What I Need to Know
This module will help you in understanding the basic concepts of electric charges
and fields. The topics covered by this module are electric charges. At the end of this
module, you should be able to:
1. Describe using a diagram charging by rubbing and charging by induction
STEM_GP12EM-IIIa-1;
2. Explain the role of electron transfer in electrostatic charging by rubbing
STEM_GP12EMIIIa-2;
3. Describe experiments to show electrostatic charging by induction
STEM_GP12EM-IIIa-3;
4. Calculate the net electric force on a point charge exerted by a system of point
charges STEM_GP12EM-IIIa-6;
5. Describe an electric field as a region in which an electric charge experiences a
force STEM_GP12EM-IIIa-7; and
6. Calculate the electric field due to a system of point charges using Coulomb’s law
and the superposition principle STEM_GP12EM-IIIa-10.
What’s In
In General Physics 1, your journey explored the various fundamental forces found
in nature. Gravity, one of the forces you studied, was examined in a detailed manner
and how it influences the movement of physical bodies.
This time, we will be exploring the electromagnetic force, one of nature’s
fundamental forces, which possesses both electric and magnetic force. However, we
need to know how this interaction involves particles with electric charge in
understanding this force. This could also be fundamentally represented by mass. When
an object with mass is accelerated by an applied force, objects with electric charges are
also accelerated by the presence of electric forces.
This behavior can be observed when we see lightning strikes the sky, feel the shock
from a metallic surface after scrubbing our shoes across a carpet or when lighter objects
stick with other objects such as dust clinging on a plastic paper.
Activity 1: GETTING RECHARGED!
Direction: This is to check what you have learned about electric charges and electric
fields. Circle the letter of the best answer.
1. Two unlike charges_________
A. attract each other
C. neutralize each other
B. repel each other
D. have no effect on each other
2. Which of the following is not a process of charging?
A. Induction
B. Friction
C. Conduction
D. Convection
3. Material A is positively charged. When brought near to material B, they attract.
Which of the following is true?
A. Material B is negatively charged
C. Material B is uncharged
B. Material B is positively charged
D. Both are uncharged
1
4. Material B has become positively charged after rubbing it with Material A. Which
of the
following statements is correct?
A. Material B loses protons
C. Material A loses proton
B. Material B gains electrons
D. Material A gains electron
5. What will happen when two unlike charges are brought together? They will _____
A. repel each other
C. attract each other
B. neutralize each other
D. no effect on each other
6. If you comb your hair and the comb becomes positively charged, then your hair
becomes _________.
A. positively charged
C. uncharged
B. negatively charged
D. discharged
For No. 7 & No. 8 study the given Triboelectric series, where moving up means
positive and moving down means negative.
7. Which of the following pairs has the strongest electrical force
Melgi
of attraction?
Xatzki
A. Welcru and Lokfu
C. Xatzki and Melgi
Lofku
B. Zysmu and Melgi
D. Kharmi and Xatzki
Khamri
Welcru
8. Which of the following would have a negative net charge when
Zysmu
rubbed with Kharmi?
A. Lokfu
B. Welcru
C. Xatzki
D. Melgi
9. A negatively charged rod is brought near a metal can that rests on a wooden box.
You touch the opposite side of the can momentarily with your finger. If you remove
your finger before removing the rod, what will happen to the can?
A. It will be discharged
C. It will become negatively charged
B. it will become positively charged
D. Its charge will remain as it was
10. Which of the following can be attracted by a positively charged object?
A. Another positively charged object
C. A neutral object
B. Any object
D. No other object
What’s New
Activity 2: WATER BENDING
Direction: This activity will help you acquire real-life concepts of static electricity.
Materials You Need: 3 Styrofoam cups (you can also use 2 paper cups and an inflated
balloon) and a toothpick. Also, this experiment will also require water and someone with
dry and clean hair.
1. Prepare the set-up by pushing the toothpick at the bottom side of the cups. Leave the
toothpick to produce a gentle drop of water after filling the cup. Hold the cup directly
over the second cup. Fill the cup (with a toothpick) with water and check if it is leaking
steadily.
2. Observe the flow of water from the top cup to the cup below. What are your
observations? ____________________________________________________
3. Rub the third cup on someone with dry hair several times. (This process will help
you in acquiring electrical charges).
