric field
b. Electric potential energy
d. Work
8. What is voltage?
a. The amount of electric potential energy per one coulomb of charge
b. The number of volts per coulomb
c. The amount of 1 coulomb of charge per unit of potential energy
d. The amount of electric potential energy per volt
9. If the electrical potential energy between two equal charges quadruples,
describe the change in the distance between the particles.
a. The distance was quadrupled.
b. The distance was halved.
c. The distance was not changed.
d. The distance was quartered.
10. Which of the following is not true regarding electric potential?
a. The positive terminal of a battery has higher electric potential than the
negative terminal.
b. Electric potential can be expressed with units of Volts or Joules per
Coulomb.
c. When a positive charge moves from a region of low potential to high
potential, the electric field does positive work on the charge.
d. A negative charge moving from low potential to high potential will
accelerate.
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Learning Competency
Relate the electric potential with work, potential energy and electric field
(STEM_GP12EM-IIIb-15)
Activity 1: THE CAUSE TO FLOW!
Directions: Give at least 4 examples of a phenomenon and what causes it flow.
the flow of…
is caused by a difference in…
1. Water in a hose
pressure
2.
3.
4.
5.
Activity 2: IT IS COMPLICATED!
Directions: Identify the relationship of the following paired quantities.
1. Electric Potential and Work done
2. Electric Potential and Electric Potential Energy
3. Displacement/distance and Electric Potential
4. Electric Field Strength and Electric Potential
5. Electric/Electrostatic Force and amount of Electric Charge
Activity 3: CHOOSE YOUR BET!
Directions: Choose one (1) from the panels you want to perform or answer.
What object do you associate
Make a slogan consisting of the
Potential difference? Why
words; Electric potentials, Work,
(Essay)
Energy
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Compose a short song with a
Make a short poem about Electric
familiar tune about Electric
potential, Work and Energy
potential.
REFLECTION
1.I learned that________________________________________________________
_______________________________________________________________
_______________________________________________________
2.I enjoyed most on ____________________________________________________
_______________________________________________________________
_______________________________________________
3.I want to learn more on ________________________________________________
_______________________________________________________________
_______________________________________________
105
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REFERENCES
D. (2018a, August 22). Work and Electric Potential Energy. Retrieved February 13,
2021,
from
https://howtomechatronics.com/learn/electricity/work-electric-potential-
energy/
Elert, G. (2015, February 24). Electric Potential –. Retrieved February 14, 2021, from
https://physics.info/electric-potential/
Khan Academy. (2015, September 9). Electric potential (article). Retrieved February 14,
2021,
from
https://www.khanacademy.org/test-prep/mcat/physical-
processes/electrostatics-1/a/electric-potential
Libretexts. (2020a, November 5). 6.2: Electric Flux. Retrieved January 20, 2021, from
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physi
cs_(OpenStax)/Map%3A_University_Physics_II__Thermodynamics_Electricity_and_Magnetism_(OpenStax)/06%3A_Gauss’s_Law/6.0
2%3A_Electric_Flux
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ANSWER KEY:
Quiz Time
2. C
2. D
3. C
4. D
5. C
6. A
7. A
8. A
9. D
10. C
Activity 1 Answers vary
Activity 2
1. The potential difference between two points in an electric field is defined as
amount of work done in moving a unit charge from one point to other point.
Potential difference or Electric Potential = Work Done/ Quantity of Charge
moved. (Direct proportion)
2. Electric Potential is directly proportional to the electric potential Energy. In this
way, the electric potential at any point in the electric field is the electric potential
energy of a unit positive charge at that point.
3. Moving towards and away from the charge results in change of potential; the
relationship between distance and potential is inverse.
4. The relationship between electric potential and field is similar to that between
gravitational potential and field in that the potential is a property of the field
describing the action of the field upon an object.
5. Electrostatic/electric force is directly related to the charge of each object. So, if
the charge of one object is doubled, then the force will become two times
greater.
Activity 3 Answers vary
Prepared by:
JOHN DAVID B. MEDRANO
APARRI EAST NATIONAL HIGH SCHOOL
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GENERAL PHYSICS 2
Name: ___________________________________ Grade Level: _____________
Date: ____________________________________ Score: __________________
LEARNING ACTIVITY SHEET
ELECTRIC POTENTIAL OF CONTINUOUS CHARGE DISTRIBUTION
Background Information for the Learners (BIL)
Introduction to Continuous Charge Distribution
With the help of Coulomb’s Law and Superposition Principle, we can easily
find out the electric field due to the system of charges or discrete system of
charges. The word discrete means every charge is different and has the existence
of its own. Suppose, a system of charges having charges as q 1 , q 2, q 3……. up to
q n. We can easily find out the net charge by adding charges algebraically and net
electric field by using the principle of superposition. This is because:
•
Discrete system of charges is easier to solve
•
Discrete system of charges do not involve calculus in calculations
Image 1: A system in which charge is distributed over a conductor, is
called
continuous
charge
distribution
system
taken
from
https://www.askiitians.com/iit-jee-electrostatics/continuous-charge-distribution/
If the charge distribution is continuous, the potential at a point P can be
found by summing over the contributions from individual differential elements of
charge dq.
Consider the charge distribution shown in Image 1. Taking infinity as our reference
point zero potential, the electric potential at P due to dq is
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𝑑𝑉 =
1 𝑑𝑞
4𝜋𝜀0 𝑟
Summing over contributions from all differential elements, we have
𝑉=
1
𝑑𝑞
∫
4𝜋𝜀0
𝑟
But how to calculate electrostatics terms in continuous charge system? For
an example if there is a rod with charge q, uniformly distributed over it and we wish
to find the electric field at some distance ‘r’ due it. It would be illogical and irrelevant
to simply add electric field using principle of superposition as the charge is
uniformly distributed over the rod. So we take a small element of the ro d and
integrate it with proper limits.
We consider element, based on how density of charge is centered on the
material or object. If the charge is uniformly distributed over the surface of the
conductor, then it is called Surface Density. If the charge varies linearly along the
length of the conductor, then it is called Linear Charge Density. And if the charge
changes with volume of the conductor, then it is called Volume Charge Density.
What is Continuous Charge Distribution?
Image 2: Types of Charge Distribution
The continuous charge distribution system is a system in which the charge
is uniformly distributed over the conductor. In continuous charge system, infinite
numbers of charges are closely packed and have minor space between them.
Unlikely from the discrete charge system, the continuous charge distribution is
uninterrupted and continuous in the conductor. There are three types of the
continuous charge distribution system.
•
Linear Charge Distribution
•
Surface Charge Distribution
•
Volume Charge Distribution
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Linear Charge Density
When the charge is non-uniformly distributed over the length of a conductor, it is
called linear charge distribution. It is also called linear charge density and is
denoted by the symbol λ (Lambda).
Mathematically linear charge density is
𝜆=
𝑑𝑞
𝑑𝑙
The unit of linear charge density is C/m. If we consider a conductor of length ‘L’
with surface charge density λ and take an element dl on it, then small charge on it
will be
𝑑𝑞 = 𝜆 𝑑𝑙
So, the electric field on small charge element dq will be
𝑘𝑑𝑞
𝑟2
𝑘𝜆𝑑𝑙
𝑑𝐸 = 2
𝑟
𝑑𝐸 =
To calculate the net electric field we will integrate both sides with proper limit, that
is
𝐿
∫ 𝑑𝐸 = ∫
0
𝑘𝜆𝑑𝑙
𝑟2
𝑘 𝐿
∫ 𝑑𝐸 = 2 ∫ 𝜆 𝑑𝑙
𝑟 0
Surface Charge Density
When the charge is uniformly distributed over the surface of the conductor,
it is called Surface Charge Density or Surface Charge Distribution. It is denoted
by the symbol σ (sigma) symbol and is the unit is C / m 2.
It is also defined as charge/ per unit area. Mathematically surface charge
density is
𝜎=
𝑑𝑞
𝑑𝑠
where dq is the small charge element over the small surface ds.
The electric field due to small charge at some distance ‘r’ can be evaluated as
𝑘𝑑𝑞
𝑟2
𝑘𝜎𝑑𝑠
𝑑𝐸 = 2
𝑟
𝑑𝐸 =
Integrating both sides with proper limits we get
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𝑠
∫ 𝑑𝐸 = ∫
0
𝑘𝜎𝑑𝑠
𝑟2
𝑘 𝑠
∫ 𝑑𝐸 = 2 ∫ 𝜎 𝑑𝑠
𝑟 0
Volume Charge Density
When the charge is distributed over a volume of the conductor, it is
called Volume Charge Distribution. It is denoted by symbol ρ (rho). In other
words charge per unit volume is called Volume Charge Density and its unit is
C/m 3. Mathematically, volume charge density is
𝑑𝑞
𝑑𝑣
where dq is small charge element located in small volume dv. To find total charge
we will integrate dq with proper limits. The electric field due to dq will be
𝜌=
𝑑𝑞 = 𝜌 𝑑𝑣
𝑘𝑑𝑞
𝑟2
𝑘 𝜌 𝑑𝑣
𝑑𝐸 =
𝑟2
𝑑𝐸 =
Integrating both sides with proper limits we get
𝑣
∫ 𝑑𝐸 = ∫
0
𝑣 𝜌 𝑑𝑣
𝑟2
𝑘 𝑣
∫ 𝑑𝐸 = 2 ∫ 𝜌 𝑑𝑣
𝑟 0
Examples:
1. ELECTRIC POTENTIAL OF A LINE CHARGE
An infinite line charge has linear charge density = −1.mC 0.2 Calculate the electric
potential at a point on a line perpendicular to the line charge, at a distance of 3.0 m from
the line charge. Assume that the electric potential of the line charge is zero at the
perpendicular distance of 4.0 m.
SOLUTION: Note that the potential difference between two points a and b due to an
infinite line charge is given as
𝑣𝑎 − 𝑣𝑏 =
𝜆
𝑟𝑎
ln ( )
2𝜋𝜀0
𝑟𝑏
We get:
10−6 𝐶
3.0 𝑚
𝑚
) = 𝟏. 𝟎𝟗𝟑 𝒙 𝟏𝟎𝟒 𝑽
𝑣𝑎 − 0 =
ln (
−12
10 𝐹
4.0 𝑚
2 𝑥 3.14 𝑥 (8.85 𝑥 𝑚 )
2.0 𝑥
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2. ELECTRIC POTENTIAL DUE TO CHARGED CONDUCTING SPHERE
Two charged spherical conductors of radius r1 = 0.8 cm and r2 = 0.2 cm are separated
by a distance much larger than 10 cm. These spheres are connected by a conducting
wire and a total of 60 nC charge is placed on one of the spheres. (a) Calculate the charge
on each sphere. b) Calculate the electric potential of each sphere at a point on their
surfaces.
SOLUTION: Since the charged conducting sphere is connected through a conducting
wire to the uncharged sphere, the 60 nC charge will redistribute between the two sphere
in such a manner so that both sphere have same electric potential. Let the final charge
be q1 (on the larger sphere) and q2 on the smaller sphere.
a.
1
𝑞1
4𝜋𝜀0 𝑟1
=
1
𝑞2
4𝜋𝜀0 𝑟2
𝑟
= 𝑞2 = 𝑟2 𝑞1
1
𝑞1 +
𝑟2
𝑞 = 60 𝑛𝐶
𝑟1 1
𝑟1
80 𝑐𝑚
) 𝑥 60 𝑛𝐶 =
𝑞1 = (
𝑥 60 𝑛𝐶 = 𝟒𝟖 𝒏𝑪
𝑟1 + 𝑟2
10 𝑐𝑚
𝑞2 = (60 𝑛𝐶 − 48 𝑛𝐶) = 𝟏𝟐 𝒏𝑪
So,
b. 𝑉1 =
1
48 𝑛𝐶
4𝜋𝜀0 (8.0 𝑥 10−2 𝑚)
= 5.4 𝑘𝑉
𝑉2 =
1
12 𝑛𝐶
4𝜋𝜀0 (2.0 𝑥 10−2 𝑚)
= 5.4 𝑘𝑉
Learning Competency:
Determine the electric potential function at any point due to highly symmetric
continuous charge distribution (STEM_GP12EM-IIIc-17)
Learning Activity #1: FACT or BLUFF
Directions: State TRUE if the statement is correct and BLUFF if otherwise.
1. When the charge is uniformly distributed over the surface of the conductor,
it is called Volume Charge Density.
2. When the charge is non-uniformly distributed over the length of a conductor,
it is called non-linear charge distribution.
3. In general, for determining the electric potential of a continuous charge
distribution, we first calculate the potential due to a small element of the
charge distribution and then integrate this expression over appropriate limits
to include the effect of total charge in it.
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4. Charge per unit volume is called Volume Charge Density.
5. In continuous charge system, infinite numbers of charges are closely packed
and have minor space between them.
Learning Activity #2: MATCH ME!
Directions: Match the items from column a to its corresponding formula in Column B.
Column A
1. Linear Charge Density
Column B
A. 𝑑𝑉 =
2. Surface Charge Density
B. 𝑑𝐸 =
3. Volume Charge Density
C. 𝑣𝑎 − 𝑣𝑏 = 2𝜋𝜀 ln (𝑟𝑎 )
1
𝑑𝑞
4𝜋𝜀0 𝑟
𝑘 𝜌 𝑑𝑣
𝑟2
𝜆
𝑟
0
4. Electric Potential of a line charge
D. 𝑑𝐸 =
5. Continuous Charge distribution
E. 𝑑𝐸 =
𝑏
𝑘𝜎𝑑𝑠
𝑟2
𝑘𝜆𝑑𝑙
𝑟2
Learning Activity #3: PROBLEM SOLVING
Directions: Solve the following problems accurately with complete solution. (5 points
each)
1. The linear charge density of an infinite line charge is 0.3 10 .m C − −16  Assuming
that the electric potential at a perpendicular distance of 5.0 m from the wire is zero,
calculate the potential at the perpendicular distance of 6.0 m.
2. The radius and surface charge density of a uniformly charged spherical shell are
20 cm and ,m C0.3 −2  respectively. Calculate the electric potential at a distance
(a) 40 cm and (b) 15 cm from the center of the shell.
3. An isolated solid sphere of aluminum having radius 7.0 cm is at a potential of 500
V. Calculate the number of electrons which have been removed from the sphere
to raise it to this potential.
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REFLECTION
1.I learned that ________________________________________________________
_______________________________________________________________
___________________________________________________________
2.I enjoyed most on ____________________________________________________
_______________________________________________________________
_______________________________________________
3.I want to learn more on ________________________________________________
_______________________________________________________________
_______________________________________________
114
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References:
Padua, Alicia L. et. al, 2003, States of Equilibrium, Practical and Explorational
Physics: Modular Approach, pp. 244-254.
2018. General Physics 2, Rex Book Store, Inc. pages 2-40.
https://www.askiitians.com/iit-jee-electrostatics/continuous-charge-distribution/
115
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Answers key
Learning Activity #1: KEY TERMS
1.
2.
3.
4.
5.
BLUFF
BLUFF
FACT
FACT
FACT
Learning Activity #2: CRITICAL THINKING
1.
2.
3.
4.
5.
E
D
B
C
A
Learning Activity #3: Problem Solving
1. Vp = -9.8 x 103 V
2. a. V = 8.4 x 104 V , b. 6.7 x 104 V
3. n = 2.4 x 108
Prepared by:
JOLLY MAR D. CASTANEDA
Baggao National Agricultural School Sta. Margarita Annex
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GENERAL PHYSICS 1
Name: ________________________________
Date: ______________
Grade :________________________________
Score: _____________
LEARNING ACTIVITY SHEETS
ELECTRIC FIELD
Background Information for the Learners
Electric
field
strength
is a
vector
quantity; it has both magnitude and direction.
The magnitude of the electric field strength is
defined in terms of how it is measured. Let's
suppose that an electric charge can be
denoted by the symbol Q. This electric charge
Figure 1
creates an electric field; since Q is the source
of the electric field, we will refer to it as the source
https://www.google.com/url?sa=i&ur
charge. The strength of the source charge's
l=https%3A%2F%2Fwww.physicsclassr
electric field could be measured by any other
oom.com%2Fclass%2Festatics%2FLess
charge placed somewhere in its surroundings. The
on-4%2FElectric-Field.
charge that is used to measure the electric field
strength is referred to as a test charge since it is used to test the field strength. The test
charge
has
a
of charge denoted
symbol q.
When
quantity
by
the
placed
within the electric field, the test charge will experience an electric force - either attractive
or repulsive. As is usually the case, this force will be denoted by the symbol F. The
magnitude of the electric field is simply defined as the force per charge on the test charge.
or
The standard metric units on electric field strength arise from its definition. Since
electric field is defined as a force per charge, its units would be force units divided by
charge units. In this case, the standard metric units are Newton/Coulomb or N/C.
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In the above discussion, you will
note that two charges are mentioned the source charge and the test charge.
Two charges would always be necessary
to encounter a force. In the electric world,
it takes two to attract or repel. The
equation for electric field strength (E) has
one of the two charge quantities listed in
it. Since there are two charges involved,
we will have to be ultimately careful to use the correct charge quantity when computing
the electric field strength. The symbol q in the equation is the quantity of charge on the
test charge (not the source charge). Recall that the electric field strength is defined in
terms of how it is measured or tested; thus, the test charge finds its way into the equation.
Electric field is the force per quantity of charge on the test charge.
A new equation that defines electric field strength in terms of the variables that
affect the electric field strength. To do so, we will have to consider the Coulomb's law
equation. Coulomb's law states that the electric force between two charges is directly
proportional to the product of their charges and inversely proportional to the square of
the distance between their centers. When applied to our two charges - the source charge
(Q) and the test charge (q) - the formula for electric force can be written as:
If the expression for electric force as given by Coulomb's law is substituted for force in
the above E =F/q equation, a new equation can be derived as shown below.
118
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The Direction of the Electric Field Vector
The magnitude of the electric field vector is calculated as the force per charge on
any given test charge located within the electric field. The force on the test charge could
be directed either towards the source charge or directly away from it. The precise
direction of the force is dependent upon whether the test charge and the source charge
have the same type of charge (in which repulsion occurs) or the opposite type of charge
(in which attraction occurs). To resolve the dilemma of whether the electric field vector
is directed towards or away from the source charge, a convention has been established.
The worldwide convention that is used by scientists is to define the direction of the
electric field vector as the direction that a positive test charge is pushed or pulled
when in the presence of the electric field. By using the convention of a positive test
charge, everyone can agree upon the direction of E. Given this convention of a positive
test charge, several generalities can be made about the direction of the electric field
vector. A positive source charge would create an electric field that would exert a repulsive
effect upon a positive test charge. Thus, the electric field vector would always be directed
away
from
http://www.physicsclassroom.com/Class/estatics/u8l4c1.gif
charged
positively
objects.
On
the other hand, a positive test charge would be attracted to a negative source charge.
Therefore, electric field vectors are always directed towards negatively charged objects.
Figure 2
http://www.physicsclassroom.com/Class/estatics/u8l4c2
.gif
119
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We can represent electric potentials (voltages) pictorially, just as we drew pictures
to illustrate electric fields. Of course, the two are related., which shows an isolated
positive point charge and its electric field lines. Electric field lines radiate out from a
positive charge and terminate on negative charges. While we use blue arrows to
represent the magnitude and direction of the electric field, we use green lines to
represent places where the electric potential is constant. These are
called equipotential lines in two dimensions, or equipotential surfaces in three
dimensions. The term equipotential is also used as a noun, referring to an
equipotential line or surface. The potential for a point charge is the same anywhere
on an imaginary sphere of radius r surrounding the charge.
Figure 3. An isolated point charge Q with its electric field lines in blue and equipotential
lines in green. The potential is the same along each equipotential line, meaning that no
work is required to move a charge anywhere along one of those lines. Work is needed to
move a charge from one equipotential line to another. Equipotential lines are
perpendicular
to
electric
field
lines
in
every
case.
https://nigerianscholars.com/assets/uploads/2018/10/Figure_20_04_01a.jpg
It is important to note that equipotential lines are always perpendicular to electric
field lines. No work is required to move a charge along an equipotential, since ΔV = 0.
Thus the work is:
W = −ΔPE = −qΔV = 0.
Work is zero if force is perpendicular to motion. Force is in the same direction as
E, so that motion along an equipotential must be perpendicular to E. More precisely,
work is related to the electric field by:
W = Fd cos θ = qEd cos θ = 0.
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Note that in the above equation, E and F symbolize the magnitudes of the electric field
strength and force, respectively. Neither q nor E nor d is zero, and so cos θ must be 0,
meaning θ must be 90º. In other words, motion along an equipotential is perpendicular
to E.
One of the rules for static electric fields and conductors is that the electric field
must be perpendicular to the surface of any conductor. This implies that a conductor is
an equipotential surface in static situations. There can be no voltage difference across
the surface of a conductor, or charges will flow. One of the uses of this fact is that a
conductor can be fixed at zero volts by connecting it to the earth with a good
conductor—a process called grounding. Grounding can be a useful safety tool. For
example, grounding the metal case of an electrical appliance ensures that it is at zero
volts relative to the earth.
These are some points to consider when dealing with electric field:
•
•
•
•
Electric field lines always extend from a positively charged object to a negatively
charged object, from a positively charged object to infinity, or from infinity to a
negatively charged object.
Electric field lines never cross each other.
Electric field lines are most dense around objects with the greatest amount of
charge.
At locations where electric field lines meet the surface of an object, the lines are
perpendicular to the surface.
Learning Competency
Infer the direction and strength of electric field vector, nature of the electric field sources,
and electrostatic potential surfaces given the equipotential lines STEM_GP12EMIIIc-18
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ACTIVITY 1: TRUE OR FALSE
Directions: Write TRUE if the statement is correct and FALSE if the statement is
incorrect on the space provided before the number.
______1. Electric field is a scalar quantity.
______2. Electric field doesn’t have direction.
______3. The unit for charge is coulomb.
______4. There’s an attraction and repulsion between charges.
______5. There are two types of charges: positive and negative.
______6. Like charges attract.
______7. Unlike charge repel.
______8. Attraction between charges depends upon their nature.
______9. Source of electric field is the test charge.
______10. Newton/Coulomb is the unit for electric field.
ACTIVITY 2: PROBLEM SOLVING
Directions: Solve the following problems
1. What is the electric field due to a point of 15μC at a distance of 1 meter away
from it?
2. What is the electric field strength at a distance of 10 cm from a charge of 2 μC?
3. What is the electric field strength at a distance of 130 cm from a charge of 5.2 μC?
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123
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ACTIVITY 3: CROSSWORD PUZZLE
Directions: Supply the missing word either horizontally or vertically to complete the
empty boxes. Used the meaning of the words below to guide you in answering the
crossword puzzle.
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REFLECTION
1.I learned that ________________________________________________________
_______________________________________________________________
___________________________________________________________
2.I enjoyed most on ____________________________________________________
_______________________________________________________________
_______________________________________________
3.I want to learn more on ________________________________________________
_______________________________________________________________
_______________________________________________
125
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REFERENCES
“Ch. 1 Introduction to Science and the Realm of Physics, Physical Quantities, and Units
- College Physics.” Accessed February 14, 2021.
https://openstax.org/books/college-physics/pages/1-introduction-to-science-andthe-realm-of-physics-physical-quantities-and-units.
“Work and Energy | Physics Library | Science.” Khan Academy. Khan Academy.
Accessed February 14, 2021.
https://www.khanacademy.org/science/physics/work-and-energy.
“Electric Charge.” Electric charge - Energy Education. Accessed January 23, 2021.
https://energyeducation.ca/encyclopedia/Electric_charge.
126
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ANSWER KEY
ACTIVITY 1: TRUE OR FALSE
1.
2.
3.
4.
5.
F
F
T
T
T
6. F
7. F
8. T
9. F
10. T
ACTIVITY 2: PROBLEM SOLVING
1. 3.6 x 104 N/C.
2. 1.8 X 105 N/C.
3. 1.35 X 105 N/C.
ACTIVITY 3: CROSSOWRD PUZZLE
1. Attraction
2. Repulsion
3. Work
4. Force
5. Equipotential
6. Vector
7. Test Charge
8. Source Charge
9. Positive
10. Negative
Prepared by:
CHARLES C. DAQUIOAG
Sanchez Mira School of Arts and Trades
127
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GENERAL PHYSICS 2
Name: ____________________________
Date: _____________________________
Grade Level: ______________
Score: ____________________
LEARNING ACTIVITY SHEET
Calculting Electric Field in the Region
Background Information for the Learners (BIL)
An electric field exists in the region of space around a charged object or a source
charge. When another charged enters this electric field, it will experience and electric
force. The strength of the electric field will simply refer to electric field intensity. Electric
field is defined as the force that a test charge will experience when placed at that point.
Physicists use a unit positive charge as the test charge in defining an electric field. This
test charge and the electric field are usually represented by q0 and E.
The electric field produced by a point source charge q can be obtained using
Coulomb’s law. The electric field at any point is given by the equation:
E = FE / q0
where E is the electric field, FE is the electric force, and q0 is the test charge.
To calculate the electric field at any point at a disatnce r in space from a point
charge q, imagine a test charge q0 placed at that point. The magnitude of the electric
force on q0 is:
FE = k |qq0| / r2
Thus, the magnitude of the electric field due to the point charge is;
E = FE / q0 = k |q| / r2
It follows that E has the unit of newton/coulomb (N/C). Like electric force, electric
field is also a vector quantity and has the same direction as the electric force on a positive
charge placed at a point. The electric field also follows the superposition pronciple.
Sample Problems:
1. Calculate the magnitude and direction of the electric field 0.45 m from a +7.85x10 -9 C
point charge.
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Given: q = 7.85 x 10-9 C
r = 0.45 m
Solution:
E = k |q| / r2
= [ 9 X 109 N·m2/C2 ] [ (+7.85x10-9 C) / (0.45 m)2 ]
= 348.89 N/C = 350 N/C
2. An electron is placed on a uniform electric field that is directed downward and has a
magnitude of 5 N/C. Find the magnitude and direction of the force experienced by the
electron.
Given: E = 5 N/C, directed downward
q = 1.602 x 10 -19
Solution:
The direction of the force is upward because the electron is negatively charged.
Only the absolute value of the charge of the electron is considered.
FE = E |q|
= (5 N/C) | -1.602 X 10-19 C |
= 8 X 10-19 N
LEARNING COMPETENCY:
Calculate the electric field in the region given a mathematical function describing its
potential in a region of space (STEM_GP12EMIIIc-20)
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Activity 1: Try To Solve Me!
Directions: Solve the following problems systematically. Write your solutions on the
space provided.
1. A tiny ball weighs 0.0055 kg and carries a charge of +3.25 x 10-6 C. What electric field
(magnitude and direction) is needed for the ball to remain suspended in air?
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
_________________________________________________
2. Calculate the magnitude and direction of the electric field 0.50 m from a +6.25 x 10 -9
C point charge.
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
_________________________________________________
Activity 2: Science Connections
Directions: Answer the following problems logically. Write your answers on the space
provided.
1. Relate the concept of electric fields to the sixth sense of sharks and rays.
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
https://images.app.goo.gl/cS8hHJGcWZvpSHQn9
130
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Activity 3: Fact or Bluff
Directions: Write FACT if the statement is true, otherwise write BLUFF if it is false. Write
your answer on the space before each item.
___________1. The higher the electric field, the higher the electric force.
___________2. The electric field is directly proportional to the value of the test
charge.
___________3. The magnitude of the electric field due to the point charge can be
calculated using the formula, E= FE q0
___________4. Electric force has a unit of N/C
___________5. Electric force is a vector quantity.
REFLECTION:
1. I learned that ________________________________________________________
_____________________________________________________________________
_________________________________________________________________
2. I enjoyed most on_____________________________________________________
_____________________________________________________________________
_________________________________________________________________
3. I want to learn more on_________________________________________________
_____________________________________________________________________
_________________________________________________________________
131
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REFERENCES
https://images.app.goo.gl/cS8hHJGcWZvpSHQn9
Silverio,Angelina.”Exploring Life Through Science Series: General Physics 2.” In
Teachers Wraparound Edition. Quezon City, Phoenix Pulishing House, Inc., 2017
132
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ANSWER KEY
ACTIVITY 1:
1. 1.66 x 104 N/C, directed upward.
2. 112.50 N/C
ACTIVITY 2:
1. Sharks and rays have the ability to detect electric fields in their surroundings. This
ability is called electroreception and is considered sharks and rays’ sixth sense.
Sharks and rays can locate prey hidden beneath the sand of the bottom of the ocean
because of the electric field generated by the muscle contraction of the prey.
ACTIVITY 3:
1. F
2. B
3. B
4. B
5.F
Prepared by:
Kimberly Anne C. Pagdanganan
Licerio Antiporda Sr. National High School - Dalaya Extension
133
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GENERAL PHYSICS 1
Name: ________________________________
Grade: ________________________________
Date: ______________
Score: ____________
LEARNING ACTIVITY SHEETS
ELECTRIC POTENTIAL ENERGY AND ELECTRIC POTENTIALS
ELECTRIC POTENTIAL ENERGY
Background Information for the Learners
Electric potential energy is possessed by an object by the virtue of two elements,
those being, the charge possessed by an object itself and the relative position of an object
with
respect
to other electrically charged objects. The magnitude of electric
potential depends 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 some amount of energy
which is defined as the electric potential energy. For any charge, the electric potential is
obtained by dividing the potential energy by the quantity of charge. When an electrostatic
force acts between two or more charged particles within a system of particles, we can
assign an electric potential energy U to the system. If the system changes its
configuration from an initial state i to a different final state f, the electrostatic force does
work W on the particles. We then know that the resulting change ∆U in the potential
energy of the system is
∆U =Uf - Ui= -W.
(Equation 1)
As with other conservative forces, the work done by the electrostatic force is path
independent. Suppose a charged particle within the system moves from point i to point f
while an electrostatic force between it and the rest of the system acts on it. Provided the
rest of the system does not change, the work W done by the force on the particle is the
same for all paths between points i and f.
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Sample Problem 1:
Electrons are continually being knocked out of air
molecules in the atmosphere by cosmic-ray particles
coming in from space. Once released, each electron
experiences an electrostatic force due to the electric field
that is produced in the atmosphere by charged particles
already on Earth. Near Earth’s surface the electric field has
the magnitude E=150 N/C and is directed downward. What
Figure 1 The image shows
is the change ∆U in the electric potential energy of a
the direction of movement
released electron when the electrostatic force causes it to
of the electron.
move vertically upward through a distance d=520 m?
The change U in the electric potential energy of the electron is related to the work
W done on the electron by the electric field. (∆U = -W) gives the relation.
An electron in the atmosphere is moved upward through displacement by an
electrostatic force due to an electric field E.
A. The work done by a constant force F on a particle undergoing a displacement d
is:
W = F x d (Equation 2)
B. The electrostatic force and the electric field are related by the force equation F=q
E:, where here q is the charge of an electron (-1.6x10-19 C)
W= qE x d = qEd cos θ
W = (-1.6x10-19 C) (150 N/C) (520 m) cos 180⁰
W = 1.248 X 10 -14 J
C. This result tells us that during the 520 m ascent, the electric potential energy of
the electron decreases by 1.248 X 10 -14 J.
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Sample Problem 2:
An electron is accelerated from rest through a potential difference 12 V. What is
the change in electric potential energy of the electron?
Given:
Electron (e) = -1.60 x 10-19 Coulomb
Electric potential = voltage (V) = 12 Volt
Unknown: The change in electric potential energy of the electron (ΔPE)
ΔPE = q V
= (-1.60 x 10-19 C)(12 V)
= -19.2 x 10-19 Joule
ELECTRIC POTENTIALS
The potential energy of a charged particle in an electric field depends on the
charge magnitude. However, the potential energy per unit charge has a unique value at
any point in an electric field. For an example of this, suppose we place a test particle of
positive charge 1.60 x 10-19 C at a point in an electric field where the particle has an
electric potential energy of 2.40 x 10-19 J.
Thus, the potential energy per unit charge, which can be symbolized as U/q, is
independent of the charge q of the particle we happen to use and is characteristic only
of the electric field we are investigating. The potential energy per unit charge at a point
in an electric field is called the electric potential V (or simply the potential) at that point
(Equation 3)
The electric potential difference ∆V between any two points i and f in an electric field is
equal to the difference in potential energy per unit charge between the two points:
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(Equation 4)
Using Equation 1 to substitute –W for ∆U in Equation 4, we can define the potential
difference between points i and f as:
(Equation 5)
The potential difference between two points is thus the negative of the work done by the
electrostatic force to move a unit charge from one point to the other. A potential difference
can be positive, negative, or zero, depending on the signs and magnitudes of q and W.
The SI unit for electric potential is the joule per coulomb. This combination occurs so
often that a special unit, the volt (abbreviated V), is used to represent it. Thus,
1 volt = 1 joule per coulomb
Finally, we can now define an energy unit that is a convenient one for energy
measurements in the atomic and subatomic domain: One electron-volt (eV) is the energy
equal to the work required to move a single elementary charge e, such as that of the
electron or the proton, through a potential difference of exactly one volt. The magnitude
of this work is q ∆V; so
1 eV = e(1 V)
= (1.60 x 10-19 ) (1 J/C)
Learning Competency
Solve problems involving electric potential energy and electric potentials in contexts such
as, but not limited to, electron guns in CRT TV picture tubes and Van de Graaff
STEM_GP12EMIIIc-22
137
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ACTIVITY 1- PROBLEM SOLVING
Directions: Answer the following set of problem. The scoring is being provided for you
before the questions.
Given
Solution
Final Answer
2 pts
5 pts
3 pts
Sample Problem: An electron is accelerated from rest through a potential
difference 24 V. What is the change in electric potential energy of the electron?
Known :
The charge on an electron (e) = -1.60 x 10-19 Coulomb
Electric potential = voltage (V) = 12 Volt
Wanted: The change in electric potential energy of the electron (ΔPE)
Solution :
ΔPE = q V
= (-1.60 x 10-19 C)(12 V)
= -19.2 x 10-19 Joule
The minus sign indicates that the potential energy decreases.
1. Mr. Aguban is checking the difference in energy between a car battery and a
motorcycle battery in moving a certain amount of charge. A 12 V motorcycle
battery can move 6,000 C of charge, and a 12 V car battery can move 80,000 C
of charge. How much energy does each deliver?
2. If 10 J of work is needed to shift 25 C of charge from one place to another. The
potential difference between the places should be ?
3. Mr. Usabal uses two charged parallel plates and try to calculate the potential
energy between the plates. The separation between the plates is 5 cm and the
magnitude of the electric field between the plates is 650 Volt/meter. What is the
change in potential energy of the proton when accelerated from the positively
charged plate to the negatively charged plate.
4. The SMSAT automotive team is assembling a car for the school and they are
trying to work with different watts of headlight on a 24 V battery car. When a
24 V car battery runs a single 50 W headlight, how many electrons pass through
it each second?
138
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ACTIVITY 2 - TRUE OR FALSE
Directions: Write the word TRUE if the statement is correct and FALSE if the statement
is incorrect. Write your answer on the space provided before the number.
_____1. Electron is a positively charge particle.
_____2. The direction of electric potential depends on the amount of work done in
moving the object from one point to another against the electric field.
_____3. The work done by the electrostatic force is path independent.
_____4. The potential energy of a charged particle in an electric field depends on the
charge magnitude.
_____5. The unit for electric potential is Volts.
_____6. A potential difference can be positive, negative, or zero.
_____7. Volt is different unit from joule per coulomb.
_____8. When an object is moved against the electric field it loss some amount of
energy.
_____9. The charge possessed by an object itself is sometimes referred to as the
potential energy.
_____10. Proton is a negatively charged particle.
REFLECTION
1. I learned that ___________________________________________________
_______________________________________________________________
_______________________________________________________
2. I enjoyed most on _______________________________________________
_______________________________________________________________
_______________________________________________
3. I want to learn more on ___________________________________________
_______________________________________________________________
_______________________________________________
139
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REFERENCES
Admin. “Electric Potential Energy - Formula, Definition, Solved Examples.” BYJUS.
BYJU'S, February 26, 2021. https://byjus.com/jee/electric-potential-energy/.
https://physics.gurumuda.net/electric-potential-energy-problems-and-solutions.htm
“Electric Charge.” Electric charge - Energy Education. Accessed January 23,
2021. https://energyeducation.ca/encyclopedia/Electric_charge.
Halliday, David, Resnick, Robert, & Walker, Jearl. Fundamentals of Physics. 6th ed.New
York: John Wiley & Sons Inc, 2001
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ANSWER KEY
ACTIVITY 1 – Problem Solving
1. (– 7.2 X 104) and (9.6 X 105)
3. 32.5 V and 5.2 X 10-18
2. (– 7.2 X 104) and (9.6 X 105)
4. 1.302 x 1019
ACTIVITY 2 – True or False
1.
3.
5.
7.
9.
F
T
T
F
T
2.
4.
6.
8.
F
T
T
T
F
Prepared by:
CHARLES C. DAQUIOAG
Sanchez Mira School of Arts and Trade
141
NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 2
Name: ____________________________
Date: _____________________________
Grade Level: ______________
Score: ____________________
LEARNING ACTIVITY SHEET
Deducing the Effects of Capacitors
Background Information for the Learners (BIL)
Capacitors
One important element in an electric circuit is a capacitor. A capacitor is a device
for storing charges. The standard symbols for a capacitor are shown in figure 2-6.
There are several types of capacitors. One of the simplest types of capacitors
consists of two equally but oppositely charged parallel conducting plates separated from
each other by a thin sheet of insulating material or dielectric. When connected
to a source of charge, such as baterry, the positive terminal of the source removes
electrons from the plate connected to it and transfers them to the other plate. As a result,
the two plates are equally but oppositely charged. Figure 2-7 shows the basic parts of a
parallel plate capacitor.
A capacitor is usually named after the dielectric material used. Common dielectric
materials used in a capacitor are mica, glass, air, ceramic and paper.
(Source: Silverio,Angelina.”Exploring Life Through Science Series: General Physics 2.” In Teachers Wraparound Edition.
Quezon City, Phoenix Pulishing House, Inc., 2017)
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Capacitance
Capacitance is the ability of a capacitor to store charges. The capacitance C of a
capacitor is mathematically defined as the ratio of the amount of charge q in one plate to
the potential difference V between the plates. In symbols,
C = q/v
The SI unit of capacitance is the farad (F) nmaed after Michael Faraday. Note that
1 farad is equal to 1 coulumb per volt.
The capacitance of a parallel plate capacitor is affected by the following factors:
1. The area of plates. The bigger the area of the plates, the greater the capacitance.
2. The distance between the plates. The closer the plates to each other, the greater
the capacitance.
3. The insulating material or dielectric between them. The capacitance is determined
in terms of the material’s permittivity constant
є
– the higher the
є,
the greater
the capacitance.
The dependence of the capacitance of a parallel plate capacitor on the factors cited
above is mathematically expressed as:
C = є A/d
where A is the area of one plate, d is the
distance between the plates, and є is
the permittivity of the inslutaing material
or dielectric. Table 2-1 lists the permittivity
of some common dielectrics. Sometimes,
the relative permittivity or dielectric constant
of the dielectric material is given instead of
its permittivity. The relative permativitty or
dielectroc constant єR is the ratio of
the permittivity є of the dielectric to the
permittivity є0 of a vacuum or air.
єR
=
є/ є0
Noter that єR has no unit. Also, the relative permittivity is greater than or equal to
one.
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Rewriting the Eq. (2.8) using Eq. (2.9),
C = є A/d = єR є0 A/d
Note that C0 = є0A/d (capacitance with air or vacuum as the dielectric). Therefore,
C = єR C0
Inserting a dielectric other than air or vacuum increases the capacitance to an
amount equal to єR times its original value.
There is a limit to the potential difference between the plates of the capacitor.
When the maximum potential difference is exceeded, the dielectric becomes a conductor,
allowing the flow of charges. These movinf charges form sparks or discharge. This
condition is known as dielectric breakdown. Lightning is an example
of a dielectric breakdown.
Fig 2-10. Clouds act like a huge capacitor in the sky
with air as the dielectric. The upper portion of clouds .
is positivly charged and the lower portion is negatively
charged. The dielectric breakdown of air results to lightning .
(Source: https://images.app.goo.gl/giCK9Qp7P1G549vZA)
LEARNING COMPETENCY:
Deduce the effects of simple capacitors (e.g. parallel plate, spherical, cylindrical) on the
capacitance, charge and potential difference when the size, potential difference or charge
is changed. (STEM_GP12EM-IIId-23)
144
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Activity 1: Solve it!
Directions: Solve the following problems systematically. Write your answer on the space
provided.
1. A capacitor consists of two square metal plates, each measuring 5.00x10 -2 m on a
side. In between the plates is a sheet of mica measuring 1.00x10 -4 m thick. (a) What
is the capacitance of this capacitor? If the charge in one plate is 2.00x10 -8 C, what is
the (b) potential difference and (c) electric field between the plates?
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
_________________________________________________
2. The capacitance of a parallel plate air capacitor is 350.0 µF. When a sheet of a
dielectric is inserted between the plates, the capacitance increases to 2100.0 µF.
