Republic of the Philippines
Department of Education
Regional Office IX, Zamboanga Peninsula
12
GENERAL PHYSICS 2
2 n d Semester - Module 3
CAPACITANCE AND DIELECTRICS
Name of Learner:
Grade & Section:
Name of School:
General Physics 2 – Grade 12 (STEM)
Support Material for Independent Learning Engagement (SMILE)
Module 3: Capacitance and Dielectrics
First Edition, 2021
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Development Team of the Module
Writer:
Jeovanny A. Marticion
Editor:
Zyhrine P. Mayormita
Reviewers:
Leo Martinno O. Alejo, Zyhrine P. Mayormita
Layout Artist:
Chris Raymund M. Bermudo
Management Team: Virgilio P. Batan Jr.
- Schools Division Superintendent
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Nur N. Hussien
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Ronillo S. Yarag
- Education Program Supervisor, LRMS
Zyhrine P. Mayormita
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Leo Martinno O. Alejo
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What I Need to Know
This module will help you learn about capacitors and dielectrics. At the end
of this module, you should be able to:
1.
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_GP12EMIIId-23)
2. Calculate the equivalent capacitance of a network of capacitors
connected in series/parallel (STEM_GP12EM-IIId-24)
3.
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)
4.
Determine the potential energy stored inside the capacitor given the
geometry and the potential difference across the capacitor
(STEM_GP12EM-IIId-26)
5.
Describe the effects of inserting dielectric materials on the
capacitance, charge, and electric field of a capacitor (STEM_GP12EMIIId-29)
6. Solve problems involving capacitors and dielectrics in contexts such
as, but not limited to, charged plates, batteries, and camera flash
lamps. (STEM_GP12EM-IIId-30)
What’s In
Devices used for communication, photography, and high-energy
accelerators rely on one of the most important electric devices in modern times,
capacitors. It is a device composed of conductors separated by an insulator or
a vacuum. A potential difference is produced between the conductors (equal
magnitude with opposite signs). Aside from the potential difference, an electric
field is also formed between the conductors.
The energy stored by capacitors is similar to the mechanical energy
stored in any physical body. The energy stored is delivered to a circuit when it
discharges. The presence of an insulator produces polarization which helps in
charge distribution within the material. It also increases the capacitance of a
device.
1
What’s New
Activity 1. How Defibrillator Works.
Direction: Below is a comic strip that shows analogous examples of how a
defibrillator works. Take a look at these visual strips and answer the questions
after analyzing the strips.
Image Source: https://www.howequipmentworks.com/defibrillator/
2
(1) In your own words, what is the main function of a defibrillator?
(2) What scenario or situation was the defibrillator compared to? Give
evidence.
(3) What do we need if we want our defibrillator to work effectively?
(4) What similar experience do you have which could also be compared to a
defibrillator?
What Is It
Capacitance
Capacitors are devices formed from two conductors separated by an
insulator. Examples of insulators are plastic, liquid gel, paper, mica, ceramic, or
even air. These materials are called dielectrics. Both conductors have an equal
magnitude of charges with opposite signs. Thus, the net charge of the device is
zero. Their ability to store energy makes it useful when it produces potential
differences across the plates. They act like rechargeable batteries. The insulating
component of capacitors blocks the flow of direct current.
The presence of the electric field found between the plates is directly
proportional to the charge Q present in the conductors. Therefore, the potential
difference, V ab, is also directly proportional to charge Q. The more charges
present, the intensity of the electric field between plates increases, and potential
difference increases. However, the ratio between charge and potential
3
difference remains the same. This ratio represents the constant for any
capacitor known as the capacitance:
where
C is the capacitance expressed in
or Farad (F)
Q is the charge expressed in Coulombs (C); and
V is the potential difference between conductors expressed in
voltage (V)
Capacitors are represented by the symbol:
These devices are mainly used in communication
such as radio
transmitters and receivers. These are also used in regulating the outputs of
flash units in our cameras.