4. Hold the cup (rubbed against dry hair) near the water stream without getting the cup
wet. What happened to the water flow? _______________________________________
_______________________________________________________________Then, slowly move
the cup away from the stream and observe. Did you observe changes? ___________ If
yes, describe the change. _________________________________________________________
2
____________________________________________________________________________________
5.
Try other objects (such as balloon, paper cup or any material) aside from the
Styrofoam cup and rub it on a dry and clear hair. What objects have you tried?
_____________________________Which of the objects have changed the flow of water?
_______________________________________________________________________________
What Is It
Electric Charges
The main building block of matter is composed of atoms and molecules. Its
properties are primarily influenced by the electrically charged particles – proton,
electron, and neutron. The table below shows the properties of the charged particles in
terms of mass, charge, and location.
Particle
Mass
(in terms of kg)
proton
electron
9.1093897 x 10-31 kg
1.6726231 x 10-27 kg
neutron
1.6749286 x 10-27 kg
Charge
(in terms of Coulomb
(C))
Location in atom
+1.60217733 x 10-19 C
-1.60217733 x 10-19 C
nucleus
outside nucleus
none
nucleus
Materials contain a huge amount of positively charged particles called protons and
negatively charged particles called electrons. When there is an equal number of protons
and neutrons in a matter, the body is electrically neutral. In making a body negatively
charged, electrons are added to a body. On the other hand, a positively charged body
removes electrons.
Number of
Number of
protons
= electrons
negatively
charged
positively
charged
The transfer of electrons from one body to another proves the law of conservation
of charges. These charges are neither created nor destroyed. According to the principle
of conservation of charges, the sum of electric charges of a body within a closed system
is always constant.
In understanding electrostatic interactions, we have to keep in mind the following
conventions:
1. Any charged object can attract a neutral object.
3
2. Unlike charges attract
3. Like charges repel.
Charging Objects
Materials that allow the movement of electrons from one region to another
are called conductors of electricity, while materials that do not allow the flow of electrons
are called insulators. The majority of the metals are conductors, while nonmetals are
insulators. The electrons can move while protons and neutrons are bound to remain
fixed in the positive nuclei.
Charging objects could happen through induction and conduction. These
charging processes can be demonstrated through an electroscope. The electroscope is
composed of a metal knob, metal rod, glass container, and foil (leaf). The small metal
foils are hung at the end of the metal rod. This should freely move since they open after
being charged.
When a charged object is placed near the metal knob, this causes the foil to open
up since they are being repelled by the presence of excess charges. The foils drop down
when the charged object is placed away from the electroscope.
Metal knob
Glass
Leaves (foil)
Charging by Conduction
When a negatively charged rod touches
the neutrally charged metal knob, the
knob attracts the electrons making the
leaves negatively charged.
When a positively charged rod touches
the neutrally charged metal knob, the
rod attracts the electrons making the
leaves positively charged.
4
Charging by Induction
When
a
negatively
charged rod is placed
near the knob, the
charges
undergo
polarization.
Positive
charges are near the
knob while negative
charges stay away from
the knob. This makes
the foil open up.
Electrons
are
repelled to the Earth
when you touch the
knob or when the
ground
wire
is
connected.
This
makes
the
electroscope
positively charged.
When you touch the
knob with a finger or
attach it with ground
wire, the electrons
from
Earth
move
towards the knob
making it negatively
charged.
When a positively
charged rod is placed
near the knob, the
charges
undergo
polarization. Negative
charges are near the
knob while positive
charges stay away
from the knob. This
makes the foil open
up.
Electric Force
Charles Augustin de Coulomb (1736-1806) used a torsion balance in studying
gravitational interaction. He studied the attractive and repulsive forces between charges.
He found out that the magnitude of force decreases when the distance of separation
between the charges increases. This is shown by Coulomb’s law as expressed in the
equation below:
|𝑸𝟏 𝑸𝟐 |
𝑭=𝒌
𝒓𝟐
where k is proportionality constant 9 x 109
𝑁𝑚2
,
𝐶2
Q1 and Q2 and point charges expressed
in Coulomb (C), r is the distance of separation of two charges expressed in meter (m),
and F is the electrostatic force between the two charges expressed in Newton (N). In SI
units, k is not usually written but as
1
4𝜋∈0
𝐶2
where ∈0 = 8.854 𝑥 109 𝑁𝑚2 . This actually
complicates the formula but could somehow help when you encounter other formulas.