What is the permittivity of the dielectric?
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
_________________________________________________
145
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Activity 2: Concept Test
Directions: Answer the following questions logically. Write your answers on the space
provided.
1. Two identical parallel plate capacitors are shown in an end-view in Figure A. Each
has a capacitance of C. If the two are joined together at the edges as in Figure B,
forming a single capacitor, what is the final capacitance? What will happen to its area?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________
Activity 3: Fact or Bluff
Directions: Write FACT if the statement is true, otherwise write BLUFF if it is false.
1. A capacitor is a device for releasing charges.
2. One of the simplest types of capacitors consists of two equally but oppositely charged
perpendicular conducting plates separated from each other by a thin sheet of
insulating material
3. A capacitor is usually named after the dielectric material used.
4. The bigger the area of the plates, the lesser the capacitance.
5. The closer the plates to each other, the greater the capacitance.
146
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Activity 4: Picture Analysis
Directions: Identify which factor affects the capacitance of a parallel plate capacitor in
each situation. Write a short explanation on how each factor affects its capacitance.
1. _______________________________
_______________________________
_______________________________
_______________________________
_______________________________
_______________________________
2. _______________________________
_______________________________
_______________________________
_______________________________
_______________________________
_______________________________
3. _______________________________
_______________________________
_______________________________
_______________________________
_______________________________
_______________________________
REFLECTION:
1. I learned that ______________________________________________________
_____________________________________________________________________
_________________________________________________________________
2. I enjoyed most on___________________________________________________
_____________________________________________________________________
_________________________________________________________________
3. I want to learn more on_______________________________________________
_____________________________________________________________________
_________________________________________________________________
147
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REFERENCES
https://images.app.goo.gl/Y1bvB8gRrTGpmJpi8
https://images.app.goo.gl/mqgnj4sDgCtwBCsU6
https://images.app.goo.gl/iuVaZSaAWDZodzwL9
Silverio,Angelina.”Exploring Life Through Science Series: General Physics 2.” In
Teachers Wraparound Edition. Quezon City, Phoenix Pulishing House, Inc., 2017
148
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ANSWER KEY
ACTIVITY 1:
1. a. 1.2x10-9 F
b. 16.7 V
c. 16.7x105 V/m
ACTIVITY 2:
2. 2C. Area is doubled
ACTIVITY 3:
1. BLUFF
2. BLUFF
3. FACT
4. BLUFF
5. FACT
ACTIVITY 4:
1. The distance between the plates. The closer the plates to each other, the greater the
capacitance.
2. The area of plates. The bigger the area of the plates, the greater the capacitance.
3. The insulating material or dielectric between them. The capacitance is determined in
terms of the material’s permittivity constant є – the higher the є, the greater the
capacitance.
Prepared by:
Kimberly Anne C. Pagdanganan
Licerio Antiporda Sr. National High School - Dalaya Extension
149
NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 1
Name: ________________________________
Grade: ________________________________
Date: ______________
Score: ____________
LEARNING ACTIVITY SHEETS
CAPACITORS IN SERIES AND IN PARALLEL
Background Information for the Learners
When there is a combination of capacitors in a circuit, we can sometimes replace
that combination with an equivalent capacitor—that is, a single capacitor that has the
same capacitance as the actual combination of capacitors. With such a replacement, we
can simplify the circuit, affording easier solutions for unknown quantities of the circuit.
CAPACITORS IN SERIES
“In series” means that the capacitors are wired serially, one after the other, and
that a potential difference V is applied across the two ends of the series. The total
capacitance is less than any one of the series capacitors’ individual capacitances. If two
or more capacitors are connected in series, the overall effect is that of a single
(equivalent) capacitor having the sum total of the plate spacings of the individual
capacitors. An increase in plate spacing, with all other factors unchanged, results in
decreased capacitance.
When a potential difference V is applied across several capacitors connected in
series, the capacitors have identical charge q.The sum of the potential differences across
all the capacitors is equal to the applied potential difference V.
We can explain how the capacitors end up with identical charge by following a
chain reaction of events, in which the charging of each capacitor causes the charging of
the next capacitor. We start with capacitor 3 and work upward to capacitor 1. When the
battery is first connected to the series of capacitors, it produces charge -q on the bottom
plate of capacitor 3. That charge then repels negative charge from the top plate of
capacitor 3 (leaving it with charge +q).The repelled negative charge moves to the bottom
plate of capacitor 2 (giving it charge -q). That charge on the bottom plate of capacitor 2
then repels negative charge from the top plate of capacitor 2 (leaving it with charge +q)
to the bottom plate of capacitor 1 (giving it charge -q). Finally, the charge on the bottom
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plate of capacitor 1 helps move negative
charge from the top plate of capacitor 1 to the
battery, leaving that top plate with charge +q.
Here
are
two
important
points
about
capacitors in series:
1. When charge is shifted from one
capacitor to another in a series of
capacitors, it can move along only one
route, such as from capacitor 3 to
capacitor 2 in Fig. 1. If there are
additional routes, the capacitors are
not in series.
2. The battery directly produces charges on only the two plates to which it is
connected (the bottom plate of capacitor 3 and the top plate of capacitor 1 in Fig.
1). Charges that are produced on the other plates are due merely to the shifting of
charge already there. Charges that are produced on the other plates are due
merely to the shifting of charge already there. For example, in Fig. 1, the part of
the circuit enclosed by dashed lines is electrically isolated from the rest of the
circuit. Thus, the net charge of that part cannot be changed by the battery— its
charge can only be redistributed.
When we analyze a circuit of capacitors in series, we can simply say;
Capacitors that are connected in series can be
replaced with an equivalent capacitor that has the
same charge q and the same total potential
difference V as the actual series capacitors.
To derive an expression for Ceq in Fig. 1, we first use
equation on potential difference for each capacitor.
Figure
1.(a)
Three
capacitors
connected in series to battery B.The
battery
maintains
potential
difference V between the top and
bottom
plates
combination.
of
(b)The
the
series
equivalent
capacitor, with capacitance Ceq,
replaces the series combination.
The total potential difference V due to the battery is the sum of these three potential
differences.Thus,
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The equivalent capacitance is then; or
CAPACITORS IN PARALLEL
In figure 2, shows an electric
circuit in which three capacitors are
connected in parallel to battery B.This
description has little to do with how the
capacitor plates are drawn. Rather, “in
parallel” means that the capacitors are
directly wired together at one plate and
directly wired together at the other plate,
and that the same potential difference V
is applied across the two groups of
wired-together
plates.
Thus,
each
capacitor has the same potential difference V, which produces charge on the capacitor.
(In Fig. 2, the applied potential V is maintained by the battery.) In general, When a
potential difference V is applied across several capacitors connected in parallel, that
potential difference V is applied across each capacitor.The total charge q stored on the
capacitors is the sum of the charges stored on all the capacitors.
We can simply say that capacitors connected in parallel can be replaced with an
equivalent capacitor that has the same total charge q and the same potential difference
V as the actual capacitors.
Figure 2b shows the equivalent capacitor (with equivalent capacitance Ceq) that has
replaced the three capacitors (with actual capacitances C1,C2, and C3) of Fig. 2a.
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Figure 2. (a) Three capacitors connected
To derive an expression for Ceq in Fig.
in parallel to battery B.The battery
2b, we first use Eq. q=CV to find the
maintains potential difference V across its
charge on each actual capacitor:
terminals and thus across each capacitor.
(b)
q1 = C1V, q2 = C2V, and q3 = C3V.
The
equivalent
capacitor,
with
capacitance Ceq, replaces the parallel
combination.
The total charge on the parallel combination of Fig. 2a is then;
q = q1 = q2 = q3 = (C1 + C2 + C3)V.
The equivalent capacitance, with the same total charge q and applied potential difference
V as the combination, is then;
SAMPLE PROBLEM:
A. The Grade 12 Binggas is working with capacitance they connected three
capacitances in series to find the how well they hold charges. Find the total
capacitance for three capacitors connected in series by the stem students given
their individual capacitances to be 1 µF, 5 µF, and 8 µF.
Given: C1=1 µF
C2=5 µF
C3=8 µF
1
CTotal
=
CTotal = ??
1
1 µF
+
1
5 µF
+
1
8 µF
=
1
1.325
CTotal = 0.755 µF
B. Five capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 6 μF, C4 = 5 μF, C5 = 10 μF, are
connected in series and parallel. Determine the capacitance of a single capacitor
that will have the same effect as the combination.
Given:
C1 = 2 μF
C2 = 4 μF
C5 = 10 μF
C3 = 6 μF
C4 = 5 μF
Ceq= ???
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First; we are going to get the capacitance of C2 and C3 that are connected in
parallel. The equivalent capacitance;
C2,3 = C2 + C3
C2,3 = 4 μF + 6 μF
C2,3 = 10 μF
Next, we are going to used the formula in series connection to get the equivalent
capacitance of the circuit.
1
1
CTotal
CTotal =
=
2
1
+
10
1
+
5
1
+
10
9
=
10
1.1 μF
Learning Competency
Calculate the equivalent capacitance of a network of capacitors connected in
series/parallel STEM_GP12EM-IIId-24
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ACTIVITY 1: LOOP A WORD
DIRECTIONS: Look for the different concepts regarding the topic capacitance in series
and parallel on the pool of letters below.
D
V
C
P
V
S
R
T
P
S
F
A
L
N
E
R
B
A
C
A
P
A
C
I
T
O
R
I
R
A
T
H
R
C
A
A
R
A
C
E
I
E
I
R
L
C
D
I
S
V
S
R
A
L
U
T
T
O
O
D
V
C
E
I
C
A
R
L
S
F
A
L
T
F
N
E
T
F
T
A
V
S
E
C
H
A
R
G
E
B
D
Q
E
S
G
C
S
R
K
B
A
T
T
E
R
Y
P
O
1. Charge
6. Battery
2. Capacitance
7. Circuit
3. Series
8. Combination
4. Parallel
9. Voltage
5. Capacitor
10. Plates
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ACTIVITY 2: SOLVE
DIRECTIONS: Answer the following set of problem. The scoring is being provided for
you before the questions.
The link below will redirect you to the websites that contains a sample problem with
solution.
https://www.physicstutorials.org/home/electrostatics/capacitors-in-series-and-parallel.
Given
Solution
Final Answer
2 pts
5 pts
3 pts
1. Mr. Catenza, the electronics teacher of SMSAT is explaining the
capacitance through the used of circuit for his students. The circuit that he
showed them is given below having the capacitance as follow; C1=60µF,
C2=20 µF, C3=9 µF and C4=12 µF respectively. Calculate the total
capacitance of the capacitor in the circuit.
C3
C4
A
B
C1
C
C2
2. Five capacitors, C1 = 3 μF, C2 = 6 μF, C3 = 8 μF, C4 = 7 μF, C5 = 12 μF,
are connected in series and parallel. Determine the capacitance of a single
capacitor that will have the same effect as the combination.
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ACTIVITY 3: TRUE OR FALSE
DIRECTIONS: Write True if the statement is correct and False if the statement is
incorrect on the space provided before the number.
________1. In parallel connection different potential difference V is applied across the
two groups of wired-together plates. False
________2. Capacitors that are connected in series cannot be replaced with an
equivalent capacitor that has the same charge q and the same total potential difference
V as the actual series capacitors. False
________3.
________4.
Unit for capacitance is V. False
“In parallel” means that the capacitors are directly wired together at one
plate and directly wired together at the other plate, and that the same potential difference
V is applied across the two groups of wired-together plates. True
________5.
In series connection, the total capacitance is less than any one of the
series capacitors’ individual capacitances. True.
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REFLECTION
1. I learned that ___________________________________________________
_______________________________________________________________
_______________________________________________________
2. I enjoyed most on _______________________________________________
_______________________________________________________________
_______________________________________________
3. I want to learn more on ___________________________________________
_______________________________________________________________
_______________________________________________
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ANSWER KEY
ACTIVITY 1 – CROSSWORD PUZZLE
D
V
C
P
V
S
R
T
P
S
F
A
L
N
E
R
B
A
C
A
P
A
C
I
T
O
R
I
R
A
T
H
R
C
A
A
R
A
C
E
I
E
I
R
L
C
D
I
S
V
S
R
A
L
U
T
T
O
O
D
V
C
E
I
C
A
R
L
S
F
A
L
T
F
N
E
T
F
T
A
V
S
E
C
H
A
R
G
E
B
D
Q
E
S
G
C
S
R
K
B
A
T
T
E
R
Y
P
O
ACTIVITY 2 – PROBLEM SOLVING
1. 8 μF
2. 1.584 Μf
ACTIVITY 2 – TRUE OR FALSE
1.
2.
3.
4.
FALSE
FALSE
TRUE
TRUE
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REFERENCES
https://physics.gurumuda.net/electric-potential-energy-problems-and-solutions.htm
“Electric Charge.” Electric charge - Energy Education. Accessed January 23,
2021. https://energyeducation.ca/encyclopedia/Electric_charge.
Halliday, David, Resnick, Robert, & Walker, Jearl. Fundamentals of Physics. 6th
ed.New York: John Wiley & Sons Inc, 2001
Administrator. “Capacitors in Series and Parallel.” Accessed February 26, 2021.
https://www.physicstutorials.org/home/electrostatics/capacitors-in-series-andparallel.
“Capacitors in Series and Parallel – Problems and Solutions: Solved Problems in
Basic Physics.” Gurumuda.Net, November 13, 2018.
https://physics.gurumuda.net/capacitors-in-series-and-parallel-problems-and-
Prepared by:
CHARLES C. DAQUIOAG
Sanchez Mira School of Arts and Trade
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GENERAL PHYSICS 2
Name: ____________________________
Date: _____________________________
Grade Level: _____________
Score: __________________
LEARNING ACTIVITY SHEET
Combination of Capacitors
Background Information for the Learners (BIL)
Capacitors may be connected in series or in parallel. Figure 2-11 shows these
connections.
(Source: Silverio,Angelina.”Exploring Life Through Science Series: General Physics 2.” In Teachers Wraparound
Edition. Quezon City, Phoenix Pulishing House, Inc., 2017)
Referring to figure 2-11a, the series combination of capacitors is characterized
by only one path for charge transfer through terminals A and B. All the series
capacitors acquire the same charge. The charges in each capacitor are equivalent,
and are all equal to the total charge in the combination. But because they have
different capacitances, the potential differences between the plates of the capacitor
are different. In summary, the following relationships apply for capacitors in series.
a. Charge: qtotal = q1 = q2 =q3 = ....= qn
b. Potential differences:
Vtotal = V1 + V2 + V3 + .... + Vn
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c. Capacitance:
Using Eq. (2.7) and the above relationships between charges and volatages,
For parallel capacitors, there are several paths for the transfer of
charges through the voltage terminals A and B. Since the capacitors are
connected to the same terminals A and B, then the potential differences
between their plates are equivalent, and are equal to Vtotal.
In summary, the following relationships apply for capacitors in parallel.
a. Charge:
qtotal = q1 + q2 + q3 + .. +qn
b. Potential difference:
Vtotal = V1 = V2 = V3 = ... Vn
c. Capacitance:
CtotalVtotal = C1V1 + C2V2 +C3V3 + ..CnVn
Ctotal = C1 + C2 + C3 + ... +Cn
LEARNING COMPETENCY:
Determine the total charge, the charge on, and the potential difference across each
capacitor in the network given the capacitors connected in series/parallel .
(STEM_GP12EM_IIId-25)
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Activity 1: Try To Solve Me!
Directions: Solve the following problems systematically. Write your answers on the
space provided.
1.A parallel plate capacitor is made up of two plates, each having an area of 8.0x10 4
m2 and separated from each other by 5.0mm. Half of the space between the plates
is filled with glass and the other with mica. Find the capacitance of this capacitor.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
______
2.Two capacitors with 2.0 F and 3.0 F capacitance, respectively, are connected in
series and subjected to a total potential difference of 100V. Find the (a) total
capacitance, (b) charge stored in each capacitor, and (c) potential difference across
each capacitor.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
______
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Activity 2: Brain Twister
Directions: Answer the following problems logically. Write your answers on the
space provided.
1.If you wish to store a large amount of energy in a capacitor bank, would you connect
capacitors in series or parallel? Explain.
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
Activity 3: Choose The Best
Directions: Choose the letter of the correct answer. Write your answer on the space
before each item.
_________1. In the below figure for capacitors connected in parallel, what is the
correct statement and formula?
a. The Total charge will be the sum of the
charges in each of three capacitors
QT=Q1+Q2+Q3
b. Charge in each capacitor is the same
QT=Q1=Q2=Q3
c. Charge in each capacitor is different but current in each capacitor are the
same
d. The total charge will be QT=Q1- Q2 - Q3
_________2. Two capacitors C1=5 μF and C2= 2 μF are connected in parallel to a
20 V DC power supply as shown in the figure given below: Calculate the total
capacitance of the circuit?
a. 0.7 μF
b. 1.42 μF
c. 7 μF
d. 3 μF
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_________3. Which of the following is true for capacitors connected together in
parallel?
a. All capacitors in parallel have the same current through them
b. All capacitors in parallel store the same amount of charge
c. All capacitors in parallel must have the same capacitance
d. All capacitors in parallel have the same voltage across their plates
_________4. Which of the following is true for capacitors connected in series?
a. Capacitors in series all store the same amount of charge
b. Capacitors in series all have the same voltage across their plates
c. Capacitors in series all have the same capacitance
d. Capacitors in series must have a lower capacitance than capacitors in
parallel
_________5. What is the total capacitance when two capacitors C1 and C2 are
connected in series?
a. (C1+C2)/C1C2
c. C1C2/(C1+C2)
b. 1/C1+1/C2
d. C1+C2
Activity 4: Fact or Bluff
Directions: Write FACT if the statement is true, otherwise write BLUFF if it is false.
Write your answer on the space before each item.
___________1. If you connect two different capacitors in series and charge them up,
both of them will have equal voltage.
___________2. Capacitors in parallel have a higher total value than any individual
capacitor.
___________3. To find the total capacitance of capacitors in parallel, sum the values
of the individual capacitors.
___________4. In a series capacitive circuit the smallest capacitor will have the
largest voltage drop across it.
___________5. When a capacitor is charging, current will flow through its dielectric.
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REFLECTION:
1. I learned that
__________________________________________________________________
__________________________________________________________________
________________________________________________________________
2. I enjoyed most on__________________________________________________
__________________________________________________________________
__________________________________________________________________
__
3. I want to learn more on______________________________________________
__________________________________________________________________
__________________________________________________________________
__
166
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REFERENCES
https://quizizz.com/admin/quiz/5dbd4405efadc6001b54a183/pop-quiz-series-andparallel-capacitors
https://www.indiabix.com/electronics/capacitors/126002
Silverio,Angelina.”Exploring Life Through Science Series: General Physics 2.” In
Teachers Wraparound Edition. Quezon City, Phoenix Pulishing House, Inc., 2017
167
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ANSWER KEY
ACTIVITY 1:
2. a. 19.4X10-12 F
3. a. 1.2 F
b. 120 C
c. V cross 2.0 F capacitor is 60 V
V cross 3.0 F capacitor is 40 V
ACTIVITY 2:
3. Parallel.
ACTIVITY 3:
2. A
2. A
3. B
4. A
5.C
2. F
3. F
4. F
5.B
ACTIVITY 4:
1. B
Prepared by:
Kimberly Anne C. Pagdanganan
Licerio Antiporda Sr. National High School - Dalaya Extension
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Capacitors in Series and Parallel
Part 2
Background Information for the Learners (BIL)
A capacitor is a device used to store electrical charge and electrical energy. It
consists of at least two electrical conductors separated by a distance. The space
between capacitors may simply be a vacuum, and, in that case, a capacitor is then
known as a “vacuum capacitor.” However, the space is usually filled with an insulating
material known as a dielectric. The amount of storage in a capacitor is determined
by a property called capacitance.
The capacitance value of a
capacitor is measured in farads
(F), units named for English
physicist Michael Faraday (1791–
Figure 01
1867).
Michael Faraday was an English
scientist who contributed to the
study of electromagnetism and
electrochemistry.
A farad is a large quantity of capacitance. Most household electrical devices
include capacitors that produce only a fraction of a farad, often a thousandth of a
farad (or microfarad, µF) or as small as a picofarad (a trillionth, pF).
Supercapacitors, meanwhile, can store very large electrical charges of
thousands of farads.
How to increase capacitance
Capacitance can be increased when:
•
•
•
A capacitor's plates (conductors) are positioned closer together.
Larger plates offer more surface area.
The dielectric is the best possible insulator for the application.
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Capacitors have applications ranging from filtering static from radio reception
to energy storage in heart defibrillators. Typically, commercial capacitors have two
conducting parts close to one another but not touching, such as those in (Figure 2).
Most of the time, a dielectric is used between the two plates. When battery terminals
are connected to an initially uncharged capacitor, the battery potential moves a small
amount of charge of magnitude 𝑄 from the positive plate to the negative plate. The
capacitor remains neutral overall, but with charges
and residing on opposite
plates.
Both capacitors shown here were initially uncharged before being connected
to a battery. They now have charges of
and (respectively) on their plates. (a)
A parallel-plate capacitor consists of two plates of opposite charge with
area A separated by distance d. (b) A rolled capacitor has a dielectric material
between its two conducting sheets (plates).
Figure 2. Cross section of a capacitor
Capacitors with different physical characteristics (such as shape and size of
their plates. See Figure 3) store different amounts of charge for the same applied
voltage 𝑉 across their plates. The capacitance 𝐶 of a capacitor is defined as the ratio
of the maximum charge 𝑄 that can be stored in a capacitor to the applied voltage 𝑉
across its plates. In other words, capacitance is the largest amount of charge per volt
that can be stored on the device:
𝐶=
𝑄
𝑉
Figure 3. Capacitors of different sizes and shapes.
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ENERGY STORED IN CAPACITORS
The energy stored in a capacitor can be expressed in three ways:
2
2
Ecap= QV = CV = Q
2C
2
where Q 2is the charge,
V is the voltage, and C is the capacitance of the capacitor.
The energy is in joules for a charge in coulombs, voltage in volts, and capacitance
in farads.
DID YOU KNOW?
In a defibrillator, the delivery of a large charge in a short burst to a set of
paddles across a person’s chest can be a lifesaver. The person’s heart attack might
have arisen from the onset of fast, irregular beating of the heart—cardiac or
ventricular fibrillation. The application of a large shock of electrical energy can
terminate the arrhythmia and allow the body’s pacemaker to resume normal patterns.
Today it is common for ambulances to carry a defibrillator, which also uses an
electrocardiogram to analyze the patient’s heartbeat pattern. Automated external
defibrillators (AED) are found in many public places (Figure 4). These are designed
to be used by lay persons. The device automatically diagnoses the patient’s heart
condition and then applies the shock with appropriate energy and waveform. CPR is
recommended in many cases before use of an AED.
Figure 4. Automated external defibrillators are found in many public places. These portable units
provide verbal instructions for use in the important first few minutes for a person suffering a cardiac
attack. (credit: Owain Davies, Wikimedia Commons)
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Sample Problems:
A) A heart defibrillator delivers 4.00×102J of energy by discharging a capacitor initially
at 1.00×104V. What is its capacitance?
Solution:
We are given Ecap and V, and we are asked to find the capacitance C. Of the
three expressions in the equation for Ecap, the most convenient relationship is
Ecap =
CV2
2
Solving this expression for C and entering the given values yields
𝐶 = 2𝐸𝑐𝑎𝑝=
𝑉2
2(4.00 x 102 J) = 8.00 x 10-6 𝑭 or 8.00 𝒖𝑭
(1.00 x 104 𝑉)2
B) What is the capacitance of a capacitor that stores 12 μC of charge when
connected to a 6 V battery?
Solution:
𝐶 =
𝑄_
𝑉
=
12 μC
6V
= 2 μC
Learning Competency:
Determine the potential energy stored inside the capacitor given the
geometry and the potential difference across the capacitor (STEM_GP12EM-IIId26).
Activity 1: Warming Up!
Directions: Answer concisely the questions below.
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1. Look at the picture above. Which capacitor do you think will hold the most
charge? Why?
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________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
_____
2. Describe simply how a capacitor works. What can a capacitor be compared to?
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
_____
Activity 2 : Let’s Get Charged!
Directions: View the videos about capacitance, potential energy in capacitors, and
voltage on the internet and answer the guide questions that follow. Use the links
below.
•
•
https://www.youtube.com/watch?v=SIU_9SMd5q0
https://www.youtube.com/watch?v=u-jigaMJT10
Guide Questions:
A capacitor (parallel plate) is charged with a battery of constant voltage. Once
the capacitor reaches maximum charge, the battery is removed from the circuit.
1. Describe the charge on the plates if the plates were pushed closer together.
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
_____
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2. Describe what happens to the capacitance of the capacitor if both the plates were
moved closer together.
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
_____
3. Relate the voltage to that of the capacitance given both plates were moved closer
together in the capacitor.
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
________________________________________________________________
_____
Activity 3 : Problem Solving
Directions: Solve the following problems. Show all the steps you made in the
procedure. Use separate answer sheet(s) for your solutions.
1. It is desired that 5.8 μC of charge be stored on each plate of a 3.2- μC capacitor.
What potential difference is required between the plates?
2. A capacitor of 0.75 uF is charged to a voltage 16 V. What is the magnitude of the
charge on each plate of capacitor?
3. In a typical defibrillator, a 175-uF capacitor is charged until the potential difference
is 2240V.
(a) What is the magnitude of the charge on each plate of the final capacitor?
(b) find the energy stored in the charged-up defibrillator.
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4. In open heart surgery, a much smaller amount of energy will defibrillate the heart.
(a) What voltage is applied to the 8.00 μF capacitor of a heart defibrillator that
stores 40.0 J of energy?
(b) Find the amount of stored charge.
5. (a) What is the energy stored in the 10.0 μF capacitor of a heart defibrillator
charged to 9.00 × 103 V?
(b) Find the amount of stored charge.
REFLECTION:
1. I learned that
__________________________________________________________________
__________________________________________________________________
________________________________________________________________
2. I enjoyed most on__________________________________________________
__________________________________________________________________
__________________________________________________________________
__
3. I want to learn more on______________________________________________
__________________________________________________________________
__________________________________________________________________
__
175
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References
Ling, S.,Moebs, W. and Sanny, J. “Capacitors and Capacitance”. University
of Physics Vol. 2. Accessed January 21, 2021.
https://opentextbc.ca/universityphysicsv2openstax/chapter/capacitor
s-and-capacitance/#CNX_UPhysics_25_01_Battery
Mazur,G.A, “Digital Multimeter Principles”. American Technical Publishers.
Accessed February 1, 2021.
https://www.fluke.com/en-ph/learn/blog/electrical/what-iscapacitance#:~:text=The%20capacitance%20value%20of%20a,a%2
0large%20quantity%20of%20capacitance.
OpenStax College. “ Energy Stored in Capacitors”. College Physics.
Accessed January 21, 2021
: http://cnx.org/contents/031da8d3-b525-429c-80cf
6c8ed997733a/College_Physics.
Yenka, “ Capacitors Activity.” https://yenka.com/en Accessed February 1,
2021
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Answer Key
Activity 1. Warming Up!
1. E, because it is the biggest.
2. The capacitor fills up with charge, something like a bucket filling with water
(answers may vary)
Activity 2. Let’s Get Charged!
1. The charge deposited on the plates does not change when the battery is removed
and thus the charge and the charge density remains the same as the plates are
moved closer together.
2. Since the area of the capacitor is not changing, any decrease in plate separation
will cause an increase in the capacitance.
3. Because C= Q/V, and the charge does not change, an increase in capacitance
implies a decrease in voltage.
Activity 3. Problem solving
1. 1.8 V
2. 1.2 x 10-5 C
3. a) 0.392 C
b) 439 J
4. a) 3.16 kV
b) 25.3 mC
5. a) 405 J
b) 90.0 mC
Prepared by:
ALDRIN GRAGEDA
Pattao National High School
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GENERAL PHYSICS 2
Name: ___________________________
Date: ____________________________
Grade Level: __________
Score: _______________
LEARNING ACTIVITY SHEET
EFFECTS OF INSERTING DIELECTRIC MATERIAL ON THE
CAPACITANCE, CHARGE, AND ELECTRIC FIELD OF A
CAPACITOR.
Background Information for the Learners (BIL)
Before we proceed to discuss the effects of dielectric materials on the
capacitance, charge, and electric field of a capacitor we will review first the definition
of such terms to bridge us into a new concept.
Capacitor stores electrical energy in an electric field. It is created out of two
metal plates and an insulating material called dielectric. Metal plates are placed very
close each other, in parallel, but the dielectric sits between them to make sure they
don’t touch. The two metal plates (which act as the conductor) must carry equal
charges but opposite sign resulting to zero total charge of the capacitor. The dielectric
on the other hand is any insulating material that act as insulator and it serves three
purposes;
1. to keep the conducting plates from coming in contact, allowing for smaller
plate separations and therefore higher capacitances;
2. to increase the effective capacitance by reducing the electric field strength,
which means you get the same charge at a lower voltage; and
3. to reduce the possibility of shorting out by sparking (more formally known as
dielectric breakdown) during operation at high voltage.
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To visualize a capacitor, think of a sandwich with a spread in between. The two sliced
of loaf acts as metal plates while the spread as dielectric. The figures below illustrate
a capacitor.
www.google.com/search?q=capacitor+in++aplates&tbm...
Capacitance is the amount of charge a capacitor can store per unit of potential
difference and is always a positive quantity. It is always a positive quantity because
potential difference increases linearly with the stored charge.
It can further be defined as the ratio of the magnitude of the charge on either
conductor to the magnitude of the potential difference between the conductor, and in
equation form it can be expressed as;
We have already defined capacitor, dielectric and capacitance, we now
proceed to describe the effects of inserting dielectric materials on the capacitance,
charge, and electric field of a capacitor through some various examples.
For an instance, if we place a dielectric between two-parallel charged metal
plates with an electric field pointing from right to left. The positive nuclei of the
dielectric will move with the field to the right and the negative electrons will
move against the field to the left. Field lines start on positive charges and end on
negative charges, so the electric field within each stressed atom or molecule of the
dielectric points from left to right (as what is shown in the figure) — opposite the
external field from of the two metal plates. The electric field is a vector quantity and
when two vectors point in opposite directions you subtract their magnitudes to get
the resultant. The two fields don't quite cancel in a dielectric as they would in a metal,
so the overall result is a weaker electric field between the two plates.
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https://physics.info/dielectrics/#:~:text=Capacitance
Defining an electric field, it is the gradient of electric potential commonly known
as voltage and can be expressed in an equation form as;
And if we go back to the definition of capacitance, it is the ratio of charge to
voltage and expressed as;
Thus, introducing a dielectric into a capacitor decreases the electric field, which
decreases the voltage, which increases the capacitance.
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A capacitor with a dielectric store the same charge as one without a dielectric,
but at a lower voltage. Therefore, a capacitor with a dielectric in it is more effective.
To describe the effect of inserting dielectric materials on the capacitance
consider this example—an air is filled with 50 micro Farad capacitor has a voltage of
10V. An insulator with dielectric constant of 4.0 is inserted between the plates.
Calculate the capacitance, original charge and current charge on the capacitor given
that the new value for the voltage across is equal to 2.5V.
In dealing with any solving problem always start reading the problem with
utmost understanding while noting down the given and unknown of it.
For the problem above, know that the original capacitor has no dielectric
material between the two metals, only air with dielectric constant of approximately 1.
Other given in the problem are capacitance (which is equal to 50 micro Farad) and
the voltage across the plates (which is equal to 10V). But, if we insert a dielectric
material in between these two plates, what will happen to its capacitance? Will it
increase or decrease?
Going back to the problem, we know that the dielectric constant for the
insulator is 4. Giving us the value of “k” to be 4. The capacitance is directly
proportional to “k”, so if the value of “k” increases and so is the value of C, which in
this case the capacitance. To calculate the new capacitance (with dielectric inserted
between), multiply the original capacitance with the constant value of dielectric “k”
which is equal to 4 for this problem. The original capacitance was 50 micro farad and
the dielectric value is 4, we get 200 micro Farad for the value of new capacitance.
Notice that, the problem proved that, whenever the value of k increases, the value of
C also increases.
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To get the original charge of then capacitor, just multiply the original value of
capacitance and voltage. That is, 50 micro farad times 10V we get 500 micro
coulombs (that is the charge on the original capacitance). If we calculate the new
charge, we use the same formula as getting the original capacitance only that we will
change the values of the variables. The new capacitance is equal to 200 micro farad
while the new voltage is equal to 2.5V, multiplying these two we get the final value of
charge which is also equal to 500 micro Farad. Base on this result we can infer that,
increasing the dielectric will have no effect on the charge.
For the second example, we will describe the effect of dielectrics on the
electric field of a capacitor.
A capacitor is composed of metal plates separated by an air gap of 1mm has
a voltage of 15V. Calculate the electric field inside the capacitor and calculate the
new electric field if a material with dielectric constant of 3.0 is placed in between the
two metals.
To calculate the electric field is simply the voltage divided by the distance so,
15V over 0.001m which results to 15,000 V/m.
Moving on the second question, remember that if the value of dielectric “k”
increases the value of electric field decreases. To calculate the electric field, divide
the initial electric field with the value of “k” and that would be 15,000 over 3 resulting
to 5000V/m.
For the third example, consider the problem below followed by the complete
solution presented numerically.
A 40uF capacitor with an air gap of 2mm is connected across a 12V battery. Calculate
the charge, electric field, and potential energy stored in this capacitor.
We know that if we add an insulator and if we increase the dielectric constant of
capacitor, the capacitance increases, and the voltage decrease and the charge stays
the same while the electric filed and potential energy stored in a capacitor decreases.
To calculate the problem, the given are;
C=40F
V=12V
d=2mm
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To get the electric field, we use the formula;
Q= CV
Inserting the values; Q= (40uF) (12V)
Q= 480uC
To get the electric field, we use the formula;
E=
V
d
Inserting the values,
12V
E=
.002m
E= 6000V/m
To get the potential energy stored in the capacitor, we use the formula;
Inserting the values,
U=
1
2
U=
1
2
(480uF) (12V)
U= 2880uJ or 2.88mJ
Learning Competency
Describe the effects of inserting dielectric materials on the capacitance, charge, and
electric field of a capacitor. (STEM_GP12EMIId-29)
ACTIVITY 1: Fill in the Blanks
Directions: Choose the correct word that fits the blank
To answer this activity, choose from the following terms:
DIRECTLY
INVERSELY
DECREASES
CAPACITOR
DIRECTLY
DECREASE
INCREASES
Dielectric is any insulating material that acts as insulator and when inserted into
a _____________ it results with a direct effect with the charge, capacitance and
electric field. The capacitance is ___________ proportional to the value of dielectric
material. This means that, when the dielectric material decreases in value, the
capacitance would ____________ in value. The electric field on the other hand is
______________ proportional to the value of insulating material and it
_____________ as the value of dielectric material increases. And Lastly for the effect
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QV
of dielectric on charge, it ____________ with an increased value of insulating
material and we say that they are___________ proportional with each other.
ACTIVITY 2: Calculate the Following Problems
Directions: Solve the following problems
1. What is the capacitance of a parallel plate capacitor with metal plates, each
of area 1.00 m2, separated by 1.00 mm?
2. What charge is stored in this capacitor if a voltage of 3.00 × 103 V is applied
to it?
3. What charge is stored in a 180 μF capacitor when 120 V is applied to it?
4. What voltage must be applied to an 8.00 nF capacitor to store 0.160 mC of
charge?
5. What is the capacitance of a large Van de Graaff generator’s terminal, given
that it stores 8.00 mC of charge at a voltage of 12.0 MV?
REFLECTION:
1. I learned that
__________________________________________________________________
__________________________________________________________________
________________________________________________________________
2. I enjoyed most on__________________________________________________
__________________________________________________________________
__________________________________________________________________
__
3. I want to learn more on______________________________________________
__________________________________________________________________
__________________________________________________________________
__
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REFERENCES:
Cutnell and Johnson. Introduction to Physics. John Wiley and Sons, Inc. 2013.
Douglas Giancoli. Physics 1 and 2. 6th Edition, Pearson Education, 2014.
OpenStax: College Physics
Physics_serway.pdf
Thomson_-_Physics_For_Scientists_And_Eng.pdf
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ANSWER KEY:
Activity 1:
Dielectric is any insulating material that acts as insulator and when inserted into a
CAPACITOR it results with a direct effect with the charge, capacitance and electric
field. The capacitance is DIRECTLY proportional to the value of dielectric material.
This means that, when the dielectric material decreases in value, the capacitance
would DECREASE in value. The electric field on the other hand is INVERSELY
proportional to the value of insulating material and it DECREASES as the value of
dielectric material increases. And Lastly for the effect of dielectric on charge, it
INCREASES with an increased value of insulating material and we say that they are
DIRECTLY proportional with each other.
Activity 2:
1. 8.85uF
2.
26.6uC
3. 21.6uC
4. 20.0kV
5. 662pF
Prepared by:
ANGELIKA TORRES
Sta. Ana Fishery National High School
186
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GENERAL PHYSICS 2
Name: ____________________________
Date: _____________________________
Grade Level: _________
Score: ______________
LEARNING ACTIVITY SHEET
Capacitors and Dielectrics
Background Information for the Learners (BIL)
The Capacitance of a Capacitor
A parallel plate capacitor consists of two parallel metal plates placed near one
another but not touching. This type of capacitor is only one among many. In general,
a capacitor consists of two conductors of any shape placed near one another without
touching. For a reason that will become clear later on, it is common practice to fill the
region between the conductors or plates with an electrically insulating material called
a dielectric. A capacitor stores electric charge. Each capacitor plate carries a charge
of the same magnitude, one positive and the other negative. Because of the charges,
the electric potential of the positive plate exceeds that of the negative plate by an
amount V. Experiment shows that when the magnitude q of the charge on each plate
is doubled, the magnitude V of the electric potential difference is also doubled, so q
is proportional to V. These variables express this proportionality with the aid of a
proportionality constant C, which is the capacitance of the capacitor.
The magnitude Q of the charge on each plate of a capacitor is directly
proportional to the magnitude V of the potential difference between the plates and C
which is the capacitance, given by the equation
Q = CV
SI Unit of Capacitance: coulomb/volt = farad (F)
This equation shows that the SI unit of capacitance is the coulomb per volt
(C/V). This unit is called the farad (F), named after the English scientist Michael
Faraday (1791–1867).
Calculating Capacitance: Capacitors in Vacuum
We can calculate the capacitance of a given capacitor by finding the potential
difference Vab between the conductors for a given magnitude of charge Q and the
equation C =Q/Vab. For now we’ll consider only capacitors in vacuum; that is, we’ll
assume that the conductors that make up the capacitor are separated by empty
space.
The simplest form of capacitor consists of two parallel conducting plates, each with
area separated by a distance d that is small in comparison with their dimensions.
When the plates are charged, the electric field is almost completely localized in the
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region between the plates. The field between such plates is essentially uniform, and
the charges on the plates are uniformly distributed over their opposing surfaces. We
call this arrangement a parallel-plate capacitor.
We found that E = σ/ϵO, where σ is the magnitude (absolute value) of the
surface charge density on each plate. This is equal to the magnitude Q of the total
charge on each plate divided by the area A of the plate, or so the field magnitude can
be expressed as σ = Q/A so the field magnitude can be expressed as
𝐸=
σ
ϵ₀
=
𝑄
ϵ₀A
The field is uniform and the distance between the plates is so the potential
difference (voltage) between the two plates is
Vab = Ed =
𝑄𝑑
ϵ₀𝐴
From this we see that the capacitance of a parallel-plate capacitor in vacuum is
𝑄
𝐴
C = = ϵO
(capacitance of a parallel-plate capacitor in vacuum)
𝑉
𝑑
If A is in square meters and d in meters, C is in farads. The units of ϵO are C2 / Nm
= C2 / J thus, the units of ϵO can be expressed as; C2 / Nm2 or F/m
ϵO = 8.85 x 10-12 F/m
Sample problems:
1. The parallel plates of a 1.0-F capacitor are 1.0 mm apart. What is their area?
Given: C = 1.0 F
ϵO = 8.85 x 10-12 F/m
-3
d = 1.0 x 10 m
A=?
Solution:
(1.0𝐹)( 1.0 𝑋 10−3 𝑚)
𝐶𝑑
𝐴=
=
= 1.1 x 10-8 m2
𝜖₀
8.85 𝑥 10−12 𝐹/𝑚
2. The plates of a parallel-plate capacitor in vacuum are 5.0 mm apart and 2.00m 2 in
area. A10.0-kV potential difference is applied across the capacitor. Compute for
a. capacitance;
b. charge on each plate
c. magnitude of the electric field between the plates.
a. Given: A= 2.00 m2
ϵO = 8.85 x 10-12 F/m
d = 5.0 mm
C=?
𝐴
2.00 m2
Solution: 𝐶 = 𝜖₀
= 8.85 x 10 − 12 F/m
= 3.54 x 10-9 F
5.0 𝑥 10−3 𝑚
𝑑
b. Given: C = 3.54 x 10-9 F
V = 10,000 V
Q=?