Figure 1. Practical uses of capacitors
Image source: https://tinkersphere.com/wireless-modules-for-raspberry-pi-and-arduino/485-rf-link-transmitter-receivermodule-pair-433mhz-or-315mhz-for-arduino-raspberry-pi-b-b.html
The most commonly studied capacitors is
composed of two parallel conducting plates. These
are separated by a distance which is very small in
contrast with their true dimensions. The region in
between them has a uniform field and charges are
uniformly distributed. This is known as parallel
plate capacitors.
The electric field in this case can be
expressed as:
where
Q is the charge and
𝟄0 is the electric constant as
presented in Module 1 which is
equal to
or also
known as permittivity of space
4
d
+Q
-Q
Va
Vb
The potential difference (recall this lesson from Module 2: Electric
Potential) between plates a and b is expressed as:
Manipulating formula
divide both sides by V ab
divide both sides by d
The formula for the capacitance
of parallel plate capacitors
where A is area in terms of m 2 , d
is the distance in terms of meter
(m), 𝟄0 is permittivity of space in
terms of
Example 1:
The plates of a parallel plate capacitor are separated at 1.0 mm apart.
What is the capacitance in air if the plate is 10 cm 2 ? What is the charge in
each plate if a potential difference of 100 V is applied?
A What is/are given?
A = 10 cm 2 ; d = 1.0 mm; V = 100 V
B What is asked?
(a) C = ? (b) Q = ?
C Are the units
No, distance should be converted from mm to m.
consistent with the
Thus, 1.0 mm = 1.0 x 10 -3 m and
formula?
( )
E
What strategy must
be employed?
F
Solution
( )
We use the capacitance formula for parallel
plate
capacitors.
Then,
we
use
the
capacitance formula to derive charge Q.
(
)
( ) ( )( )
5
G
What is the
conclusion?
Therefore, the capacitance of the parallel plate capacitor
is
and the charge of the plate with 100 V is
Capacitors in Series and Parallel Connection
Suppose the capacitors were arranged as shown in Figure 2 below.
Two capacitors were connected between points a and b where a constant
potential difference is maintained. In this connection, both capacitors
must have the same magnitude of charge Q. The net charge between the
right plate of C 1 and the left plate of C2 must be zero. The same net charge
must be observed with the remaining plates. Thus, the capacitors are in
series connection where the magnitude of charge for each plate is the
same.
C1
c
C2
b
a
Figure 2. Capacitors in Series Connection
The potential difference for each
capacitor
Combining the potential difference
( )
Divide both sides by Q
algebraically,
Hence, the equivalent capacitance of
the capacitors in series connection.
The reciprocal of the equivalent
capacitance of a series combination
is equal to the sum of the reciprocals
of individual capacitors.
We may also extend this to any
number of capacitors arranged in
series connection.
6
Suppose the capacitors were arranged as shown in Figure 3 below.
Two capacitors were connected in parallel with points a and b where a
constant potential difference is maintained. The right plates are connected
together forming equipotential surfaces and the remaining plates formed
another. In a parallel arrangement, the potential difference for each
capacitor is the same.
a
C2
C1
Vab
b
and
( )
( )
The magnitude of charge for each capacitor
Thus, the total charge is the sum of individual
charges of the capacitor
The common factor is V since the value of
potential difference are the same for both
capacitors
Divide both sides by V
Take note:
The equ ivalent capacitance of a parallel
combination
is the
su m
of individu al
capacitance.
We may also extend this to any number of
capacitors arranged in parallel connections.
Example 2:
Suppose you have two capacitors 6.0 µF and 3.0 µF connected in series
and parallel with points a and b whose potential difference is 18 V, what is
the equivalent capacitance, charge, and potential difference for each
capacitor.
A What is/are given?
C 1 = 6.0 µF; C 2 = 3.0 µF; V ab = 18 V
B What is asked?
Ceq=?; Q and V for each capacitor
C Are the units
No, C 1 and C 2 must be converted to F
consistent with the
Thus, C 1 = 6.0 x 10 -6 F and C 2 = 3 x 10 -6 F
formula?
E What strategy must
We use the formula for series and parallel
7
be employed?