If test charges are placed at some angle with respect to other charges, this
involves computing the x and y components of forces. Recall your previous lessons on
vector resolution and trigonometric identities.
Refer to the diagram below:
F
F
Fy
Ɵ
5
Fx
Solving for x component
∑
Solving for y component
𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃
𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥 + ⋯ + 𝐹𝑛𝑥
Solving for magnitude of resultant force
𝐹𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 = √(∑
2
𝐹𝑥 ) + (∑
𝐹𝑦 )
𝐹𝑦 = 𝐹𝑠𝑖𝑛𝜃
∑
𝐹𝑦 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥 + ⋯ + 𝐹𝑛𝑦
Solving for the direction of resultant force
∑
𝜃=(
∑
2
𝐹𝑦
𝐹𝑥
)
The next examples will help you understand the application of Coulomb’s Law.
Please prepare your scientific calculator and notebook.
Example 1:
Two charges lie on positive x-axis. Charge A (2.0 x 10-9 C) is 2.0 cm from the origin and
Charge B is 4.0 cm from the origin. (-3.0 x 10-9 C). What is the total force exerted by
these two charges on Charge C (5.0 x 10-9 C) located at the origin?
A What is/are
QA = 2.0 x 10-9 C at 𝑟𝐴 = 2.0 𝑐𝑚
given?
QB = -3.0 x 10-9 C at 𝑟𝐵 = 4.0 𝑐𝑚
QC = 5.0 x 10-9 C at 𝑟𝐶 = (0,0)
B What is asked? F = ? at Q3
C Are the units
No, distance of separation, r, given should be converted
consistent with from cm to m.
the formula?
Thus, r1 = 0.02 m and r2 = 0.04 m.
How will you
Q
Q
Q
x
D draw the
0.01 m
0.03 m
0.04 m
problem?
0.02 m
A
C
B
0.02 m
0..04 m
E
What strategy
must be
employed?
In solving the total force experienced by Q C from QA and
QB, forces must be computed individually: FAonC and FBonC.
The vector sum of these forces will determine the net force
exerted by QA and QB on QC.
F Solution
Solving for FAonC:
𝐹𝐴𝐶
Solving for FBonC:
|𝑄𝐴 𝑄𝐶 |
=𝑘
𝑟𝐴 2
𝑁𝑚2 |(2 𝑥10−9 𝐶)(5 𝑥 10−9 𝐶)|
= 9 𝑥 10
(
)
(0.02 𝑚)2
𝐶2
𝑁𝑚2 |(1 𝑥10−17 𝐶 2 )|
= 9 𝑥 109 2 (
)
𝐶
4 𝑥 10−4 𝑚2
= 9 𝑥 109 𝑁(2.5𝑥 10−14 )
𝐹𝐴𝐶 = 2.25 𝑥 10−4 𝑁
The direction of this force lies along the
negative x-component since like charges
repel. Thus, - 2.25 𝑥 10−4 𝑁
9
𝐹𝐵𝐶 = 𝑘
|𝑄𝐵 𝑄𝐶 |
𝑟𝐵 2
𝑁𝑚2 |(−3 𝑥10−9 𝐶)(5 𝑥 10−9 𝐶)|
= 9 𝑥 10
(
)
(0.04 𝑚)2
𝐶2
𝑁𝑚2 |(1.5 𝑥10−17 𝐶 2 )|
= 9 𝑥 109 2 (
)
𝐶
1.6 𝑥 10−3 𝑚2
= 9 𝑥 109 𝑁(9.375𝑥 10−15 )
𝐹𝐵𝐶 = 8.44 𝑥 10−5 𝑁
The direction of this force lies along the
positive
x-component
since
unlike
charges attract. Thus, 8.44 𝑥 10−5 𝑁
9
6
QC
∑
G
FBon
QB
𝐹 = 𝐹𝐴𝐶 + 𝐹𝐵𝐶 = (−2.25 𝑥 10−4 𝑁) + (8.44 𝑥 10−5 𝑁) = −1.41 𝑥 10−4 𝑁
What is the
conclusion?
Therefore, the magnitude of total force experienced by QC from
QA and QB is 1.406 x 10-4 N directed to the left.
Example 2:
Two point charges are located in xy coordinate system. A charge 2.00 𝑥 10−9 𝐶 is
located at (0,4.00 cm) and the other charge −3.00 𝑥 10−9 𝐶 is located at (3.00 cm,
4.00 cm). If the third charge 5.00 𝑥 10−9 𝐶 is placed at origin, find the resultant force
at the third charge.