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Solution: Q = CV = (3.54 x 10-9 F) (10,000 V) = 3.54 X 10-5 C
c. Given: Q = 3.54 X 10-5 C
ϵO = 8.85 x 10-12 F/m
Solution: E=
𝑄
ϵ₀A
=
A = 2.00 m2
E=?
3.54 X 10−5 C
F
(8.85 x 10−12m)(2.00 m2)
= 2.00 X 10-6 N/C
Dielectrics
Most capacitors have a nonconducting material, or dielectric, between their
conducting plates. A common type of capacitor uses long strips of metal foil for the
plates, separated by strips of plastic sheet such as Mylar. A sandwich of these
materials is rolled up, forming a unit that can provide a capacitance of several
microfarads in a compact package. Placing a solid dielectric between the plates of a
capacitor serves three functions.
First, it solves the mechanical problem of maintaining two large metal sheets
at a very small separation without actual contact.
Second, using a dielectric increases the maximum possible potential
difference between the capacitor plates. Any insulating material, when subjected to
a sufficiently large electric field, experiences a partial ionization that permits
conduction through it. This is called dielectric breakdown. Many dielectric materials
can tolerate stronger electric fields without breakdown than can air. Thus using a
dielectric allows a capacitor to sustain a higher potential difference and so store
greater amounts of charge and energy.
Third, the capacitance of a capacitor of given dimensions is greater when
there is a dielectric material between the plates than when there is vacuum.
When the charge is constant, Q₀= C₀V₀ = CV and C/C₀ = V₀/VIn this case, thus
𝑉=
𝑉₀
𝐾
(when Q is constant)
With the dielectric present, the potential difference for a given charge Q is reduced
by a factor K. The dielectric constant K is a pure number. Because C is always
greater than C₀, K is always greater than unity. Some representative values of K are
given below;
Values of Dielectric Constant K at 20 C
Material
k
Vacuum
1
Air (1 atm)
1.00059
Air (100 atm)
1.0548
Teflon
2.1
Polyethylene
2.25
Benzene
2.28
Mica 3
3–6
Mylar
3.1
Material
Polyvinyl chloride
Plexiglas®
Glass
Neoprene
Germanium
Glycerin
Water
Quartz
k
3.18
3.40
5.6
6.70
16
42.5
80.4
3.8
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No real dielectric is a perfect insulator. Hence there is always some leakage current
between the charged plates of a capacitor with a dielectric.
Induced Charge and Polarization
When a dielectric material is inserted between the plates while the charge is
kept constant, the potential difference between the plates decreases by a factor K.
Therefore the electric field between the plates must decrease by the same factor. If
E₀ is the vacuum value and E is the value with the dielectric, then
𝐸=
𝐸₀
𝐾
(when Q is constant)
The capacitance when the dielectric is present is given by
𝐴
𝐴
𝐶 = 𝐾𝐶0 = 𝐾𝜖0 𝑑 = 𝜖 𝑑 (parallel-plate capacitor, dielectric between plates)
Sample Problems:
1. Electronic circuitry enables the computer to detect the change in capacitance,
thereby recognizing which key has been pressed. The separation of the plates is
normally 5.00 x 10-3 m but decreases to 0.150 x 10-3 m when a key is pressed. The
plate area is 9.50 x 10-5 m2, and the capacitor is filled with a material whose dielectric
constant is 3.50. Determine the change in capacitance that is detected by the
computer.
Given: A = 9.50 x 10-5 m2
d = 0.150 x 10-3 m
K = 3.50
ϵ₀ = 8.85 x 10-12 F/m
𝐴
F
9.50 x 10−2 m2
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝐶 = 𝐾𝜖 0 𝑑 = (3.50) ( 8.85 x 10−12 m) 0.150 x 10−3 𝑚 = 19.0 𝑥10−12 𝐹
2. A parallel plate capacitor is filled with an insulating material with a dielectric
constant of 2.6. The distance between the plates of the capacitor is 2.0 x 10 -4 m. Find
the plate area if the new capacitance is 3.4µF.
Given: K = 2.6
C = 3.4 X 10-6 F
d = 2.0 x 10-4 m
ϵ₀ = 8.85 x 10-12 F/m
A=?
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝐴 =
𝐶𝑑
𝐾𝜖₀
=
(3.4𝑥10−6 𝐹)(2.0𝑥10−4 𝑚)
(2.6)(8.85𝑥10−12 𝐹/𝑚
= 29.6𝑚2
Learning Competency
Solve problems involving capacitors and dielectrics in context, such as, but not
limited to charged plates, batteries and camera flashlamps. (STEM_GP12EM-IIId-30)
190
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Activity 1: Study & Solve
Directions: Analyze the given problems below. Write your complete solution.
1. A parallel plate capacitor has square plates 7.5 cm on a side, separated by 0.29
mm. The capacitor is charge to 12 V, then disconnected from the charging power
supply.
a. Calculate the capacitance of this capacitor.
b. What is the total charge on each plate?
c. What is the electric field between the plates?
2. Calculate the capacitance and total charge if the following dieletric materials is
inserted on the parallel plate capacitor in problem # 1. ( The same given for problem
#1)
a. sheet of glass
b. polyethylene
c. quartz
Note: The table for K is given above for your reference.
Activity 2: Conceptual Analysis
Directions: Analyze the given situations below. Describe and explain briefly the
concept being asked.
An empty capacitor is connected to a battery and charged up. The capacitor is then
disconnected from the battery, and a slab of dielectric material is inserted between
the plates. Explain how each of the following would change.
a. Capacitance
_____________________________________________________________
b. Voltage
_____________________________________________________________
c. Electric Field
_____________________________________________________________
d. The charge on the surface of the insulator
_____________________________________________________________
2. If you want to increase the capacitance of a parallel plate capacitor, which of the
following dielectrics will you insert to yield the greatest value. Explain your answer.
a. polyvinyl chloride
b. mylar
c. neoprene
Answer:
_____________________________________________________________
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Activity 3: Test Your Understanding
Directions: Read and analyze the statements carefully. Write only the letter of your
choice.
1. Which of the following is TRUE about the capacitance in a parallel plate
capacitor?
A. It is proportional to the plate area.
B. It is proportional to the charge stored.
C. It is independent of any material inserted between the plates.
D. It is proportional to the potential difference of the plates.
2. Which of the following situations increases the ability of capacitors to store
more charges?
A. Increasing the distance between metal plates.
B. Decreasing the area of each plate.
C. Inserting a conducting material within the capacitor.
D. Decreasing the distance between metal plates.
3. What will happen if you pull the plates of an isolated charged capacitor
apart?
A. Capacitance will increase.
B. Potential difference will increase.
C. It does not affect the potential difference.
D. Potential difference will decrease.
4. What happens when an insulating material is placed between the plates of a
capacitor?
I.
The electric field becomes weaker due to the opposing force of the
polarized material.
II. The voltage increases.
III. The capacitance increases.
A. I and II
B. I and III C. II and III
D. I, II and III
5. If the area of the plates of a parallel-plate capacitor is doubled while the
spacing between the them is halved, what will happen on its capacitance?
A. C will be doubled
B. C will increase by four times
C. C will decrease by ¼
D. C will not change
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REFLECTION:
1. I learned that
__________________________________________________________________
__________________________________________________________________
________________________________________________________________
2. I enjoyed most on__________________________________________________
__________________________________________________________________
__________________________________________________________________
__
3. I want to learn more on______________________________________________
__________________________________________________________________
__________________________________________________________________
__
193
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References:
Kernion et.al, Physics prep,https://.physics-prep.com/index.php/about-us
Young,Hugh D, “University Physics”, Pearson Education,Inc.2012
Cutnell, Johnson, “Physics”, 8th Edition, John Wiley & Sons, Inc.,2009
194
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Answer Key
Activity 1
1.
a. C= 1.7 x 10-10 F
b. Q= 2.1 x 10-10 C
c. E= 4.1 x 104 Vm-1
2. Glass, C= 9.5 x 10-10 F
Q= 1.2 x 10-8 C
Quartz, C= 6.5x10-10F
Q= 4.8X10-9C
Polyethylene, C=3.9x10-10F
Q= 8.0X10-9 C
Activity 2
1.
a. The insertion of dielectric will always increase the capacitance.
b. Since Q = CV and q is fixed, the potential difference V across the
plates must decrease in order for q to remain unchanged. The
amount by which the potential difference decreases from the
value initially established by the battery depends on the
dielectric constant of the slab.
c.Since voltage is directly related to electric field based from the
equation, then a decrease in voltage would also mean a
decrease in electric field. The electric field will be minimized due
to the polarization of the material.
d.A surface charge will be created on the surface of the insulator due
to the rearrangement of molecules. The insertion of the
dielectric will cause polarization on the insulating material.
2. C. Neoprene, the higher the value of the dieletric the greater will be the increase
in the capacitance of the material because it is directly related.
Activity 3
1. A
2. D
3.A
4. D
5. B
Prepared by:
MARJOHN C. ADDURU
PATTAO NATIONAL HIGH SCHOOL
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GENERAL PHYSICS 2
Name:___________________________
Grade Level:____________
Date:____________________________
Score:__________________
LEARNING ACTIVITY SHEET
CONVENTIONAL CURRENT AND ELECTRON FLOW
Background Information for the Learners (BIL)
In this lesson, we study the flow of electric charges, recall that whenever there
is a net flow of charge through some region, an electric current is said to exist. It is
instructive to draw an analogy between water flow and current. In many localities it
is common practice to install low-flow showerheads in homes as a water
conservation measure. We quantify the flow of water from these and similar devices
by specifying the amount of water that emerges during a given time interval, which is
often measured in liters per minute. On a grander scale, we can characterize a river
current by describing the rate at which the water flows past a location.
But for this topic, we shall focus particularly on differentiating the conventional
current and electron flow. Look at the figure below;
https://www.codrey.com/dc-circuits/conventional-current-vs-electron-current
Notice that, for conventional current the direction of the arrow is from positive
terminal to negative terminal. While on electron flow the opposite is shown. From
this, we define conventional current as charge per unit time transported from the
positive terminal of a source to the negative terminal of the source. It behaves as if
positive charges carriers cause current to flow. Thus, we may assume that the flow
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of electrons is the flow of protons in the opposite direction. The simplest way to think
about this is to pretend as if the movement of positive charges carriers constituted
current flow.
Electron flow out of the negative terminal through the circuit into the positive
terminal of the source. It is what happens in the circuit, but both conventional current
and electron flow are used. To distinguish the two, the opposite of what happens in
conventional current is the electron flow.
CONVENTIONAL CURRENT
ELECTRON FLOW
Floyd, 1989, Principles of Electric Floyd, 1990, Principles of Electric
Circuits, 5th edition, Conventional
Circuits, 4th edition, Electron Flow
Current Version.
Version.
Although, it makes no difference which way current is flowing if it is used consistently.
The direction of current flow does not affect what the current does. The difference
between conventional current flow and electron flow in no way effects any real-world
behavior or computational results. In general, analyzing an electrical circuit yields
results that are independent of the assumed direction of current flow. Conventional
current flow is the standard that most all the world follows.
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http://web.engr.oregonstate.edu/~traylor/ece112/beamer_lectures/elect_flow_vs_co
nv_I.pdf
Learning Competency:
Distinguish between conventional current and electron flow. (STEM_GP12EM-IIId32)
Activity 1: Compare and Contrast
Directions: Complete the table below, on the left column describe what an electron
flow is, on the rightmost column write a description of the conventional current, and
in the middle column write a brief description of what the two may have in common.
Electron Flow
Similarities
Conventional Current
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Activity 2: Label Us
Directions: Label the directions of both electron flow and conventional flow in this
simple circuit
Activity 3: Multiple Choice
Directions: Choose the best correct answer.
1. Which one of the following statements is true?
A. Using conventional current flow, we picture current leaving a voltage
source's positive terminal and entering its negative terminal.
B. Using electron current flow, we picture current leaving a voltage source's
positive terminal and entering its negative terminal.
2. A person using electron current flow would imagine current in this circuit
flowing
A. Clockwise
B. Counter-clockwise
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3. Which of the following describes the electron charge moves from the
negative (surplus) side of the battery to the position (deficiency) side?
A. Electron Flow
B. Current Flow
4. Which of the following describes the electron charge moves from the positive
(surplus) side of the battery to the negative (deficiency) side?
A. Electron Flow
B. Current Flow
Activity 4: True or False
Directions: Write T if the statement is true and F if the statement is false.
1. Electron flow states that electron flows from positive terminal to negative
terminal.
2. Conventional current states that electric charge flows from positive to
negative terminal.
3. Conventional current behaves as if positive charges carriers cause
current to flow.
4. Electron flow is the standard that most all the world follows.
5. The direction of current flow affects what the current does.
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REFLECTION:
1. I learned that
__________________________________________________________________
__________________________________________________________________
________________________________________________________________
2. I enjoyed most on__________________________________________________
__________________________________________________________________
__________________________________________________________________
__
3. I want to learn more on______________________________________________
__________________________________________________________________
__________________________________________________________________
__
201
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REFERENCES:
Cutnell and Johnson. Introduction to Physics. John Wiley and Sons, Inc. 2013.
Douglas Giancoli. Physics 1 and 2. 6th Edition, Pearson Education, 2014.
Jewet and Serway.Physics for Scientists and Engineers.6 th Edition, Thomson
Brooks/Cole, 2004
OpenStax: College Physics
Physics_serway.pdf
Thomson_-_Physics_For_Scientists_And_Eng.pdf
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ANSWER KEY:
Activity 1 and Activity 2: Answers may be stated in various ways.
Activity 3: A, B, A, B
Activity 4: F, T, T, F, F
Prepared by:
ANGELIKA TORRES
Sta. Fishery National High School
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GENERAL PHYSICS 2
Name: ____________________________Grade Level: _________
Date: _____________________________Score:______________
LEARNING ACTIVITY SHEET
ELECTRIC CURRENT
BACKGROUND INFORMATION FOR THE LEARNERS
Many practical applications and devices are based on the principles of static
electricity, but electricity was destined to become an inseparable part of our daily
lives when scientists learned how to produce a continuous flow of charge for relatively
long periods of time using batteries. The battery or voltaic cell was invented in 1800
by Italian physicist Alessandro Volta. Batteries supplied a continuous flow of charge
at low potential, in contrast to earlier electrostatic devices that produced a tiny flow
of charge at high potential for brief periods. This steady source of electric current
allowed scientists to perform experiments to learn how to control the flow of electric
charges in circuits. Today, electric currents power our lights, radios, television sets,
air conditioners, computers, and refrigerators. They ignite the gasoline in automobile
engines, travel through miniature components making up the
chips of
microcomputers, and provide the power for countless other invaluable tasks.
There are requirements that must be met to establish an electric circuit for a
charge to flow known as current. It is not enough that there is simply a closed
conducting loop; the loop itself must extend from the positive terminal to the negative
terminal of the electrochemical cell. When this is met, then charge will flow through
the external circuit. It is said that there is a current - a flow of charge. As a physical
quantity, current is the rate at which charge flows past a point on a circuit. The
current in a circuit can be determined if the quantity of charge Q passing through a
cross section of a wire in a time t can be measured. The current is simply the ratio
of the quantity of charge and time.
Current is a rate quantity. There are several rate quantities in physics. For
instance, velocity is a rate quantity - the rate at which an object changes its position.
Mathematically, velocity is the position change per time ratio. Acceleration is a rate
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quantity - the rate at which an object changes its velocity. Mathematically,
acceleration is the velocity change per time ratio. And power is a rate quantity - the
rate at which work is done on an object. Mathematically, power is the work per time
ratio. In every case of a rate quantity, the mathematical equation involves some
quantity over time. Thus, current as a rate quantity would be expressed
mathematically as
Current =
I
=
∆𝑸
∆𝒕
Note that the equation above uses the symbol I to represent the quantity
current.
The standard metric unit for current is the ampere. Ampere is often shortened
to Amp and is abbreviated by the unit symbol A. A current of 1 ampere means that
there is 1 coulomb of charge passing through a cross section of a wire every 1
second.
1 ampere = 1 coulomb / 1 second
To determine the amount of electrical charge that flows in a circuit, you need
to know the current flow and how long it flows for. The equation is:
charge in coulombs=current in amperes×time in seconds
Charge =∆Q = I∆t
Learning Competency
Apply the relationship charge= current x time to new situation of to solve
related problems (STEM_GP12EM-IIIe-33)
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ACTIVITY 1: LIGHT ME UP!
Materials: small light bulb, dry cell, bare copper wire
Directions: Find the four different arrangements of the three items that would result
in the formation of an electric circuit that would light the bulb.
Answer the following:
Q1. What four arrangements would result in the successful lighting of the
bulb? Draw the four diagrams.
Q2. What does each of the four arrangements have in common that wouldlead
us into an understanding of the requirement of an electric circuit?
ACTIVITY 2: TAKE THE CURRENT VALUE!
Directions: Compute for what is being asked.
1. What is the current flowing when
(a) 20 C flow by in 4 s
(b) 120 C flow by in 2 minutes
(c) 2400 C flow by in 5 minutes
(d) 20 C flow by in 100 s
2. How much charge has passed when
(a) a current of 2A flows for 5 s
(b) a current of 25 mA flows for 150 s
(c) a current of 3A flows for 5 minutes
(d) a current of 5 MA flows for 5 ms
3. How long does it take when
(a) 15 C passes through 15 A
(b) 250 C passes through 8 kA
(c) 48 C passes through 4 mA
(d) 720 passes through 2.4 x 106 A
4. Directions: Complete the following table with complete solutions.
1
2
3
4
5
CHARGE
50 C
CURRENT
0.2A
0.2 C
120 C
12A
50mA
TIME
5s
2 minutes
300s
25s
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ACTIVITY 3: TAKE CHARGE OF IT!
Directions: Compute the following problems and show your complete solutions.
1. A defibrillator is used during a heart attack to restore the heart to its normal
beating pattern. A defibrillator passes 18 A of current through the torso of a
person in 2.0 ms. (a) How much charge moves during this time? (b) How many
electrons pass through the wires connected to the patient?
2. An especially violent lightning bolt has an average current of 1.26 X 10 3 A
lasting 0.138 s. How much charge is delivered to the ground by the lightning
bolt?
3. The large window air conditioner in Anita Breeze's room draws 11 amps of
current. The unit runs for 8.0 hours during the course of a day. Determine the
quantity of charge that passes through Anita's window AC during these 8.0
hours.
4. Over the course of an 8 hour day, 3.8x104 C of charge pass through a typical
computer (presuming it is in use the entire time). Determine the current for
such a computer.
5. Determine the amount of time that the following devices would have to be
used before 1.0x106 C (1 million Coulombs) of charge passes through them.
a.LED night light(I=0.0042 A)
b. Incandescent night light (I=0.068 A)
c. 60-Watt incandescent light bulb (I=0.50 A)
d. Large bathroom light fixture (I=2.0 A)
ACTIVITY 4: TAKE TIME TO REALIZE!
Directions: Analyze the situation below and answer the following questions.
1. Handheld calculators often use small solar cells to supply the energy required
to complete the calculations needed to complete your next physics exam. The
current needed to run your calculator can be as small as 0.30 mA. How long
would it take for 1.00 C of charge to flow from the solar cells? Can solar cells
be used, instead of batteries, to start traditional internal combustion engines
presently used in most cars and trucks?
2. Circuit breakers in a home are rated in amperes, normally in a range from 10
amperes to 30 amperes, and are used to protect the residents from harm and
their appliances from damage due to large currents. A single 15-amp circuit
breaker may be used to protect several outlets in the living room, whereas a
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single 20-amp circuit breaker may be used to protect the refrigerator in the
kitchen. What can you deduce from this about current used by the various
appliances?
REFLECTION:
1. I learned that
____________________________________________________________
_________________________________________________________________
_____________________________________________________.
2. I enjoyed most on
____________________________________________________________
_________________________________________________________________
_____________________________________________________.
3.I want to learn more on
_____________________________________________________________
__________________________________________________________________
___________________________________________________.
208
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References:
✓ http://tsigaridisjunior.weebly.com/uploads/8/6/0/4/8604600/current_problems
.pdf
✓ https://www.physicsclassroom.com/class/cuircuits/Lesson-2/Electric-Current
✓ https://www.physicsclassroom.com/calcpad/circuits/problems
✓ https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Univer
sity_Physics_(OpenStax)/Map%3A_University_Physics_II__Thermodynamics_Electricity_and_Magnetism_(OpenStax)/09%3A_Current
_and_Resistance/9.02%3A_Electrical_Current#:~:text=and%20circuit%20br
eakers.-,Figure%209.2.,the%20area%20A%20each%20second.
✓ https://sciencing.com/calculate-coulombs-2645.html
✓ https://www.schoolphysics.co.uk/age1114/Electricity%20and%20magnetism/Current%20electricity/text/Charge_and
_current/index.html
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ANSWER KEY
ACTIVITY 1: LIGHT ME UP!
1. Diagrams should be associated with the picture:
2. Answers may vary.
There must be a closed conducting path that extends from the positive
terminal to the negative terminal. It is not enough that there is simply a
closed conducting loop; the loop itself must extend from the positive terminal
to the negative terminal of the electrochemical cell.
ACTIVITY 2: TAKE THE CURRENT VALUE!
1. (a) 5 A
(b) 1 A
(c) 8 A
(d) .2 A
2. (a) 10 C
(b) 3.75 C
(c) 900 C
(d) 25 C
3. (a) 1 s
(b) 3.125 x 10-2 s
(c) 12,000 s
(d) 3 x 10-4 s
4.
CHARGE
1
2
CURRENT
10 A
24 C
6 x 10-4 A
3
4
5
TIME
10 s
1.25 C
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ACTIVITY 3: TAKE CHARGE OF IT!
1. (a) ∆Q = I∆t
= (18 A) (2.0x10-3 s)
= 3.60 x 10-2 C
(b) no.of e- = 3.60 x 10-2 C / 1.6 x10-19 C
= 2.25 x 1017 e2. ∆Q = I∆t
= (1.26x103 A) (0.138 s)
= 173.88 C
3. ∆Q = I∆t
= (8.0 hr x 60 min/hr x 60 s/min) (11 A)
= (28,800 s) (11 A)
= 3.17x105 C
4. I
5. (a)
=
∆𝑄
∆𝑡
= 3.8 x 104 C
28,800 s
= 1.319 A
∆𝑡 =
∆𝑄
𝐼
= 1 x 106 C
.0042 A
= 2.4x108 sec = 6.6x104 hr = 2.8x103 d = 7.5 yr
(b)
∆𝑡 =
∆𝑄
𝐼
= 1 x 106 C
.068 A
= 1.5x107 sec = 4.1x103 hr = 170 d
(c)
∆𝑡 =
∆𝑄
𝐼
= 1 x 106 C
.50 A
= 2.0x106 s = 560 hr = 23 d
(d)
∆𝑡 =
∆𝑄
𝐼
= 1 x 106 C
.2.0 A
= 5.0x105 s = 140 hr = 5.8 d
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ACTIVITY 4: TAKE TIME TO REALIZE!
1. The time for 1.00 C of charge to flow would be 3.33 X 10 3 s
Answers may vary.
The calculator takes a very small amount of energy to operate, unlike the
truck’s starter motor. There are several reasons that vehicles use batteries
and not solar cells. Aside from the obvious fact that a light source to run the
solar cells for a car or truck is not always available, the large amount of current
needed to start the engine cannot easily be supplied by present-day solar
cells. Solar cells can possibly be used to charge the batteries. Charging the
battery requires a small amount of energy when compared to the energy
required to run the engine and the other accessories such as the heater and
air conditioner. Present day solar-powered cars are powered by solar panels,
which may power an electric motor, instead of an internal combustion engine.
2. Answers may vary.
The total current needed by all the appliances in the living room (a few lamps,
a television, and your laptop) draw less current and require less power than
the refrigerator.
Prepared by:
JACKSON B. CASIBANG
SOLANA FRESH WATER FISHERY SCHOOL
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
ELECTRICAL RESISTANCE
BACKGROUND INFORMATION FOR THE LEARNERS
Electrical Resistance
An electron traveling through the wires and loads of the external circuit
encounters resistance. Resistance is the hindrance to the flow of charge. For an
electron, the journey from terminal to terminal is not a direct route. Rather, it is a
zigzag path that results from countless collisions with fixed atoms within the
conducting material. The electrons encounter resistance - a hindrance to their
movement. While the electric potential difference established between the two
terminals encourages the movement of charge, it is resistance that discourages it.
The rate at which charge flows from terminal to terminal is the result of the combined
effect of these two quantities.
Variables Affecting Electrical Resistance
The flow of charge through wires is often compared to the flow of water
through pipes. The resistance to the flow of charge in an electric circuit is analogous
to the frictional effects between water and the pipe surfaces as well as the resistance
offered by obstacles that are present in its path. It is this resistance that hinders the
water flow and reduces both its flow rate and its drift speed. Like the resistance to
water flow, the total amount of resistance to charge flow within a wire of an electric
circuit is affected by some clearly identifiable variables.
First, the total length of the wires will affect the amount of resistance. The
longer the wire, the more resistance that there will be. There is a direct relationship
between the amount of resistance encountered by charge and the length of wire it
must traverse. After all, if resistance occurs as the result of collisions between charge
carriers and the atoms of the wire, then there is likely to be more collisions in a longer
wire. More collisions mean more resistance.
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Second, the cross-sectional area of the wires will affect the amount of resistance.
Wider wires have a greater cross-sectional area. Water will flow through a wider pipe
at a higher rate than it will flow through a narrow pipe. This can be attributed to the
lower amount of resistance that is present in the wider pipe. In the same manner, the
wider the wire, the less resistance that there will be to the flow of electric charge.
When all other variables are the same, charge will flow at higher rates through wider
wires with greater cross-sectional areas than through thinner wires.
A third variable that is known to affect the resistance to charge flow is the
material that a wire is made of. Not all materials are created equal in terms of their
conductive ability. Some materials are better conductors than others and offer less
resistance to the flow of charge. Silver is one of the best conductors but is never used
in wires of household circuits due to its cost. Copper and aluminum are among the
least expensive materials with suitable conducting ability to permit their use in wires
of household circuits. The conducting ability of a material is often indicated by
its resistivity. The resistivity of a material is dependent upon the material's electronic
structure and its temperature. For most (but not all) materials, resistivity increases
with increasing temperature. The table below lists resistivity values for various
materials at temperatures of 20 degrees Celsius.
Resistivity
Material
(ohm•meter)
Silver
1.59 x 10-8
Copper
1.7 x 10-8
Gold
2.2 x 10-8
Aluminum
2.8 x 10-8
Tungsten
5.6 x 10-8
Iron
10 x 10-8
Platinum
11 x 10-8
Lead
22 x 10-8
Nichrome
150 x 10-8
Carbon
3.5 x 10-5
Polystyrene
107 - 1011
Polyethylene
108 - 109
Glass
1010 - 1014
Hard Rubber
1013
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As seen in the table, there is a broad range of resistivity values for various
materials. Those materials with lower resistivities offer less resistance to the flow of
charge; they are better conductors. The materials shown in the last four rows of the
above table have such high resistivity that they would not even be considered to be
conductors.
How Temperature Changes Resistance
Although the resistance of a conductor changes with the size of the conductor (e.g.
thicker wires have less resistance to current flow than thinner wires), the resistance
of a conductor also changes with changing temperature. This may be expected to
happen because, as temperature changes, the dimensions of the conductor will
change as it expands or contracts.
However, materials that are classed as conductors tend to increase their
resistance with an increase in temperature. insulators however are liable to
decrease their resistance with an increase in temperature. Materials used for
practical insulators (glass, plastic etc.) only exhibit a marked drop in their resistance
at very high temperatures. They remain good insulators over all temperatures they
are likely to encounter in use.
These changes in resistance cannot therefore be explained by a change in
dimensions due to thermal expansion or contraction. In fact for a given size of
conductor the change in resistance is due mainly to a change in the resistivity of the
material, and is caused by the changing activity of the atoms that make up the
material.
Variation In Resistance Of A Material At Different Temperatures:
(1)
Increase in resistance of a conductor is directly proportional to original
resistance.
R  R1..............(a)
(2)
Change in resistance is directly proportional to change in temperature.
R  T………….(b)
Combining (a) and (b)
R  R1 T
R = (constant) R1T
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Here constant = 
R = RT
Where a = temperature coefficient
As R = R2 - R1
And
T = T2 -T1
We get
R2 – R1 = R1 (T2 – T1)
R2 = R1 + R (T2 – T1)
R2 = R1 (1 + ( – T1))
Where,
R2 = Resistance of conductor at temperature T2
R1 = Resistance of conductor at reference temperature T1
α = Temperature coefficient of resistance at reference temperature
T2 = Temperature of conductor in degrees Celsius
T1 = Temperature reference that αis specified at for the conductor material
When T1 = 0 and T2 = t
Rt = R0 [1 + (t – 0)]
Rt = R0 (1 + t)
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Temperature Coefficient of Resistance Table
The table below gives the temperature coefficient of resistance for a variety of
substances
SUBSTANCE
Aluminium
Antimony
Bismuth
Brass
Cadmium
Cobalt
Constantan (Alloy)
Copper
Gold
Carbon (Graphite)
Germanium
Iron
Lead
Manganin
Molybdenum
Nichrome
Nickel
Platinum
Silicon
Silver
Tantalum
Tin
Tungsten
Zinc
TEMPERATURE COEFFICIENT °C(20°C)
4.3 x 10-3
(18°C - 100°C)
4.0 x 10-3
4.2 x 10-3
~1.0 x 10-3
4.0 x 10-3
7 x 10-5
3.3 x 10-3
4.0 x 10-3
3.4 x 10-3
-5.6 x 10 -6
-4.8 x 10-2
5.6 x 10-3
3.9 x 10-3
~2 x 10-5
4.6 x 10-3
1.7 x 10-4
5.9 x 10-3
3.8 x 10-3
-7.5 x 1024
4.0 x 10-3
3.3 x 10-3
4.5 x 10-3
4.5 x 10-3
3.6 x 10-3
Learning Competency:
Describe the effect of temperature increase on the resistance of a metallic conductor.
STEM_GP12EM-IIIe-35
217
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ACTIVITY 1: GETTING TO KNOW!
Directions: Analyze the following statements and answer the questions being asked.
1. Given two lengths of metal wire, which one will have the least electrical
resistance: one that is short, or one that is long? Assume all other factors are
equal (same metal type, same wire diameter, etc.).
2. Given two lengths of solid metal wire with round cross-sections, which one will
have the least electrical resistance: one that is small-diameter, or one that is
large-diameter? Assume all other factors are equal (same metal type, same
wire length, etc.).
3. What happens to resistance when temperature increases?
4. Why does resistance of metal increase with temperature?
5. Why is resistance directly proportional to temperature?
RUBRICS FOR SCORING
A. (5 points) Outstanding – student responses far exceed what is expected
B. (4 points) Very good – information is factually accurate and offers extra
supporting facts.
C. (3 points) Good – Sufficiently developed information with adequate
elaboration or explanation.
D. (2 points) Fair – Limited content with inadequate elaboration or explanation
E. (1 point) Poor – the ideas are not clear.
ACTIVITY 2: IT’S ALL ON THE TABLE!
Directions: Examine the following specific resistance table for various metals:
Metal type
Zinc (very pure)
Tin (pure)
Copper (pure annealed)
Copper (hard-drawn)
Copper (annealed)
Platinum (pure)
Silver (pure annealed)
Nickel
Steel (wire)
Iron (approx. pure)
Gold (99.9 % pure)
Aluminum (99.5 % pure)
ρ in Ω · cmil / ft @ 0 oC
34.595
78.489
9.390
9.810
9.590
65.670
8.831
74.128
81.179
54.529
13.216
15.219
ρ in Ω · cmil / ft @ 24 oC
37.957
86.748
10.351
10.745
10.505
71.418
9.674
85.138
90.150
62.643
14.404
16.758
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1. Of the metals shown, which is the best conductor of electricity?
2. Which is the worst?
3. What do you notice about the resistivity of these metals as temperature is
increased from 0oC to 24oC?
4. Re-arrange the table to show resistivity from least to greatest.
ACTIVITY 3: I RESIST!
Directions: Calculate the resistance of each of these conductors, given their
resistance at the reference temperature (R1 @ T1), and their present temperatures
(T2):
1.
2.
3.
4.
5.
Copper ; R1 = 200 Ω @ T1= 20oC ; T2 = 45oC ; R2 = ?
Aluminum ; R1 = 1,250 Ω @ T1 = 20oC ; T2 = 100oC ; R2 = ?
Iron ; R1 = 35.4 Ω @ T1 = 20oC ; T2 = −40oC ; R2 = ?
Nickel ; R1 = 525 Ω @ T1 = 20oC ; T2 = 70oC ; R2 = ?
Gold ; R1 = 25 kΩ @ T1 = 20oC ; T2 = 65oC ; R2 = ?
ACTIVITY 4: THE VALUE OF UNKNOWN
Directions: Compute for what is being asked:
1. A length of copper wire (α = 0.004041 at 20 o C) has a resistance of 5 ohms
at 20 degrees Celsius. Calculate its resistance if the temperature were to
increase to 50 degrees Celsius.
2. A platinum resistance thermometer uses the change in R to measure
temperature. Suppose R1 = 50 Ω at T1=20 ºC, what is final resistance when
Temperature increases to 50.0 ºC?
3. A platinum resistance thermometer has a resistance R1 = 50.0 Ω at T1=20
ºC. The thermometer is immersed in a vessel containing melting tin, at which
point R increases to 91.6Ω. What is the melting point of tin?
4. Resistance of a conductor is 1.72 Ω at a temperature of 20ºC.Find the
resistance at 0ºC and 100ºC. Given the coefficient of resistivity is a =
0.00393.
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REFLECTION:
1. I learned that
____________________________________________________________
_________________________________________________________________
_____________________________________________________.
2. I enjoyed most on
____________________________________________________________
_________________________________________________________________
_____________________________________________________.
3. I want to learn more on
_____________________________________________________________
__________________________________________________________________
___________________________________________________.
220
NOTE: Practice personal hygiene protocols at all times
References:
https://courses.physics.ucsd.edu/2010/Fall/physics1b/documents/LECT11_212216.
pdf
https://www.citycollegiate.com/electricityXIchp13c.htm
https://www.electronics-notes.com/articles/basic_concepts/resistance/resistanceresistivity-temperature-coefficient.php
https://www.allaboutcircuits.com/textbook/direct-current/chpt-12/temperaturecoefficient-resistance/
https://www.allaboutcircuits.com/worksheets/temperature-coefficient-of-resistance/
https://www.embibe.com/study/examples-on-calculations-on-temperaturecoefficient-of-resistivity-concept
221
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ANSWER KEY
ACTIVITY 1: GETTING TO KNOW!
1. The short wire will have less electrical resistance than the long wire.
2. The large-diameter wire will have less electrical resistance than the smalldiameter wire.
3. (Answers may vary.) As electrons move through a metal conductor, some
collide with atoms, other electrons or impurities. These collisions
cause resistance and generate heat. Heating the metal conductor causes
atoms to vibrate more, which in turn makes it more difficult for the electrons
to flow, increasing resistance.
4. (Answers may vary.) The resistance of a conductor increases with
an increase in temperature because the thermal velocity of the free
electrons increase as
the temperature
increases.
This
results
in increase in number of collisions between the free electrons.
5. (Answers may vary.) Resistance of a conductor is directly proportional
to temperature. With the increase in temperature, vibrational motion of
the atoms of conductor increases. Due to increase in vibration, probability
of
collision
between
atoms
and
electrons
increases.
As
a
result, resistance of conductor increases.
ACTIVITY 2: IT’S ALL ON THE TABLE!
Resistivity from least to greatest:
Metal type
ρ in Ω · cmil / ft @ 0
oC
ρ in Ω · cmil / ft @ 24
oC
Silver (pure annealed)
Copper (pure annealed)
Copper (annealed)
Copper (hard-drawn)
Tin (pure)
Gold (99.9 % pure)
Aluminum (99.5 % pure)
Zinc (very pure)
Iron (approx. pure)
Platinum (pure)
Nickel
Steel (wire)
8.831
9.390
9.590
9.810
78.489
13.216
15.219
34.595
54.529
65.670
74.128
81.179
9.674
10.351
10.505
10.745
86.748
14.404
16.758
37.957
62.643
71.418
85.138
90.150
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ACTIVITY 3: I RESIST
1. 220.2 Ω
2. 1.681 kΩ
3. 23.35 Ω
4. 679 Ω
5. 29.18 kΩ
ACTIVITY 4: THE VALUE OF UNKNOWN
1. 5.606 Ω
2. 55.9 Ω
3. T = 232 ºC
4. (A) 1.58 Ω (B) 2.26 Ω
Prepared by:
JACKSON B. CASIBANG
SOLANA FRESH WATER FISHERY SCHOOL
223
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
RESISTIVITY AND CONDUCTIVITY
BACKGROUND INFORMATION FOR THE LEARNERS
We know what the electric wires are made of; they are either made of copper
or aluminum. We also know that gold and silver would have been a better choice,
had they been cheaper. However, the question is what makes them suitable for
the same? Why don’t we see lead wires? The answer lies in the very basic property
of these materials, namely their resistivity and electrical conductivity. This article
explains the essential concepts of resistivity and electrical conductivity. Resistivity
and conductivity are actually two sides of the same coin, if you understand one, you
will get the other as well.
Resistivity
The resistivity of a material is the measure of its property due to which it
opposes the flow of electrons through it.
The flow of electrons leads to current flow, therefore if a material opposes the
flow of electrons, the current that can pass through it is limited. Some materials
oppose the flow more than others. This is due to the varied atomic structure of
different materials.
In order to understand resistivity better, first revise the concept of ohm’s law.
OHMS LAW
The ohm’s law states that when a voltage (a potential difference) is applied
across a conductor, current starts to flow through it. This current is directly
proportional to the voltage.
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Ohms law can be written as:
V α I; where V = voltage, I= Current through the conductor
Or V= RI
Here, R = Constant, known as Resistance.
This resistance restricts the amount of current or we can say that it restricts
the amount of electron flow. This means that the conductor resists the current flow
to some extent.
SI UNIT OF RESISTANCE
R = ohm denoted by Ω. = Volt/Amp.
Factors affecting Resistance:
The resistance of a material is its ability to oppose the flow of electrons through
it. It depends on the physical dimensions; that is its length and area of the cross
section. Other factors include temperature and type of material used to make the
conductor.
1. Length. Resistance is directly proportional to length, L. With a decrease in
length, the resistance of the material to the flow of electron decreases. Hence
a decrease in length would decrease the resistance.
2. Area. When its area of cross-section is increased, the flow of charges[current]
increase, meaning the opposition to the movement of charges decreases,
hence there is a decrease in resistance. Thus, increasing the area of crosssection decreases the resistance. Also, when the area of cross section is
decreased, the space available for the electrons to flow decreases, and
therefore the opposition to the electron flow increases, meaning there is an
increase in the resistance.
Resistivity Equation
After analyzing the two cases, we know that the resistance is directly
proportional to length and inversely proportional to A. Mathematically, it can be
represented as
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What about the other two factors, the temperature, and type of material? For
this, proportionality constant is introduced in the equation. This constant is called
the resistivity. Since for a given material at a given temperature, the resistivity is
constant, it is taken as the proportionality constant here.
It is denoted by a Greek letter, rho, ρ. The resistance, R, now can be written
in terms of the following
The Relevance of Resistivity
It is the measure of the ability of how much a material can oppose the flow of
electrons.
Knowing the resistivity of a material, we can choose it accordingly, for different
uses. An insulator such as a glass has a high resistivity, whereas that of a good
conductor like copper has a very low resistivity, and hence is suitable for making
connecting wires.
Factors affecting Resistivity
Since ρ is defined for a particular material, so the type of material is one factor
that affects it. It is due to the fact that different materials have different atomic and
molecular arrangements. If we had to increase the resistivity of that material we
could just increase the amount of material used. But then it is not that common a
practice, since it would make the conductor nothing but bulky.
Another factor that affects the resistivity is the temperature. Every material
has a temperature coefficient, and resistivity is dependent on it. It is due to this
reason, whenever the resistivity of a material is given, the temperature at which it
was measured is specified. If the temperature is not specified, it is assumed to be at
the room temperature.
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A positive temperature coefficient implies that the resistivity increases with
temperature. Whereas a negative coefficient implies that it decreases with
temperature. Metals such as copper, have positive temperature coefficient whereas
that of semiconductors like Silicon, have negative temperature coefficient.
Electrical Conductivity
The electrical conductivity of a material is the measure of the property of a
material due to which it allows the electrons to flow through it.
From the definition, it is clear, that electrical conductivity is actually the
opposite of resistivity. There is another parameter called conductance, in an
electric circuit. It is the measure of how much the current can be conducted through
the conductor. The electric current, as we know is the flow of electrons. In simple
terms, it is the opposite of resistance. If resistance is the factor that restricts the
current in an electric circuit, conductance is the factor that allows the current to flow.