F
combination of capacitors
Solution
For series combination of C 1 and C 2 , we solve first for the equivalent
capacitance.
(
()
)
Then, we solve for the charge for given the potential difference and
computed equivalent capacitance. The computed value of charge will be
the charge for each capacitor C 1 and C 2 . Recall that charges for each
capacitor in a series combination are the same.
(
)(
)
Solve for the potential difference in each capacitor C 1 and C 2 using the
given magnitude of capacitance and the charge:
For parallel combination of C 1 and C 2 , we solve first for the equivalent
capacitance.
The potential difference for C 1 and C 2 is equal to 18 V. Since parallel
capacitors have the same magnitude of potential difference.
G
Solve for the charge for each capacitor C 1 and C 2.
(
)(
)
(
)(
)
What is the
Therefore, in a series connection C 1 and C2 has an
conclusion?
equivalent capacitance of , a potential difference of 6 V
and 1 2 V, respectively and a unfirom charge of
.
In a parallel connection, C 1 and C2 has an equivalent
capacitance of , a potential difference of 18 V and
magnitude of charge , respectively.
8
The Energy of a Charged Capacitor
A capacitor can be similar to a spring. Suppose a charge was removed
from the positive plate and released towards the negative plate. Then, the
charge lost potential energy equivalent to V. This is also equal to the work done
(W = ΔQV) on a charge due to electric field present between the plates (recall the
concepts of electric potential energy from the previous Module 2: Electric
Potential).
However, the sum of the potential energy is not just QV since potential is
directly proportional to the remaining charges in the plate. Hence, transferring
these charges to the other plate would decrease the potential. Discharging the
plate by transferring all charges to the other plate, the decrease of potential
energy is the average potential of the process. This is expressed as:
Since , then
(
)
and , then
( )
This expression is similar to the potential energy stored in a stretched spring
where k is the spring constant and x is the displacement. The
constant k is analogous to 1 / C in the potential energy of the capacitor. The
work done in stretching the spring is similar to the work done in supplying
energy to the capacitor.
This shows a direct relationship between stored energy within the
capacitor and the electric field between its plates. We could understand this by
finding the energy density or the energy per volume.
Since , then
(
Recall that , then
9
)
where is the permittivity of space, C is the capacitance and E is the electric
field.
Example 3:
Fibrillation is a condition where the heart loses its ability to pump blood
effectively leading to heart attacks. Electric shock can be used to correct this
behavior. Appropriate shocks can be administered using
a 10µF capacitor
which was charged with a potential difference of 6000V.
(a) How much energy is released in the current pulse?
(b) How much charge passes through the patient’s body?
What is/are given?
What is asked?
Are the units
consistent with the
formula?
How will you
visualize the
problem?
C 1 = 10 µF; V ab = 6000 V
U = ?; Q = ?
No, C 1 must be converted to F
Thus, C 1 = 10.0 x 10 -6 F
E
What strategy must
be employed?
We use the formula for potential energy and
charge.
F
Solution
(a) ( )( ) ( )
A
B
C
D
G
(B) ( )( )
What
is
conclusion
the
Therefore, the energy stored with 6000 V on a
capacitor is 180 J while the charge it produced is
0.06 C.
Dielectric Constant
Suppose a slab of an insulator was inserted between plates of a
capacitor. Although this could not conduct electric current, this could also
respond to the electric field. The electrons in the molecules of most substances
are not uniformly distributed. These molecules are called polar molecules. These
molecules have one end as positively charged and the other end as negatively
charged. Without an electric field, these polar molecules are randomly
arranged. If an electric field is present, the molecules align opposite to the field.
In the case of nonpolar molecules, an electric field can disturb their
arrangement since they have symmetric charge distributions. The figure below
shows the electrons of polar and nonpolar molecules when an electric field is
present or not.
10
Figure 2. Polar and nonpolar molecules with and without electric field
Image Source:
https://www.google.com/search?q=polar+molecules+nonpolar+molecules+dielectric+constant&sxsrf=ALeKk03rOlnKsOSX1RgxVEbKZcaQ
MXTwKg:1614824231919&source=lnms&tbm=isch&sa=X&ved=2ahUKEwj2iN-FyZXvAhWaMd4KHROB6kQ_AUoAXoECBAQAw&biw=1280&bih=658#imgrc=EUPaaboX6AqYNM
We can also demonstrate this phenomenon using a sensitive electrometer.