A What is/are
Q1 = 2.0 x 10-9 C at 𝑟1 = (0,4)
given?
Q2 = -3.0 x 10-9 C at 𝑟2 = (3,4)
Q3 = 5.0 x 10-9 C at 𝑟3 = (0,0)
B What is asked? F = ? at Q3
C Are the units
No, the distance of separation, r, given should be
consistent with converted from cm to m.
the formula?
Thus, r1 = (0 m, 0.04 m), r2 = (0.03 m, 0.04 m) and
r3 = (0,0 m)
D How would you
Solving for unknown (c=?)
draw the
Q1
using Phytagorean
Q2
theorem:
problem?
0.03 m
b = 0.03 m
(opposite)
0.04 m
a = 0.04
(adjacent)
Ɵ
36.87°
c=?
0.05 m
(hypotenuse)
Solving for unknown (Ɵ = ?)
using trigonometric identities:
Q3
Locating the individual forces through a diagram
Q1
Q2
F2on3
x-component
F2on3
36.87°
y-component
F2on3
F2on3
F1on3
Q3
F1on3
E
What strategy
must be
employed?
The forces experienced by Q3 from Q1 and Q2 must be computed
individually. However, x and y components must be determined
first. For instance, F2on3 has x and y components. The x
component will be calculated by multiplying the magnitude of
F2on3 by sinƟ. On the other hand, y component will be calculated
by multiplying the magnitude of F2on3 by cosƟ. For F1on3, x
component is zero since it lies along y component.
7
F Solution
(1) Solving for F1on3:
𝐹13
(2) Solving for F2on3:
|𝑄1 𝑄3 |
=𝑘
𝑟1 2
𝐹23 = 𝑘
𝑁𝑚2 |(2 𝑥10−9 𝐶)(5 𝑥 10−9 𝐶)|
(
)
(0.04 𝑚)2
𝐶2
𝑁𝑚2 |(1 𝑥10−17 𝐶 2 )|
= 9 𝑥 109 2 (
)
𝐶
4 𝑥 10−3 𝑚2
= 9 𝑥 109 𝑁(6.25𝑥 10−15 )
𝐹13 = 5.625 𝑥 10−5 𝑁
The direction of this force lies along the
negative y-component since like charges
repel. Thus, - 5.625 𝑥 10−4 𝑁
= 9 𝑥 109
𝑁𝑚2 |(−3 𝑥10−9 𝐶)(5 𝑥 10−9 𝐶)|
(
)
(0.05 𝑚)2
𝐶2
𝑁𝑚2 |(1.5 𝑥10−17 𝐶 2 )|
= 9 𝑥 109 2 (
)
𝐶
2.5 𝑥 10−3 𝑚2
= 9 𝑥 109 𝑁(6 𝑥 10−15 )
= 5.4 𝑥 10−5 𝑁
= 9 𝑥 109
𝐹23
The magnitude of the force from Q2
towards Q3 is 5.4 𝑥 10−5 𝑁
(3) Solving for x, y components 𝐹23𝑥 =
(4) Solving for magnitude and direction of
the resultant force
5.4 𝑥 10−5 𝑁(𝑠𝑖𝑛 36.87°) = 3.24 𝑥 10−5 𝑁
𝐹23𝑦 = 5.4 𝑥 10−5 𝑁(𝑐𝑜𝑠 36.87°) = 4.32 𝑥 10−5 𝑁
𝐹13𝑥 = 0𝑁 (since F13 lies along y-axis)
−4
𝐹13𝑦 = − 5.625 𝑥 10 𝑁
F13
F23
Sum
x component
y component
0𝑁
− 5.625 𝑥 10−4 𝑁
3.24 𝑥 10−5 𝑁
3.24 𝑥 10−5 𝑁
|𝑄2 𝑄3 |
𝑟1 2
𝐹𝑅 = √(∑
4.32 𝑥 10−5 𝑁
−1.31 𝑥 10−5 𝑁
𝜃=(
2
𝐹𝑥 ) + (∑
𝐹𝑦 )
2
= √(3.24 𝑥 10−5 𝑁)2 + (−1.31 𝑥 10−5 𝑁)2
𝐹𝑅 = 3.49 𝑥 10−5 𝑁
∑
∑
𝐹𝑦
𝐹𝑥
−1.31 𝑥 10
) = 𝑡𝑎𝑛−1 (
−5
−5
3.24 𝑥 10
𝑁
𝑁
) = −21.94° 𝑜𝑟 338°
G
What is the
Therefore, the magnitude and direction of the resultant
conclusion?
force is 3.49 𝑥 10−5 𝑁 , 338°.