Conductance is often represented as G = 1/R.
Factors affecting Conductivity and Conductance:
Since conductance is just the opposite of resistance, the factors that have an
effect on conductance remain the same as that of resistance, but how they affect
changes, it just becomes opposite.
1. Length , L: As length increases the conductance decreases.
G α 1/L
2. Area of the cross-section , A: As area of cross-section increases the
conductance increases.
GαA
Thus: G α A/L
Conductivity Equation
Here a proportionality constant, named as conductivity is introduced.
It is denoted by the symbol sigma, Ϭ and is the reciprocal of resistivity.
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Ϭ = 1/ρ
This knowledge of conductivity helps to determine the good and bad
conductors of electricity. If the conductivity of a material is high enough, it is a good
conductor of electricity, so it allows more charge to flow through it.
Also like resistivity, conductivity also depends on temperature and is specified
at a particular temperature.
Metals like copper, aluminum, gold, and silver have high conductivity in the
range of MS/m, making them good conductors of electricity and suitable for making
electrical wires.
Table of Resistivity and Conductivity at 20°C
Material
Silver
Copper
Annealed copper
Gold
Aluminum
Calcium
Tungsten
Zinc
Nickel
Lithium
Iron
Platinum
Tin
Carbon steel
Lead
Titanium
Grain oriented electrical
steel
Manganin
Constantan
Stainless steel
Mercury
Nichrome
GaAs
ρ (Ω•m) at 20 °C
Resistivity
1.59×10−8
1.68×10−8
1.72×10−8
2.44×10−8
2.82×10−8
3.36×10−8
5.60×10−8
5.90×10−8
6.99×10−8
9.28×10−8
1.0×10−7
1.06×10−7
1.09×10−7
(1010)
2.2×10−7
4.20×10−7
4.60×10−7
σ (S/m) at 20 °C
Conductivity
6.30×107
5.96×107
5.80×107
4.10×107
3.5×107
2.98×107
1.79×107
1.69×107
1.43×107
1.08×107
1.00×107
9.43×106
9.17×106
1.43×10−7
4.55×106
2.38×106
2.17×106
4.82×10−7
4.9×10−7
6.9×10−7
9.8×10−7
1.10×10−6
5×10−7 to 10×10−3
2.07×106
2.04×106
1.45×106
1.02×106
9.09×105
5×10−8 to 103
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Carbon (amorphous)
Carbon (graphite)
Carbon (diamond)
Germanium
Sea water
Drinking water
Silicon
Wood (damp)
Deionized water
Glass
Hard rubber
Wood (oven dry)
Sulfur
Air
Paraffin wax
Fused quartz
PET
Teflon
5×10−4 to 8×10−4
2.5×10−6 to 5.0×10−6 //basal
plane
3.0×10−3 ⊥basal plane
1×1012
4.6×10−1
2×10−1
2×101 to 2×103
6.40×102
1×103 to 4
1.8×105
10×1010 to 10×1014
1×1013
1×1014 to 16
1×1015
1.3×1016 to 3.3×1016
1×1017
7.5×1017
10×1020
10×1022 to 10×1024
1.25 to 2×103
2 to 3×105 //basal
plane
3.3×102 ⊥basal plane
~10−13
2.17
4.8
5×10−4 to 5×10−2
1.56×10−3
10−4 to 10-3
5.5×10−6
10−11 to 10−15
10−14
10−16 to 10-14
10−16
3×10−15 to 8×10−15
10−18
1.3×10−18
10−21
10−25 to 10−23
https://www.chegg.com/homework-help/questions-and-answers/table-302-resistivity-conductivityconducting-materials-conductivity-resistivity-material--q20357273
Learning Competency:
Describe the effect of ability of a material to conduct current in terms of resistivity
and conductivity. (STEM_GP12EM-IIIe-36)
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ACTIVITY 1: TRUE OF FALSE
Directions: State whether the each of the following statement is true of false.
1.The materials, which do not allow electric current to pass through them easily, are poor
conductors of electricity.
2.For metal conductors the electrical resistivity decrease with rise in temperature.
3.The electrical resistivity of a material is the measure of the property of a material due to
which it allows the electrons to flow through it.
4.Wires with different length and shape made with the same material have different
resistivity.
5.Wires with different length and shape made with the same material have the same
conductivity and resistivity, but different resistance.
6.Both conductivity and resistivity are temperature dependent.
7. Good conductors will have high value of resistivity.
8.The resistivity of a material is the measure of its property due to which it opposes
the flow of electrons through it.
9.Good conductor like copper has a very low resistivity
10.Changing the amount of material used changes the resistivity of that material.
ACTIVITY 2: WHICH IS RESISTIVE AND CONDUCTIVE?
Directions: Answer the following questions:
1. Which wire materials have lower resistances, thicker or thinner wires?
Higher resistances, longer or shorter wires?
2. Which metal has lower resistivity and is a better conductor of electricity?
Copper or Iron?
3. How resistance of a wire is being affected with its length?
4. Is copper the best metal conductor?
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ACTIVITY 3: LEAD! LEAD! LEAD!
Directions: Applying what you learned on this topic, answer the question below:
Why lead being a metal is a bad conductor of electricity and not used in
electric wires?
RUBRICS FOR SCORING
F. (10 points) Outstanding – student responses far exceed what is expected
G. (8 points) Very good – information is factually accurate and offers extra
supporting facts.
H. (6 points) Good – Sufficiently developed information with adequate
elaboration or explanation.
I. (4 points) Fair – Limited content with inadequate elaboration or explanation
J. (2 point) Poor – the ideas are not clear.
REFLECTION:
1. I learned that
____________________________________________________________
_________________________________________________________________
_____________________________________________________.
2. I enjoyed most on
____________________________________________________________
_________________________________________________________________
_____________________________________________________.
3. I want to learn more on
_____________________________________________________________
__________________________________________________________________
___________________________________________________.
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References:
✓ https://www.thoughtco.com/table-of-electrical-resistivity-conductivity-608499
✓ https://courses.lumenlearning.com/physics/chapter/20-3-resistance-andresistivity/
✓ https://www.circuitstoday.com/resistivity-electrical-conductivity
✓ https://www.education.com/science-fair/article/resistivity-iron-conductelectricity-copper/
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ANSWER KEY
ACTIVITY 1: TRUE OR FALSE
1. TRUE
2. TRUE
3. FALSE
4. FALSE
5. TRUE
6. TRUE
7. FALSE
8. TRUE
9. TRUE
10. TRUE
ACTIVITY 2: WHICH IS RESISTIVE AND CONDUCTIVE?
1. Thicker wires will have lower resistances, but longer wires will have higher
resistances.
2. Copper has a lower resistivity and is a better conductor of electricity than
iron.
3. The resistance of a wire increases with length. Because resistance is the
property of a material that resist electron flow, it makes sense that the longer
the length of a material is, the more resistance it will have.
4. Copper is a better conductor than iron, which means current can flow easier
(with less resistance) through copper. This is an inherent property of a
material.
ACTIVITY 3: LEAD! LEAD! LEAD!
-
Answers may vary.
-
Lead being a metal is a bad conductor of electricity. This is because it
readily reacts with the atmosphere to form lead oxide which does not
allow electricity to pass through it. Lead is a soft, silvery white or grayish
metal. Also, it is a highly toxic metal and a very strong poison.
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Prepared by:
JACKSON B. CASIBANG
SOLANA FRESH WATER FISHERY SCHOOL
234
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GENERAL PHYSICS 2
Name: ______________________________
Grade: ______________________________
Date: ________________
Score: ________________
LEARNING ACTIVITY SHEETS
RESISTANCE AND RESISTIVITY
Background Information for the Learner (BIL)
Is there an instance in your life where your mother scolds and blames you
because of the high electric bill in your house? If yes, have you ever wondered what
could have been the reason why she scolds you? Is it maybe because you keep
charging your cellphones, computer, tablets and your PSP? Or maybe because you
keep on watching television all day and the electric fan is also on. Well maybe it is
high time for you to roam around your house and see if the following scenarios is
seen in your home.
1. Do you use extension wires when you recharge your gadgets or when you
use your appliances?
2. Is your refrigerator or any electrical appliances situated in a warm place or
where the rays of the sun can reach it all day?
3. Is the electrical wirings in your home has narrow diameters?
If your answer to these three questions are all YES, then you are not the only one
to be blamed of the high electric bill but your whole family, including your mother who
scolds you. If you want to know the reason why, then go over this Leaning Activity
Sheet and be informed and be educated.
Consider Figure 1. If the same potential difference V is applied across a metallic
conductor and a block of wood, different current values will result. The property of
the material that enters here is the resistance.
Resistance is the opposition a material offers to current.
The symbol for resistance is R. All materials offer some resistance
to current but the amount of resistance differs from each other.
There are high resistance and low resistance materials. More
energy is required to move electrons through high resistance
materials.
The unit used to specify the amount of resistance is the ohm,
represented by the symbol Ω. An ohm is defined as the amount of
resistance that allows 1A of current to flow when the voltage is 1V.
It can also be defined as the amount of resistance of a column of
mercury 106.3cm in length, with a cross-sectional area of 1mm2,
and a temperature of 00C.
Applying a known potential V between the ends of the Figure 1. (a) A metallic
conductor and measure the resulting current i. The ratio of V to i is
material, (b) A block of
the resistance R, or
wood
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𝑅=
𝑉
𝑖
Eq. 1
If V is in volts and i in amperes, R is in ohms.
𝑣𝑜𝑙𝑡
1
= 1 𝑜ℎ𝑚 = 1Ω
𝑎𝑚𝑝𝑒𝑟𝑒
The Greek letter Ω is used for the word ohm.
Resistance of an object depends on four factors (see Table 1): 1) length, 2)
cross-sectional area, 3) resistivity of a material and 4) temperature. The amount of
resistance of an object is directly proportional to its length because the electrons will
encounter greater opposition as they flow through the wire, and inversely proportional
to its cross-sectional area because the electrons will have an easier path to travel.
Resistance then is mathematically defined as
𝐿
𝑅 ∝𝐴
Eq. 2
where R is the resistance, L is the length and A is the cross-sectional area of a
material.
Consider a cylindrical conductor of cross-sectional area A and length L. When
potential difference V is applied between its ends, a current i will result. Since both
ends of the conductor are equipotential surfaces, the electric filed intensity j will be
constant for all points in the conductor.
𝑉
Eq. 3
𝐸=
𝐿
𝑖
Eq. 4
𝑗=
𝐴
Hence,
𝑉
Eq. 5
𝐸 𝐿
𝜌= =
𝑖
𝑓
𝐴
or
𝑉 𝐴
Eq. 6
𝜌= ●
𝑖 𝐿
But
𝑉
Eq. 7, therefore
=𝑅
𝑖
𝐿
Eq. 8
𝑅=𝜌
𝐴
Rewriting equation 8 gives
𝐴
Eq. 9
𝜌=𝑅
𝐿
If A is a unit area and L is a unit length, then resistivity ρ is numerically equal
to R. The resistivity is numerically equal to the resistance offered by a conductor of
unit length and unit cross section to the passage of a current with the current flowing
in a direction perpendicular to the cross section. See Table 2 for the resistivity values
of some materials.
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Table 1. Factors that Affect Resistance
Factor
Less Resistance
Length
Greater Resistance
L1
L2
A1
A2
Copper
Aluminum
Cross-sectional Area
Type of Material
Temperature
Consider a situation when a variable potential difference V between the ends
of a 100-meter coil of no. 22 copper wire. For each applied potential difference, the
current i plotted against V (see Fig. 2).
The resulting straight line suggest
that the resistance of this conductor is
constant no matter what applied voltage
is used to measure it. This important
results hold for all metallic conductors and
is known as Ohm’s Law. This was
discovered and formulated by Georg Simon
Ohm. Mathematically, Ohm’s law is defined
as
𝑉
𝑅=
𝑖
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Example 1. Given an electric silver wire, find the resistance if it is 0.8m long and it
has a diameter of 1.21mm at 200C.
Given:
Length = L
0.8m
1𝑚
Diameter = d
0.21𝑚𝑚 𝑥 1 𝑥 103 𝑚𝑚 = 1.21 𝑥 10−3 m
Radius = r
Resistivity value of silver =ρ
6.05 x 10-4 m
1.59 𝑥 10−8 Ω●𝑚
Solution: From equation 8
𝐿
0.8𝑚
𝑅 = 𝜌 = (1.59𝑥 10−8 Ω●𝑚)
= 0.01Ω
(𝜋)(6.05 𝑥 10−4 𝑚)2
𝐴
Example 2. Karl Louie is trying to work on a project. In this case, he want to use a
glass rod as an insulator. The rod’s diameter is 10.54 mm and has a
resistivity value of 1* 109 Ω●m. How long must be the rod to offer a
resistance of 12.149Ω?
Given:
Diameter = d
Radius = r
Ρ
Resistance = R
1𝑚
10.54𝑚𝑚 ∗ 1∗103 𝑚𝑚 = 1.21 ∗ 10−2 m
5.22*10-3 m
1* 109 Ω●𝑚
12.149Ω
Solution
𝑅𝐴 (12.149 Ω)[(𝜋) ∗ (5.22 ∗ 10−3 𝑚)2 ]
𝐿=
=
= 3.31 ∗ 10−4 𝑚 𝑜𝑟 0.33𝑚𝑚
𝜌
1 ∗ 109 Ω●𝑚
Table 2. Resistivity of Some Materials
Resistance
Nichrome
Platinum
Iron
Tungsten
Aluminum
Gold
Copper
Silver
Glass
Quartz
Germanium
Silicon (pure)
Resistivity at 200C (Ω-m)
100 x 10-8
10.6 x 10-8
9.71 x 10-8
5.65 x 10-8
2.65 x 10-8
2.24 x 10-8
1.72 x 10-8
1.59 x 10-8
107 to 1010
7.5 x 1017
5.0 x 10-1
3.0 103
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Learning Competency:
Apply the relationship of the proportionality between resistance and the length
and cross-sectional area of a wire to solve problems. STEM_GP12EM-IIIe-37
ACTIVITY 1: CHOOSING THE BEST!
Directions: Choose the letter of the best answer.
1. Consider a wire of length L, cross-sectional-area A and a resistance R. A
second wire at the same temperature has a length 2L and a cross1
sectional area of 2 𝐴.What is the resistance of the second wire?
a. R
b. 2R
1
c. 2 𝑅
d. 4R
2. Xylene works in an electric company. She always recommends that longer
extension cords should have a thicker diameter. Her recommendation is
made because resistance is __________________.
a. directly with length and inversely with cross-sectional area
b. inversely with length and directly with cross-sectional area
c. directly with both length and cross-sectional area
d. inversely with both length and cross-sectional area
3. What is the resistance of a wire when its length is reduced by half?
a. Halved
b. quartered
c. doubled
d. quadrupled
4. Leejun is working on his physics project when his mother told him to wash
the dishes. He left the circuit he was working and forgot to remove the
source causing the wire to become hot. What happens to resistance when
temperature increases?
a. Decreases b. increases d. remains the same
d. unknown
5. A wire is always present in a complete circuit. Which graphs best
represents the relationship between the wire’s length and its resistance?
a.
b.
c.
d.
6. Which change decreases the resistance of a piece of copper wire?
a. An increase in the wire’s length
b. An increase in the wire’s resistivity
c. A decrease in the wire’s temperature
d. A decrease in the wire’s diameter
7. When an incandescent light bulb is turned on, its thin wire filament heats
up quickly. As the temperature of this wire filament increases, its electrical
resistance __________________.
a. Decreases b. increases c. remains the same
d. unknown
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8. A metal conductor is used in a circuit. The electrical resistance provided
by the conductor could be increased by __________________.
a. Decreasing the length of the conductor
b. Decreasing the applied voltage in the circuit
c. Increasing the temperature of the conductor
d. Increasing the cross-sectional area of the conductor
9. The diagrams below represents four pieces of copper wire at 200C.
Assume that temperature is the same for all of the four wires.
The piece of wire that has the greatest
resistance is _________________.
a.
Wire 1
b.
Wire 2
c.
Wire 3
d.
Wire 4
10. The electrical resistance of a metallic conductor is inversely proportional
to its ________________.
a. Temperature b. length
c. cross-sectional area
d. resistivity
ACTIVITY 2. FILL ME UP!
Directions: Fill in the blanks with the correct word to complete the paragraph. You
may choose from the table below.
Less
Travels
Increases
Resistance
Wires
Material
Short
Cross-sectional area
Thick
Length
First
Low
Insulators
Long
Difficult
Current
Temperature
Low
Ohm
Four
The amount of ______________ in a circuit also depends on the
______________ of the material through which it ______________. Resistance is
the measure of how ______________ it is for charges to flow through a material. The
greater the resistance, the ______________ current there is for a given voltage. The
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unit of measure of resistance is the ______________. There are ______________
factors that determine the resistance of a wire, or any object. The ______________
is the ______________ from which the wire is made. ______________ have high
resistance, and conductors have ______________ resistance. The second factor is
______________. ______________ wires have more resistance than
______________ wires. The third factor is ______________. Thin ______________
have more resistance than ______________ wires. The fourth factor is the
______________ of the wire. The electrical resistance of most materials
______________ as temperature increases.
ACTIVITY 3. CRITICAL THINKING!
Directions: Solve the following problems. Show your complete solution.
1. A copper wire 1mm in diameter and L m long has a resistance of 0.5 ohm.
Find the resistance of another copper wire at the same temperature if the
second wire is (a) 2mm in diameter and L m long, (b) 3mm in diameter and
0.5L m long.
2. Gauge no. 36 copper wire has a diameter of 0.058 m. Find the length of no.
36 copper wire which will give a resistance of 0.0005 ohms.
3. What is the resistance of gauge no. 10 copper wire (diameter of 0.026 m)
which is 1km long at 200C?
4. A cable consist of 10 strands of gauge no. 10 copper wire with a diameter of
0.5m . What is the resistance of 5km of this cable at 200C?
5. A copper wire has a resistance of 200 ohms. A second copper wire with twice
cross-sectional area and the same length. What is the resistance of the
second wire?
PREFORMANCE TASK I. REAL WORLD EXAMPLE – THE CIRCUIT MALL
A. Scenario
Imagine you are in line to enter a brand new shopping mall and you are very
much excited because after months of waiting, you will now be able to enter
shopping malls after the pandemic. The stores comprising the mall are very
popular. These stores include Converse, Gucci, Jansport, Fendi and all other
popular brands, and so the demand to enter is very high. Security guards only
allow one person through the doors at a time. As a result, the line and the wait
are quite long. To add to the chaos, the waiting crowd is made mostly of frequent
shoppers who are growing very impatient and holding up the line with their
constant complains. Together, these factors affect the flow of customers into the
shopping mall, just as electric resistance affects the flow of charges through a
current.
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B. Creative applications.
1. What makes less resistance favorable in a circuit?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
________________________________________________________
2. If the guard formed two adjacent lines to allow two people into the mall at a
time, we would expect the mall to be filled up faster. If we encased these two
lines within a wire, the wire would be wider and shorter than the original. In
terms of flowing electros why are wider and shorter wires favorable over
narrow, long ones?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
________________________________________________________
3. Compare the impatient shoppers to insulator in a circuit. What do insulators
do to current flow? Are insulators necessarily bad for the circuit? (Hint: How
they are helpful to humans?) Which type of material is commonly used as an
insulator?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
________________________________________________________
4. The type of customers (patient or impatient) the line is made up of will
determine how orderly or chaotic the shopping mall is. Most wires are made
up of what type of material and why? Name some examples of this material
that are ideal for wires.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
________________________________________________________
5. Let’s say security manages to calm down impatient shoppers with coupon
offers. The environment becomes less heated and so the line becomes
orderly. Therefore, do you think cooler wires have less resistance than warmer
wires? Why or why not?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
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_____________________________________________________________
________________________________________________________
Reflection
1. I learned that
_____________________________________________________________
_____________________________________________________________
__
2. I enjoyed most on
_____________________________________________________________
_____________________________________________________________
__
3. I want to learn more on
_____________________________________________________________
_____________________________________________________________
__
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References:
Halliday, D & Resnick (2007). Fundamentals of Physics (10th ed., pp 698). Quad
Graphics, USA.
Ling, Samuel and Sanny, Jeff (2005). University Physics Volume 2 (pp. 406-415).
Rice University, Texas, USA.
Serway, Reynard and Vuille, Chris (2005). College Physics (pp. 598-602). Boston
USA.
Retrieved from http://phet.colorado.edu/en/simulation/battery-resistor-circuit on
January 26, 2021.
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Answer Key
Activity 1.
1. a
2. a
3. a
4. b
5. b
6. c
7. b
8. c
9. b
10. c
Activity 2.
The amount of current in a circuit also depends on the resistance of the
material through which it travels. Resistance is the measure of how difficult it is for
charges to flow through a material. The greater the resistance, the less current there
is for a given voltage. The unit of measure of resistance is the ohm. There are four
factors that determine the resistance of a wire, or any object. The material is the first
from which the wire is made. Insulators have high resistance, and conductors have
low resistance. The second factor is length. Long wires have more resistance than
short wires. The third factor is cross-sectional area. Thin wires have more
resistance than thick wires. The fourth factor is the temperature of the wire. The
electrical resistance of most materials increases as temperature increases.
Activity 3.
1. a. 5.47*10-3Ω; b. 1.22 *10-3Ω
2. 75.80m
3. 0.03Ω
4. 4.38*10-4Ω-m
5. 100Ω
Prepared by:
GRACE ANN M. CALIBOSO – AGCAOILI
David M. Puzon Memorial National High School
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GENERAL PHYSICS 2
Name: ______________________________
Grade: ______________________________
Date: ________________
Score: ________________
LEARNING ACTIVITY SHEETS
OHMIC AND NON-OHMIC MATERIALS
Background Information for the Learner
Ohm’s Law, discovered and named after Georg Simon Ohm, states the
relationship between voltage, current and resistance of a conductor. This is important
in designing electrical and electronic circuits in order to ensure that the voltages and
currents in the components stay within specs. Just about any component that is
capable of carrying current is considered to be a conductor, it is just a matter of
whether the conductor is ohmic or non-ohmic. The main distinguishing feature
between these two is determined by Ohm’s Law.
Under constant physical conditions, the potential difference is proportional to
the current (𝑉 ∝ 𝐼), and is given by Ohm’s Law:
𝑉=𝐼𝑅
Fig. 1. A graph of Ohm’s Law
OHMIC CONDUCTORS
An ohmic conductor would have a linear relationship between
the current and the voltage (see Fig. 1). A good example of an ohmic
conductor is the resistor. The voltage drop across a resistor directly
correlated to the current that is flowing through it. But, this is only
true when the resistor is kept within the temperature range that is
rated for.
As more current flows through a resistor, it generates more
and more heat. This heat, when it becomes excessive, can cause Fig. 2. A graph showing the
the resistor to become non-ohmic and the resistance would also
relationship between V and I
increase. Even ordinary wires are also considered as Ohmic
conductors. These ordinary wires still have resistance but are often designed to be
extremely low to minimize losses.
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NON-OHMIC CONDUCTORS
Non-ohmic conductors do not follow Ohm’s Law and have their own
characteristics. This happens when the voltage increase but the current will not
increase in proportion as shown in Figure 3.
There are a number of examples of
non-Ohmic conductors, including bulb
filaments and semiconductors like diodes and
transistors.
Fig. 3. I-V curve of non-ohmic
conductor (headlamp bulb and I-V curve
of anon-ohmic conductor (diode)
A diode (see Fig. 4) provides a near
constant voltage drop even if you vary the current, so it does not follow Ohm’s Law.
The opposite happens in a light bulb filament; even as you increase the voltage
significantly, it only allows a certain amount of current to pass through.
Fig. 4. A diode as a semiconducting
device that allows current flow only if the
diode is forward biased, which means
that the anode is positive and the cathode
is negative
Even if non-ohmic conductors do not follow
Ohm’s Law, they have their own specialized uses that aid greatly in electrical and
electronic circuits. Incandescent light bulbs have been lighting our home for more
than a century and semiconductors have made a lot of things possible. Almost all
electronic gadgets like phones, computers and even ordinary watches and remotes
uses semiconductors.
I-V CURVE
The I-V Characteristic Curves, which is short for Current-Voltage
Characteristic Curves or simply I-V curves of an electrical device or component, are
a set of graphical curves which are used to define its operation within an electrical
circuit. It shows the relationship between the current flowing through an electronic
device and the applied voltage across its terminals.
It is generally used as a tool to determine and understand the basic
parameters of a component or device which can also be used mathematically model
its behavior within an electronic circuit.
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I-V CURVE OF AN OHMIC CONDUCTOR
The -V characteristic curves define the resistive
element, in the sense that if any voltage value is applied to
the resistive element, the resulting current is directly
obtainable from the I-V characteristics. As a result, the power
dissipated by the resistive element can also be determined
from the I-V curve.
Fig. 5. I-V Characteristic Curve of an
Ohmic Conductor
If the voltage and current are positive in nature, then the I-V characteristic curves will
be positive in quadrant I, if the voltage and current are negative in nature then the
curve will be displayed in quadrant III as shown in figure 5.
In a pure resistance, the relationship between voltage and current is linear and
constant at a constant temperature, such that current I is proportional to the potential
difference V times the constant of proportionality.
I-V CHARACTERISTIC CURVES OF DIODES
Fig. 6. I-V Characteristic Curve of A Non-Ohmic Conductor
(Source: General Physics 2 Quexbook)
When the diode is forward biased, anode positive with respect to the cathode,
a forward or positive current passes through the diode and operates in the top right
quadrant of its I-V Curve as shown in figure 6. Starting at the zero intersection, the
curve increases gradually into the forward quadrant but the forward current and
voltage are extremely small.
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When the forward voltage exceeds the diodes P-N junction internal barrier
voltage, which for silicon is about 0.7 volts, avalanche occurs and the forward current
increases rapidly for a very small increase in voltage producing a non-linear curve.
The “knee” point on the forward curve.
Likewise, when the diode is reversed biased, cathode positive with respect to
the anode, the diode blocks current except for an extremely small leakage current,
and operates in the lower left quadrant (se Fig. 6) of its I-V Characteristic Curve. The
diode continues to block current flow through it until the reverse voltage across the
diode becomes greater than its breakdown voltage point resulting in a sudden
increase in reverse current producing a fairly straight line downward curve as the
voltage losses control. This reverse breakdown voltage point is used to good effect
in Zener diodes.
Learning Competency
Differentiate ohmic and non-ohmic materials in terms of their I-V curves.
STEM_GP12EM-IIIe-38
ACTIVITY 1. CHOOSING THE BEST!
Directions: Choose the letter of the best answer.
1. It states the relationship between current, voltage, and resistance of a
conductor.
a. Faraday’s Law
c. Ohm’s Law
b. Len’z law
d. Le Chatelier Principle
2. What is the main difference between ohmic and non-ohmic materials?
a. Non-ohmic material changes in temperature.
b. Ohmic material varies in magnitude.
c. Ohmic material follows Ohm’s Law
d. Non-ohmic material follows Ohm’s Law
3. In an ohmic conductor, the relationship of current and voltage is
___________.
a. Non-linear
c. linear
b. Curve
d. scattered
4. In a non-ohmic material, the relationship of current and voltage is
___________.
a. Non-linear
c. linear
b. Curve
d. scattered
5. Resistors are example of _____________.
a. Ohmic material
c. both ohmic and non-ohmic
b. Non-ohmic
d. neither ohmic nor non-ohmic
6. Diodes are examples of ______________.
a. Ohmic material
c. both ohmic and non-ohmic
b. Non-ohmic
d. neither ohmic nor non-ohmic
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7. What happens as more current flows through a resistor?
a. It explodes
c. It cools down
b. It dissipates power
d. It generates more heat
8. Why does diodes do not follow Ohm’s Law?
a. Because it provides constant current all the time
b. Because it provides a constant drop
c. Because it automatically varies according to its temperature
d. Because it can act like a switch
9. Non-ohmic materials are useful in society nowadays.
a. True
b. false
c. partly true
d. partly false
10. Why is Ohm’s Law important in designing electrical and electronic circuits?
a. To ensure that heat would be dissipated equally throughout the circuit
b. To ensure that resistance will be maximize as current flows through the
circuit
c. To ensure that voltage and current in components stay within specs
d. None of these choices
ACTIVITY 2. CLASSIFY ME.
Directions: Write O if the device and/or material is Ohmic and NO if it Non-Ohmic
1. _____ Transistors
6. ____ Metals
2. _____ Incandescent light bulbs
7. ____ Electrolytes
3. _____ Potentiometer
8. ____ Nichrome
4. _____ Transmission lines
9. ____ Thyristors
5. _____ Vacuum tubes
10. ____ Copper wire
ACTIVITY 3. PLEASE COMPLETE ME!
Directions: Comparison Table Between Ohmic and Non-Ohmic Conductors.
Complete the table by filling up the needed information. Refer to the parameter of
comparison on Column 1.
Parameter of
Ohmic Conductors
Non-Ohmic Conductor
Comparison
Basic Definition
Relationship between
current and voltage
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The slope between
current and voltage
Effect of temperature
variations
ACTIVITY 3. COMPARE ME NOT!
Directions: Compare and contrast Ohmic and Non-Ohmic Materials using a Venn
Diagram
Ohmic Materials
Non-Ohmic Materials
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ACTIVITY 4. REAL LIFE APPLICATION
Write an essay about the importance and uses of ohmic and non-ohmic materials to
you, to industry and to the society as a whole. Your essay will be graded using the
rubric below.
Level
Outstanding
9-10
Good
7-8
Fair
6
Poor
4-5
Very Poor.
10-Point Rubric
Description
Well written and very organized.
Excellent grammar mechanics.
Clear and concise statements.
Excellent effort on the presentation of
details.
Demonstrate a thorough understanding
of the topic.
Writes fairly clear. Good grammar
mechanics.
Good presentation and organization.
Sufficient effort and detail.
Minimal
effort.
Good
grammar
mechanics.
Fair presentation and few supporting
details.
Somewhat unclear. Shows little effort.
Poor grammar mechanics. Confusing
and choppy, incomplete sentences. No
organization of thoughts.
Lacking effort. Very poor grammar
mechanics. Does not address topic and
thoughts are very unclear.
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Reflection
1.I learned that
_____________________________________________________________
_____________________________________________________________
__
2.I enjoyed most on
_____________________________________________________________
_____________________________________________________________
__
3.I want to learn more on
_____________________________________________________________
_____________________________________________________________
__
253
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References
Retrieved from http://www.sciencedirect.com/science/article/pii/002230937290227X
on January 20, 2021.
Retrieved from http://www.tandfonline.com/doi/abs/1.01080/14786445008560976
on January 20,2021
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Answer Key
Activity 1
1. C
2. C
3. C
4. A
5. A
6. B
7. D
8. B
9. A
10. C
Activity 2
1. O
2. NO
3. O
4. O
5. NO
6. O
7. NO
8. O
9. NO
10. O
Activity 3
Parameter of
Comparison
Basic Definition
Relationship between
current and voltage
Ohmic Conductors
Non-Ohmic Conductor
Ohmic conductors follow
Ohm’s law, which implies
that the resistance of the
conductors
remains
constant
on
varying
current and voltage.
Non-ohmic conductors do
not follow Ohm’s law,
which
means
the
resistance
of
the
conductor
varies
on
sharing current, voltage
and temperature.
In non-ohmic conductors,
the current and voltage
are
not
directly
proportional
to
one
another, that is, current
and voltage have a nonlinear
relationship
between them.
In ohmic conductors, the
current and voltage are
directly proportional to
each other, that is, there
is a linear relationship
between current and
voltage.
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The slope between
current and voltage
The
slope
between
current and voltage in
ohmic conductor is a
straight line.
The slope between the
current and the voltage in
non-ohmic conductor is
not straight but a curved
line.
Effect of temperature
variations
Conductors follow Ohm’s
law when the temperature
is in the range for which
the conductor has been
made
but
as
the
temperature increases,
ohmic conductors also
behave as non-ohmic
conductors.
In non-ohmic conductors,
the resistance of the
conductors
varies
according to the variation
in the temperature.
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Activity 3.
Ohmic Materials
•
•
•
•
Non-Ohmic Materials
Follows Ohm’s law
Resistance
remains
unchanged on changing
current and voltage
Current and voltage are
directly proportional to
each other (linear
relationship)
I-V graph, the slope is a
straight line
•
•
•
•
•
•
Does not follow Ohm’s
law
Resistance changes on
changing current, voltage
and temperature
Current and voltage are
not directly proportional
(non-linear relationship)
I-V graph, the slope is a
curve line
They are both conductor
They both dissipate heat
Prepared by:
GRACE ANN M. CALIBOSO – AGCAOILI
David M. Puzon Memorial National High School
257
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GENERAL PHYSICS 2
Name: ___________________________________
Date: _____________
Grade Level & Strand: _____________________
Score: ____________
LEARNING ACTIVITY SHEETS
ELECTROMOTIVE FORCE AND POTENTIAL DIFFERENCE
Background Information for The Learners
You would be saddened if you found a battery-operated toy car, but you cannot play
with it for it has no battery. The toy car needs a battery that gives an electromotive force
(also called emf) to maintain a potential difference across its circuit while it is running.
Potential difference and emf are two concepts often used interchangeably. However, these
two concepts are not the same. In this material, you will learn how to distinguish an
electromotive force from the potential difference.
The Water-Pressure Analogy
The best way to understand an
electromotive force (emf) and a potential
difference in a circuit is to think of a waterpressure
analogy.
Consider
the
two
connected tanks filled with water of different
heights, as shown in Figure 1. Tank A, in
which water is at a greater height, has more
potential (pressure) than tank B with lower
water
height.
Because
of
the
potential
difference between the two tanks, water will
flow from tank A to tank B. Note that water
flows by itself from higher potential to lower
potential.
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The water stops flowing when the two tanks have the same amount of water, as
displayed in Figure 2. In this case, there is no (zero) difference in the potential of the two
tanks. Also, note that water cannot flow by itself from lower potential to higher potential.
In figure 3, a pump is used and does
the work to drive back the water from tank B
to tank A. So, tank A will always be at a higher
potential than tank B, letting the water
continue flowing from tank A to B. And this is
how an electric circuit works.
The representations used in the analogy are as follows:
1. The water tank system is like an electric circuit, which provides a path for
electrons (charges) to flow.
2. The water characterizes the charges flowing in an electric circuit.
3. The two tanks denote the two distinct points across an electric circuit - the
positive terminal (at which the electric potential is high) and the negative
terminal (at which the electric potential is low).
4. The pump portrays the source’s emf, a driving force to move the charges from
a lower potential to a higher potential region.
The potential difference between two tanks is like the potential difference across two
points in a circuit.
Consider Figure 4. When a potential difference due to an electric field is
present between two distinct points (a and b), charges flow across it. It flows from a
higher electric potential to a lower electric potential. (Hint: Current is the amount of
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charges flowing in a conductor per unit time). However, the electric field between
these points is not enough to sustain the current in the circuit. It needs a driving force
to maintain the flowing of charges from a lower potential to a higher potential region.
This driving force sustains the potential difference across a circuit. And this driving
force is called emf - induced by a battery or any source of energy.
Difference Between Electromotive Force and Potential Difference
Electromotive force is the amount of energy supplied by the source to each coulomb
of charge. In simpler words, it is the energy converted to electrical energy per unit of charge.
Mathematically, it defined as the work done by the source to drive a unit charge around the
circuit. In symbols,
𝜀=
𝑊
𝑄
where: ε (Greek letter epsilon) denotes emf in volt (V)
W denotes the work done in joule (J)
Q denotes charge in coulomb (C)
Potential difference is the amount of energy used by one coulomb of charge in
moving from one point to another point in a circuit. It is the electrical energy converted to
other forms of energy per unit charge. Mathematically, it is defined as the work done to drive
a unit charge across two points in the circuit. In symbols,
𝑉=
𝑊
𝑄
where: V denotes potential difference in volt (V)
W denotes the work done in joule (J)
Q denotes charge in coulomb (C)
Example 1: Consider the circuit on the right.
The battery has an emf of 12 V. It implies that the
battery supplies 12 J of energy to each coulomb of
charge as the charge travels from the positive terminal to the negative terminal through an
external circuit. Also, it suggests that 12 J of work is done in driving a unit of charge around
the circuit.
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Example 2: Consider the circuit below.
A 12-volt supply applies across the total resistance
of the circuit. The potential difference between any
two points, says A and B, is the energy used by one
coulomb of charge in moving from point A to point
B. Using Ohm’s Law, the potential difference between point A and B is 7 volts. It indicates
that the work done to move a charge from point A to B is 7 J. Also, it means that 7 J of
energy is converted to other forms of energy like sound, light, and heat as it passes across
a 7 Ω resistor.
The comparison chart below shows some of the other contrasting points of emf and
potential difference.
Electromotive force
Potential Difference
It is independent of the resistance in the It is directly dependent on the resistance
circuit.
of the circuit.
It transmits current throughout the circuit.
It transmits current between two points.
It is greater than the potential difference It is always less than the maximum value
between any two points.
of emf.
It is the cause.
It is the effect.
It remains constant.
It does not remain constant.
It is the maximum voltage that the source It is less than the maximum voltage that
can transfer.
the source delivers.
It gains energy.
It loses energy.
It is present even when no current is It is zero in the absence of current.
drawn through the battery.
Learning Competency
Differentiate emf of a source and potential difference across the circuit.
(STEM_GP12EMIIIe-40)
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ACTIVITY 1: LET’S CROSS THE PUZZLE
Directions: Read the clues to complete the crossword. The words are related to
emf and potential difference.
ACROSS
DOWN
1. electromotive force
2. the potential is lower at this terminal of the
3. symbol of electromotive force
battery
6. it denotes the amount of work done
4. the potential difference when no charges
8. it is represented by water in the analogy
flow in the circuit
10. SI unit for potential difference
5. SI unit of charge
7. the electric potential at the positive terminal
of a battery
9. an example of a source
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ACTIVITY 2: LET’S MAKE A CHOICE
Directions: Choose the best answer from the options. Write the letter of your choice in the
space provided.
_____ 1. The _____ is the measure of energy that it gives to each coulomb of charge,
whereas the _____ is the amount of energy used by the one coulomb of charge.
a. emf, potential difference
c. emf, emf
b. potential difference, emf
d. potential difference; charge
_____ 2. It is the work done by a battery to drive an electron to move around the circuit.
a. V
c. W
b. Q
d. ε
_____ 3. In the absence of emf, charges move from a _____ potential region to a _____
potential region.
a. lower; higher
c. high; high
b. higher; lower
d. low; low
_____ 4. It is the energy converted to light and heat as the charges travel across a lightbulb.
a. Emf
c. potential difference
b. work done by the source
d. chemical energy
_____ 5. The potential difference is zero when the current across the circuit is ______.
a. high
c. zero
b. low
d. equal to emf
ACTIVITY 3: LET’S FIND THE LIE
Directions: Write True if the statement is correct; otherwise, change the underlined word or
phrase to make it right. Write the answer in the space provided before the
statement.
_____________ 1. Current flows from one point to another point in the circuit when the
potential difference between the two points is zero.
_____________ 2. Both emf and potential difference are measured by volts.
_____________ 3. Potential difference is energy and emf is force.
_____________ 4. If we say that the voltage applied to the radio is 120 volts, we mean that
the emf between the two electrical contacts on the lightbulb is 120 volts.
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_____________ 5. The emf exists in the circuit even when the current does not flow in the
circuit.
ACTIVITY 4: LET’S APPLY LEARNINGS
Directions: Read, understand, and perform the tasks by applying what you learned.
1. Make a Venn Diagram comparing emf of a source and potential difference across
the circuit.
2. How much work is done in moving one coulomb of charge from 9 V potential to a
point where the potential is 5 V?
3. The battery supplies a maximum voltage of 12 V. When it is connected to an external
circuit, the voltage measured by the voltmeter across the circuit is only 10 V. Is this
possible? Explain your answer.
Reflection
1. I learned that ______________________________________________________
_______________________________________________________________________
_______________________________________________________________
2. I enjoyed most on __________________________________________________
_______________________________________________________________________
_______________________________________________________________
3. I want to learn more on ______________________________________________
_______________________________________________________________________
_______________________________________________________________
264
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REFERENCES
Circuit Globe. “Difference Between Electromotive Force and Potential Difference”.
Accessed February 2,
2021.
https://circuitglobe.com/difference-betweenelectromotive-force-and-potential-difference.html
Electrical
Genius.
“Voltage”.
Accessed
February
https://electricalg4u.blogspot.com/2015/06/voltage.html
2,
2021.
Moore, Thomas A. Six Ideas that Shaped Physics, Unit E: Electric and Magnetic Fields are
Unified. 2nd ed. New York: Mc Graw Hill, 2003.
Physics About. “What Is the Difference Between emf and Potential Difference”. Accessed
February 2, 2021. https://physicsabout.com/difference-emf-potential-difference/
Serway, Raymond A. and Jewette, John W. Jr. Physics for Scientists and Engineers with
Modern Physics. 6th ed. Singapore: Thomson Learning Asia, 2004.