The figure below shows a dielectric inserted between plates. Readings of its
potential difference are shown by the electrometers. When we remove the
dielectric, the potential difference returns to its original value. This shows that
the charges in the plates did not change. The measure of how effective the
substance in reducing the electric field is called dielectric constant. This is
represented by symbol K. For the capacitor, it should reduce the electric field
and its potential difference. This implies increasing the capacitance since
potential difference and capacitance is inversely proportional
( ).
Figure 3. The electrometer reads the potential difference before and after the dielectric
is inserted between the plates.
Image source: https://www.google.com/search?q=effect+of+dielectric+using+electrometer&tbm=isch&ved=2ahUKEwi7 -YLXy5XvAhUODpQKHcOQDEAQ2cCegQIABAA&bih=658&biw=1280#imgrc=fDTK5Fei4vcChM
The dielectric can also be expressed as the ratio of the final capacitance C
to its initial capacitance C0.
11
It could also be the ratio of initial potential V0 difference to its final potential
difference V.
On the other hand,
This implies that E becomes smaller when the dielectric is present. The
decreasing electric field is due to the smaller surface charge density. The
induced charge appears on each surface of the dielectric. The induced surface
charges are due to the distribution of charges within the dielectric, a property
called polarization. Table 1 below shows the dielectric constant K at 20°C.
Table 1. Dielectric constant K at 20°
Material
Vacuum
Glass
Mica
Mylar
Neoprene
K
1
5-10
3-6
6.70
3.40
Material
Teflon
Germanium
Strontium titanate
Water
Glycerin
K
2.1
16
310
80.4
42.5
Material
Benzene
Air (1 atm)
Air (100 atm)
Plexiglass
Polyethylene
K
2.28
1.00059
1.0548
2.25
3.18
Source: sears, F., Zemansky, M. and Young, H. College Physics 7 t h Edition
In deriving the expression for the induced charge, capacitance and
potential energy stored in capacitors with dielectric:
Solving for the induced charge
Rearranging the formula
From the previous section,
Recall that
Algebraic solution
Get the common factor
( )
Thus, the expression for the induced
charge on a capacitor with dielectric
12
Solving for capacitance
Permittivity of dielectric
In terms of the permittivity of dielectric,
we can express the electric field.
The capacitance with dielectric.
Since
Recall that
Thus, the expression for capacitance
in terms of permittivity of dielectric,
area, and distance of separation.
Solving for potential energy
In terms of potential energy stored in
capacitors, we recall the formula in
terms of permittivity of space and
electric field.
We can express the energy stored with
the dielectric constant.
Recall that
Thus, the expression of potential
energy stored in capacitor in terms of
electric field and permittivity.
Example 5:
The parallel plates have an area of 2000 cm 2 and are separated at 1.00 cm
apart. The original potential difference between them is 3000 V and it decreased
to 1000 V when a sheet of dielectric was inserted. What is the
(a) Original capacitance
(b) The magnitude of the charge on each plate
(c) Capacitance C after dielectric is inserted
(d) Dielectric constant K
(e) Permittivity of dielectric
(f) The magnitude of induced charge
(g) Original electric field
(h)Electric field after dielectric is inserted
A What is/are given?
A = 2000 cm 2 ; d = 1.0 cm;
V0 = 3000 V; V = 1000 V
B What is asked?
(a)C0 = ? (b) Q =? (c) C = ? (d) K = ?
(e) (f) induced charge (g) E0 =? (h) E=
C Are the units
No, area should be converted from cm 2 to m 2
consistent with the
(
)
formula?
13
E
F
What strategy must
be employed?