You can verify the magnitude and direction of the resultant force using the
graphical method for vector analysis.
Electric Fields
You were introduced to the behavior of electric charges and how these charges
produce attractive and repulsive forces. Aside from these forces, it also creates an
electric field E. The electric field of charge Q is the space surrounding the charge. It also
exerts a force F on any test charge q placed within that region.
8
The electric field is represented by the equation below:
𝑬=
𝑁
𝑭
𝒒
where E is the electric field expressed in 𝐶 , F is the electric force expressed in newton
(N), and q is the charge expressed in coulombs (C). If q is positive, the direction of E is
the direction of F. On the other hand, the force on a negative charge is opposite to the
direction of the E.
F
E
F
We can also calculate E given the magnitude and position of all charges
|𝑄 𝑄 |
𝐹
involved. Since 𝐹 = 𝑘 1 2 2 , we substitute this formula in 𝐸 = . Thus,
𝑟
𝑞
𝒒
𝑬=𝒌 𝟐
𝒓
Like dealing with charge situated at some angles, you can similarly perform the
same using the trigonometric and vector resolutions concepts. Refer to the diagram
below. Suppose we have a negative charge at xy plane. We want to calculate the electric
field at point P:
E
E
Ey
P
Ɵ
Ex
Solving for x component
Solving for y component
𝐸𝑥 = 𝐸𝑐𝑜𝑠𝜃
𝐸𝑦 = 𝐸𝑠𝑖𝑛𝜃
∑
𝐸𝑥 = 𝐸1𝑥 + 𝐸2𝑥 + 𝐸3𝑥 + ⋯ + 𝐸𝑛𝑥
Solving for magnitude
resultant force
𝐹𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 = √(∑
2
𝐸𝑥 ) + (∑
of
∑
𝐸𝑦 = 𝐸1𝑥 + 𝐸2𝑥 + 𝐸3𝑥 + ⋯ + 𝐸𝑛𝑦
the Solving for the direction of
resultant force
2
∑
𝐸𝑦
)
𝜃=(
∑
𝐸𝑦 )
𝐸𝑥
the
Example 3:
A point charge 𝑞1 = +6.00 𝑥 10−9 𝐶 is at the point 𝑥 = 0.800 𝑚, 𝑦 = 0.600 𝑚 and a second
point charge 𝑞2 = −2.00 𝑥 10−9 𝐶 is at the point 𝑥 = 0.800 𝑚, 𝑦 = 0 𝑚. Calculate the
magnitude and direction of the resultant electric field at the origin due to these
charges.
9
A
B
C
D
What is/are
given?
What is asked?
Are the units
consistent with
the formula?
How would you
draw the
problem?
q1 = +6.00.0 x 10-9 C at 𝑟1 = (0.800 𝑚, 0.600 𝑚)
q2 = -2.00.0 x 10-9 C at 𝑟2 = (0.800 𝑚, 0 𝑚)
E = ? at (0,0)
Yes, given values have correct SI units.
Solving for unknown
(c=?) using Phytagorean
theorem:
Q1
c=?
1m
(hypotenuse)
Ɵ
53°
0.6 m
b = 0.03 m
(opposite)
Solving for unknown (Ɵ = ?)
using trigonometric
identities:
37°
Q2
0.8 m
Review field of
charges
a = 0.04
(adjacent)
Locating the individual electric fields at (0,0):
E2
E1
53°
E1y
53°
E1x
E
What strategy
must be
employed?
The electric fields experienced in origin (0,0) must be
computed individually. Furthermore, x and y components
must be solved for each E1 and E2. We then calculate the
magnitude and direction of the resultant electric field.