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ANSWER KEY
ACTIVITY 1
1. emf
2. higher
3. negative
4. charges
5. volts
6. epsilon
7. battery
8. joule
9. zero
10. joule
ACTIVITY 2
1. a
2. d
3. b
4. c
5. c
ACTIVITY 3
1. zero
2. True
3. energy
4. potential difference
5. True
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ACTIVITY 4
1. (Answers may vary. See
sample Venn Diagram)
2. 4 joules
3. Yes, it is possible. The 10-V
voltage measured in a voltmeter is the potential difference across the circuit.
The potential difference is dependent on the resistance of the circuit and is
always lesser than the maximum value of emf. 10 V potential difference is less
than the 12 V emf.
Prepared by:
Techie Gammad-Vera Cruz
Amulung National High School
267
NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 2
Name: ____________________________________Grade Level: _________
Date: ______________________________________Score:______________
LEARNING ACTIVITY SHEET
ENERGY AND POWER IN ELECTRIC CIRCUITS
Background Information for the Learners (BIL)
Power is associated by many people with electricity. Knowing that power is
the rate of energy use or energy conversion, what is the expression for electric
power? Consider a lamp connected to a pair of batteries as shown in figure1 , treat
the connecting wires as ideal conductors with no resistance and so virtually no
potential difference across them but full potential difference of the battery appears
across the lamp. The magnitude of the current in the circuit is constant , the condition
known as direct current.
As the current passes through the lamp , charges are moving from a higher
potential to a lower one. Energy is being lost from the battery and converted in the
filament of the lamp into heat and light. The amount of energy released by a charge
q as it falls through the potential V across the lamp is W = qV. Since the potential is
constant , the rate at which energy is released , or the power P is
P=
=
∆𝑊
∆𝑡
∆ q( V)
∆𝑡
∆q
P = 𝑉 ∆𝑡
But
∆q
∆𝑡
Figure 1
is the current so P = IV
where :
P = power in watts, W
I = current in amperes, A
V= voltage in volts , V
Q = charge in coulombs, C
W= work in joules, J
t = time in seconds , s
The SI unit of power is watt, W , based from the units derived from
P = IV
=
=
𝐜𝐨𝐮𝐥𝐨𝐦𝐛
𝐬𝐞𝐜𝐨𝐧𝐝
joule
second
𝐣𝐨𝐮𝐥𝐞
. 𝐜𝐨𝐮𝐥𝐨𝐦𝐛
= watt
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Take note that 1 watt, W = 1 joule/second, J /s
In a resistor, the energy dissipated appears as thermal energy. This effect is used
in appliances such as electric stoves, hair dryers and heaters. In an incandescent
lamp, the energy delivered to the filament raises its temperature so high that light is
emitted. In other circuit elements, the energy may take on different forms. For
example, the energy may appear as mechanical work done by a motor, as sound
from a loudspeaker, or as stored chemical energy in a battery when the battery is
being recharged. Conversion from electrical to mechanical is never 100% efficient.
The difference appears as heat.
When an electric current pass through a resistor, electrical energy is irreversibly
transformed to thermal energy. And so we can write another equation for power that
is P = I 2 R . Another equation relating power to the resistance and voltage across
an electric device is P = V2 / R . These equations known as Joule’s law.
Let us have examples on how to solve problems involving power in electric
circuits.
Example 1. A typical designed to operate on a 120 – V household circuit is rated at
1500 W . What is the resistance of the dryer?
Given :
V = 120 V
P = 1500 W
Find: R
Solution :
P
I= V
=
R=
1500 W or 1500 J/s
120 V or J/C
𝑪
𝐈 = 𝟏𝟐. 𝟓 𝒔 𝒐𝒓 𝑨
=
V
I
120 V
12.5 A
𝐑 = 𝟗. 𝟔𝟎 Ω
Example 2. A piece of wire has a resistance of 30 Ω . How much power is dissipated
in the wire if it carries a current of 0.50 A?
Given :
R = 30 Ω
I = 0.50 A
Find: P
Solution :
P = I2 R
= (0.50𝐴)2 (30 Ω)
𝐏 = 𝟕. 𝟓𝐀𝟐 Ω 𝐨𝐫 𝟕. 𝟓 𝐖
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The Cost of Electricity
The more electric appliances you use and the longer they are left on, the higher
your
electric bill. This familiar fact is based on the relationship between energy and power.
You pay for the energy used. Since P = E/ t, we see that the energy dissipated in a
circuit is the product of the power and time, E= Pt and the energy used by a device
using power P for a time interval t .For example, the more light bulbs burning, the
greater P used; the longer they are on, the greater t is. The energy unit on electric
bills is the kilowatt-hour (kW ⋅ h), consistent with the relationship E = Pt.. It is easy to
estimate the cost of operating electric appliances if you have some idea of their power
consumption rate in watts or kilowatts, the time they are on in hours, and the cost per
kilowatt-hour for your electric utility.
Study the example below on how energy is dissipated in the different appliances
at home.
Example 3. In a stairwell of a ten- storey building there are two continuously burning
75- W safety lamps for each floor?
a. What is the total energy in kilowatt-hours used in 1 year?
b. What will it cost to use the lamps for a year if the cost of electricity is
5.014/ kWh?
Given :
P total use = 2 lamps/floor x 10 floors x 75W/ lamp = 1,500 W𝑥
1𝑘𝑊
1000 𝑊
= 1.5
kW
Cost of energy/kWh = 5.014/kWh
Find:
a. Total energy consumed in a year
b. Cost for a year of operation
Solution :
𝑎. W = E = Pt
365days
24 hours
= 1.5 kW(1year) ( 1 year ) ( 1 day )
𝐄 = 𝟏. 𝟑𝒙𝟏𝟎𝟒 𝒌𝑾𝒉
b. Cost = Energy consumed x Cost of energy/kWh
= 1.3𝑥104 𝑘𝑊ℎ x 5.014/kWh
Cost for
a year of
operation = 𝐏 𝟔𝟓, 𝟏𝟖𝟐
For you to better understand the concept of electric power and energy in electric
circuits , do the following activities :
Learning Competency :
Given an emf source connected to a resistor , determine the power supplied or
dissipated by each element in a circuit (STEM_GP12V-EM - IIIf-47)
270
NOTE: Practice personal hygiene protocols at all times
ACTIVITY 1: POWER EQUATIONS AND UNITS CHALLENGE
Directions : Write the formula of electric power based on the description in the
table below :
Description of Power in words
Formula of Power
1.Power = current x time
2.Power = current squared x resistance
3.Power = voltage squared x resistance
4. Show that the units of A2 Ω are watts as implied by the equation P= I2 R
5. Show that the units 1V2/Ω = 1W as implied by the equation P = V2/ R.
.
ACTIVITY 2 : CALCULATING ENERGY AND POWER IN ELECTRIC CIRCUITS
Directions : Read, understand and analyze each of the problems very carefully.
Then, solve and show your complete solutions. Encircle your final
answer. (5points each)
1. What is the power consumption of an electric iron if its resistance is 13.1 Ω
and it operates on a household circuit with a voltage of 120 V?
2. A flashlight lamp connected to a battery that provides 1.4 V draws a current
of 0.10 A . What electric power is used by the lamp?
3. The label on a toaster reads 800 W at 120 V . How much current does it draw?
4. A 100 Ω resistor is rated at 1.00 W maximum power capacity.
a. What is the maximum voltage that can be applied across the resistor
without exceeding its maximum power rating?
b. What is the current at this voltage ?
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NOTE: Practice personal hygiene protocols at all times
5. Find the power dissipated in each of these extension cords: (a) an extension
cord having a 0.0600 Ω resistance and through which 5.00 A is flowing; (b) a
cheaper cord utilizing thinner wire and with a resistance of 0.300 Ω.
6. A drilling machine operates with an electric power consumption of 840W. How
much does it cost to operate the machine continuously for eight hours if the
electricity costs P 5.012 per kWh ?
7. A 150 W street lamp is operated for 12 hours a day. How much energy does
it take to operate the lamp for 30 days ? Express your answer in kilowatt hours
.
8. How much does it cost to operate a 100- W lamp hours a day for 30 days if
electricity costs 12. 45 / kWh ?
9. How many joules of energy are necessary to run a washing machine for 30
minutes if it is hooked up to a 220-V line and has a resistance of 10 ohms?
10. A 800 C flow through a flashlight with 5 ohms of resistance and is on for 30
minutes. How much power was used?
ACTIVITY 3 : POWER SAVING !!!
Directions : Make a poster on how you can save energy and power on a piece of
A4 size bond paper. Below is the rubric for evaluating your output.
CATEGORY
Graphics
Clarity
Graphics
Originality
Graphics
Relevance
-
–
–
4
3
2
1
Graphics are all in
focus
and
the
content
easily
viewed
and
identified from 6 ft.
away.
Several
of
the
graphics used on
the poster reflect a
exceptional degree
of student creativity
in their creation
and/or display.
Most graphics are
in focus and the
content
easily
viewed
and
identified from 6 ft.
away.
Most
graphics
are in focus and
the content is
easily
viewed
and
identified
from 4 ft. away.
Many graphics are
not clear or are too
small.
One or two of the
graphics used on
the poster reflect
student creativity
in their creation
and/or display.
The graphics are
made by the
student, but are
based on the
designs or ideas
of others.
No graphics made
by the student are
included.
All graphics are
related to the topic
and most make it
easier
to
understand.
All
borrowed graphics
have a source
citation.
All
graphics
relate to the
topic.
Most
borrowed
graphics have a
source citation.
Graphics do not
relate to the topic
OR
several
borrowed graphics
do not have a
source citation.
All graphics are
related to the topic
and make it easier
to understand. All
borrowed graphics
have
a
source
citation.
272
NOTE: Practice personal hygiene protocols at all times
Several items of
importance on
the poster are
clearly labeled
with labels that
can be read from
at least 3 ft.
away.
All but 1 of the
required
elements
are
included on the
poster.
Student
can
accurately
answer
about
75%
of
questions
related to facts in
the poster and
processes used
to create the
poster.
3-4
accurate
facts
are
displayed on the
poster.
The poster is
acceptably
attractive though
it may be a bit
messy.
Labels
All
items
of
importance on the
poster are clearly
labeled with labels
that can be read
from at least 3 ft.
away.
Almost all items of
importance on the
poster are clearly
labeled with labels
that can be read
from at least 3 ft.
away.
Required
Elements
The poster includes
all
required
elements as well as
additional
information.
All
required
elements
are
included on the
poster.
Knowledge
Gained
Student
can
accurately answer
all questions related
to facts in the poster
and processes used
to create the poster.
Student
can
accurately answer
most
questions
related to facts in
the poster and
processes used to
create the poster.
At least 7 accurate
facts are displayed
on the poster.
5-6 accurate facts
are displayed on
the poster.
Attractiveness
The
poster
is
exceptionally
attractive in terms of
design, layout, and
neatness.
The
poster
is
attractive in terms
of design, layout
and neatness.
Title
Title can be read
from 6 ft. away and
is quite creative.
Title can be read
from 6 ft. away and
describes content
well.
Title can be read
from 4 ft. away
and
describes
the content well.
Mechanics
Capitalization and
punctuation
are
correct throughout
the poster.
There is 1 error in
capitalization
or
punctuation.
There are 2
errors
in
capitalization or
punctuation.
Content
Accuracy
–
Labels are too
small to view OR
no important items
were labeled.
Several required
elements
were
missing.
Student appears to
have insufficient
knowledge about
the
facts
or
processes used in
the poster.
Less
than
3
accurate facts are
displayed on the
poster.
The
poster
is
distractingly
messy or very
poorly designed. It
is not attractive.
The title is too
small and/or does
not describe the
content of the
poster well.
There are more
than 2 errors in
capitalization
or
punctuation.
273
NOTE: Practice personal hygiene protocols at all times
Reflection
Complete this statement:
I learned that
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
____________
I enjoyed most on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
____________
I want to learn more on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
____________
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REFERENCES:
Jones E. , Childers , R. . Contemporary College Physics ( 2nd Edition ) : AddisonWesley Publishing Company Inc. pp. 500-504
Walker , J. , Halliday, R. , & Resnick D. (2011) . Fundamentals of Physics ( 9th
Edition): Hoboken, NJ : John Wiley & Sons Inc. pp. 605-606
https://courses.lumenlearning.com/physics/chapter/20-4-electric-power-and-energy/
https://www.google.com/search?q=energy+and+power+in+electric+circuits&source
=lnms&tbm=isch&sa=X&ved=2ahUKEwi2osHzibTuAhXGUt4KHUF3DNYQ_AUoAX
oECBAQAw&biw=1366&bih=657#imgrc=VBtBl9Pvq3d
https://intl.siyavula.com/read/science/grade-11/electric-circuits/11-electric-circuits03
https://drive.google.com/file/d/1UBn5AnlVhN8o63YBRGCWKtTlWGT3WHRe/view
275
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Answer Key
ACTIVITY 1: POWER EQUATIONS AND UNITS CHALLENGE
Description of Power in words
1.Power = current x time
Formula of Power
P= IV
2.Power = current squared x resistance
P= I2 R
3.Power = voltage squared x resistance
P = V2 / R
4. A2 Ω are watts as implied by the equation P= I2 R
= A2 Ω
= (A) ( A )(V / A)
= coulomb / second x joule /coulomb
Watt = joule/second
5. 1V2/Ω = 1W as implied by the equation P = V2/ R
= V.V / V/A
= V.V . A/V
= (V )( A)
= (joule /coulomb) ( coulomb / second )
1 watt = 1 joule/second
ACTIVITY 2 : CALCULATING ENERGY AND POWER IN ELECTRIC CIRCUITS
1. P = 1099. 24 W
2. P = 0.14 W
3. P = 6.67 A
4. V= 10 V ; I = 0.1 A
5. a. P = 1.50 W
b. P = 7.50 W
6. E= 6.72 kWh cost = P 33.68
7. E = 54 kWh
8. E = 72 kWh
cost = 896 .40
9. E= 17.42 x107 joules
10. P = 0.968 W
ACTIVITY 3: POWER SAVING !!!
Answers vary
Prepared by:
FE S. CAGUMBAY
ANDARAYAN NATIONAL HIGH SCHOOL
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NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score:______________
LEARNING ACTIVITY SHEET
CURRENT, RESISTANCE AND RESISTIVITY
Background Information for the Learners (BIL)
Electric Current
I=
∆q
∆t
Electric current is defined to be the rate at which charge flows. A large
current, such as that used to start a truck engine, moves a large amount of charge in
a small time, whereas a small current, such as that used to operate a hand-held
calculator, moves a small amount of charge over a long period of time. In equation
form, electric current I is defined to be :
where: Δq - the amount of charge passing through a given area
Δt - time (In previous lessons ,initial time is often taken to be zero, in which
case Δt = t .)
The SI unit for current is the ampere (A), named for the French Physicist AndréMarie Ampère (1775–1836). Since I =
∆q
∆t
, we see that an ampere is one coulomb
per second:
1 A = 1 C/s .
How do you calculate electric current ? Study the examples below on how to
compute for electric current.
Example 1
What is the current involved when a truck battery sets in motion 720 C of charge
in 4.00 s while starting an engine?
Given : q = 720 C
t = 4.00 s
Find :I
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NOTE: Practice personal hygiene protocols at all times
Solution : I =
∆q
∆t
720 C
= 4.00 s
I = 1.80 A
Example 2
How long does it take 1.00 C of charge to flow through a handheld calculator if a
0.300-mA current is flowing
∆𝑡 =
=
∆𝑞
𝐼
1.00 C
0.300−3
𝑐/𝑠
∆𝑡 = 3.33𝑥103 𝑠 s
Ohm’s Law: Resistance and Simple Circuits
What drives current? We can think of various devices—such as batteries,
generators, wall outlets, and so on—which are necessary to maintain a current. All
such devices create a potential difference and are loosely referred to as voltage
sources. When a voltage source is connected to a conductor, it applies a potential
difference V that creates an electric field. The electric field in turn exerts force on
charges, causing current.
The current that flows through most substances is directly proportional to the
voltage V applied to it. The German Physicist Georg Simon Ohm (1787–1854) was
the first to demonstrate experimentally that the current in a metal wire is directly
proportional to the voltage applied:
I∝V
This important relationship is known as Ohm’s law. It can be viewed as a causeand-effect relationship, with voltage the cause and current the effect.
If voltage drives current, what impedes it? The electric property that impedes
current is called resistance R . Collisions of moving charges with atoms and
molecules in a substance transfer energy to the substance and limit current.
Resistance is defined as inversely proportional to current, or
I ∝ 1/ R
Thus, for example, current is cut in half if resistance doubles. Combining the
relationships of current to voltage and current to resistance gives
I = V/ R
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This relationship is also called Ohm’s law. Ohm’s law in this form really defines
resistance for certain materials The many substances for which Ohm’s law holds are
called ohmic. These include good conductors like copper and aluminum, and some
poor conductors under certain circumstances.
Ohmic materials have a resistance R that is independent of voltage V and current
I . An object that has simple resistance is called a resistor, even if its resistance is
small. The unit for resistance is an ohm and is given the symbol Ω .
Rearranging I = V/R gives R = V/I , and so the units of resistance are 1 ohm = 1
volt per ampere:
1 Ω = 1V/ A
Figure 2 shows the schematic for a simple circuit. A simple circuit has a single
voltage source and a single resistor. The wires connecting the voltage source to the
resistor can be assumed to have negligible resistance, or their resistance can be
included in R .
Figure 1 .A simple electric circuit in which a closed path for current to flow is supplied by conductors (usually
metal wires) connecting a load to the terminals of a battery, represented by the red parallel lines. The zigzag
symbol represents the single resistor and includes any resistance in the connections to the voltage source.
An example on how to apply Ohm’s law is shown below :
Example 3
A voltage of 10 volts is placed across a 500 ohm resistor. Calculate the amount
of current that will flow in the circuit
Given :
V = 10 V
R = 500 Ω
Find : I
Solution
:
I=
=
V
R
10 V
500Ω
I = 0.02 A
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Resistance and Resistivity
Material and Shape Dependence of Resistance
The resistance of an object depends on its shape and the material of which it is
composed. The cylindrical resistor in Figure 3 is easy to analyze, and, by so doing, we can
gain insight into the resistance of more complicated shapes. As you might expect, the
cylinder’s electric resistance R is directly proportional to its length L , similar to the
resistance of a pipe to fluid flow. The longer the cylinder, the more collisions charges will
make with its atoms.
The greater the diameter of the cylinder, the more current it can carry (again similar to
the flow of fluid through a pipe). In fact, R is
inversely proportional to the cylinder’s cross-sectional area A .
Figure2 A uniform cylinder of length L and cross-sectional area A . Its resistance to the flow of current is similar to the
resistance posed by a pipe to fluid flow. The longer the cylinder, the greater its resistance. The larger its cross-sectional
area A , the smaller its resistance.
For a given shape, the resistance depends on the material of which the object is
composed. Different materials offer different resistance to the flow of charge. We
define the resistivity ρ of a substance so that the resistance R of an object is directly
proportional to ρ . Resistivity ρ is an intrinsic property of a material, independent of its
shape or size. The resistance R of a uniform cylinder of length L , of cross-sectional area
A , and made of a material with resistivity ρ , is
R=
Where : R – resistance of material in ohms, Ω
L – length of the conductor in meters, m
A –cross sectional area in square meters , m2
ρ-resistivity in Ω ⋅ m
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Table 1. Resistivities ρ of Various materials at 20ºC
Resistivity at 20 C
ρ(Ω⋅m)
Material
Conductors
Silver
1.59×10−8
Copper
1.72×10−8
Gold
2.44×10−8
Aluminum
2.65×10−8
Tungsten
5.6×10−8
Iron
9.71×10−8
Platinum
10.6×10−8
Steel
20×10−8
Lead
22×10−8
Manganin (Cu, Mn, Ni alloy)
44×10−8
Constantan (Cu, Ni alloy)
49×10−8
Mercury
96×10−8
Nichrome (Ni, Fe, Cr alloy)
100×10−8
Semiconductors
Carbon (pure)
3.5×105
Carbon
(3.5 − 60)×105
Germanium (pure)
600×10−3
Germanium
(1 − 600)×10−3
Silicon (pure)
2300
Silicon
0.1–2300
Insulators
Amber
5×1014
Glass
109 – 1014
Lucite
>1013
Mica
1011 – 1015
Quartz (fused)
75×1016
Rubber (hard)
1013 – 1016
Sulfur
1015
Teflon
>1013
Wood
108 – 1011
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For most metals , the resistivity increases with increasing temperature as
shown in the figure below. For some materials, over narrow ranges, the change in
resistivity is approximately proportional to the change in temperature. Let’s work out
an example to understand resistance and resistivity.
Figure 3. Resistivities of some metals as a function of temperature
Example 4 :
What is the electric resistance of an iron wire 0.50 m long with a diameter
of 1.3m if the resistivity of iron is 9.71×10−8 Ω ⋅ m
Given : L = 0.50 m
d =1.3 m
resistivity of iron is 9.71×10−8 Ω ⋅ m
Find :
R
Solution : Solve first for the cross sectional area of the wire by using the
formula
A=
A=
=
πd2
4
𝜋( 1.33 𝑥 10 −3 𝑚)2
4
A = 1.33 𝑥 10−6 𝑚2
Then, find resistance
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R=
= 9.71×10−8 Ω ⋅ m ( 0.50 m )
1.3 x 10-3 m2
R = 0.037 Ω
Learning Competency:
Solve problems involving current , resistivity , resistance ,and Ohm’s law in contexts
such as , but not limited to, batteries and bulbs, household wiring and selection of
fuses(STEM_GP12V-EM - IIIf-47)
ACTIVITY 1: NAME ME !
Directions : Identify the word/ s that best describes the following statement . Write
your answer on the blank provided before each number.
_________ 1. The amount of charge flowing through a particular area in a unit of
time.
_________ 2. The electric property that impedes current.
_________ 3. SI unit of electric current.
_________ 4. The electric property that drives current.
_________ 5. A measure of the resistance of a given size of a specific material to
electrical conduction.
_________ 6. It states that the current flowing through a conductor is directly
proportional to the potential difference and inversely proportional to
the resistance of the circuit.
_________ 7. He is a German physicist who discovered Ohm’s law.
_________ 8. The symbol for resistivity.
__________ 9. The mathematical expression of Ohm’s law.
__________10.It is the relationship of electrical resistance to the length of a
conductor.
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ACTIVITY 2: OHM’S LAW CHALLENGE
Directions : Using Ohm’s Law, calculate the missing value (V, I, or R) for each of
the following circuits. Show your complete solutions for each item.(3points each)
CURRENT, I
( Ampere , A )
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
4
3
0.2
0.0015
9
6
0.30
9.16
VOLTAGE, V
Volts, V
1000
RESISTANCE,R
Ohms, Ω
100
30
25
110
220
90
55
3.33
120
18
9.60
ACTIVITY 3: RESISTANCE VS RESISTIVITY
Directions: Give the difference between resistance and resistivity in tabular form.
DIFFERENCE IN TERMS OF…
RESISTANCE
1.
2.
3.
4.
5.
RESISTIVITY
MEANING
FORMULA
SYMBOL
UNIT
FACTORS
THAT
WILL
AFFECT THE PHYSICAL
QUANTITY
ACTIVITY 4 : CURRENT , RESISTANCE AND RESISTIVITY WORD PROBLEMS
Directions : Read , understand and analyze each of the problems very carefully.
Then solve , and show your complete solutions . ( 5points each )
1. During open-heart surgery, a defibrillator can be used to bring a patient out of
cardiac arrest. The resistance of the path is 500Ω and a 10.0-mA current is needed.
What voltage should be applied?
2. Calculate the effective resistance of a pocket calculator that has a 1.35-V battery
and through which 0.200 mA flows.
3. The diameter of 10-gauge copper wire is 8.252 mm. Find the resistance of a 1.00km length of such wire used for power transmission.
284
NOTE: Practice personal hygiene protocols at all times
4. What is the resistance of a 20.0-m-long piece of 12-gauge copper wire having a
2.053-mm diameter?
5. Calculate the resistance of a piece of a 20- gauge copper wire 2 m long . The cross
sectional area of the wire is 0.5176mm2
6. Compute the resistance of a hardened copper rod 2 meters long and 8 mm in
diameter if the resistivity of the material is 1.756 x 10 -8 ohm-meters.
7. A 0.500-meter length of wire with a cross-sectional area of 3.14 x 10 -6 meters
squared is found to have a resistance of 2.53 x 10-3 ohms. According to the resistivity
chart, from what material is the wire made?
8. The resistance of a uniform copper wire 50.0 meters long and 1.15 mm in diameter
is 0.830 ohms at 20° C. What is the resistivity of the copper at this temperature?
9. At 20° C, 33 meters of copper wire has a resistance of 0.639 ohms. What is the
resistance of 165 meters?
10. A manufacturer recommends that the longer the extension cord used with an
electric drill, the thicker (heavier gauge) the extension cord should be. Why is this the
recommendation made by the manufacturer ?
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Reflection
Complete this statement:
I learned that
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
___
I enjoyed most on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
___
I want to learn more on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
___
286
NOTE: Practice personal hygiene protocols at all times
REFERENCES:
Jones E. , Childers , R. . Contemporary College Physics ( 2nd Edition ) : AddisonWesley Publishing Company Inc. pp. 500-504
Walker , J. , Halliday, R. , & Resnick D. (2011) . Fundamentals of Physics ( 9th
Edition): Hoboken, NJ : John Wiley & Sons Inc. pp. 605-606
https://courses.lumenlearning.com/physics/chapter/20-4-electric-power-and-energy/
https://www.google.com/search?q=energy+and+power+in+electric+circuits&source
=lnms&tbm=isch&sa=X&ved=2ahUKEwi2osHzibTuAhXGUt4KHUF3DNYQ_AUoAX
oECBAQAw&biw=1366&bih=657#imgrc=VBtBl9Pvq3d
https://intl.siyavula.com/read/science/grade-11/electric-circuits/11-electric-circuits03
https://drive.google.com/file/d/1UBn5AnlVhN8o63YBRGCWKtTlWGT3WHRe/view
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Answer Key :
ACTIVITY 1: NAME ME!
1. Current
2. Resistance
3. Ampere, A
4. Voltage
5. Resistivity
6. Ohm’s law
7. Georg Simon Ohm
8. ρ
V
9.I = R
10. inversely proportional
ACTIVITY 2: OHM’S LAW CHALLENGE
CURRENT, I
( Ampere , A )
1.
10
2.
4
3.
3
4.
0.2
5.
4
6.
0.0015
7.
9
8.
6
9.
0.30
10.
0.04
VOLTAGE, V
Volts, V
1000
120
75
110
220
90
30
120
18
24
RESISTANCE,R
Ohms, Ω
100
30
25
550
55
60 , 000
3.33
20
60
560
ACTIVITY 3 : RESISTANCE VS RESISTIVITY
1. The resistance is the property of the material which obstructs the flow of
current, whereas the resistivity gives the resistance of the material which has
fixed dimension.
2. The resistance is the ratio of the length and cross-section area of the
conductor, whereas the resistivity of the material is the ratio of the product of
the resistance and area to the length of the conductor.
3. The resistance is represented by the symbol R whereas the resistivity is
represented by the symbol ρ.
4. The SI unit of the resistance is ohm, and the SI unit of resistivity is ohm-meter.
5. The resistance of the material depends on the length, cross-section and area
of conductor whereas the resistivity depends on the nature and temperature
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of the material. The inverse of the resistivity is known as the conductivity of
the material.
ACTIVITY 4: CURRENT , RESISTANCE AND RESISTIVITY WORD PROBLEMS
1. V= 5V
2. R= 6750 Ω
3. R= 0.354 Ω
4. R= 0.104 Ω
5. R= 6.65 x10-2 Ω.
6. R= 6.99 x10-4 Ω.
7. ρ= 1.59 x10-8 Ω.m - silver
8. ρ= 6.99 x10-4 Ω.m
9. R= 3.51 Ω.
10. This recommendation is made by the manufacturer because resistance of a
wire varies directly with length and inversely with cross-sectional area .
Prepared by:
FE S. CAGUMBAY
ANDARAYAN NATIONAL HIGH SCHOOL
289
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GENERAL PHYSICS 2
Name: ____________________________Grade Level: _________
Date: _____________________________Score:______________
LEARNING ACTIVITY SHEET
DEVICES FOR MEASURING CURRENTS AND VOLTAGES
Background Information for the Learners (BIL)
Electricians usually use ammeters and voltmeters to help them understand what
is happening in a circuit . What is the use of an ammeter and a voltmeter? Do you
know how to operate these devices?
Voltmeters
A voltmeter as shown in figure 1 is an instrument that measures the difference in
electrical potential between two points in an electric circuit. An analog voltmeter
moves a pointer across a scale in proportion to the circuit’s voltage; a digital voltmeter
provides a numerical display. Any measurement that can be converted to voltage
can be displayed on a meter that is properly calibrated; such measurements include
pressure, temperature, and flow.
Figure 1
In order for a voltmeter to measure a device’s voltage, it must be connected in
parallel to that device. This is necessary because objects in parallel experience the
same potential difference.
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Voltmeter in Parallel: Figure 2 (a) To measure the potential difference in this series circuit, the voltmeter
(V) is placed in parallel with the voltage source or either of the resistors. Note that terminal voltage is
measured between points a and b. It is not possible to connect the voltmeter directly across the EMF without
including its internal resistance, r. (b) A digital voltmeter in use
291
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Ammeters
An ammeter measures the electric current in a circuit. The name is derived from
the name for the SI unit for electric current, amperes (A).
In order for an ammeter to measure a device’s current, it must be connected in
series to that device. This is necessary because objects in series experience the
same current. They must not be connected to a voltage source — ammeters are
designed to work under a minimal burden, (which refers to the voltage drop across
the ammeter, typically a small fraction of a volt).
Ammeter in Series: Figure 3 An ammeter (A) is placed in series to measure current. All of the current in
this circuit flows through the meter. The ammeter would have the same reading if located between points
d and e or between points f and a, as it does in the position shown. (Note that the script capital E stands
for EMF, and r stands for the internal resistance of the source of potential difference. )
There are two types of ammeters. They are known as analog and digital ammeters
(see figure 4)
a. Analog ammeter
b. Digital ammeter
Figure 4
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How to read an analog ammeter:
As in any measurement, you should always read the smallest division on your
scale, and then estimate the next digit.
For Example:
Current measurement: 13.4 A or 13.5 A
NOTE: Everyone will not likely obtain exactly the same
measurement. Therefore, as long as your estimate for the
last digit is reasonable it will be considered to be correct.
Figure 5
How to read a digital ammeter :
Unlike the analog ammeter, once you have a measurement on your digital
ammeter, it is very simple to read. All you need to do is read the number on the
display screen and use the correct units, depending on the dial setting you have
chosen. Some of the meters in automobile dashboards, digital cameras, cell phones,
and tuner-amplifiers are voltmeters or ammeters.
Often a single meter is packaged so that, by means of a switch, it can be made
to serve as either an ammeter or a voltmeter—and usually also as an ohmmeter,
designed to measure the resistance of any element connected between its terminals.
Such a versatile unit is called a multimeter.
Figure 6. A multimeter
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Using a multimeter to measure voltage:
A multimeter can measure either AC or DC quantities.
The following symbols are used to distinguish between the two:
AC
DC
• Plug one wire into the black COM socket.
• Plug another into the red V socket.
• Select the 20V DC range by turning the dial to the ‘20’ mark next to the ‘V ’ symbol.
(It is good practice to set the meter on a range that is much higher than the reading
you are expecting. Then you can refine the measurement by choosing a lower range
that suits the voltage you find.)
• Plug the two wires into the sockets at the ends of the component under
investigation.
• Press the red ON/OFF switch when you are ready to take a reading.
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Using a multimeter to measure current:
A multimeter can measure either AC or DC quantities. The following symbols are
used to distinguish
between the two:
AC
DC
•
Plug one wire into the black COM socket.
•
Plug another into the red mA socket.
•
Select the 200mA DC range by turning the dial to
the ‘200m’ mark next to the ‘A
’
symbol.
• Break the circuit where you want to measure the current, by removing a
link, and then plug the two wires in its place.
• Press the red ON/OFF switch when you are ready to take a reading.
For more information on how to use a mutimeter visit
https://www.youtube.com/watch?v=qZH98XN2Suw
https://www.youtube.com/watch?app=desktop&v=hgTgx_h5QOk
Learning Competency :
Operate devices for measuring currents and voltages (STEM_GP12V-EM - IIIe-45)
ACTIVITY 1: TRUE OR FALSE
Directions : Write TRUE if the statement is correct, and FALSE if otherwise.
1. A voltmeter is connected in parallel with a device to measure its voltage.
2. An ammeter is connected in series with a device to measure its current.
3. The resistance of a voltmeter is smaller than the resistance of any circuit
element across which the voltmeter is connected.
4. The resistance of the ammeter is very much bigger than other resistances in
the circuit.
5. A multimeter is a tool that is capable of measuring two or more electrical
values.
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ACTIVITY 2: LABEL ME …
Directions : Label the diagram by using the words in the box below .
5.
6.
1.
2.
7.
3.
4.
8.
A.C. voltage ranges
off switch
D. C. current ranges
display screen
D. C. voltage ranges
range selector
diode tester
continuity tester
resistance ranges
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ACTIVITY 3: VOLTAGE – CURRENT PLACEMENT
Directions: Read each of the questions carefully. Choose the letter of the best
answer. Write the letter only.
1. You are to connect an ammeter in such a way that you will be able to
directly read the current intensity running through resistor R1. Which of
the diagrams below illustrates the way the ammeter should be
connected?
A)
C)
B)
D)
R1
R1
R2
R2
2. You have to connect a voltmeter to determine the potential difference
across the terminals of a resistor in a simple circuit. In which diagram below
is the voltmeter properly connected?
A)
C)
B)
D)
3. How must the ammeter and the voltmeter be connected?
A)
C)
A
B)
Appliance
V
D)
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For nos. 4-5 , give the correct reading on the ammeter.
4.
Current measurement:
5.
Current measurement:
ACTIVITY 4 : IGNITE YOUR UNDERSTANDNG…
Directions : Answer briefly the following questions : ( 10 points )
1. How do you set up an ammeter in a circuit ?
2. What happens when the ammeter is connected in parallel with the lamp?
3. Why do the problems occur when an ammeter is connected in parallel with
the lamp?
298
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REFLECTION:
1. I learned that ________________________________________________
_____________________________________________________________
_____________________________________________________________
2. I enjoyed most on ____________________________________________
_____________________________________________________________
_____________________________________________________________
3. I want to learn more on _________________________________________
______________________________________________________________
__________________________________________________________
299
NOTE: Practice personal hygiene protocols at all times
REFERENCES:
Walker , J. , Halliday, R. , & Resnick D. (2011) . Fundamentals of Physics ( 9th
Edition): Hoboken, NJ : John Wiley & Sons Inc. p. 720
https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuitconstruction-kit-dc_en.html
https://www.yenka.com/activities/Using_Ammeters_and_Voltmeters__Activity/#:~:text=Electricians%2C%20and%20other%20people%20who,water%20
behind%20a%20dam%20pushes)
https://www.theautomationstore.com/using-a-multimeter-voltmeter-ammeter-andan-ohmmeter/
https://www.ck12.org/physics/ammeters-and-voltmeters/lesson/Ammeters-andVoltmeters-PHYS/
https://flexbooks.ck12.org/cbook/ck-12-physics-flexbook2.0/section/16.4/primary/lesson/ammeters-and-voltmeters-phys
https://www.youtube.com/watch?v=qZH98XN2Suw
https://www.tes.com/teaching-resource/the-digital-multimeter-6449605
300
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ANSWER KEY
ACTIVITY 1 : TRUE OR FALSE
1. TRUE
2. TRUE 3. FALSE
4. FALSE
5. TRUE
ACTIVITY 2: LABEL ME …
1.
2.
3.
4.
5.
6.
7.
8.
AC voltage
Range selector
Resistance range
Continuity tester
Display screen
D.C. voltage
D.C. current range
Diode selector
ACTIVITY 3 : Voltmeter – Ammeter placement
1.
2.
3.
4.
5.
A
A
B
10.5 A
6.5 A
ACTIVITY 4 : IGNITE YOUR UNDERSTANDING ...
How to set up an ammeter:
1. In order for an ammeter to accurately measure current, it needs to be placed
in series in the circuit. To begin, you should first set up your circuit as
necessary, without including the ammeter. The steps for placing an ammeter
in a circuit are as follows:
Step 1: Break the circuit in the desired location. This involves physically
disconnecting a wire or component to make a place for the ammeter in the
circuit.
Step 2: Connect the red and black leads to the ammeter. The black lead will
always go to the negative/grounded terminal. The red lead will go to the
positive terminal on the ammeter.
Step 3: Connect the ammeter to your circuit. The black lead should always be
connected to the side of the circuit that is closest to the negative terminal of
the power source. The red lead should always be connected to side of the
301
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circuit that is closest to the positive terminal of the power source. If you fail to
do this correctly, the ammeter may be damaged.
2. It has a negative value .
3. Short Circuit occurs when the narrator places the ammeter in parallel with the
lamp.
Prepared by :
Fe S. Cagumbay
Andarayan National High School
302
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GENERAL PHYSICS 2
Name: ____________________________Grade Level: _________
Date: _____________________________Score:______________
LEARNING ACTIVITY SHEET
Circuit Symbols and Circuit Diagrams
Background Information for the Learners (BIL)
Electric circuits, whether simple or complex, can be described in a variety of ways.
An electric circuit is commonly described with mere words. Saying something like
"A light bulb is connected to a dry cell" is a sufficient amount of words to describe a
simple circuit. Upon hearing (or reading) the words, a person becomes familiar and
quickly imagine the picture of the circuit in their mind.
Another means of describing
a circuit is to simply draw it. Such drawings provide a quicker mental picture of the
actual circuit .
An example is shown below:
Describing Circuits with Words
Describing Circuits with
Drawings
"A circuit contains a light bulb and a
1.5-Volt D-cell."
A final means of describing an electric circuit is by use of conventional circuit
symbols to provide a schematic diagram of the circuit and its components. Some
circuit symbols used in schematic diagrams are shown below.
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Fuse
Connecting wires
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Electric circuits are represented by schematic diagrams. Schematics are very
useful in visualizing the main features of a circuit. A single schematic diagram can
represent a wide variety of situations.
As an illustration of the use of electrical symbols in schematic diagrams, consider
the following examples :
Example 1:
Description with Words: Three D-cells are placed in a battery pack to power a circuit
containing three light bulbs.
Using the verbal description, one can acquire a mental picture of the circuit being
described. This verbal description can then be represented by a drawing of three
cells and three light bulbs connected by wires. Finally, the circuit symbols presented
above can be used to represent the same circuit. Note that three sets of long and
short parallel lines have been used to represent the battery pack with its three Dcells. And note that each light bulb is represented by its own individual resistor
symbol. Straight lines have been used to connect the two terminals of the battery to
the resistors and the resistors to each other.
The above circuits presumed that the three light bulbs were connected in such a
way that the charge flowing through the circuit would pass through each one of the
three light bulbs in consecutive fashion. The path of a positive test charge leaving
the positive terminal of the battery and traversing the external circuit would involve a
passage through each one of the three connected light bulbs before returning to the
negative terminal of the battery. The diagram shows a series connection where all
the components are connected end-to-end, forming a single path for current to flow.
Example 2:
Description with Words: Three D-cells are placed in a battery pack to power a circuit
containing three light bulbs.
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Using the verbal description, one can acquire a mental picture of the circuit being
described. But this time, the connections of light bulbs is done in a manner such that
there is a point on the circuit where the wires branch off from each other. The
branching location is referred to as a node. Each light bulb is placed in its own
separate branch. These branch wires eventually connect to each other to form a
second node. A single wire is used to connect this second node to the negative
terminal of the battery.
These two examples illustrate the two common types of connections made in
electric circuits. When two or more resistors are present in a circuit, they can be
connected in series or in parallel. Example 1 is a series circuit where all the
components are connected end-to-end, forming a single path for current flow while
example 2 is a parallel circuit where all the components are connected across each
other , forming two sets of electrically common points.
Example 3:
The diagram shows a fixed resistor ,in series with an ammeter, variable resistor and
a bulb. A voltmeter is connected across the battery and fixed resistor while the switch
is left open in the circuit.
Learning Competency :
Draw circuit diagrams with power sources ( cell or battery ) , switches, lamps ,
resistors, ( fixed and variable) ,fuses ,ammeters and voltmeters (STEM_GP12V-EM
- IIIf-47)
306
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ACTIVITY 1: CIRCUIT SYMBOLS AND CIRCUIT DIAGRAMS
CHALLENGE
Directions : Read and understand each of the questions very carefully. Then,
choose the letter of the correct answer. Write the letter only.
1. Which of the following shows the electrical symbol of a dry cell ?
A.
B.
C.
D.
2.
What are the two types of circuit connections ?