Solution
Distance should be converted from cm to m. Thus,
1.0 cm = 1.0 x 10 -2 m
We use the previous formaul presented under
dielectric constant section
(a) ( ) ( )
(b) ( )( )
(c)
(d)
(e) ( )
(f) ( ) ( ) ( )
(g)
(h)
What’s More
Activity 2: Qualitative Problems. Direction:
Answer the following questions.
(1) A student claims that capacitors in parallel combinations have greater
equivalent capacitance compared to the magnitude of either the capacitor and
less than either the capacitors when connected in series combination. How will
you agree or disagree with the claim?
(2) A capacitor was made of aluminum foil strips which were separated by Mylar
films. The capacitor was subjected to too much voltage and produced holes in
Mylar films. The capacitance was still found to have the same magnitude but
the voltage was lesser. Why do you think this happened?
(3) Is there any kind of material that when inserted between parallel plate
capacitors would reduce its capacitance?
14
What I Have Learned
Activity 3: Quantitative Problem.
Direction: Write your answers on a separate sheet of paper. You may also
consult your Physics teacher.
(1) Compute the equivalent capacitance of the network.
(2)A 300 µF capacitor separated by a distance of 4.0 mm was charged to a
potential difference of 250 V. What is the energy density in its region in terms of
Joules/m 3 ?
(3)Two parallel plates have equal and opposite charges. When space was
evacuated, the electric field was 2.80 x 105 V/m. When it was filled with a
dielectric material, the electric field was 1.20 x 105 V/m. What is the dielectric
constant? What is the material?
Criteria
Physics
Approach
3
Approach is
appropriate and
complete
2
Approach contains
minor errors
Procedure
Mathematical and
logical procedures
are clear,
complete and
connected
Diagrams and
symbols used are
appropriate and
complete
Description
0
Solution doesn’t
indicate an
approach
Mathematical and
logical procedures
are missing/contain
errors
1
Some of the
concepts and
principles are
missing or
inappropriate
Most of the
mathematical
and logical
procedures
Parts of the
diagrams and
symbols contain
errors
Most of the parts
of the diagrams
and symbols are
not useful
The entire
visualization is
wrong or did
not include
visualization.
15
All procedures
are incomplete
and contain
errors
What I Can Do
Activity 4. Building Concept Map.
Direction: Create a concept map from what you have learned in this module.
You can use words, terms, phrases, or formulas in connecting these
concepts. Refer to the scoring guide below:
Legible (easy to read)
Accurate (concepts
were used
accurately)
Complete (sufficient
number of relevant
concepts and
relationships)
Sophisticated
(finding meaningful
connections between
relevant concepts)
No (0-1)
Many inaccuracies
(0-2)
Yes (2)
A few inaccuracies
(3-4)
No inaccuracies (5)
Limited use of
concepts
(0-2)
Some use of
concepts
(3-4)
Sufficient number
of concepts
(5)
Little or
none
(0-1)
Few
meaningful
connections
made (2-4)
Some
meaningful
connections
made (5-7)
Meaningful and
original
insights
demonstrated
(8)
Mueller’s Classroom Concept Rubric
Assessment
Direction: Read each item and write the letter of your choice in the space
provided.
1.If the value you computed has an SI unit of , the quantity could be
a.potential difference
b.dielectric constant
c.electric field
d.capacitance
2.A charge was removed from one of the plates, the capacitance of the
capacitors
a.decreases
b.stays the same
c.increases
d.it is halved
3.The energy of a charged capacitors could be found in
a.plates
b.potential difference
c.charges
d.electric field
4.The Farad is not equivalent to which of the following units:
a.
b.
16
c.
d.
5.When a slab of insulating material is placed between the plates of a
charged capacitor, the electric field becomes
a.less
b.similar
c.greater
d.depends on the situation
6.A capacitor has a charge of 0..02 C when connected with a 0 V battery.