F Solution
(1) Solving for F1on3:
(2) Solving for E2:
|Q1 |
E1 = k 2
r1
2 |(6
x10−9 C)|
Nm
= 9 x 109 2 (
)
(1 m)2
C
Nm2 |(6 x10−9 C)|
= 9 x 109 2 (
)
C
1m2
= 9 x 109 N(6x 10−9 )
N
E1 = 54 C (outward direction)
(3) Solving for x, y components E1x =
N
54 C (sin 53°)
E1y = 54
= −43.126
N
(cos 53°)
C
N
C
(along –x-axis)
|𝑄2 |
𝑟2 2
2 |(−2
𝑥10−9 𝐶)|
𝑁𝑚
= 9 𝑥 109 2 (
)
(0.8 𝑚)2
𝐶
𝑁𝑚2 |(−2 𝑥10−17 𝐶)|
= 9 𝑥 109 2 (
)
𝐶
0.64𝑚2
= 9 𝑥 109 𝑁(−3.125 𝑥 10−17 )
𝑁
= 28.125 𝑁 𝐶
𝐸2 = 𝑘
𝐹23
The electric field is directed towards the
positive x-axis since it is a negative
charge (inward direction of field).
(4) Solving for magnitude and direction
of the resultant force
𝐹𝑅 = √(∑
N
= −32.498 (along – y axis)
C
N
E2x = 28.125
C
2
𝐹𝑥 ) + (∑
𝐹𝑦 )
2
= √(−15.001 𝑁/𝐶)2 + (−32.498 𝑁/𝐶)2
𝐹𝑅 = 35.79 𝑁/𝐶
10
N
∑
𝜃=(
∑
E2y = 0 C since it lies along x-axis
x component
E1
E2
Sum
G
N
−43.126
C
N
28.125 C
N
−15.001 C
What is the
conclusion?
y component
-32.498
0
N
C
−32.4898
𝐹𝑦
𝐹𝑥
) = 𝑡𝑎𝑛−1 (
= 65.22°
−32.498 𝑁/𝐶
)
−15.001 𝑁/𝐶
N
C
Therefore, the magnitude and direction of the resultant force
𝑁
is 35.79 𝐶 , 65.22°.
You can verify the magnitude and direction of the resultant force using a
graphical method for vector analysis.
Electric Flux
The relationship between electric charge and electric field was also formulated
alternatively in Gauss’s law (Karl Friedrich Gauss 1777-1855). This is logically
equivalent to Coulomb’s law, but this was easier to use in finding electric field for
symmetrical charge distribution.
Gauss’s law general statement is expressed as
𝜙𝐸 = 𝐸𝐴cosƟ
where is the electric flux
𝑁𝑚2
,
𝐶
E is the magnitude of electric field expressed in
𝑁
𝐶
, Ɵ is
the angle between the normal line of the surface and electric field lines. and A is the
area of a given surface expressed in m2.
Electric flux refers to the amount of electric field lines penetrating a given
surface. The electric flux is maximum if the electric field lines are perpendicular to the
surface or parallel to the normal line of the surface. The electric flux is zero when the
electric field line is parallel to the surface or perpendicular to the normal line of the
surface. At a given angle, the electric flux is directly proportional to the component of
the electric field lines.
Example 4:
A point charge 𝑞 = 8.00 𝑥 10−9 𝐶 is at the center of the cube with sides of length 0.200
m. What is the electric flux through one of the six faces of the cube?
A
B
What is/are
given?
What is asked?
q = +8.00 x 10-9 C ; 𝑠 = 0.200 𝑚
𝞍 = ? at one of the faces of the cube
11
C
D
Are the units
consistent with
the formula?
How would you
draw the
problem?
Yes, given values have correct SI units.
c = 0.141
c = 0.283
0.200 m
0.200 m
0.200 m
E
F
What strategy
must be
employed?
Solution
Hence, the distance from the charge to the surface of the cub
is 0.141 m
Using Gauss’s law, substitute E with formula for Electric Field
𝒒
𝑬 = 𝒌 𝒓𝟐
2
−9
𝑞
𝑁𝑚 |(8 𝑥10 𝐶)|
9
𝐸 = 𝑘 2 = 9 𝑥 10 2 (
)
(0.141 𝑚)2
𝑟
𝐶
9
= 9 𝑥 10
G
What is the
conclusion?