A. series circuit and parallel circuit
B. simple circuit and series circuit
C. network
D. parallel and simple circuit
3.Which of the following symbols represent an AC (alternating current )source ?
A.
C.
B.
D.
4.
, the symbol shows a switch that is ______.
A. on
B. off
C. open
D. close
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5. Which of the following best describes a bulb connected in a dry cell with
a closed switch and an ammeter that will measure the flow of current in the
circuit?
A.
B.
C.
D.
ACTIVITY 2: CHECK YOUR UNDERSTANDING…
Directions :Use circuit symbols to construct schematic diagrams for the following
circuits ( 5 points each )
1. A simple circuit which consists of a single cell, light bulb and switch are placed
together such that the switch can be opened and closed to turn the light bulb on.
2. Two light bulbs , one fixed resistor with an open switch are connected in series
with a dry cell.
3.
A load connected in series with a closed switch with three dry cells and a
fuse.
4. Four dry cells, fixed resistor , a variable resistor and switch are placed together
in a series circuit such that the switch is closed to turn the light bulb on. An
ammeter is connected in series with the circuit and a voltmeter across the fixed
resistor to measure the voltage.
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ACTIVITY 3: SKETCH ME MORE
Directions : Draw the circuit diagram or the schematic diagrams of the following
circuits . ( 3 points each )
PICTURE
Schematic Diagram
1.
2.
3.
4.
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5.
310
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Reflection
Complete this statement:
I learned that
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
___
I enjoyed most on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
___
I want to learn more on
311
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REFERENCES:
Jones E. , Childers , R. . Contemporary College Physics ( 2nd Edition ) : AddisonWesley Publishing Company Inc. p.505.
https://www.physicsclassroom.com/class/circuits/Lesson-4/Circuit-Symbols-andCircuit-Diagrams
https://www.google.com/search?q=Three+Dcells+are+placed+in+a+battery+pack+to+power+a+circuit+containing+three+light+
bulbs.&tbm=isch&ved=2ahUKEwi7yaqZ86LuAhVJ
https://www.google.com/search?q=complete+schematic+diagram+of+fixed+resistor
s+with+a+fuse+%2C+ammeter+%2C+switch+and+a+voltmeter&tbm=isch&ved=2a
hUKEwi7jM
312
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ANSWER KEY
ACTIVITY 1: CIRCUIT SYMBOLS AND CIRCUIT DIAGRAMS
CHALLENGE
1.D
2.A
3.D
4.B &C
5.A
ACTIVITY 2: CHECK YOUR UNDERSTANDING…
1.
2.
3.
4.
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ACTIVITY 3: SKETCH ME MORE …
1.
2.
3.
4.
5.
Prepared by:
FE S. CAGUMBAY
ANDARAYAN NATIONAL HIGH SCHOOL
314
NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 2
Name: _______________________________
Date: ______________
Grade Level & Strand: __________________
Score: _____________
LEARNING ACTIVITY SHEETS
THE EQUIVALENT RESISTANCE, CURRENT, AND VOLTAGE OF
RESISTORS CONNECTED IN SERIES AND PARALLEL CIRCUITS
Background Information for the Learners
A circuit (or an electric circuit) is an arrangement of
conductors to form a path along which electrons can flow.
A simple circuit, as shown in Figure 1, is a path that
contains the three components necessary to have a
functioning
electric
circuit
-a
power
source, the
conducting wires, and a resistor. A power source can
either be an alternating current (AC) source or a direct
current (DC) source like a battery or a cell. The conducting
wires connect the power source to the resistor. And a resistor is any device that uses
electric energy such as lightbulb, speaker, etc. However, most circuits have more
than one resistor. Just think of the bulbs in Christmas lights or the lightbulbs in a
household. The resistors are connected in one of the two circuit types, the series or
the parallel. These two connections differ in characteristics, in terms of the voltage,
the current, and the resistance across the circuit. Below is a table summarizing the
different properties of these two types of connection.
Properties
Description
Series Connection
Parallel Connection
The resistors are arranged in a The resistors are connected by
single path. The electrons pass forming branches. Each branch
through the first resistor, then gives a separate path for the flow
the next resistors, and then flow of electrons. Thus, if one resistor
back to the battery. When one is damaged, the other resistors
resistor is damaged, it results in are not affected as they have
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a break in the circuit, and the separate
electrons cease to flow.
pathways
for
the
electric current.
Illustration
Current
The current has only one path The total current in the circuit
through the circuit. Thus, the equals the sum of the currents in
current passing through all the its parallel branches. This sum
resistors in the circuit is the equals the current in the battery
same. In symbols,
or
ITotal = I1 = I2 = I3….
other
voltage
source.
In
symbols,
ITotal = I1 + I2 + I3….
Voltage
The total voltage across the Each resistor is independently
circuit
divides
among
each connected to the same power
resistor in the circuit. So, the supply. Therefore, the voltage is
sum of the "voltage drops" the same across each resistor. In
across each resistor is equal to symbols,
the total voltage supplied by the
source. In symbols,
VTotal = V1 + V2 + V3….
Resistance
VTotal = V1 = V2 = V3….
The current is resisted by all the As
the
number
of
parallel
resistors in the circuit. So, the branches is increased, the overall
total resistance to current in the resistance
circuit
is
the
sum
of
of
the
circuit
is
the decreased. The total resistance
individual resistances along the of the circuit is always lesser than
circuit path. In symbols,
the least resistance in the circuit.
In symbols,
𝟏
RTotal = R1 + R2 + R3….
𝐑 𝐓𝐎𝐓𝐀𝐋
=
𝟏
𝟏
𝟏
+
+
…
𝐑𝟏 𝐑𝟐 𝐑𝟑
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Example
Three resistors are connected in Three resistors are connected in
series with a 9-V battery, as parallel with a 10-V battery, as
shown in the figure below.
shown in the figure below.
Find for the following:
Find for the following:
a) the total resistance
a) the total resistance
b) the currents through each
b) the voltage drops across
resistor
each resistor
c) the voltage drops across
c) the currents through each
each resistor
resistor
Solution:
Solution:
a) RTOTAL = R1 + R2 + R3
a)
= 5 Ω + 5 Ω + 10 Ω
1
=
RTOTAL
R1
+
1
R2
1
+
1
1
R3
1
=4Ω + 4Ω + 2Ω
= 20 Ω
=
b) ITOTAL = I1 = I2 = I3
V
1
9V
0.25
Ω
+
0.25
Ω
+
0.5
Ω
1
=Ω
= RTOTAL = 20 Ω
To get the total resistance,
= 0.45 A
get the reciprocal of
TOTAL
c) Use
Ohm’s
determine
drops
the
Law
to
1
RTOTAL
.
RTOTAL = 1Ω/1 = 1 Ω
voltage
b) VTOTAL = V1 = V2 = V3 = 10 V
each
c) Use Ohm’s Law to quantify
across
resistor.
for the individual currents
V1 = V2 = I1R1
through each resistor.
= (0.45 A) (5 Ω)
V
I1 = I2 = R1 =
1
= 2.25 V
V3= I3R3
V
I3 = R3 =
3
10 V
2Ω
10 V
4Ω
= 2.5 A
=5A
= (0.45 A) (10 Ω)
To check if the sum of
= 4.5 V
currents equals the current
To check if the sum of the
voltage drops equals the
in
the
battery
or
other
voltage source:
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V
voltage supplied by the
ITOTAL = RTOTAL =
TOTAL
battery: VTOTAL = V1 + V2 +
10 V
1Ω
=
10 A
V3 = 2.25 V + 2.25 V + 4.5
ITOTAL = I1 + I2 + I3 = 2.5 A +
V = 9 V.
2.5 A + 5 A = 10 A
Learning Competency
Evaluate the equivalent resistance, current, and voltage in a given network of
resistors connected in series and/or parallel. (STEM_GP12EMIIIg-48)
Activity 1: Let’s Appraise our Knowledge
Directions: Select the letter of the choice that correctly answers each question.
1. Which of the following quantities do devices connected in series have in
common?
A. Current
B. Voltage
C. Resistance
D. Both a & b
2. Which of the following quantities connected in parallel have in common?
A. Current
B. Voltage
C. Resistance
D. Both a & b
3. What is the current on the battery if the current through one lamp is 2A in a
series connection?
A. half, 1 A.
C. more than 2 A.
B. 2 A.
D. cannot be calculated.
4. What is the current on the battery if the current through one lamp is 2A in a
parallel connection?
A. half, 1 A.
C. more than 2 A.
B. 2 A.
D. cannot be calculated.
5. Why should household appliances be connected in parallel with the voltage
source?
I. To increase the resistance of the household circuit.
II. To impress upon each appliance the same voltage as the power supply.
A. I only
B. II only
C. Both I & II
D. Neither I & II
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6. In the circuit shown, which two bulbs are in series?
A. A and B
B. C and D
C. B and C
D. None of the bulbs are connected in
series.
7. In the circuit shown in number 6, which two bulbs will have the same amount
of voltage drop?
A. A and B
B. C and D
C. B and C
D. A and D
8. What happens to the intensity or the brightness of the lamps connected in
series as more and more lamps are added?
A. increases
C. remains the same
B. decreases
D. cannot be predicted
9. What happens to the intensity or the brightness of the lamps connected in
parallel as more and more lamps are added?
A. increases
C. remains the same
B. decreases
D. cannot be predicted
10. As you add more resistors in a series circuit, what happens to its total
resistance?
A. increases
C. remains the same
B. decreases
D. cannot be predicted
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Activity 2: Let’s Analyze Circuits
Directions: Calculate the total resistance, total current, individual currents, and
individual voltage drops across the given circuits.
1.
2.
ACTIVITY 3: Let’s Be Logical
Directions: Rank the bulbs of equal resistances according to intensity or brightness
of light, from highest to lowest. Make models using current and/or
voltage to support your answer.
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Reflection
1. I learned that ______________________________________________________
__________________________________________________________________
__________________________________________________________________
__
2. I enjoyed most on __________________________________________________
__________________________________________________________________
__________________________________________________________________
__
3. I want to learn more on ______________________________________________
__________________________________________________________________
__________________________________________________________________
__
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REFERENCES
“Are Christmas Lights in Series or Parallel?”. Accessed February 26, 2021.
https://www.wired.com/2014/12/christmas-lights-series-parallel/
Halliday, David, Resnick, Robert, & Walker, Jearl. Fundamentals of Physics. 6th ed.
New York: John Wiley & Sons Inc, 2001.
Hewitt, Paul G. Conceptual Physics. 10th ed. United States of America: Pearson
Addison-Wesley, 2006.
Serway, Raymond A. and Jewette, John W. Jr. Physics for Scientists and Engineers
with Modern Physics. 6th ed. Singapore: Thomson Learning Asia, 2004.
“Simple Circuit Project for Kids to Make”. Accessed February 26, 2021.
https://cubscoutideas.com/9187/simple-circuit-project/
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ANSWER KEY
ACTIVITY 1
1.
2.
3.
4.
5.
a
b
b
c
b
6. a
7. b
8. b
9. c
10. a
ACTIVITY 2
1. RT = 10.5 Ω
IT = I1 = I2 = I3 = 0.857 A
V1 = 2.1425 V
V2 = 2.995 V
V3 = 3.8565 V
2. RT = 1.96 Ω
VT = V1 = V2 = V3 = 9 V
I1 = 1.8 A
I2 = 1.5 A
I3 = 1.29 A
ACTIVITY 3
Rank
Rank
1st
2nd
3rd
4th
-
Bulb
A
C
B & D (tie)
E & F (tie)
Prepared by:
Techie Gammad-Vera Cruz
Amulung National High School
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GENERAL PHYSICS 2
Name: _____________________________
Date: _______________
Grade: ____________________________
Score: _____________
LEARNING ACTIVITY SHEETS
Kirchhoff’s Loop and Junction Rules
Background Information for Learners
You had just learned resistors connected in series and parallel in the previous
learning activity sheet. And if you remember, you can solve resistance, current and
voltage in series, parallel and a combination of series-parallel circuits. Now, how
about if there are more complex circuits which cannot be reduced to either series
and/or parallel? How would you analyze and solve such problems? You will
understand that using the concept Kirchhoff’s Rule.
Key Terms
Junction – the intersection of three or more wires or pathways in a circuit. Typically
represented by a dot on a circuit diagram. And also called a node.
J1
J2
Two junctions J1 and J2 that are represented by dots and the current pathways
highlighted.
Branch – a path connecting two nodes.
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B1
B2
Two branches B1 and B2 in a circuit each highlighted in a different color.
Kirchhoff’s First Rule (Junction Rule)
This rule states that the total current entering a junction is equal to the current
leaving the junction. In other words, the sum of all currents entering and leaving the
junction must be equal to zero. That is,
Iin = Iout
or
Iin + Iout = 0
Consider the diagram below;
Based from the diagram, the junction or node is the point of intersection of the
currents. The currents entering the junction or node are I1, I2 and I3, while the currents
leaving the node are I4 and I5.
By Kirchhoff’s junction rule, the equation can be formulated as;
I 1+ I 2 + I 3 = I 4 + I 5
or
I1+ I2 + I3 + (-I4 + - I5) = 0
I1+ I2 + I3 -I4 - I5 = 0
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Kirchhoff’s Second Rule (Loop Rule)
This rule states that the sum of the voltages around a closed loop must add
up to zero. Mathematically, it can be written as;
∑
𝑉=0
Conventions for Loop Rule
The following conventions apply in determining the sign of delta V or the potential
drop across circuit elements. The direction of travel is the direction that we choose
to proceed around the loop. See figure below;
Each of these resistors and voltage sources is traversed from a to b.
(a) When moving across a resistor in the same direction as the current flow,
subtract the potential drop.
(b) When moving across a resistor in the opposite direction as the current flow,
add the potential drop
(c) When moving across a voltage source from the negative terminal to the
positive terminal, add the potential drop.
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(d) When moving across a voltage source from the positive terminal to the
negative terminal, subtract the potential drop.
Consider a simple loop below,
The labels A, B, C and D serves as reference while R is the resistance across each
wire. The loop rotation is indicated which will be a guide to track the voltage
differences around the circuit.
Applying Kirchhoff’s Loop Rule, the sum of the voltage drop around the loop
is equal to zero. Thus, the equation is formulated as;
VAB + VBC + VCD + VDA = 0
Let’s try another diagram below with two nodes to fully understand Kirchhoff’s loop
rule.
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I1
+
Va
R1
I1
I1
+
A
Vb
I2
B
R2
I3
Vc
I3
+
R3
I3
The circuit has two nodes A and B and also two loops with a counter clockwise
direction. There are three branches through the circuit a) at the top b) middle and c)
bottom. This serves as reference that there is current flowing in each branch as
indicated. Applying the loop rule and convention for loop rules in this situation
generates the following equations.
Loop 1:
-Va + Vb + I2R2 + I1R1 = 0
Loop 2:
-Vb + Vc + I3R3 – I2R2 = 0
Procedure for Applying Kirchhoff’s Rules
1. Assume all voltage sources and resistances are given. (If not label them V1,
V2 ..., R1, R2 etc.)
2. Label each branch with a branch current and its direction (if not indicated).
3. Apply junction rule at each node.
4. Applying the loop rule for each of the independent loops of the circuit.
5. Solve the equations by substitutions/linear manipulation.
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Example 1. Express the currents in the junction ‘a’ as an equality.
I4
a
I2
I1
I5
I3
Solution: Applying the junction rule, we obtain the equation
I2 + I3 + I4 = I1 + I5
or
I2 + I3 + I4 - I1 – I5 = 0
Example 2. Suppose the current entering the junction in example 1 is 4A each. What
is the value of the current leaving the junction?
Solution: The junction rule for this case is:
I2 + I3 + I4 = I1 + I5
Since the current entering the junctions are I2, I3 and I4 which is equal to 4A. Then,
solving the current leaving the junction:
I2 + I3 + I4 = I1 + I5
4A + 4A+ 4A = I1 + I5
12A = I1 + I5
Therefore, the current leaving the junction is 12A which is distributed to I1 and I5.
Example 3. Calculate the current flowing through each resistor.
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The circuit has two nodes A and B and two loops 1 and 2. Applying the Kirchhoff’s
rule from the problem,
Junction A:
I1 + I 2 = I 3
eq. 1
Junction B:
I3 = I 1 + I 2
eq. 2
Loop 1:
10 – 10I1 – 40I3 = 0
eq. 3
Loop 2:
20 – 20I2 – 40I3 = 0
eq. 4
Solving I1 from eq. 1 gives,
I1 = I3 – I2
eq. 5
Substituting eq. 5 from eq. 3, we get
10 – 10(I3 – I2) - 40I3 = 0
10 – 10I3 + 10I2 - 40I3 = 0
10 – 50I3 + 10I2 = 0
10I2 = 50I3 -10
I2 = (50I3 – 10)/10
I2 = 5I3 – 1
eq. 6
Substituting eq. 6 from eq. 4,
20 – 20(5I3 – 1) – 40I3 = 0
20 – 100I3 + 20 – 40I3 = 0
40 – 140I3 = 0
-140I3 = -40
I3 = -40/-140
I3 = 0.29A
Solving I2 in eq. 6 when I3 = 0.29A, we get
I2 = 5(0.29A) – 1
I2 = 1.45A – 1
I2 = 0.45A
Lastly, solving I1 when I2 = 0.45A and I3 = 0.29A gives,
I1 = I3 - I2
I1 = 0.29A – 0.45A
I1 = -0.13A (the negative sign indicates wrong direction of current I 1)
Therefore, the currents across the circuit are I1 = -0.13A, I2 = 0.45A and I3 = 0.29A.
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Example 4. Determine the current flowing through each of the resistors. Find the
voltage across the resistors.
I1
+
R1 = 2Ω
6V
I1
I1
I2
+
B
9V
A
R2 = 6Ω
I3
I3
18V
+
R3 =3Ω
I3
Again, the circuit consists of two junction A and B and two loops (loop 1 and loop 2).
The direction of current is also indicated.
By Kirchhoff’s rule,
Junction A:
I2 + I 3 = I 1
eq. 1
Junction B:
I1= I2 + I3
eq. 2
Loop 1:
9 + 6I2 + 2I1 – 6 = 0
eq. 3
Loop 2:
18 + 3I3 – 6I2 – 9 = 0
eq. 4
Solving for I3 in eq. 1,
I3 = I 1 - I2
eq. 5
Substituting eq. 5 from eq. 4,
18 + 3(I1 – I2) -6I2 – 9 = 0
9 + 3I1 – 3I2 – 6I2 = 0
3I1- 9I2 = -9
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3I1 = 9I2 – 9
I1 = (9I2 – 9)/3
I1 = 3I2 – 3
eq. 6
Substituting eq. 6 from eq. 3, we get
9 + 6I2 + 2(3I2 – 3) – 6 = 0
3 + 6I2 + 6I2 – 6 = 0
12I2 – 3 = 0
12I2 = 3
1
I2 = 3/12 = 3A
1
Solving for I1 from eq. 6 when I2 = 3A,
1
I1 = 3I2 – 3 = 3(3) – 3 = -2A
1
Then, solving I3 from eq. 5 when I1 = - 2A and I2 = 3A,
1
7
I3 = I1 - I2 = -2A - 3A = − 3A = -2.33A
The negative sign from I1 and I3 only indicates the wrong direction of currents. Thus,
1
7
the current through the circuit are I1 = 2A, I2 = 3A and I3 = 3A.
The Voltage across the resistor R1, R2 and R3 are;
I1R1 = 2A(2Ω) = 4V
I2R2 = (1/3A)(6Ω) = 2V
I3R3 =s (7/3A)(3 Ω) = 7V
Learning Competency
Calculate the current and voltage through and across circuit elements using
Kirchhoff’s loop and junction rules (at most 2 loops only) (STEM_GP12EMIIIg-49)
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Activity 1. STOP! Kirchhoff’s JUNCTION Rules.
Directions: Formulate and solve unknowns using the junction rule.
1. What is the amount of current from the figure below?
2. Identify the node from the circuit and write the junction rule for each
node.
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Activity 2. LETS TRAVEL AROUND Kirchhoff’s LOOP.
Directions: Write down the equation using for each loop presented in each diagram.
1. Diagram 1
2. Diagram 2:
I2
I1
I3
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Activity 3. PROBLEM SOLVING using Kirchhoff’s Rule.
Directions: Solve the following problems and show your complete solution. (10
points each)
1. Analyze the circuit and calculate the current and voltage through each
resistor.
2.
A
I2
I1
I3
B
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3. Calculate the current and voltage through each of the resistors.
I1
B
I3
I2
A
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Reflection
1. I learned that
__________________________________________________________________
__________________________________________________________________
________________________
2. I enjoyed most on
__________________________________________________________________
__________________________________________________________________
_____________________________________
3. I want to learn more on
__________________________________________________________________
__________________________________________________________________
______________________________________
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References:
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_P
hysics_(OpenStax)/Book%3A_University_Physics_II__Thermodynamics%2C_Electricity%2C_and_Magnetism_(OpenStax)/10%3A_Dire
ct-Current_Circuits/10.04%3A_Kirchhoff's_Rules
https://web.iit.edu/sites/web/files/departments/academic-affairs/academic-resourcecenter/pdfs/Kirchhoff_s_Circuit_Laws.pdf
https://openpress.usask.ca/physics155/chapter/6-3-kirchhoffs-rules/
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Answer Key:
ACTIVITY 1.
1. I1 + 15A = 15A + 30A + 20A
I1 = 65A – 15A
I1 = 50A
2. The nodes are the points C and F.
Node C:
I1 + I 3 = I 2
Node F:
I2 = I 1 + I 3
ACTIVITY 2.
1. Loop 1:
Loop 2:
2. Loop 1:
Loop 2:
V1 –I1R1 – I2R2 = 0
-V3 + V2 + I2R2 + I3R3 = 0
-2I2 + I3 – 7 = 0
-2I2 – 4I1 – 28 = 0
ACTIVITY 3.
1. Applying Kirchhoff’s Rule:
Node A:
I3 = I 1 + I 2
eq. 1
Node B:
I1 + I 2 = I 3
eq. 2
Loop 1:
28 + 4I1 – 2I2 = 0
eq. 3
Loop 2:
7 - I3 - 2I2 = 0
eq. 4
Solving the equations simultaneously;
Substituting eq. 1 from eq. 4, we obtain
7 – (I1 + I2) - 2I2 = 0
7 - I1 - I2 - 2I2 = 0
-3I2 – I1 = -7
I1 = 7- 3I2
eq. 5
Substituting eq. 5 from eq. 3, we get
28 + 4I1 – 2I2 = 0
28 + 4(7- 3I2) – 2I2 = 0
28 + 28 -12I2 – 2I2 = 0
-14 I2 = -56
I2 = -56/-14 = 4A
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Substituting I2 = 4A from eq. 4, we obtain,
7 - I3 – 2(4) = 0
- I3 - 1 = 0
I3 = -1A
Solving for I1 from eq. 1 when I2 = 4A and I3 = -1A,
I3 = I 1 + I 2
I1 = I 3 – I2
I1 = -1A – 4A = -5A
Note: the negative sign indicates that we have chosen the wrong direction of
current.
Voltage across each resistor:
V1 = I1R1 = 5A(4Ω) = 20v
V2 = I2R2 = 4A(2 Ω) = 8v
V3 = I3R3 = 1A(1 Ω) = 1v
2. By Applying Kirchhoff’s rule
Node A:
I2 = I 1 + I 3
eq. 1
Node B:
I1 + I3 = I2
eq. 2
Loop 1:
10 – 2I2 – 20 – 4I2 = 0
eq. 3
Loop 2:
-30 +4I2 + 5I3 = 0
eq. 4
Solving the equations simultaneously
Solving I1 from eq. 1 gives,
I1 = I 2 - I3
eq. 5
Substituting eq. 5 form eq. 3, we obtain
-2(I2 - I3) - 4I2 = 10
-6I2 + 2I3 = 10
I3 = 3I2 + 5
eq. 6
Solving I2 substitute eq. 6, from eq. 4,
4I2 + 5(3I2+5) = 30
4I2 + 15I2 +25 = 30
19I2 = 30-25
I2 = 5/19A = 0.26A
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Solving I3 from eq. 6 when I2 = 5/19A gives,
4(5/19) + 5I3 = 30
5I3 = 30 – 20/19
I3 = 110/19 A = 5.789A
Then, solving for I1 using eq. 5 when I2 = 5/19A and I3 = 110/19 A, we get
I1 = I 2 – I3
I1 = 110/19 – 5/19 = 105/19A = 5.526A
The current through each resistors are I1= 105/19A, I2 = 5/19A and I3 = 110/19 A.
The voltage through each resistor are also as follows;
V1 = I1R1 = 5.526(2) = 11.05v
V2 = I2R2 = 0.26(4) = 1.04v
V3 = I3R3 = 5.789(5) = 28.945v
Prepared by:
Silverio V. Macarilay
Magalalag National High School
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
R-C CIRCUITS
Background Information for the Learners (BIL)
An RC circuit is one containing a resistor R and a capacitor C. The capacitor
is an electrical component that stores electric charge. RC circuit that employs a DC
(direct current) voltage source. The capacitor is initially uncharged. As soon as the
switch is closed, current flows to and from the initially uncharged capacitor. As charge
increases on the capacitor plates, there is increasing opposition to the flow of charge
by the repulsion of like charges on each plate.
Figure 1. (a) A circuit with an initially uncharged capacitor. Current flows in the
direction shown (opposite of electron flow) as soon as the switch is closed. Mutual
repulsion of like charges in the capacitor progressively slows the flow as the capacitor
is charged, stopping the current when the capacitor is fully charged and Q = C ⋅ emf.
(b) A graph of voltage across the capacitor versus time, with the switch closing at
time t = 0. (Note that in the two parts of the figure, the capital script E stands for
emf, q stands for the charge stored on the capacitor, and τ is the RC time constant.)
Voltage on the capacitor is initially zero and rises rapidly at first, since the
initial current is a maximum. Figure 1(b) shows a graph of capacitor voltage versus
time (t) starting when the switch is closed at t = 0. The voltage approaches emf
asymptotically, since the closer it gets to emf the less current flows. The equation for
voltage versus time when charging a capacitor C through a resistor R, derived using
calculus, is
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V = emf(1 − e−t/RC) (charging),
where V is the voltage across the capacitor, emf is equal to the emf of the DC voltage
source, and the exponential e = 2.718 … is the base of the natural logarithm. Note
that the units of RC are seconds. We define
τ = RC
where τ (the Greek letter tau) is called the time constant for an RC circuit. As noted
before, a small resistance R allows the capacitor to charge faster. This is reasonable,
since a larger current flows through a smaller resistance. It is also reasonable that
the smaller the capacitor C, the less time needed to charge it. Both factors are
contained in τ = RC. More quantitatively, consider what happens when t = τ = RC.
Then the voltage on the capacitor is
V = emf (1 − e−1) = emf (1 − 0.368) = 0.632 ⋅ emf.
This means that in the time τ = RC, the voltage rises to 0.632 of its final value. The
voltage will rise 0.632 of the remainder in the next time τ. It is a characteristic of the
exponential function that the final value is never reached, but 0.632 of the remainder
to that value is achieved in every time, τ. In just a few multiples of the time constant τ,
then, the final value is very nearly achieved, as the graph in Figure 1(b) illustrates.
Discharging a Capacitor
Discharging a capacitor through a resistor proceeds in a similar fashion, as Figure
2 illustrates. Initially, the current is I0 =V0RI0=V0R, driven by the initial voltage V0 on
the capacitor. As the voltage decreases, the current and hence the rate of discharge
decreases, implying another exponential formula for V. Using calculus, the
voltage V on a capacitor C being discharged through a resistor R is found to be
V = V0e−t/RC(discharging).
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Figure 2. (a) Closing the switch discharges the capacitor C through the resistor R.
Mutual repulsion of like charges on each plate drives the current. (b) A graph of
voltage across the capacitor versus time, with V = V0 at t = 0. The voltage decreases
exponentially, falling a fixed fraction of the way to zero in each subsequent time
constant τ.
The graph in Figure 2(b) is an example of this exponential decay. Again, the
time constant is τ = RC. A small resistance R allows the capacitor to discharge in a
small time, since the current is larger. Similarly, a small capacitance requires less
time to discharge, since less charge is stored. In the first-time interval τ = RC after
the switch is closed, the voltage falls to 0.368 of its initial value, since V = V0 ⋅ e−1 =
0.368V0.
During each successive time τ, the voltage falls to 0.368 of its preceding value.
In a few multiples of τ, the voltage becomes very close to zero, as indicated by the
graph in Figure 2(b). Now we can explain why the flash camera in our scenario takes
so much longer to charge than discharge; the resistance while charging is
significantly greater than while discharging. The internal resistance of the battery
accounts for most of the resistance while charging. As the battery ages, the
increasing internal resistance makes the charging process even slower. (You may
have noticed this.)
The flash discharge is through a low-resistance ionized gas in the flash tube
and proceeds very rapidly. Flash photographs, such as in Figure 3, can capture a
brief instant of a rapid motion because the flash can be less than a microsecond in
duration. Such flashes can be made extremely intense. During World War II,
nighttime reconnaissance photographs were made from the air with a single flash
illuminating more than a square kilometer of enemy territory. The brevity of the flash
eliminated blurring due to the surveillance aircraft’s motion. Today, an important use
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of intense flash lamps is to pump energy into a laser. The short intense flash can
rapidly energize a laser and allow it to reemit the energy in another form.
Figure 3. This stop-motion photograph of a rufous hummingbird (Selasphorus rufus)
feeding on a flower was obtained with an extremely brief and intense flash of light
powered by the discharge of a capacitor through a gas. (credit: Dean E. Biggins, U.S.
Fish and Wildlife Service)
EXAMPLE 1: INTEGRATED CONCEPT PROBLEM: CALCULATING
CAPACITOR SIZE—STROBE LIGHTS
High-speed flash photography was pioneered by Doc Edgerton in the 1930s,
while he was a professor of electrical engineering at MIT. You might have seen
examples of his work in the amazing shots of hummingbirds in motion and a drop of
milk splattering on a table. To stop the motion and capture these pictures, one needs
a high-intensity, very short pulsed flash, as mentioned earlier in this module.
Suppose one wished to capture the picture of a bullet (moving at 5.0 × 102
m/s) that was passing through an apple. The duration of the flash is related to
the RC time constant, τ. What size capacitor would one need in the RC circuit to
succeed, if the resistance of the flash tube was 10.0 Ω? Assume the apple is a sphere
with a diameter of 8.0 × 10–2m.
Strategy
We begin by identifying the physical principles involved. This example deals
with the strobe light, as discussed above. Figure 2 shows the circuit for this probe.
The characteristic time τ of the strobe is given as τ = RC.
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Solution
We wish to find C, but we don’t know τ. We want the flash to be on only while
the bullet traverses the apple. So we need to use the kinematic equations that
describe the relationship between distance x, velocity v, a nd time t:
x = v/t or t=x/v
The bullet’s velocity is given as 5.0 × 102 m/s, and the distance x is 8.0 × 10-2 m The
traverse time, then, is
t= x/v
t= 8.0x10-2 m / 5.0x102 m/s
t= 1.6x10-4 s
We set this value for the crossing time t equal to τ. Therefore,
C=t/R
=1.6×10-4 s/10.0 Ω
=16 μ F
(Note: Capacitance C is typically measured in farads, F, defined as Coulombs per
volt. From the equation, we see that C can also be stated in units of seconds per
ohm.)
Discussion
The flash interval of 160 μs (the traverse time of the bullet) is relatively easy
to obtain today. Strobe lights have opened up new worlds from science to
entertainment.
RC Circuits for Timing
RC circuits are commonly used for timing purposes. A mundane example of
this is found in the ubiquitous intermittent wiper systems of modern cars. The time
between wipes is varied by adjusting the resistance in an RC circuit. Another
example of an RC circuit is found in novelty jewelry, Halloween costumes, and
various toys that have battery-powered flashing lights. (See Figure 4 for a timing
circuit.)
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A more crucial use of RC circuits for timing purposes is in the artificial
pacemaker, used to control heart rate. The heart rate is normally controlled by
electrical signals generated by the sino-atrial (SA) node, which is on the wall of the
right atrium chamber. This causes the muscles to contract and pump blood.
Sometimes the heart rhythm is abnormal and the heartbeat is too high or too low.
The artificial pacemaker is inserted near the heart to provide electrical signals to the
heart when needed with the appropriate time constant. Pacemakers have sensors
that detect body motion and breathing to increase the heart rate during exercise to
meet the body’s increased needs for blood and oxygen.
Figure 4. (a) The lamp in this RC circuit ordinarily has a very high resistance, so that
the battery charges the capacitor as if the lamp were not there. When the voltage
reaches a threshold value, a current flows through the lamp that dramatically reduces
its resistance, and the capacitor discharges through the lamp as if the battery and
charging resistor were not there. Once discharged, the process starts again, with the
flash period determined by the RC constant τ. (b) A graph of voltage versus time for
this circuit.
Learning Competency:
Solve problem involving the calculation of currents and potential difference in circuits
consisting of batteries, resistors and capacitors. (STEM_GP12EM-IIIg-51)
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ACTIVITY 1. CALCULATING TIME: RC CIRCUIT IN A HEART DEFIBRILLATOR
Directions: Analyze the problem and solve what is required of it.
A heart defibrillator is used to resuscitate an accident victim by discharging a
capacitor through the trunk of her body. A simplified version of the circuit is seen in
Figure 2. (a) What is the time constant if an 8.00-μF capacitor is used and the path
resistance through her body is 1.00 × 103 Ω? (b) If the initial voltage is 10.0 kV, how
long does it take to decline to 5.00 × 102 V?
Strategy
Since the resistance and capacitance are given, it is straightforward to multiply them
to give the time constant asked for in part (a). To find the time for the voltage to
decline to 5.00 × 102 V, we repeatedly multiply the initial voltage by 0.368 until a
voltage less than or equal to 5.00 × 102 V is obtained. Each multiplication
corresponds to a time of τ seconds.
Solution for (a)
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Solution for (b)
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Discussion
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ACTIVITY 2. CHECK YOUR UNDERSTANDING
Directions: Answer the following questions.
1. When is the potential difference across a capacitor an emf?
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ACTIVITY 3. PROBLEMS AND EXERCISES
Directions: Compute the following problems involving the calculations of
currents and potential difference.
1. The timing device in an automobile’s intermittent wiper system is based on
an RC time constant and utilizes a 0.500-μF capacitor and a variable resistor.
Over what range must R be made to vary to achieve time constants from 2.00
to 15.0 s?
2. The duration of a photographic flash is related to an RC time constant, which
is 0.100 μs for a certain camera. (a) If the resistance of the flash lamp is
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0.0400 Ω during discharge, what is the size of the capacitor supplying its
energy? (b) What is the time constant for charging the capacitor, if the
charging resistance is 800 kΩ?
3. After two time constants, what percentage of the final voltage, emf, is on an
initially uncharged capacitor C, charged through a resistance R?
4. A heart defibrillator being used on a patient has an RC time constant of 10.0
ms due to the resistance of the patient and the capacitance of the defibrillator.
(a) If the defibrillator has an 8.00-μF capacitance, what is the resistance of the
path through the patient? (You may neglect the capacitance of the patient and
the resistance of the defibrillator.) (b) If the initial voltage is 12.0 kV, how long
does it take to decline to 6.00 × 102 V?
REFLECTION
1. I learned that
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__________________________________________________________________
__________________________________________________________________
__________________________________________________
2. I enjoyed most on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________
3. I want to learn more on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
_________________________________________________.
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REFERENCES
LUMEN PHYSICS: DC Circuits containing Resistors and Capacitors
https://courses.lumenlearning.com/physics/chapter/21-6-dc-circuits-containingresistors-and-capacitors/
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Answer Key
Activity 1
Solution for (a)
The time constant τ is given by the equation τ = RC. Entering the given values for
resistance and capacitance (and remembering that units for a farad can be expressed as
s/Ω) gives
τ = RC = (1.00 × 103 Ω) (8.00 μF) = 8.00 ms.
Solution for (b)
In the first 8.00 ms, the voltage (10.0 kV) declines to 0.368 of its initial value. That is:
V = 0.368 Vo = 3.680 × 103 V at t = 8.00 ms.
(Notice that we carry an extra digit for each intermediate calculation.) After another 8.00 ms,
we multiply by 0.368 again, and the voltage is
V'=0.368 V
=(0.368)(3.680×103 V)
=1.354×103 V at t=16.0 ms
Similarly, after another 8.00 ms, the voltage is
V′′=0.368 V′= (0.368)(1.354×103 V)
=498 V at t=24.0 ms
Discussion
So after only 24.0 ms, the voltage is down to 498 V, or 4.98% of its original value. Such
brief times are useful in heart defibrillation, because the brief but intense current causes a
brief but effective contraction of the heart. The actual circuit in a heart defibrillator is slightly
more complex than the one in Figure 2, to compensate for magnetic and AC effects that will
be covered in Magnetism.
Activity 2
CHECK YOUR UNDERSTANDING:
1. Only when the current being drawn from or put into the capacitor is zero.
Capacitors, like batteries, have internal resistance, so their output voltage is
not an emf unless current is zero. This is difficult to measure in practice so we
refer to a capacitor’s voltage rather than its emf. But the source of potential
difference in a capacitor is fundamental and it is an emf.
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Activity 3
PROBLEMS AND EXERCISES:
1. range 4.00 to 30.0 MΩ
2. 2. (a) 2.50 μF (b) 2.00 s
3. 3. 86.5%
4. 4. (a) 1.25 kΩ (b) 30.0 ms
Prepared by:
ARNOLD C. TEODORO
ANDARAYAN NATIONAL HIGH SCHOOL
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
ELECTRICITY AND MAGNETISM
Background Information for the Learners (BIL)
The one of the major difference between the magnetic and electric field is that
the electric field induces around the static charge particle which is either negative or
positive, whereas the magnetic field produces around the poles (i.e., the north and
south pole) of the magnet. Some other differences between them are explained
below in the form of comparison chart
Comparison Chart
Basis For
Comparison
Definition
Electric Field
Magnetic Field
It is the force around the
The region around the
electrical charge particle.
magnetic where poles exhibit
a force of attraction or
repulsion.
Unit
Volt/meter or Newton/coulomb
Tesla, (Newton × Second)
/(Coulomb × Meter)
Symbol
E
B
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Basis For
Electric Field
Magnetic Field
Magnetometer
Electrometer
Pole
Mono pole
Dipole
Electromagnetic
It is perpendicular to the
It is perpendicular to the
Field
magnetic field.
electric field.
Field
Vector
Vector
Field Line
Induces on a positive charge
Generate at north pole and
and terminate on a negative
terminate at the south pole.
Comparison
Formula
Measuring
Instrument
charge
Loop
Electric field lines do not form a
Magnetic line forms a closed
closed loop.
loop.
Type of charge
Negative or positive charge.
North or south pole.
Force
Repulsion force on like charges
Repulsion force on like poles
and attraction force on unlike
and attraction force on unlike
charges.
poles.
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Basis For
Electric Field
Magnetic Field
Dimension
Exist in two dimensions
Remain in three dimensions
Work
Field can do work (the speed
Magnetic field cannot do
and direction of particles
work (speed of particles
changes)
remain constant)
Comparison
Learning Competency
Differentiate electric interactions from magnetic interactions (STEM_GP12EMIIIh-54)
ACTIVITY 1. WHAT IS AN ELECTROMAGNET?
Directions: Indicate whether the statement is true or false. If false, change the
identified word or phrase to make the statement true.
1.
A conductor is a material that doesn't allow electrons to flow through it
easily. _________________________.
2.
A lightning bolt occurs when billions of protons are transferred at the same
time. _________________________.
3.
The rearrangement of electrons on a neutral object caused by a nearby
charged object is called charging by induction. _________________________.
4.
Voltage difference is measured in amperes. _________________________.
5.
The unit used to measure current is the volt. _________________________.
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B. Multiple Choice: Identify the choice that best completes the statement or answers
the question.
6.
Which of the following is a device designed to open an overloaded circuit
and prevent overheating?
a. circuit breaker
7.
b. magnet
c. resistor
d. transformer
Current that does NOT reverse direction is called ____.
a. alternating
b. circuit current
c. direct current
d. magnetic
current
8.
current
The location of the strongest magnetic forces is the ____.
a. electromagnets b. magnetic
c. magnetic fields d. magnetic poles
domains
9.
Current that reverses direction in a regular pattern is called ____.
a. alternating
b. circuit current
c. direct current
d. magnetic
current
10.
current
The region around a magnet where the magnetic forces act is the ____.
a. electromagnetic b. magnetic
pole
11.
Objects that keep their magnetic properties for a long time are called ____.
domains
energy
d. temporary
magnets
magnets
c. electrical energy to mechanical
energy
b. electrical energy to chemical
energy
d. mechanical energy to electrical
energy
The function of a generator is to change ____.
a. chemical energy to electrical
energy
c. electrical energy to mechanical
energy
b. electrical energy to chemical
energy
14.
c. permanent
The function of an electric motor is to change ____.
a. chemical energy to electrical
13.
d. magnetic pole
domain
a. electromagnets b. magnetic
12.
c. magnetic field
d. mechanical energy to electrical
energy
The current that flows in an electric circuit carries ____.
a. chemical
energy
b. mechanical
c. thermal energy d. electrical
energy
energy
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15.