The capacitance is
a.1 µF
b.4 µF
c.2 µF
d.40 µF
7. A 50 µF has a potential difference of 8 V. What is its charge?
a.4 x 10 -3 C
b.6.25 x 10 - 5 C
c.4 x 10 -4 C
d.6.25 x 10 -6 C
8.A parallel plate capacitor has an energy of 2.5 J. It must be placed in a
potential difference of
a.150 V
b.350 V
c.500 V
d.0.25 MV
9.Two 50 µF capacitor are connected in series. The equivalent
capacitance of the combination is
a.25 µF
b.100 µF
c.50 µF
d.200 µF
10. Two 50 µF capacitor are connected in parallel. The equivalent
capacitance of the combination is
a.25 µF
b.100 µF
c.50 µF
d.200 µF
11.Which of the following portrays the electric field of a parallel plate
capacitor?
17
12. The capacitance of a parallel-plate capacitor depends on
I.plate area
II. distance of separation
III.voltage
a.II only
b.III only
c.I and II only
d.II and III only
13. A parallel-plate capacitor has a capacitance C. What is the new
capacitance if a dielectric with K = 4 is inserted between its plates?
a.
b.
c.C
d.4 C
14.What is the new capacitance if the plate separation is doubled?
18
a.
b.
c.C
d.2 C
15.What can dielectric do for the capacitors?
a.decrease the value of C
b.decrease the intensity of electric field
c.increase the value of C
d.increase the intensity of electric field
Additional Activities
Activity 5. Social Context.
Direction: The community is a rich repository of learning opportunities for
sources of magnetic forces and fields. Choose one from the following suggested
activities in understanding the importance and utilization of electric potential in
our daily lives:
1.
2.
Conduct simulations on capacitors and dielectric.
From this, write a short reflection. Scan the QR code to
gain access to the simulations.
Making your capacitor. You will need the following
materials: aluminum foil, paper, scissors, 2 AA
batteries, 2 office fasteners, 2 paper clips, tape, and a
multimeter
Step 1 . Cut out the strip of wax paper about 3 inches wide. Cut a piece of
aluminum foil about inches wide.
Step 2. Cut two squares from the aluminum foil strip. Cut the paper so it is
about 0.5-inch wider than the aluminum foil at the top and bottom. Cut
the paper so it is 4 times that of the width of the aluminum foil.
Step 3. Bend the other end of the paper clip until it becomes straight
leaving the other part in its original form.
Step 4. Place the aluminum foil near the end of the paper. Use a little piece
of tape to secure the paper clip. Tape the curved end of the paper clip in the
middle of the aluminum foil and make sure the straight portion of the clip
is placed outside the wax paper.
Step 5. Fold the wax paper so the aluminum foils are covered on each side
with paper.
19
Step 6. Place the second foil directly over the same location as the other foil
square. Use tape on top to secure it in place. Do step 4 but this time the
paper clip must be on the other side opposite the first paper clip.
Step 7. Fold the remaining length of paper over the top of the foil square so
it is not exposed. You can have the option of rolling the sandwiched foil and
paper.
Step 8. Attach your home-made capacitor to your battery.
Step 9. After some time, measure the amount of voltage present by dialing
the multimeter to microvolts. Make sure the probes are properly inserted in
the negative and positive ports.
Step 10. Check if the multimeter has a reading of voltage stored.
Step 1 1 . Prepare a reflection paper on the experience you had while
making the capacitor. You may do some revisions or changes and share
your observations after applying the revisions or changes.
.
Capture the QR code and visit this site if you are on the right track.
20
Answer Key General Physics 2 Module 3
References
Printed Resources
Sears, F., Zemansky, M. and Young, H. (1992). College Physics 7 t h Edition. Addison-Wesley
Publishing Company
Zitzewits, Haase and Harper (2013). PHYSICS Principles and Problems. The MAcGraw-Hill
Companies, Inc.
Online References
Digital Vector Studios (2021). How to Make a Capacitor. Retrieved last March 2, 2021, from
https://sciencewithkids.com/Experiments/Energy-Electricity-Experiments/how-tomake-a-capacitor.html
How Equipment Works (n.d.). How defibrillators work explained simply. Retrieved last March
1 , 2021, from https://www.howequipmentworks.com/defibrillator/
Physics Prep. (2020). Virtual Activity: Capacitor Lab. Retrieved last March 3, 2021, from
https://www.physics-prep.com/index.php/virtual-activity-capacitor-lab-2
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