0.200 m
𝑁𝑚2
𝐶2
|(8 𝑥10−9 𝐶)|
𝑁𝑚2
9
(4.02 𝑥 10−7 )
(
) = 9 𝑥 10
2
2
0.0199 𝑚
𝐶
𝐸 = 3,621.55 𝑁/𝐶
𝐴 = 𝑠 2 = (0.200 𝑚)2 = 0.04 𝑚2
𝑁
𝜙𝐸 = 𝐸𝐴𝑐𝑜𝑠Ɵ = 3,621.55 (0.04 𝑚2 ) = 144,862
𝐶
𝑁𝑚2
𝐶
Therefore, the electric flux at any surface is 144,862
𝑁𝑚2
𝐶
What’s More
Activity 3: QUALITATIVE PROBLEMS
Direction: Solve the following problems as directed. You may show your solution on a
separate sheet of paper
(1) Two spheres of equal mass and equal charge are separated at a distance r. (a)
Derive an expression for the quantity of charge that must be on each sphere so that
the spheres are in an equilibrium where attractive and repulsive forces are balanced.
(b) How would doubling the distance between spheres affect the expression for the
value of q from the previous problem. Explain.
_____________________________________________________________________________
_____________________________________________________________________________
____________________________________________________________________
12
(2) How would you draw the electric field lines given the test charges below:
(3) Coulomb’s law and Newton’s law of gravitation are similar in structure. Can Gauss’s
law be applied to gravitational fields? _________If so, what changes are
needed?____________________________________________________________________________
__________________________________________________________________________
What I Have Learned
Activity 4: QUANTITATIVE PROBLEMS
Direction: Write your answers on a separate sheet of paper. You may also consult
your Physics teacher.
Three charged spheres are at the positions shown in the figure.
(a) Find the net electrostatic force at sphere B.
(b) Find the net electric field at (4,-3)
4 cm
4.5µC
A
8.2 µC
B
3 cm
6.0µC
C
Scoring Rubric
Criteria
3
Physics
The approach is
Approach
appropriate and
complete
Procedure
Description
Mathematical and
logical procedures
are clear,
complete, and
connected
Diagrams and
symbols used are
appropriate and
complete
2
The approach
contains minor
errors
Mathematical
and logical
procedures are
missing/contain
errors
Parts of the
diagrams and
symbols contain
errors
13
1
Some of the
concepts and
principles are
missing or
inappropriate
Most of the
mathematical and
logical procedures
0
The solution
doesn’t indicate
an approach
Most of the parts of
the diagrams and
symbols are not
useful
The entire
visualization is
wrong or did
not include
visualization.
All procedures
are incomplete
and contain
errors
What I Can Do
Activity 5. BUILDING CONCEPT MAP
Direction: Create a concept map out from the things you have learned from this module.
You can use words, terms, phrases, or formulas in connecting these concepts. Refer to
the scoring guide below:
Mueller’s Classroom Concept Rubric
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)
Yes (2)
Many inaccuracies
(0-2)
A few inaccuracies
(3-4)
No inaacuracies (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)
Assessment
Direction: Write the letter of your choice in the space provided.
______1.The diagram below shows the behavior of the electroscope before and after a
positively charged rod is placed near the electroscope knob. This tells us that the
movement of electrons is from ______.
a.rod to leaves
b.leaves to knob
c.knob to leaves, then back to knob
d.leaves to knob, then back to leaves
_____2.A glass rod was positively charged when rubbed with a silk cloth. The net
positive charge is accumulated because the glass rod
a.loses protons
c.gains electrons
b.loses electrons
d.loses electrons
_____3.Gravitational forces and electric forces are both
a.forces with attractive and repulsive behaviors
b.indirectly proportional to the square root of separation between bodies
c.directly proportional to the product of the masses and charges
d.decreasing when the distance between two bodies is decreasing
14
_____4.At which point is the electric field strength strongest?
A
D
C
B
_____5.Three spheres were brought together. When Spheres A and B are brought
together, they attract. When spheres B and C are brought together, they also
repel. Which of the following is true?
a.Spheres A & C have same signs
c.Spheres A & C have opposite signs
b.Spheres B & C have same signs
d.Spheres A & B have similar signs
_____6.The first object has a charge of +3nC, and the second object has a charge of 6
nC. Which is true about the electric forces between these objects?
a.F1on2= 2F2on1
b.3F1on2= 6F2on1
c.-6F1on2=2F2on1
d.F1on2=F2on1
For Nos. 7 to 9, refer to the diagram below:
_____7. A positively charged rod was brought near to a metallic plate; what is the type
of charge induced along the side facing the rod?
a.positive
b.negative
c.netural
d.depends on number of charges
_____8.If the positively charged rod was touched on the neutral metallic plate, what is
the charge of the plate?