There is a repulsive force between two charged objects when
a. charges are of unlike sign.
c. charges are of like sign.
b. they have the same number of
d. they have the same number of
protons.
electrons.
ACTIVITY 2. HOW ELECTROMAGNETS WORK?
A. Matching: Match each item with the correct statement below.
a. Conductor
d. fuse
b. Insulator
e. electroscope
c. circuit breaker
____ 1.
It contains a piece of metal that melts if the current becomes too high
____ 2.
It allows electrons to move through it easily
____ 3.
It contains a piece of metal that bends when it gets hot
____ 4.
It detects the presence of electric charges
____ 5.
It does not allow electrons to move through it easily
B. Completion: Complete each statement.
Figure 1.
1.
In Figure 1, circuit ____________________ is wired in series.
2.
In Figure 1, circuit ____________________ is wired in parallel.
3.
In Figure 1, circuit ____________________ represents the way that
homes are usually wired.
4.
In Figure 1, circuit ____________________ is the type of circuit that
causes an entire string of decorative lights to go out when one of the
bulbs burn out.
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ACTIVITY 3. PRINCIPLE OF ELECTRICITY AND MAGNETISM
A. Short Answer:
1. Describe how you could quickly determine whether a string of lights is
wired in series or in parallel.
2. Why is a fuse an important device in an electrical circuit?
3. Identify the types of elements in the schematic diagram below and the
number of each type.
4. Is a current flowing in the schematic diagram below? Explain your
answer.
5. Will the magnets in the figure below attract or repel each other?
B. True/False: Indicate whether the statement is true or false.
1.
When you bring the south ends of two magnets close together, they repel
each other.
2.
The strength of an electromagnet can be increased by reducing the number
of turns on the wire coil.
3.
An electric motor is used to change mechanical energy into electrical
energy.
4.
Moving a wire through a magnetic field creates an electrical current in the
wire.
5.
Paper clips and other objects that contain iron can become temporary
magnets.
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REFLECTION
1. I learned that
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
________________________.
2. I enjoyed most on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
________________________.
3. I want to learn more on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
________________________.
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REFERENCES
•
•
Electricity and Magnetism
www.srpnet.edu.com
https://www.helpteaching.com/tests/printKey.htm?test=334359
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Answer Key
ACTIVITY 1
A. MODIFIED TRUE/FALSE
1.
F, An insulator
2.
F, electrons
3.
T
4.
F, volts
5.
F, ampere
B. MULTIPLE CHOICE
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
A
C
D
A
C
C
C
D
D
C
ACTIVITY 2:
A. MATCHING
1.
2.
3.
4.
5.
D
A
C
E
B
B. COMPLETION
1.
A
2.
B
3.
B
4.
A
ACTIVITY 3:
A. SHORT ANSWER
1. Remove one bulb. If it's a series, all the
lights will go out; if it's parallel, the
remaining lights will stay lit.
2. A fuse will stop the current if it becomes
too high for the wires to handle,
preventing fires.
3. Three resistors, one battery, and one
light bulb.
4. No, because the switch is open, so there
is not a closed-loop path for the electrons
to follow.
5. The magnets will repel each other.
B. TRUE OR FALSE
1. T
2. F
3. F
4. T
5. T
Prepared by:
ARNOLD C. TEODORO
ANDARAYAN NATIONAL HIGH SCHOOL
362
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Magnetic Flux
Background Information for the Learners (BIL)
What is Magnetic Flux?
Magnetic Flux is a measurement of the total magnetic field which passes
through a given area. It is a useful tool for helping describe the effects of the
magnetic force on something occupying a given area. The measurement of
magnetic flux is tied to the particular area chosen. We can choose to make the
area any size we want and orient it in any way relative to the magnetic field.
If we use the field-line picture of a magnetic field then every field line
passing through the given area contributes some magnetic flux. The angle at
which the field line intersects the area is also important. A field line passing
through at a glancing angle will only contribute a small component of the field
to the magnetic flux. When calculating the magnetic flux, we include only
the component of the magnetic field vector which is normal to our test area.
If we choose a simple flat surface with area A as our test area and there
is an angle θ between the normal to the surface and a magnetic field vector
(magnitude B) then the magnetic flux is,
Φ=BAcosθ
where:
Φ=
B=
A=
Magnetic Flux
Magnetic Field
Area
cosθ = angle between a perpendicular vector to the area and the magnetic
field
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How do we measure magnetic flux?
The SI unit of magnetic flux is the Weber (named after German physicist
and co-inventor of the telegraph Wilhelm Weber) and the unit has the
symbol 𝑊.
Because the magnetic flux is just a way of expressing the magnetic field
in a given area, it can be measured with a magnetometer in the same way as
the magnetic field. For example, suppose a small magnetometer probe is
moved around (without rotating) inside a 0.5 𝑚2 area near a large sheet of
magnetic material and indicates a constant reading of 5 𝑚𝑇. The magnetic flux
through the area is then (5𝑋10−3 𝑇)(0.5 𝑚2 ) = 0.0025 𝑊𝑏. In the event that
the magnetic field reading changes with position, it would be necessary to find
the average reading.
A related term that you may come across is the magnetic flux density.
This is measured in 𝑊𝑏/𝑚2 . Because we are dividing flux by area, we could
also directly state the units of flux density in Tesla. In fact, the term magnetic
flux density is often used synonymously with the magnitude of the magnetic
field.
Why is this useful?
There are a couple of reasons why the description of magnetic flux can
be more useful than that of a magnetic field directly.
1. When a coil of wire is moved through a magnetic field a voltage is
generated which depends on the magnetic flux through the area of the
coil. This is described by Faraday's law and is explored in our article
on Faraday's law. Electric motors and generators apply Faraday's law
to coils which rotate in a magnetic field as depicted in Figure 1. In this
example the flux changes as the coil rotates. The description of
magnetic flux allows engineers to easily calculate the voltage generated
by an electric generator even when the magnetic field is complicated.
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Figure 1. Simplified diagram of a rotating coil in an electric generator (public
domain).
2. Although we have thus-far only concerned ourselves with magnetic flux
measured for a simple flat test-area, we can make our test-area a
surface of any shape we like. In-fact, we can use a closed surface such
as a sphere which encloses a region of interest. Closed surfaces are
particularly interesting to physicists because of Gauss's law for
magnetism. Because magnets always have two poles there is no
possibility (as far as we know) that there is a magnetic monopole inside
a closed surface. This means that the net magnetic flux through such a
closed surface is always zero and therefore all the magnetic field lines
going into the closed surface are exactly balanced by field lines
coming out. This fact is useful for simplifying magnetic field problems.
Learning Competency
Evaluate the total magnetic flux through an open surface. STEM_GP12EMIIIh-55
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ACTIVITY 1. MAGNETIC FLUX THROUGH GIVEN AREAS
Directions: Explain the Magnetic Flux through given areas.
In the case that the surface is perpendicular to the field then the angle is zero
and the magnetic flux is simply BA, Figure 2 shows an example of a flat test area at
two different angles to a magnetic field and the resulting magnetic flux.
Figure 2: Magnetic flux through given areas (blue) oriented at an angle (left) and normal to (right) the magnetic field.
If the blue surfaces shown in Figure 2 both have equal area and the angle 𝜃
is 25 °, how much smaller is the flux through the area in Figure 2-left vs Figure 2right?
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____________
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6
6
6
6
6
5
5
6
6
6
6
6
6
6
6
5
5
6
6
6
5
5
6
5
6
6
5
5
5
5
5
6
5
5
5
4
4
5
5
6
5
5
4
4
5
3
5
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5
5
5
5
4
4
3
4
4
3
5
5
4
4
4
4
3
3
3
3
3
4
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
All numbers represent mT out of the page.
Grid lines drawn with 1 cm spacing.
Figure 3: A map of magnetic field measurements around a loop of wire (green).
Figure 3 shows a map of a non-uniform magnetic field measured near a sheet
of magnetic material. If the green line represents a loop of wire, what is the magnetic
flux through the loop?
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ACTIVITY 2. MAGNETIC FLUX AROUND CURRENT-CARRYING
WIRE
Figure 4. Magnetic flux through a loop near a straight current carrying wire.
1. Figure 4 shows a square loop of wire placed near a current carrying wire.
Using the dimensions shown in the figure, find the magnetic flux through a
coil. If you don't know how to calculate the magnetic field around a wire, review
our article on the magnetic field. Hint: it may be useful to plot the magnetic
field vs vertical distance from the wire.
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2. A square coil of side 5 cm lies perpendicular to a magnetic field of flux density
4.0 T. The coil consists of 200 turns of wire. What is the magnetic flux cutting
the coil?
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_________________________________________________________________________________
_________________________________________________________________________________
__________________________________________________________________
3. Th e coil is rotated through an angle of 90° in 0.2 second. Calculate the
magnitude of the average e.m.f. induced in the coil while it is being rotated.
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ACTIVITY 3. DETERMINING MAGNETIC FLUX
Directions: Problem Solving: Calculate the magnetic flux density of the
following:
1. Copy the following diagram and mark in the polarities of the two ends of the coil.
2. Calculate the magnetic flux density at the following places:
(a) 2 m from a long straight wire carrying a current of 3 A
(b) at the centre of a solenoid of 2000 turns 75 cm long when a current of 1.5 A
flows
Note: take µo = 4π  10-7 N A-2
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3. A solenoid of length 25 cm is made using 100 turns of wire wrapped round an
iron core. If the magnetic flux density produced when a current of 2 A is passed
through the coil is 2.5 T calculate the permeability (µ) of the core.
4. A Hall probe measures a steady magnetic field directly by detecting the effect of
the field on a slice of semiconductor material. A student sets up the circuit below
to investigate, using a Hall probe, the factors which determine the magnetic flux
density within a long solenoid.
5. A solenoid similar to that shown in the diagram has 100 turns connected in a
circuit over a length of 0.50 m. µo = 4π  10-7 N A-2 Calculate the flux density at
the centre of the solenoid when a current of 10 A flows.
REFLECTION:
1. I learned that
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________
2. I enjoyed most on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________
3. I want to learn more on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
_________________________________________________.
370
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REFERENCES
Khan Academy, Physics Library, Magnetic Flux
https://www.khanacademy.org/science/physics/magnetic-forces-andmagnetic-fields/magnetic-flux-faradays-law/a/what-is-magnetic-flux
The Physics Teacher, Electromagnetic Induction
http://www.thephysicsteacher.ie/LC%20Physics/Revision/Long%20Question
s/12.%20Electromagnetic%20Induction.pdf
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Answer Key
Activity 1
Exercise 1: the magnetic flux through the tilted area is about 9% smaller than through the
area normal to the field.
Exercise 2: 0.0123 m Wb
Activity 2
1. 8 x 10 -8 Wb
2. =0.01 Wb
3. E = 10 V
Activity 3
1.
N
S
2. (a)
At distance r from a long straight wire: Magnetic flux density (B) = moI / 2pr =
3 x 10-7 T
(b)
T
At the centre of a solenoid: Magnetic flux density (B) = moNI / L = 5.03 x 10-3
3. Magnetic flux density (B) = mNI / L = 2.5 = m x 100 x 2 / 0.25 T
Permeability of the core (m) = 2.5 x 0.25 / 100 x 2 = 0.0031 N A-2
4. Factors affecting field strength are current I and spacing of coils, N coils in length L:
B  I, B 
N
L
5. Calculation using I = 30 A, N = 100, L = 0.50 m:
B=
 o NI
4  10 −7 N A –2  100  10 A
=
= 2.5 mT
L
0.50 m
Prepared by:
ARNOLD C. TEODORO
ANDARAYAN NATIONAL HIGH SCHOOL
372
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Background Information for the Learners (BIL)
Magnetic Fields and Electric Fields
If a charged particle moves, it creates a magnetic field. Unlike the electric field,
the force lines are directed at right angles to the direction of motion. So, if electrons
are moving in a wire, there has to be a magnetic field encircling the length of the wire,
the direction of the magnetic force depending on the direction of the electron
movement. Magnetic and electric fields interact such that a changing magnetic field
creates an electric field and a changing electric field creates a magnetic field. The
magnetic field is created by the moving electric field associated with the charged
particle. Conversely, if a magnetic field moves, an electric field is generated. Moving
a magnet past a wire will create a voltage that moves the electrons in the wire.
Magnetic Field
Magnetic fields are created whenever there is a flow of electric current. A
magnetic field is one that’s created around a permanent magnetic substance or a
moving electrically charged object. Magnetic fields are measured in milliGauss (mG).
Magnetic field lines, in the case of a magnet, are generated at the north pole
and terminate on a south pole. Magnetic poles do not exist in isolation. Like in the
case of electric field lines, the magnetic field is tangent to the field lines. Charged
particles will spiral around these field lines.
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Image Source
https://www.google.com/search?q=magnetic+field+clipart&hl=en&sxsrf=ALeKk02TGby4gTyoXTyLa5Asabt_yzfr4Q:1613303207425&tbm=isch&source=iu&ictx=1&fir=JR2qGH0i43IQV
M%252CcaliXA7tNSQpPM%252C_&vet=1&usg=AI4_-kSDvANi8KfsIk8t2l0zO6Lv8K5eg&sa=X&ved=2ahUKEwjfqN_kpunuAhVSyYsBHWu9CEoQ9QF6BAgQEAE&biw=1366&bih=663#imgrc=5dVzbk1gf-KlOM
Electric Field
All matter contains electrons and protons. Electrons have a negative charge,
while protons have a positive charge. An electric field is essentially a force field that’s
created around an electrically charged particle. An electric field occurs wherever a
voltage is present. Electric fields are created around appliances and wires wherever
a voltage exists. In an electromagnetic field, the directions in which the electric and
magnetic field move, are perpendicular to each other.
A charged particle produces an electric field in all directions. This field
produces a force that is either directed away or toward the original This force attracts
oppositely charged particles and repels particles with the same charge. If the particle
moves, such as electrons in an antenna, the associated electric field moves with it.
Electric field lines are generated on positive charges and terminate on negative ones.
The field lines of an isolated charge are directly radially outward. The electric field is
tangent to these lines.
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Figure 2. Diagram of a positively-charged
Figure 3. Diagram of a negatively-charged
particle showing the lines of force created by
particle showing the lines of force created by
the electric field
the electric field.
Image credit: https://www.met.nps.edu/~psguest/EMEO_online/module2/module_2_3.html
A difference in electric field strength between two locations is called a voltage.
So, when we apply a voltage across two ends of a wire the electrons in the wire move
toward the positive voltage and away from the negative voltage. The strength of an
electric field decreases rapidly as you move away from the source. Electric fields can
also be shielded by many objects, such as trees or the walls of a building. Electric
field strength is measured in volts per meter (V/m).
An object with moving charge always has both magnetic and electric field.
Electric and Magnetic forces both affect the trajectory of a charged particles but in a
qualitatively different way.
Some of the important applications associated with the presence of the two fields
include:
•
The motion of a charged particle in electric and magnetic fields
•
Measurement of specific charge of an electron (J.J.Thompson experiment)
•
Acceleration of charged particles (cyclotron)
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The motion of a charged particle in the electric and magnetic field.
In case of motion of a charge in a magnetic field, the magnetic force is
perpendicular to the velocity of the particle. So no work is done and no change in the
magnitude of the velocity is produced (though the direction of momentum may be
changed). We shall consider the motion of a charged particle in a uniform magnetic
field. First, consider the case of v (velocity) perpendicular to B (direction of magnetic
field)
Figure 4
Image Credit
:https://www.toppr.com/guides/physics/moving-charges-and-magnetism/motion-combined-electric-magneticfields/#:~:text=The%20motion%20of%20a%20charged,the%20velocity%20of%20the%20particle.&text=The%20perpendicular%20force%2C%20q%20v%20%C3%97,perpendicular%20
to%20the%20magnetic%20field.
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The perpendicular force, q v × B, acts as a centripetal force and produces a
circular motion perpendicular to the magnetic field. If velocity has a component along
B, this component remains unchanged as the motion along the magnetic field will not
be affected by the magnetic field.
Figure 5
Image Credit :https://www.toppr.com/guides/physics/moving-charges-and-magnetism/motion-combined-electric-magneticfields/#:~:text=The%20motion%20of%20a%20charged,the%20velocity%20of%20the%20particle.&text=The%20perpendicular%20force%2C%20q%20v%20%C3%97,perpendicular%20
to%20the%20magnetic%20field.
The motion in a plane perpendicular to B is as before a circular one, thereby
producing a helical motion. However, the electric field in y-direction imparts acceleration
in that direction. The particle, therefore, acquires velocity in the y-direction and resulting
motion is a helical motion. The radius of each of the circular element and other periodic
attributes like time period, frequency and angular frequency is same as for the case of
circular motion of a charged particle in perpendicular to magnetic field. If there is a
component of the velocity parallel to the magnetic field (denoted by v2), it will make the
particle move along both the field and the path of the particle would be a helical one.
The distance moved along the magnetic field in one rotation is called pitch p.
Constant Velocity Produces A Straight Line
If a charged particle’s velocity is parallel to the magnetic field, there is no net force
and the particle moves in a straight line .Newton’s first law of motion states that if an
object experiences no net force, then its velocity is constant. If a charged particle’s
velocity is completely parallel to the magnetic field, the magnetic field will exert no
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force on the particle and thus the velocity will remain constant. A particle with
constant velocity will move along a straight line through space.
There are many cases where a particle may experience no net force. The particle
could exist in a vacuum far away from any massive bodies (that exert gravitational
forces) and electromagnetic fields. Or there could be two or more forces on the
particle that are balanced such that the net force is zero. This is the case for, say, a
particle suspended in an electric field with the electric force exactly counterbalancing
gravity. If the net force on a particle is zero, then the acceleration is necessarily zero
from Newton’s second law: F=ma. If the acceleration is zero, any velocity the particle
has will be maintained indefinitely (or until such time as the net force is no longer
zero). Because velocity is a vector, the direction remains unchanged along with the
speed, so the particle continues in a single direction, such as with a straight line.
Speed, Acceleration and Kinetic Energy
The magnetic field accelerates the charge particles by changing the direction
of velocity. The magnetic field doesn’t change the speed of the charged particles.
The reasons is that the magnetic field doesn’t affect the speed is because the
magnetic field applies a force perpendicular to the velocity. Hence the force can’t do
work on the particles. As a result, the particles can’t change the kinetic energy. So it
cannot change the speed.
Learning Competency:
Describe the motion of a charged particles in a magnetic field in terms of its speed,
acceleration cyclotron radius cyclotron frequency and kinetic energy
(STEM_GP12EM-IIIH-58)
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Activity 1: It’s time to reveal the similarities and differences.
Directions: Based on your readings about the electric field and magnetic field, give
the similarities and difference through the use of the Venn Diagram.
Electric Field
Magnetic Field
Activity 2: Be A Detective
Directions: Draw five things that can be found inside your house that uses magnetic
field and electric field and explain how does they use magnetic and electric field.
Explanation
1
2
3
4
5
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Activity 3: Think Critically
Directions: Read and analyze each item. Write the letter of the correct answer
Write A – if both of the statements are TRUE
B- if both of the statement are FALSE
C- if the first statement is TRUE and the second statement is FALSE
D- if the first statement is FALSE and the second statement is TRUE
1. A. The movement of electron creates magnetic field.
B. The magnets creates electric field.
2. A. Magnetic field are created whenever there is a flow of electric current.
B. Changing magnetic field creates electric field.
3. A. Not all matter contain proton and electron.
B. Electric fields will be created even without the presence of voltage.
4. A. The strength of an electric field decreases rapidly as you move toward the
source.
B. An object with moving charges may contain both electric and magnetic field.
5. A. Magnetic force is always perpendicular to the velocity of a particle in a
uniform magnetic field.
B. When magnetic force is perpendicular to the velocity there’s no changes in
magnitude.
6. A. The particles when in a circular motion the attributes like its radius and
angular frequency are not the same.
B. The distance move along the magnetic field in one rotation is called pitch.
7. A. A charged particle with constant velocity will move along in a circular
manner through space.
B. If a charged particles velocity is completely parallel to the magnetic force,
the velocity will remain zero as well.
8. A. Magnetic field can accelerates the charged particles by changing the
direction of velocity.
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B. Magnetic field can change the speed of the charged particles, if the
magnetic field is perpendicular to the velocity of the particles.
9. A. If the net force on a particle is zero then the acceleration is also zero.
B. Particles may experience no net force because the particles exist inside a
vacuum.
10. A. The direction of magnetic field depends to the movement of electron.
B. The direction of electric field will depend on the force line.
Activity 4: Hide and Seek
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Directions: Encircle the 10 words listed below. Word may appear across, straight, up
,down back word straight across.
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REFLECTION:
1. I learned that _________________________________________________
_____________________________________________________________
___________________________________________________________
2. I enjoyed most on _____________________________________________
_____________________________________________________________
____________________________________________________________
3. I want to learn more on _________________________________________
_____________________________________________________________
____________________________________________________________
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References
“Electric Field vs Magnetic Field”. Different. Last Modified (n.d). Accessed last
February 14 2021.
https://www.diffen.com/difference/Electric_Field_vs_Magnetic_Field#:~:text=An%2
0electric%20field%20is%20essentially,a%20moving%20electrically%20charged%2
0object.
“Magnetic Fields, Magnetic Forces and Conductors”. Lumen. Last Modified
November 19 2019. Accessed last February 14 2021.
https://courses.lumenlearning.com/boundless-physics/chapter/motion-of-a-chargedparticle-in-a-magnetic-field/
“Motion in combined electric and magnetic fields”. Toppr. Last Modified (n.d).
Accessed last February 14 2021. https://www.toppr.com/guides/physics/movingcharges-and-magnetism/motion-combined-electric-magneticfields/#:~:text=The%20motion%20of%20a%20charged,the%20velocity%20of%20th
e%20particle.&text=The%20perpendicular%20force%2C%20q%20v%20%C3%97,
perpendicular%20to%20the%20magnetic%20field.
“Basic Concept Electromagnetic Radiation Concept”. Naval Postgraduate School
Gateway to Online Learning. Last Modified January 12 2004. Accessed last
February 14 2021.
https://www.met.nps.edu/~psguest/EMEO_online/module2/module_2_3.html
384
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Answer Key
Activity 1: It’s time to reveal the similarities and differences.
Electric Field
Magnetic
Field
Electric
Magnetic Field
and
are created
Magnetic
when ever
field
there is a
affects the
current
trajectory
A force field
that created
around in an
electrically
charged
particles
of charged
Note: Answer may vary
particles
Activity 2: Be A Detective
Answer may vary
Activity 3: Think Critically
1. C
2. A
3. B
4. A
5. A
6. D
7. B
8. C
9. A
10. A
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Activity 4: Hide and Seek
Prepared by:
NASHRENE ANN A. FRONDA
ALLACAPAN VOCATIONAL HIGH SCHOOL
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Background Information for the Learners (BIL)
Magnetic Field and Force in a Current Carrying Wire
“Faith is like Electricity. You can’t see it but you can see the light.” an
inspirational quotation from Sir Gregory Dickow. Electricity is very useful most
especially in our daily life, without electricity the world seems dull and dark as if there
is no sign of life. But did you know how our electricity works? How does magnetic
force and magnetic field need to team up together just to provide us electricity? I
know that you’ve been curious enough lets go let’s discover how electricity works.
Current Carrying Conductor
A wire carrying electric current will produce a magnetic field with closed field
lines surrounding the wire. A current-carrying wire feels a force in the presence of an
external magnetic field. It is found to be
F=B i ℓ,
where
ℓ is the length of the wire,
i is the current, and
θ is the angle between the current direction and the magnetic.
An electric current will produce a magnetic field, which can be visualized as a
series of circular field lines around a wire segment. Moving charges experience a
force in a magnetic field. If these moving charges are in a wire and if the wire is
carrying a current the wire should also experience a force. A current-carrying wire
generates a magnetic field and the magnetic field exerts a force on the currentcarrying wire
Magnetic Field
In physics, the magnetic field is a field that passes through space and
which makes a magnetic force move electric charges and magnetic dipoles.
Magnetic fields are around electric currents, magnetic dipoles, and changing
electric fields .Magnetic fields can illustrated by magnetic flux lines. At all times
the direction of the magnetic field is shown by the direction of the magnetic
flux lines. The strength of a magnet has to do with the spaces between the
387
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magnetic flux lines. The closer the flux lines are to each other, the stronger the
magnet is. The farther away they are, the weaker. The flux lines can be seen
by placing iron filings over a magnet. The iron filings move and arrange into
the lines. Magnetic fields give power to other particles that are touching the
magnetic field.
Magnetic fields are produced by electric currents, which can be macroscopic
currents in wires, or microscopic currents associated with electrons in atomic orbits.
Magnetic field sources are essentially dipolar in nature, having a north and south
magnetic pole. The SI unit for magnetic field is the Tesla. A smaller magnetic field
unit is the Gauss (1 Tesla = 10,000 Gauss).
Magnetic Flux
Lines
Magnet
Sourcehttps://www.google.com/search?q=magnetic+flux+lines&tbm=isch&ved=2ahUKEwiHjNWCrrTuAhX9xIsBHXJICdMQ2cCegQIABAA&oq=magnetic+flux+lines&gs_lcp=CgNpbWcQARgAMgIIADICCAAyAggAMgQIABAYMgQIABAYMgQIABAYMgQIABAYMgQIABAYMgQIABAY
MgQIABAYOgQIABBDOgQIIxAnOgcIABCxAxBDUM6NNli9qTZgxc02aABwAHgBgAHsDIgBs1GSAREwLjMuNS40LjIuMS4wLjIuMpgBAKABAaoBC2d3cy13aX
otaW1nwAEB&sclient=img&ei=s0sNYMfmBP2Jr7wP8pClmA0&bih=663&biw=1366#imgrc=tsCxrm0d8SiaWM
Source http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfie.html
Stop. Observe the two sets of magnets.
Magnet A
Magnet B
Sourcehttps://www.google.com/search?q=two+magnet+with+the+same+pole&sxsrf=ALeKk03w5gXM0A5wWC399JgCjHiX47P0RQ:1611462904214&source=lnms&tbm=isch&sa=X&ved=2ahUKEwigs
OWP37PuAhUQrpQKHXOFAq0Q_AUoAXoECAUQAw&biw=1366&bih=663#imgrc=QAIC9-BGlJ03LM
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Magnetic Force
The magnetic force is the force of attraction or repulsion that arises between
electrically charged particles due to their motion The magnetic force between two
moving charges may be described as the force exerted upon their charge by the
magnetic field created by the other. This force causes the magnets to attract or repel
one another. Two objects containing charge with the same direction of motion have
a magnetic attraction force between them. Similarly, objects with charge moving in
opposite directions have a repulsive force between them.
Electric forces exist among stationary electric charges; both electric and
magnetic forces exist among moving electric charges. The magnetic force between
two moving charges may be described as the effect exerted upon either charge by
a magnetic field created by the other. Examples of magnetic force is a compass,
a motor, the magnets that hold stuff on the refrigerator, train tracks, and new
roller coasters. All moving charges give rise to a magnetic field and the charges
that move through its regions, experience a force. It may be positive or negative
depending on whether the force is attractive or repulsive. The magnetic force is
based on the charge, velocity and magnetic field of the object.
How to find the magnetic force?
In finding the direction of magnetic force considers two objects. The
magnitude of the magnetic force between them depends on how much charge is
in how much motion in each of the two objects and how far apart they are. The
direction of the force depends on the relative directions of motion of the charge in
each
case.
Physicists use a hand mnemonic known as Flemming’s hands rule to help
remember the direction of magnetic forces. To form the mnemonic, first make an Lshape with the thumb and first two fingers of your right hand. Then, point your middle
finger perpendicular to your thumb and index finger, like this:
389
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Sourcehttps://www.google.com/search?q=right+hand+rule+physics+magnetic+force+&tbm=isch&ved=2ahUKEwjW8r-VtbTuAhXLAaYKHTsMDPMQ2-cCegQIABAA#imgrc=-X72AhHWktCNeM
Image showing a hand in the right-hand-rule configuration
The right-hand rule is based on the underlying physics that relates magnetic fields
and the forces that they exert on moving charges it just represents an easy way to
remember the directions that things are supposed to point. Occasionally a physicist
will accidentally use their left hand, causing them to predict that the magnetic force
will point in a direction opposite the true direction!
Moving charges
When charges are sitting still, they are unaffected by magnetic fields, but as soon
as they start to move, the magnetic field pushes on them. But, the direction in
which the field pushes on charges is not the same as the direction of the magnetic
field lines. It instead looks more like this:
Sourcehttps://www.khanacademy.org/test-
prep/mcat/physical-processes/magnetism-
mcat/a/using-the-right-hand-
rule#:~:text=Moving%20charges,-
When%20charges%20are&text=We%20can%20remember%20this%20diagram,pushing%20on%20the%20moving%20charge.
Image showing Lorentz forces
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We can remember this diagram using the right-hand rule. If you point your pointer
finger in the direction the positive charge is moving, and then your middle finger in
the direction of the magnetic field, your thumb points in the direction of the
magnetic force pushing on the moving charge. When you’re dealing with negative
charges like moving electrons the force points in the opposite direction as your
thumb.
Sourcehttps://www.khanacademy.org/test-
prep/mcat/physical-processes/magnetism-
mcat/a/using-the-right-hand-
rule#:~:text=Moving%20charges,-
When%20charges%20are&text=We%20can%20remember%20this%20diagram,pushing%20on%20the%20moving%20charge.
Diagram of moving charge, magnetic force, and magnetic field line on a hand
making the right-hand rule gesture
Current in a wire
When we talk about conventional current in a wire, we’re talking about the way
positive charges move through a wire. Since we know that current is just moving
charges, the wire will also be affected by a magnetic field in the same way as a single
moving charge, but only when there is a current passing through it.
391
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Sourcehttps://www.khanacademy.org/test-
prep/mcat/physical-processes/magnetism-mcat/a/using-
the-right-hand-rule#:~:text=Moving%20charges,-
When%20charges%20are&text=We%20can%20remember%20this%20diagram,pushing%20on%20the%20moving%20charge.
The right-hand rule applied to a conductive wire
We can use the same right-hand rule as we did for the moving charges—pointer
finger in the direction the current is flowing, middle finger in the direction of the
magnetic field, and thumb in the direction the wire is pushed.
Magnetic field caused by current in a wire
Not only are moving charges affected by magnetic fields, they can also create
them. We can find the magnetic field that is caused by moving charges using a
second right-hand rule. The magnetic field made by a current in a straight wire curls
around the wire in a ring. You can find it by pointing your right thumb in the
direction of the current in the wire and curling your fingers. Your fingers will be
curled in the same direction as the magnetic field around the wire.
Sourcehttps://www.khanacademy.org/test-prep/mcat/physical-processes/magnetism-mcat/a/using-the-right-hand-rule#:~:text=Moving%20charges,When%20charges%20are&text=We%20can%20remember%20this%20diagram,pushing%20on%20the%20moving%20charge.
A magnetic field around a wire with current moving upward
It turns out that you can do the opposite of this rule to figure out the direction of the
current in a wire if you already know the direction of the magnetic field. Point your
392
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thumb in the direction of the magnetic field this time and curl your fingers just as
before. This time, the circular direction of your fingers tells you the direction of the
current that creates the magnetic field.
Sourcehttps://www.khanacademy.org/test-prep/mcat/physical-processes/magnetism-mcat/a/using-the-right-hand-rule#:~:text=Moving%20charges,When%20charges%20are&text=We%20can%20remember%20this%20diagram,pushing%20on%20the%20moving%20charge.
Magnetic Field and Forces
Differences between electric and magnetic forces:
• The Electric force acts long the direction of the electric field, where as the
magnetic force acts as perpendicular to the magnetic field.
• The electric force acts on a charged particle regardless of whether the particle
is moving, whereas the magnetic force acts on a charged particle only when
the article is in motion
• The electric force does work in displacing a charged particle, whereas the
magnetic force associated with a steady magnetic field does not work when a
particle is displaced because the force is perpendicular to the displacement.
Learning Competency:
Evaluate the magnetic force on an arbitrary wire segment place in a uniform magnetic
field (STEM_GP12EM-IIIh-60)
393
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Activity 1: Predict Observe Explain
Directions: Study the diagram below and answer the different questions.
We have here giant poles of a magnet the North Pole and the South Pole
Source https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnets-magnetic/a/what-is-magnetic-force
Diagram 1
1. What part of the magnets where the magnetic field will occur? On the side of
two poles, in between the poles. Explain your answer.
Write your answer here?
Source
Diagram 2
https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnets-magnetic/a/what-is-magnetic-force
Source https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnets-magnetic/a/what-is-magnetic-force
Diagram 3
Diagram 3 shows that in between the two magnets they put the current
carrying wires so it means that in between the two magnet poles and the
current carrying wires they are now having magnetic force on their own.
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2. What will happen to the force if the current will increase? Is it going to increase
or decrease? Why?
Write your answer here
A
B
Source https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnets-magnetic/a/what-is-magnetic-force
Diagram 4
Diagram A shows that the current carrying wire is perpendicular to the
magnetic field while the Diagram B the current carrying wires is somehow
slanting to the South Pole.
3. In the two diagrams which one has the greatest force and greatest current? Is
letter A or letter B? Why?
Write your answer here
Source https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnets-magnetic/a/what-is-magnetic-force
Diagram 5
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4. What do you think is the current and magnetic force of this diagram? Is the
current the same or not? How about its magnetic force? Is the force become
stronger or weaker?
Write your answer here
Activity 2: Fact or Bluff
Directions: Write Fact if you think the underline word/s makes the statement correct
and if bluff write the correct word/s to make the statement right.
1. A current carrying wire can generate its own magnetic field.
2. The farther the magnetic flux to one another the stronger the magnet is.
3. The SI unit for magnetic field is Newton.
4. Solenoid and the earth are the source of magnetic field.
5. The magnetic force is the attraction of electrical charges only.
6. Current carrying wire produces magnetic field while magnetic field produces
magnetic force.
7. A wire carrying electric current will produce a magnetic field with open field
lines surrounding the wire.
8. The magnetic force is based on the charged, velocity and magnetic field of the
object.
9. Train tracks and new roller coaster are examples of magnetic field.
10. Magnetic force exist among stationary electric charges.
396
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Activity 3: Electricity Info graphics
Material: Marker, Coloring Materials, Pencil, Long Bond Paper
Procedure: Create an info graphics about the importance of using of solar energy to
generate electricity. Use the rubrics below in creating your Info Graphics.
Components
Excellent
Title
Title is clear
and easily
identified
Content
Relevant
details support
the title and
topic
Appropriate
and
Outstanding
use of color,
design and
space
Free of
grammatical
error
Visual Appeals
Mechanics
Above
Average
Title are
mostly clear
and easily
identified
Most details
support the
title and topic
Average
Title are not
clear and hard
to identify
Below
Average
Title are not
clearly
identified
Few details
No details
support the title support the
and topic
title and topic
Adequate use
of use of color,
design and
space
Inappropriate
use of color,
design and
space
Less usage of
color design
an space
Nearly free of
grammatical
error
Frequent
grammatical
error
Too frequent
grammatical
errors
Activity 4: Please Help Me Find My Direction
Directions: Using the Fleming’s hand rule locate the direction of the magnetic field,
magnetic force and the current to the following experiments.
For number 1 and 2 use the diagram below to answer the questions
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Source:https://www.google.com/search?sxsrf=ALeKk00Dqqd4omAVbDWwDm8A3AEEzbd2Hg:1611638440646&source=univ&tbm=isch&q=activity+about+right+hand+
rule+magnetic+field&sa=X&ved=2ahUKEwjw6oqG7bjuAhWnF6YKHQoUDQAQjJkEegQIHhAB&biw=1366&bih=663#imgrc=nMsdoBBGizIbHM&imgdii=9RSJhKc3KHfy
0M
1. In the hand grip rule, what is the direction of the current?
a. south
b. east
c. west
d. north
2. Which of the following is the direction of the magnetic field based on the
hand grip rule?
a. clockwise
c. anti clockwise
b. counter clockwise
d. both a and b
For number 3-5 use the experiment below to answer the questions
Sourcehttps://www.google.com/search?q=left+hand+rule+for+current+carrying+conductor&tbm=isch&ved=2ahUKEwj9hoKS7bjuAhUZy4sBHaiNDbEQ2cCegQIABAA&oq=left+h
and+rule+&gs_lcp=CgNpbWcQARgDMgQIABBDMgIIADICCAAyAggAMgIIADICCAAyAggAMgIIADICCAAyAggAOggIABCxAxCDAToFCAAQsQM6BwgAELEDEEM6CggAELE
DEIMBEENQlcVqWLTdamDF_mpoAHAAeACAAc4BiAG7EJIBBjAuMTQuMZgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei=waYPYL2iKJmWr7wPqJu2iAs&bih=663
&biw=1366#imgrc=zju8hv4eK2SmxM&imgdii=XOkzp5wY4DQSYM
3. Using the left hand rule what is the direction of the magnetic field?
a. south to north
c. east to west
b. north to south
d. west to east
4. What is the direction of the force?
a. upward
b. downward
c. left
d. right
5. What is the direction of the current?
a. toward the plane
c. below the plane
b. outside the plane
d. inside the plane
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REFLECTION:
1.I learned that _________________________________________________
_____________________________________________________________
___________________________________________________________
2.I enjoyed most on _____________________________________________
_____________________________________________________________
____________________________________________________________
3.I want to learn more on _________________________________________
_____________________________________________________________
_____________________________________________________________
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References:
Website
R. Nave. “Magnetic Field”. Hyperphysics. Last Modified (n.d). Accessed on January
24 2021. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfie.html
“Magnetic Field and Magnetic Force”. Toppr. Last Modified (n.d). Accessed on
January 24 2021. https://www.toppr.com/guides/magnetic-effects-of-electriccurrent/magnetic-field-and-magnetic-force/
“What is magnetic force”. Khan Academy. Last Modified (n.d). Accessed on
January 24 2021. https://www.khanacademy.org/science/physics/magnetic-forcesand-magnetic-fields/magnets-magnetic/a/what-is-magnetic-force
“Magnetic Force”. the editors of Encyclopedia Britannica. Last Modified (n.d).
Accessed on January 24 2021 https://www.britannica.com/science/magnetic-force
“The Right Hand Rule ”. Khan Academy. Last Modified (n.d). Accessed on January
24
2021.
https://www.khanacademy.org/test-prep/mcat/physicalprocesses/magnetism-mcat/a/using-the-right-handrule#:~:text=Moving%20charges,When%20charges%20are&text=We%20can%20remember%20this%20diagram,pu
shing%20on%20the%20moving%20charge.
“Magnetic Filed and Forces”. Last Modified (n.d). Accessed on January 24 2021.
https://profiles.uonbi.ac.ke/nyangondat/files/lesson_6_magnets_charge_in_magneti
c_field.pdf
Teacher Education Agency.“Magnetic Field, Field Lines and Force”. Texas
Gateway. Last Modified (n.d). Accessed on January 24 2021.
https://www.texasgateway.org/resource/201-magnetic-fields-field-lines-andforce#:~:text=Like%20the%20electric%20field%2C%20the,magnetic%20field%20is
%20almost%20zero.
“Magnetic Field”. Simple English Wikipedia. Last Modified December 02, 2020.
Accessed on January 24 2021.
https://simple.wikipedia.org/wiki/Magnetic_field#:~:text=The%20magnetic%20field
%20is%20the,charges%20can%20make%20magnetic%20fields.&text=In%20physi
cs%2C%20the%20magnetic%20field,electric%20charges%20and%20magnetic%2
0dipoles.
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ANSWER KEY
Activity 1: Experiment Time
Possible answers. Answer might vary.
1. It is located in between the two magnet poles .Magnetic fields will found in
between the two poles. Because the magnetic field lines become denser
between the poles and the magnets experience an attractive force. As you
can see in the figure two the magnetic field is uniform. They show that
magnetic field has only one direction it goes from the North Pole to South pole
of the magnet
2. The force will also increase because the current carrying wire is just like a
magnet also that why the magnetic field around it will also affect the current
carrying wire so if you are going to decrease the current the magnetic force
will be decrease and if you will turn it into zero the magnetic force will
vanished.
3. Diagram A and B has the same amount of current but differ in force. Diagram
A has stronger force than diagram B. Because when the current carrying wire
is perpendicular to the magnetic field it means that it attain the 90 degree of
its angle the highest the force is but as the current carrying wire is decreasing
its angle the force is also decreasing.
4. The current will be remain the same but the force is weaker because the
angle of the current carrying wires and the magnetic field is parallel to one
another so the force is zero.
Activity 3: Electricity Info graphics
Activity 2: FACT OR BLUFF
Refer your work to the rubrics given
1. FACT
2. BLUFF- CLOSER
3. BLUFF-TESLA
4. FACT
5. BLUFF-ATTRACTION AD REPULSION
6. FACT
7. BLUFF-CLOSE FIELD
Activity 4:
Please Help Me Find My Direction
1. D
2. A
3. B
4. A
5. B
8. FACT
9. BLUFF- MAGNETIC FORCE
10. BLUFF- ELECTRIC FORCE
Prepared by:
Nashrene Ann A. Fronda
ALLACAPAN VOCATIONAL HIGH SCHOOL
401
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Background Information for the Learners (BIL)
Did you know that electricity is always strictly linked to magnetism? It is the result of
one of Maxwell's equations which says that flowing electric current produces a
magnetic field.
Magnetic Field and Biot- Savart Law
What is Biot- Savart Law?
The Biot Savart Law is an equation describing the magnetic field generated
by a constant electric current. It relates the magnetic field to the magnitude, direction,
length, and proximity of the electric current. Biot–Savart law is consistent with both
Ampere’s circuital law and Gauss’s theorem. The Biot Savart law is fundamental to
magnetostatics, playing a role similar to that of Coulomb’s law in electrostatics.
Biot-Savart law was created by two French physicists, Jean Baptiste Biot and
Felix Savart derived the mathematical expression for magnetic flux density at a point
due to a nearby current-carrying conductor, in 1820. Viewing the deflection of a
magnetic compass needle, these two scientists concluded that any current element
projects a magnetic field into the space around it.
But before we dig deep in the Bio Savart Law lets talk about the magnetic field.
Electric current produces a magnetic field. This magnetic field can be
visualized as a pattern of circular field lines surrounding a wire. One way to explore
the direction of a magnetic field is with a compass, as shown by a long straight
current-carrying wire in.
Hall probes can determine the magnitude of the field. Another version of the
right hand rule emerges from this exploration and is valid for any current segment—
point the thumb in the direction of the current, and the fingers curl in the direction of
the magnetic field loops created by it.
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Image Credit:
https://courses.lumenlearning.com/boundless-physics/chapter/magnetismmagnetic-fields/
and-
The Magnetic field lines around a straight conductor carrying current are concentric
circles whose centres lie on the wire. The direction of magnetic field lines can be
determined using Right-Hand Thumb Rule.
Factors affecting the strength of magnetic field around a straight current carrying
conductor
1. Magnetic field strength is directly proportional to the magnitude of current flowing
in the conductor. i.e. B∝I, greater the current in the conductor, stronger will be the
magnetic field produced.
2. Magnetic field strength is inversely proportional to the distance from the wire
i.e. B∝1r, greater the distance from the current carrying conductor, weaker will be
the magnetic field.
To calculate magnetic field of a straight current carrying conductor you may used this
formula
𝜇𝑜Ι
𝐵=
2𝜋𝑟
Where:
B is the strength of the magnetic field produce at distance
I represents is the current
r is the distance between the location of dℓ, and the location at which the magnetic
field is being calculated
μ0 is the permeability of free space which have constant value
= 4𝜋 × 10−7
Example 1.1. Calculate the magnetic field of a straight current carrying conductor.
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A current of 15 A flows north along a wire. Calculate the magnitude and direction of
the magnetic field at a point 10 cm east of the wire.
Given:
I= 15 A
Unknown: B
Use the formula:
μ0= 4𝜋 × 10−7
R= 10 cm (0.1m)
𝐵=
𝜇𝑜Ι
2𝜋𝑟
Solution:
𝐵=
(4𝜋𝑥107 )(15A)
2𝜋 (0.1𝑚)
𝐁 = 𝟑 𝐱 𝟏𝟎−𝟓 𝐓
Note: If you want to increase the strength of the magnetic field you can move closer
to the current wire or increase the current.
Learning Competency:
Evaluate the magnetic field vector at a given point in space infinetisimal current
element or a staight current carrying conductor (STEM_GP12EM-IIIH-60)
Activity 1: Reveal the mystery
Directions: Decode the word using the letters and numbers
A
N
13
1
7
14
5
20
Hint: Object that is capable of producing magnetic field and attracting unlike poles
and repelling like poles
C
R
3
21
18
18
5
Hint: The rate at which charge flows through a surface
13
G
7
1
6
E
5
14
9
E
5
14
20
12
20
9
3
4
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Hint: Vector field that describes the magnetic influence on moving electric charges,
electric currents, and magnetic materials
13
1
7
14
T
20
9
U
21
4
5
Hint: Quantity or distance
1
13
P
16
5
R
18
5
Hint: unit of electric current in the International System of Units (SI)
Activity 2: Choose the Best
Directions: Read and analyze each item. Write the correct answer.
1. What is magnetic field?
a. Defined as the magnetic flux per unit area across an area at right angles
to the magnetic field
b. A measurement of the total magnetic field which passes through a given
area
c. A region of charged particles
d. Region around a magnet where magnetic force can be experienced
2. What must happen for an electromagnet to have a magnetic field?
a. It has to be touching another magnet
b. It must be lined up with earth magnetic field
c. It must be connected to an electrical source
d. It must be heated
3. What is the path of a charge moving at right angles in a magnetic field?
a. Straight
c. Parabola
b. Circular
d. Sinusoidal
4. What is the unit magnetic field strength?
a. Tesla
c. Fared
b. Weber
d. Gauss
5. What substance is attracted to a magnet?
a. Silver
c. Water
b. Lead
d. Iron
6. Earth’s magnetic field and bar magnets both attract particles to the same
locations. Where are they?
a. All around the outside
b. The North and South Pole
c. The Middle
d. First to the middle then to the ends
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7. Which of the following would be the best way to observe and compare the
magnetic field of various magnet?
a. Ask a high school science teacher about the history of natural magnets
b. Use a reference book and read a section about electromagnetism and
magnets
c. Conduct an experiment using different types of magnets and iron fillings
d. Search for an internet web site to find experiments on electricity and
magnets
8. Ancient people discovered magnetic rocks called. Iodestone. What did they
used for?
a. To start fires
b. Compasses
c. Sculptures
d. Telephone receivers
9. What is the symbol for magnetic field strength?
a. T
b. B
c. M
d. F
10. Biot- Savart Law is consistent to what law and theorem?
a. Ampere circuital law and Gauss Theorem
b. Ampere circuital law and Newton Theorem
c. Ampere circuital law and Browns Theorem
d. Ampere circuital law and Hooke’s Theorem
Activity 3: My Own Stand
Directions: Write a simple explanation to your understanding about the question below?
“How do you think electricity generation will change in the near future?”
Activity 4: Picture Analysis
Directions: Analyze each pictures depicts and answer each questions. Solve the
following worded problem
1. Let say that the diagram has a current of 2 A and distance of 3m away from
the wire. Calculate the magnitude of the magnetic field?
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2A
3m
?
Image
Credit:
https://courses.lumenlearning.com/boundless-physics/chapter/magnetism-and-magnetic-fields/
For question number 2 use the picture below
Image Credit: https://quizizz.com/admin/quiz/5a8a5956d7c2a3002086364f/magnetic-field
2. Is the direction of the magnetic field correct? Yes or No? Why?
Write your answer here?
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For question number 3 use the picture below
Image
Credit: https://quizizz.com/admin/quiz/5a8a5956d7c2a3002086364f/magnetic-field
3. Two long parallel wires each carry the same current I in the same direction.
What is the total magnetic field at the point P midway between the wires?
Explain your answer.
REFLECTION:
1.I learned that______________________________________________________
_____________________________________________________________
_____________________________________________________________
2.I enjoyed most on ________________________________________________
____________________________________________________________
_____________________________________________________________
_
3.I want to learn more on _____________________________________________
_____________________________________________________________
_____________________________________________________________
_
408
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References:
“What is the Biot Savart Law”. Electrical4U. Last Modified October 28, 2020.
Accessed on February 02, 2020. https://www.electrical4u.com/biot-savartlaw/#:~:text=The%20Biot%20Savart%20Law%20is,proximity%20of%20the%20elec
tric%20current
“Magnetism and Magnetic Field”. Lumen Boundless Physics. Last Modified (n.d).
Accessed on February 02, 2020. https://courses.lumenlearning.com/boundlessphysics/chapter/magnetism-and-magnetic-fields/
“Magnetic Field of a Straight Current Carrying Wire”. The Organic Chemistry Tutor.
Uploaded last December 20 2017. Accessed on February 05, 2020.
https://www.youtube.com/watch?v=wZlHLHnZSRg
“What is Magnet”. BYJU’S Classes. Last Modified (n.d). Accessed on February 05,
2020.https://byjus.com/physics/magnet/#:~:text=A%20magnet%20is%20defined%2
0as,poles%20and%20repelling%20like%20poles
“Electric Current”. The Physics Hyper books. Last Modified (n.d). Accessed on
February 05, 2020. https://physics.info/electriccurrent/#:~:text=Electric%20current%20is%20defined%20as,a%20wire%2C%20for
%20example).&text=The%20unit%20of%20current%20is,Amp%C3%A8re%20(177
5%E2%80%931836)
“Magnetic Field”. Wikipedia Free Encyclopaedia. Last Modified (n.d). Accessed on
February 05, 2020.
https://en.wikipedia.org/wiki/Magnetic_field#:~:text=A%20magnetic%20field%20is%
20a,and%20to%20the%20magnetic%20field
“What is magnitude in Physicss”. Toppr. Last Modified (n.d). Accessed on February
05, 2020. https://www.toppr.com/guides/physics/fundamentals/what-is-magnitudeinphysics/#:~:text=Introduction%20to%20What%20is%20Magnitude%20in%20Physi
409
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cs%3F&text=Magnitude%20generally%20refers%20to%20the,of%20the%20object
%20while%20travelling
The Editors of Encyclopaedia Britannica. “Ampere”. Encyclopaedia Britannica. Last
Modified (n.d). Accessed on February 05, 2020.
https://www.britannica.com/science/ampere
CK-12. “Magnetic Field due to Current Carrying Conductor. CK-12.org. Last
Modified February 18 2016. Accessed on February 05, 2020.
https://www.ck12.org/book/cbse-physics-book-class-x/section/4.4/
Khan Academy. Magnetic Field Created by Current Carrying Wire Physics Khan
Academy. Last Modified (n.d). Accessed on February 05, 2020.
https://www.youtube.com/watch?v=Ri557hvwhcM
“Magnets:. Review Game Zone.Com. Academy. Last Modified (n.d). Accessed on
February 05, 2020.
https://reviewgamezone.com/mc/candidate/test/?test_id=658&title=Magnets
Hanadi Fatayerji.”Magnetic Field”. Quizizziz.com. Last Modified (n.d). Accessed on
February 05, 2020.
https://quizizz.com/admin/quiz/5a8a5956d7c2a3002086364f/magnetic-field
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ANSWER KEY
Activity 1: Reveal the mystery
1. MAGNET
2. CURRENT
3. MAGNETIC FIELD
4. MAGNITUDE
5. AMPERE
Activity 2: Choose the Best
1. D
6.
B
2. D
7.
C
3. B
8.
A
4. A
9.
A
5. A
10.
A
Activity 3: My Own Stand
Answers May Vary
Activity 4: Picture Analysis
1. Let say that the diagram has a current of 2 A and distance of 3m away from the
wire. Find the magnitude of the magnetic field?
Given:
I= 15 A
R= 10 cm (0.1m)
μ0= 4𝜋 × 10−7
Unknown: B
Use the formula:
Solution:
(4𝜋𝑥107 )(2A)
𝐵=
2𝜋 (3 𝑚)
𝐁 = 𝟏. 𝟑 𝐱 𝟏𝟎−𝟕 𝐓
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2. Is the direction of the magnetic field correct? Yes or No? Explain your
answer?
Yes because the magnetic field run from north to south (Possible answer but
answer may vary)
3. Two long parallel wires each carry carry the same current I in the same
direction. What is the total magenti field at the point P midway between the
wires?
Zero (0)
Prepared by:
Nashrene Ann A. Fronda
ALLACAPAN VOCATIONAL HIGH SCHOOL
412
NOTE: Practice personal hygiene protocols at all times
GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Magnetic Field Due to a Thin Straight Wire
Background Information for the Learners (BIL)
How much current is needed to produce a significant magnetic field, perhaps as
strong as Earth’s field? Surveyors will tell you that overhead electric power lines
create magnetic fields that interfere with their compass readings. Indeed, when
Oersted discovered in 1820 that a current in a wire affected a compass needle, he
was not dealing with extremely large currents. How does the shape of wires carrying
current affect the shape of the magnetic field created? How does the magnetic field
of a current loop compare with that of a straight wire? We can use the Biot-Savart
law to answer all of these questions, including determining the magnetic field of a
long straight wire.
Figure 1. A section of a thin, straight current-carrying wire. The independent variable
has the limits
𝜃1 and 𝜃2
Figure 1. Shows a section of an infinitely long, straight wire that carries a current .
What is the magnetic field at a point P, located a distance R from the wire?
Let’s begin by considering the magnetic field due to the current element Idx located
at the position x. Using the right-hand rule 1 from Magnetic Fields and
Lines, dx(r) points out of the page for any element along the wire. At point P,
therefore, the magnetic fields due to all current elements have the same direction.
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This means that we can calculate the net field there by evaluating the scalar sum of
the contributions of the elements. With |𝑑𝑥(𝑟)| = (𝑑𝑥)(1) sin 𝜃, we have from the
Biot-Savart law
𝐵=
μ0
∫ 𝐼 𝑠𝑖𝑛𝜃𝑑𝑥
4π 𝑤𝑖𝑟𝑒 𝑟2
The wire is symmetrical about point O, so we can set the limits of the integration from zero
to infinity and double the answer, rather than integrate from negative infinity to positive
infinity. Based on the picture and geometry, we can write expressions for
terms of
and
and
in
, namely:
𝑟 = √𝑥 2 + 𝑅 2
sin 𝜃 =
𝑅
√𝑥 2 + 𝑅 2
Substituting these expressions into, the magnetic field integration becomes
𝐵=
𝜇0 𝐼 ∞ 𝑅𝑑𝑥
∫
2πR 0 (𝑥2 +𝑅2 ) 3⁄2
Evaluating the integral yields
μ0𝐼
𝑅𝑑𝑥
2πR (𝑥 2 + 𝑅 2 ) 3⁄
2
Substituting the limits gives us the solution
𝐵=
𝐵=
μ0𝐼
2πR
The magnetic field lines of the infinite wire are circular and centered at the wire (Figure 1),
and they are identical in every plane perpendicular to the wire. Since the field decreases with
distance from the wire, the spacing of the field lines must increase correspondingly with
distance. The direction of this magnetic field may be found with a second form of the righthand rule (illustrated in Figure 2). If you hold the wire with your right hand so that your thumb
points along the current, then your fingers wrap around the wire in the same sense as B.
Figure 2 Some magnetic field lines of an infinite wire. The direction of B can be found with
a form of the right-hand rule.
The direction of the field lines can be observed experimentally by placing several small
compass needles on a circle near the wire, as illustrated in Figure 3. When there is no current
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in the wire, the needles align with Earth’s magnetic field. However, when a large current is
sent through the wire, the compass needles all point tangent to the circle. Iron filings
sprinkled on a horizontal surface also delineate the field lines, as shown in Figure 3.
Figure 3 The shape of the magnetic field lines of a long wire can be seen using (a) small
compass needles and (b) iron filings
EXAMPLE 1. CALCULATING MAGNETIC FIELD DUE TO THREE WIRES
1. Three wires sit at the corners of a square, all carrying currents of 2 amps into the
page as shown in Figure 9.2.4. Calculate the magnitude of the magnetic field at the
other corner of the square, point P, if the length of each side of the square is 1 cm.
Figure 4. Three wires have current flowing into the page. The magnetic field is determined
at the fourth corner of the square.
Strategy
The magnetic field due to each wire at the desired point is calculated. The diagonal distance
is calculated using the Pythagorean theorem. Next, the direction of each magnetic field’s
contribution is determined by drawing a circle centered at the point of the wire and out toward
the desired point. The direction of the magnetic field contribution from that wire is tangential
to the curve. Lastly, working with these vectors, the resultant is calculated.
Solution
Wires 1 and 3 both have the same magnitude of magnetic field contribution at point P
𝑚
(4𝜋𝑥10−7 𝑇. 𝐴 ) (2𝐴)
𝜇0 𝐼
𝐵1 = 𝐵2
=
= 4 𝑥 10−5 𝑇
2πR
2π(0.01 m)
Wire 2 has a longer distance and a magnetic field contribution at point P of:
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−7 𝑚
μ0𝐼 (4𝜋𝑥10 𝑇. 𝐴 ) (2𝐴)
𝐵1 = 𝐵2
=
= 3 𝑥 10−5 𝑇
2πR
2π(0.01414 m)
The vectors for each of these magnetic field contributions are shown.
The magnetic field in the x-direction has contributions from wire 3 and the x-component of
wire 2:
Bnet x= -4 x 10-5 T- (2.83 x 10 -5 T) sin 45º = - 6 x 10-5 T.
The y-component is similarly the contributions from wire 1 and the y-component of wire 2:
Bnet y = -4 x 10-5 T- (2.83 x 10 -5 T) sin 45º = - 6 x 10-5 T.
Therefore, the net magnetic field is the resultant of these two components:
Significance
The geometry in this problem results in the magnetic field contributions in the x– and ydirections having the same magnitude. This is not necessarily the case if the currents were
different values or if the wires were located in different positions. Regardless of the numerical
results, working on the components of the vectors will yield the resulting magnetic field at
the point in need.
ACTIVITY 1. CONCEPTUAL QUESTIONS
1. Using Example 1, keeping the currents the same in wires 1 and 3, what should the
current be in wire 2 to counteract the magnetic fields from wires 1 and 3 so that there
is no net magnetic field at point P?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
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__________________________________________________________________
__________________________________________
2. How would you orient two long, straight, current-carrying wires so that there is no net
magnetic force between them? (Hint: What orientation would lead to one wire not
experiencing a magnetic field from the other?)
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________
ACTIVITY 2. PROBLEM SOLVING
Compute the following problems.
1. The magnitude of the magnetic field 50 cm from a long, thin, straight wire
is 8.0μT. What is the current through the long wire?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________
__________________________________________________________________
__________________________________________________________________
______________________________________________________
2. A long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is
placed in a uniform magnetic field of magnitude 4.0×10−5T that is directed vertically
downward, what is the resultant magnitude of the magnetic field 20 cm above the
wire? 20 cm below the wire?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
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__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
______
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
______________________________________________
ACTIVITY 3. MAGNETIC FIELD IN PARALLEL WIRES
1. The accompanying figure shows two long, straight, horizontal wires that are parallel
and a distance 2a apart. If both wires carry current I in the same direction, (a) what
is the magnetic field at P1? (b) P2?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
______________________
2. Consider the area between the wires of the preceding problem. At what distance from
the top wire is the net magnetic field a minimum? Assume that the currents are equal
and flow in opposite directions.
__________________________________________________________________
__________________________________________________________________
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__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
______________________
REFLECTION
1. I learned that
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________
2. I enjoyed most on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________
3. I want to learn more on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
_________________________________________________.
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REFERENCES
OpenStax CNX® is supported by the William and Flora Hewlett Foundation, the Maxfield
Foundation, and Rice University
https://legacy-content01.cnx.org/content/m10533/1.3/
Physics by Cutnell and Johnson 8th Edition, Chapter 21, Electricity and Magnetism
Circuits, 721-729
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Answer Key
Activity 1
1. 4 amps flowing out of the page
2. You would make sure the currents flow perpendicular to one another.
Activity 2
5. 20 A
6. Both answers have the magnitude of magnetic field of 4.5×10-5 T
Activity 3
1. At P1, the net magnetic field is zero. At P2,
3𝜇9 𝐼
8𝜋𝑎
into the page.
2. The magnetic field is at a minimum at distance a from the top wire, or half-way
between the wires.
Prepared by:
ARNOLD C. TEODORO
ANDARAYAN NATIONAL HIGH SCHOOL
421
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GENERAL PHYSICS 2
Name: ____________________________
Grade Level: _________
Date: _____________________________
Score: ______________
LEARNING ACTIVITY SHEET
Magnetic Force between Two Parallel Conductors
Background Information for the Learners (BIL)
You might expect that there are significant forces between current-carrying
wires, since ordinary currents produce significant magnetic fields and these fields
exert significant forces on ordinary currents. But you might not expect that the force
between wires is used to define the ampere. It might also surprise you to learn that
this force has something to do with why large circuit breakers burn up when they
attempt to interrupt large currents.
The force between two long straight and parallel conductors separated by a
distance r can be found by applying what we have developed in preceding sections.
Figure 1 shows the wires, their currents, the fields they create, and the subsequent
forces they exert on one another. Let us consider the field produced by wire 1 and
the force it exerts on wire 2 (call the force F2). The field due to I1 at a distance r is
given to be
B1=μ0I1/2πr
Figure 1. (a) The magnetic field produced by a long straight conductor is
perpendicular to a parallel conductor, as indicated by RHR-2. (b) A view from above
of the two wires shown in (a), with one magnetic field line shown for each wire. RHR1 shows that the force between the parallel conductors is attractive when the currents
422
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are in the same direction. A similar analysis shows that the force is repulsive between
currents in opposite directions.
This field is uniform along wire 2 and perpendicular to it, and so the force F2 it exerts
on wire 2 is given by 𝐹 = 𝐼𝑙𝐵𝑠𝑖𝑛𝜃 with 𝑠𝑖𝑛𝜃 = 1
𝐹2 = 𝐼2 𝑙𝐵1
By Newton’s third law, the forces on the wires are equal in magnitude, and so we just
write F for the magnitude of F2. (Note that F1=−F2.) Since the wires are very long, it
is convenient to think in terms of 𝐹𝐼𝑙, the force per unit length. Substituting the
expression for B1 into the last equation and rearranging terms gives
𝐹 𝜇0 I1I2
=
𝑙
2πr
F/l is the force per unit length between two parallel currents I1 and I2 separated by a
distance r. The force is attractive if the currents are in the same direction and
repulsive if they are in opposite directions. This force is responsible for the pinch
effect in electric arcs and plasmas. The force exists whether the currents are in wires
or not. In an electric arc, where currents are moving parallel to one another, there is
an attraction that squeezes currents into a smaller tube. In large circuit breakers, like
those used in neighborhood power distribution systems, the pinch effect can
concentrate an arc between plates of a switch trying to break a large current, burn
holes, and even ignite the equipment. Another example of the pinch effect is found
in the solar plasma, where jets of ionized material, such as solar flares, are shaped
by magnetic forces.
The operational definition of the ampere is based on the force between currentcarrying wires. Note that for parallel wires separated by 1 meter with each carrying 1
ampere, the force per meter is
𝐹 (4π × 10 − 7 T ⋅ m/A)(1 A)2
=
= 2 × 10−7 N/m
𝑙
(2π)(1 m)
Since μ0 is exactly 4π × 10-7 T ⋅ m/A by definition, and because 1 T = 1 N/(A ⋅ m), the
force per meter is exactly 2 × 10-7 N/m. This is the basis of the operational definition
of the ampere.
Learning Competency:
Calculate the force per unit length on a current carrying wire due to the magnetic
field produced by other current-carrying wire. (STEM_GP12EM-IIIi-63)
423
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ACTIVITY 1. CHECK YOUR UNDERSTANDING
1. (a) The hot and neutral wires supplying DC power to a light-rail commuter train
carry 800 A and are separated by 75.0 cm. What is the magnitude and
direction of the force between 50.0 m of these wires?
(b)
Discuss
the
practical
consequences
of
this
force.
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
________
ACTIVITY 2. PROBLEMS AND EXERCISES
Directions: Compute the following problems involving magnetic force
between two parallel conductors.
1.A 2.50-m segment of wire supplying current to the motor of a submerged submarine
carries 1000 A and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away.
What is the direction and magnitude of the current in the other wire?
2.An AC appliance cord has its hot and neutral wires separated by 3.00 mm and
carries a 5.00-A current. (a) What is the average force per meter between the wires
in the cord? (b) What is the maximum force per meter between the wires? (c) Are the
forces attractive or repulsive? (d) Do appliance cords need any special design
features to compensate for these forces?
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3.Find the direction and magnitude of the force that each wire experiences in
Figure 5(a) by, using vector addition.
ACTIVITY 3. MAGNETIC FORCES BETWEEN TWO PARALLEL CURRENTS
Directions: Calculate the force of attraction or repulsion between two currentcarrying wires.
1. Two wires, both carrying current out of the page, have a current of magnitude
5.0 mA. The first wire is located at (0.0 cm, 3.0 cm) while the other wire is
located at (4.0 cm, 0.0 cm) as shown in Figure below. What is the magnetic
force per unit length of the first wire on the second and the second wire on the
first?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
________
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2. Two wires, both carrying current out of the page, have a current of magnitude
2.0 mA and 3.0 mA, respectively. The first wire is located at (0.0 cm, 5.0 cm)
while the other wire is located at (12.0 cm, 0.0 cm). What is the magnitude of
the magnetic force per unit length of the first wire on the second and the
second wire on the first?
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
__________
REFLECTION:
1. I learned that
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________
2. I enjoyed most on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________
3. I want to learn more on
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
_________________________________________________.
426
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REFERENCES
LUMEN PHYSICS: Magnetism
https://courses.lumenlearning.com/physics/chapter/22-10-magnetic-forcebetween-two-parallel-conductors/
Physics by Cutnell and Johnson 8th Edition, Chapter 21, Electricity and
Magnetism Circuits, 721-729
427
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Answer Key
Activity 1
1. (a) 8.53 N, repulsive (b) This force is repulsive and therefore there is never a risk that
the two wires will touch and short circuit.
Activity 2
1. 400 A in the opposite direction
2. (a) 1.67 × 10−3 N/m (b) 3.33 × 10−3 N/m (c) Repulsive (d) No, these are very small
forces
3. (a) Top wire: 2.65 × 10−4 N/m s, 10.9º to left of up (b) Lower left
wire: 3.61 × 10−4 N/m, 13.9º down from right (c) Lower right
wire: 3.46 × 10−4 N/m, 30.0º down from left
Activity 3
6.
𝐹
𝑙
= (−8𝑥10−11 𝑖 + 6𝑥10−11 𝑗) 𝑁/𝑚
7. Both have a force per unit length of 9.23×10-12 N/m
Prepared by:
ARNOLD C. TEODORO
ANDARAYAN NATIONAL HIGH SCHOOL
428
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GENERAL PHYSICS 2
Name: ________________________________ Grade Level: _________________
Date: _________________________________ Score: ______________________
LEARNING ACTIVITY SHEET
MAGNETIC FIELD FROM A CURRENT LOOP
Background Information for the Learners (BIL)
The
mathematics
of
Ampere’s
law
is
manageable when one can find a path along which the
magnetic field is constant. Unfortunately, that is not
possible in many cases such as the current loop
sketched Figure 1, which shows the pattern of field
lines. There is no closed path on which 𝐵∥ (the
component of the magnetic field along the path) is
constant, so there is no way to apply Ampere’s law to
Figure 1: Field from a current
lifted from 2018, General
⃗⃗ canloop
find 𝐵∥ in this situation. Other methods for calculating ⃗𝑩
be applied, and it is
Physics 2, Rex Bookstore Inc.
found that the field at the center of a circular current loop of radius R is
𝑩=
𝝁𝟎 𝑰
𝟐𝑹
The field at the center of the loop is perpendicular to the plane of the loop as given
by right-hand rule number 1.
Field inside a Solenoid
The pattern of a magnetic field lines produced by a circular current loop is very
similar to the field produced by a bar magnet. One handy feature of the current loop
is that its field is adjustable simply by varying the current I. One disadvantage is that
for reasonable values of I the field of a single current loop is much smaller than that
of a bar magnet, but this disadvantage can be overcome by stacking many current
loops together. Figure 2.A, shows such a stack of current loops and the pattern of
field lines they produce. This field can be understood in terms of the field of a single
current loop along with the principle of superposition.
429
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Figure 2: Field inside a solenoid lifted from 2018, General
Physics 2, Rex Bookstore Inc.
Usually, it is not practical to stack many separate current loops. Instead, a
long piece of wire is wrapped in a very high helix as shown in Figure 2.B. Such a
helical winding of wire is called a solenoid. The field lines circulate around the
outside of the solenoid, emerging from one end and re-entering at the other. Outside
⃗⃗ is very much
the solenoid the field lines “expand” to fill very large volume, so ⃗𝑩
smaller outside than inside the solenoid. For a very long solenoid, it is a good
approximation to assume the field is constant inside and zero outside. Applying the
Ampere’s law, we can solve the field inside a solenoid using this equation:
𝐵𝑠𝑜𝑙𝑒𝑛𝑜𝑖𝑑 =
𝜇0 𝑁𝐼
𝐿
This result for 𝐵𝑠𝑜𝑙𝑒𝑛𝑜𝑖𝑑 is a good approximation for solenoid whose length L
is much greater than its diameter.
Learning Competency: Evaluate the magnetic field vector at any point along the
axis of a circular current loop. (STEM_GP12EM-IIIi-64)
430
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Activity #1: Choose the Best!
Directions: Encircle the letter of the correct answer.
1. A circular loop of wire is carrying a constant current 𝐼 in a clockwise direction as
viewed from above. The current creates a magnetic field. Based on the diagram,
state the direction of the magnetic field at the center of the coil.
A. Upward
B. Downward
C. Out of the screen
D. Into the screen
E. Side of the screen
2. A bar magnet is divided in two pieces. Which of the following statements is true?
A. The bar magnet is demagnetized.
B. The magnetic field of each separated piece becomes stronger.
C. The magnetic poles are separated.
D. Two new bar magnets are created.
E. The electric field is created
3. A current-carrying wire is placed perpendicular to the page. Determine the
direction of the electric current from the direction of the magnetic field.
A. Into the page.
B. Out of the page.
C. Clockwise.
D. Counter-clockwise.
E. To the left.
4. A circular loop of wire carries a constant current of 0.9 A. The radius of the loop
is 13 mm. Calculate the strength of the magnetic field at the center of the loop. Give
your answer in teslas expressed in scientific notation to 1 decimal place. Use a value
of 4𝜋×10-7 T⋅m/A for 𝜇0.
A.
B.
C.
D.
E.
3.3 x 10-3 T
1.4 x 10-5 T
8.7 x 10-5 T
3.5 x 10 T
4.3 x 10-5 T
431
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5. A circular loop of wire of radius 50 mm carries a constant current 𝐼 A and produces
a magnetic field of strength 𝐵1 T at its center. Another circular loop of wire has a
radius of 150 mm. Given that this wire also carries a constant current of 𝐼 A, which of
the following correctly shows the relation between 𝐵2, the strength of the magnetic
field produced by the larger loop at its center, and 𝐵1?
A.
B.
C.
D.
E.
B2 = B 1
B2 = 3B1
B2 = 1/3B1
B2 = 9B1
B2 = 1/9B1
432
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Activity #2: SOLVE ME!
Directions: Solve the following problems with complete solutions. (5 points each)
1. A solenoid with length 30 cm, radius 1.2 cm, and 300 turns carries a current of
2.6 A. Find B on the axis of the solenoid (a) at the center, (b) inside the solenoid
at a point 10 cm from one end, and (c) at one end.
2. A wire loop is bent into the shape of a square with each side of length 4.5 cm.
The loop is placed horizontally on a tabletop with two of the sides oriented
north/south and two of the sides oriented east/west. A battery is connected so
that a current of 24 mA is produced around the loop; the current flows in the
clockwise direction looking from the top. What is the force produced by the
earth’s magnetic field on each section of current-carrying wire? What is the
overall torque on the loop? What would the torque be if the same length of wire
were bent into a circle instead of a square (assuming the same current)?
3. A wire of length 24 cm is bent into a square and placed flat on a table. A current
of 45 mA is passed through the wire in a counter-clockwise direction (looking
from above). What is the magnitude and direction of the resulting magnetic field
at the center of the square?
4. Two loops of wire carry the same current of 10 mA, but flow in opposite
directions. One loop is measured to have a radius of R = 50 cm while the other
has a loop of 2R = 100 cm. the distance from the loop to the point where the
magnetic field is measured is 0.25 m, and the distance from that point to the
second loop is 0.75 m. what is the magnitude of the net magnetic field at that
point?
5. A long straight wire carries a current of 4 A to the right of page. Find the
magnitude and direction of the B-field at a field at a distance of distance of 5 cm
above the wire.
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Activity #3: DESIGNING A SOLENOID
You are given the task of designing a solenoid that must produce a field in its
interior that is 20 times larger than the Earth’s field (BEarth = 5 x 10-5 T). This solenoid
is to be 10 cm long and 1.0 cm in diameter, with N = 500 turns. What current must
the wire be able to carry without failing?
Recognize the principle:
Sketch the Problem:
Identify the relationships/ Equation:
Solution:
What does it mean? :
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REFLECTION
1.I learned that ____________________________________________________
____________________________________________________________
_______________________________________________________
2.I enjoyed most on _________________________________________________
____________________________________________________________
__________________________________________________
3.I want to learn more on _____________________________________________
____________________________________________________________
__________________________________________________
435
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References:
Padua, Alicia L. et. al, 2003, States of Equilibrium, Practical and Explorational
Physics: Modular Approach, pp. 244-254.
2018. General Physics 2, Rex Book Store, Inc. pages 154-156.
Chapter 30 - - Magnetic Fields Magnetic Fields and Torque and Torque A PowerPoint
Presentation by Paul E. Tippens, 2007
https://www.askiitians.com/iit-jee-electrostatics/continuous-charge-distribution/
Fields in a loop retrieved from
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_P
hysics_(OpenStax)/Map%3A_University_Physics_II__Thermodynamics_Electricity_and_Magnetism_(OpenStax)/12%3A_Sources_of_M
agnetic_Fields/12.05%3A_Magnetic_Field_of_a_Current_Loop
436
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Answers key
Learning Activity #1: CHOOSE THE BEST!
1.
2.
3.
4.
5.
D
D
B
E
C
Learning Activity #2: SOLVE ME!
1.
2.
3.
4.
5.
a. 3.25 T, b. 3.25 T, c. 1.63 T
T1 = 2.4 x 10-9 N . m East , T2 = 3.0 x 10-9 N . m East
B = 0.85 µT
B = 5.77 x 10-9 T to the right
B = 1.60 x 10-5 T or 16 µT
Prepared by:
JOLLY MAR D. CASTANEDA
Baggao National Agricultural School Sta. Margarita Annex
437
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GENERAL PHYSICS 1
Name: __________________________ Grade Level: _____________
Date: ___________________________ Score: __________________
LEARNING ACTIVITY SHEET
ELECTRIC POTENTIAL OF CONTINUOUS CHARGE DISTRIBUTION
Background Information for the Learners (BIL)
Key Points
•
A wire carrying electric current will produce a magnetic field with closed field
lines surrounding the wire.
•
Another version of the right hand rules can be used to determine the magnetic
field direction from a current—point the thumb in the direction of the current,
and the fingers curl in the direction of the magnetic field loops created by it.
•
The Biot-Savart Law can be used to determine the magnetic field strength from
a current segment. For the simple case of an infinite straight current-carrying
𝜇0 𝐼
wire it is reduced to the form 𝐵 = 2𝜋𝑟
.
A more fundamental law than the Biot-Savart law is Ampere‘s Law, which
relates magnetic field and current in a general way. It is written in integral form
as ∮B⋅dl=μ0Ienc, where Ienc is the enclosed current and μ0 is a constant.
A current-carrying wire feels a force in the presence of an external magnetic
field. It is found to be F=Bisinθ 𝐹 = 𝐵𝐼𝑙 𝑠𝑖𝑛𝜃, where ℓ is the length of the wire, I
is the current, and θ is the angle between the current direction and the magnetic
field.
•
•
Key Terms
•
•
Biot-Savart Law: An equation that describes the magnetic field generated by
an electric current. It relates the magnetic field to the magnitude, direction,
length, and proximity of the electric current. The law is valid in the
magnetostatic approximation, and is consistent with both Ampère’s circuital law
and Gauss’s law for magnetism.
Ampere’s Law: An equation that relates magnetic fields to electric currents
that produce them. Using Ampere’s law, one can determine the magnetic field
associated with a given current or current associated with a given magnetic
field, providing there is no time changing electric field present. (𝜇0 =
4𝜋 𝑥 10−7 𝑇 . 𝑚/𝐴)
∑ 𝐵∥ ∆𝐿 = 𝜇0 𝐼𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑
438
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Learning Competency:
Solve Problems involving magnetic fields, forces due to magnetic fields and the
motion of charges and current-carrying wires in contexts such as, but not limited to,
determining the strength of Earth’s magnetic field, mass spectrometers, and
solenoids. (STEM_GP12EM-IIIi-66)
Activity #1: TRUE or FALSE
Directions: Write TRUE if the statement is correct, and if otherwise write FALSE and
underline the word/phrase that makes it incorrect.
1. The magnetic field direction is the opposite of the direction a compass
needle points, which is tangent to the magnetic field line at any given point.
2. The strength of the B-field is directly proportional to the distance between
field lines. It is exactly proportional to the number of lines per unit area
perpendicular to the lines.
3. A magnetic field line can cross another field line. The magnetic field is
unique at every point in space.
4. Magnetic field lines are continuous and unbroken, forming closed loops.
Magnetic field lines are defined to begin on the north pole of a magnet and
terminate on the South Pole.
5. The direction of the magnetic field is tangent to the field line at any point in
space. A small compass will point in the direction of the field line.
6. The strength of the field is proportional to the closeness of the lines. It is
exactly proportional to the number of lines per unit area parallel to the lines
(called the areal density).
7. Magnetic field lines can never cross, meaning that the field is unique at any
point in space.
8. Magnetic field lines are continuous, forming closed loops without beginning
or end. They go from the South Pole to the North Pole.
9. A wire carrying electric current will produce a magnetic field with closed field
lines surrounding the wire.
10. Another version of the right hand rules can be used to determine the
magnetic field direction from a current—point the thumb in the direction of
the current, and the fingers curl in the direction of the magnetic field loops
created by it.
439
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Activity #2: ESSAY
Directions: Answer the following situations comprehensively.
1. A long straight current-carrying wire is placed in a region where there is
magnetic field B that has a constant direction. If these is no magnetic force on
the wire, what can you say about the direction of B relative to the wire?
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
____________
2. A positively charged particle moves with velocity in a region where there is a
magnetic field and a nonzero magnetic force on the particle. Consider how these
three vectors are oriented with respect to one another. (a) Which pairs of these
vectors are always perpendicular? (b) Which pairs can have any angle between
them?
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
____________
3. Does a current-carrying wire placed in a magnetic field always experience a
magnetic force? Explain.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
____________
4. A square loop with constant current I is placed in a uniform magnetic field.
Explain why the total force on the loop is zero, no matter how the loop is oriented
relative to the field direction.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
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_______________________________________________________________
____________
5. A wire carrying electric current will produce a magnetic field with closed field
lines surrounding the wire. Explain.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
____________
441
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Activity #3: PROBLEM SOLVING
Directions: Solve the following accurately with complete solution.
1. A long, straight wire of length 0.75 m carries current I = 1.5 A in a region where
B = 2.3 T. If the force on the wire is 1.4 N, what is the angle between the field
and the wire?
2. A long, straight wire carries a current of 5.4 A. What is the magnitude of the
magnetic field at a distance of 10 cm from the wire?
3. Two long, straight, parallel wires of length 4.0 m carry parallel currents of 3.5
A and 1.2 A. (a) if the wires are separated by a distance of 3.5 cm, what is the
magnitude of the force between the two wires? (b) Is this force attractive or
repulsive? (c) If the currents are in opposite directions (antiparallel), how do
the answers to parts (a) and (b) change?
4. Consider two long, straight, parallel wires each carrying a current I =2.0 A,
with the currents in the same direction. (a) Find the magnetic field at one wire
produced by the other wire if the wires are separated by a distance r = 1.0
mm. Assume for simplicity that the wire diameters are much less than their
separation. (b) What is the magnitude of the magnetic force exerted by one of
these wires on a 1.0-m-long section of the other wire?
442
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REFLECTION
1.I learned that ____________________________________________________
____________________________________________________________
_______________________________________________________
2.I enjoyed most on _________________________________________________
____________________________________________________________
__________________________________________________
3.I want to learn more on _____________________________________________
____________________________________________________________
__________________________________________________
443
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References:
Padua, Alicia L. et. al, 2003, States of Equilibrium, Practical and Explorational
Physics: Modular Approach, pp. 244-254.
2018. General Physics 2, Rex Book Store, Inc. pages 2-40.
https://www.askiitians.com/iit-jee-electrostatics/continuous-charge-distribution/
444
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Answers key
Learning Activity #1: TRUE OR FALSE
1. FALSE, opposite
2. FALSE, directly
3. FALSE, can cross
4. TRUE
5. TRUE
6. FALSE, parallel
7. TRUE
8. FALSE, South Pole to the North Pole
9. TRUE
10. TRUE
Learning Activity #2: ESSAY
ANSWERS MAY VARY
Learning Activity #3: Problem Solving
1. 32.760
4. (a) 4.0 x 10-4 T , (b) 8.0 x 10-4 N
Prepared by:
JOLLY MAR D. CASTANEDA
Baggao National Agricultural School Sta. Margarita Annex
445
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