a.same
b.negative
c.positive
d.depends on number of charges
_____9.After the positively charged rod is placed near the plate, a grounded wire was
attached. What is the charge of the plate after the wire is removed?
a.negative
c.neutral
b.positive
d.depends on number of charges
_____10.The electric flux through the surface at the right is ____
η
a.zero
c.positive
b.negative
d.unknown
_____11.A test charge produced an electric field, E, at point 3 m away
from the charge. The point where the field is half its original values is located at
a.1.5 m from the charge
c.4.5 m from the charge
b.6 m from the charge
d.at the point of charge
_____12.The electric field of a body is directly related to its
a.momentum
b.kinetic energy
c.potential energy
d.charge it carries
_____13.Charge q1 (26.0 µC) and q2 (-47.0 µC) experienced an electrostatic force of 5.70
N between them at a distance of ____
a.1.39 m
b.1.93 m
c.2.14 x 10-19
d.1.464 x 10-8
_____14.How many electrons have been removed from a positively charged particle if it
has a net charge of 5 x 10-9 C?
a.5 x 10-9 electrons
c.1 x 10-8 electrons
-9
b.2.5 x 10 electrons
d.incomplete information
15
_____15.A sphere of radius 0.05 m has a charge of 2.0 nC. The charge is said to be
located at the center of the sphere. The magnitude of electric field inside the
surface is
a.zero
c.thrice the original value
b.twice the original value
d.half the original value
Additional Activities
Activity 6. SOCIAL CONTEXT
Direction: The community is a rich source for learning opportunities in electrostatics.
Choose one from the following suggested activities in understanding the importance
and utilization of electrostatics in our daily lives:
1. Ask
a
local
weather
forecaster/Science
teacher/physicist/electrician/engineer through phone calls or
interview through Zoom, Google Meet, or Messenger and inquire (with
supervision from your parents/guardian/learning facilitator) about
lightning, lightning safety, and lightning rods.
2. Conduct simulations on electrostatics, electric fields, Gauss’s law using
online sources. From this, write a short reflection. Refer to any sites
below:
- https://phet.colorado.edu/en/simulation/charges-and-fields
- https://www.falstad.com/emstatic/
- http://web.mit.edu/viz/soft/visualizations/DLIC/doc/simulations/ex
periments/electrostatics/package-summary.html
- https://javalab.org/en/category/electricity_en/static_electricity_en/
3. Suppose you are to demonstrate a trick to kids using electrostatic
charging. How will you demonstrate the trick? List down the materials
(preferably available at home), step-step procedures, and brief
explanation of how this trick works.
16
Answer Key General Physics 2 Module 1
Activity 1: Getting
Recharged
1. A
2. D
3. A
4. A
5. A
6. B
7. D
8. C
9. B
10. C
Activity 2: Water Bending
2. vertical, thin, unbroken stream of water
4. the stream of water was bent towards the cup; yes
(there should be changes); the stream went back to its
original flow
5. answers may vary; the influence of charged objects to
water bending may vary depending on their ability to
gain or lose charges
Activity 3. Qualitative Problems
Activity 4. Quantitative Problems
17
References
Printed Resources
Sears, F., Zemansky, M. and Young, H. (1992). College Physics 7th Edition. Addison-Wesley
Publishing Company
Zitzewits, Haase and Harper (2013). PHYSICS Principles and Problems. The MAcGraw-Hill
Companies, Inc.
Online References
Harmon, K. (2011). Bend water with static electricity. Scientific American.
https://www.scientificamerican.com/article/static-electricity-bring-science-home/
Java Lab (n.d.). Static Electricity Simulation. Retrieved last February 18, 2021 from
https://javalab.org/en/category/electricity_en/static_electricity_en/
Massachusetts Institute of Technology. (n.d.) Package simulations experiments electrostatics.
Retrieved last February 20, 2021 from
http://web.mit.edu/viz/soft/visualizations/DLIC/doc/simulations/experiments/electros
tatics/package-summary.html
Mueller, J. (n.d.) Concept map rubric. Retrieved last February 22, 2021 from
https://teach.its.uiowa.edu/sites/teach.its.uiowa.edu/files/docs/docs/Concept_Map_Ru
brics_ed.pdf
PHET (2021). Charges and Fields. University of Colorado. Retrieved last February 19, 2021
from https://phet.colorado.edu/en/simulation/charges-and-fields
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