Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
Electric Charge, Coulomb’s Law, Electric
Fields, and Electric Flux
Module 4
EXPECTED LEARNING OUTCOMES
4.1 Explain the role of electron transfer in electrostatic charging by rubbing.
4.2 Calculate the electric field due to a system of point charges using Coulomb’s law and the
superposition principle.
4.3 Use Gauss’s law to infer electric field due to uniformly distributed charges on long wires, spheres,
and large plates.
4.4 Solve problems involving electric charges, dipoles, forces, fields, and flux in contexts such as, but
not limited to, systems of point charges, electrical breakdown of air, charged pendulums,
electrostatic ink -jet printers.
THE BIG IDEA
As early as 600 BC, the Greeks were already aware that amber – a hardened translucent
yellowish tree resin – when vigorously rubbed with a piece of cloth, could attract nearby objects. In
1600, William Gilbert, wo served as physician to Queen Elizabeth I of England, found out many other
substances possess the same ability as that of amber when rubbed against another substance. He called
these substances electrics and this ability of amber electricity, from the Greek word electron, which
means “amber.” Subsequently, it was shown that every object acquires this ability to attract small pieces
of matter after being rubbed against another object. The object attracted is said to have acquired charge
or electrified. Because the charge is at rest, it is often referred to as static electricity. Electrostatics is
the study of all phenomena associated with electric charges at rest.
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
DEEPEN YOUR UNDERSTANDING
CONDUCTORS AND INSULATORS
Charges can exist in materials and move through them. However, the ease with which charges
move through them differs. Conductivity is the measure of the ease at which an electric charge moves
through a material.
Materials that readily allow the flow of charges through them are called conductors. Metals are
good conductors because they have plenty of free electrons that can easily move in the material.
Insulators are materials that resist the flow of charges. The conductivity of insulators is low.
Some examples of insulators are rubber, plastic, mica, paper, glass, and air.
Semiconductors are intermediate between conductors and insulators. Semiconductors are not
as conductive as metals, but they are more conducive than insulators such as rubber, plastic, and mica.
The conductivity of a semiconductor in its pure form is incredibly low. Atoms of different elements in
small amounts, (i.e., one part per million or even less) are added to pure semiconductors to improve
their conductivity. This process is referred to as doping. Semiconductors have paved the way for the
development of miniaturized electronic devices such as transistors and integrated circuits. Some
examples of semiconductors are silicon, germanium, and gallium arsenide.
Superconductors offer practically no resistance to the flow of charges below some critical
temperatures. A current in a superconductor can keep flowing without any decay. In 1911, Dutch
physicist Heike Kamerlingh Onnes discovered superconductivity by cooling mercury to a temperature
of about 4 K. Most superconductors only work at temperatures close to absolute zero. Scientists are
now focused on developing superconductors that will work at normal and high temperatures because
superconductors that work at room temperature would make everyday electricity generation and
transmission vastly more efficient in as much as there will be no power losses. The highest known
critical temperature of a superconducting material is 203 K(–70°C); the material involved is hydrogen
sulfide.
CHARGING BY RUBBING
Conductors are materials that allow
electrical charges to move from one material to
another. Conductors may be charged through
different methods – rubbing, conduction, and
induction.
An electricity-neutral body can gain a
charge by rubbing or friction. Consider two
different uncharged bodies. Because of the
difference in their material composition, the
Fig. 4.1 Both the glass rod and silk cloth become electrically
nuclei of their atoms pull their electrons with
charged after rubbing them together.
different strengths. Rubbing these two bodies will
force their atoms to interact with one another, resulting in the “ripped-off” electrons are then transferred
to the other body. After rubbing, one of the bodies will have more electrons, and the other one will have
fewer. Thus, both will now be electrically charged.
An example of charging by rubbing is shown in Figure 4.1. A glass rod acquires a positive
charge after being rubbed with silk cloth, and silk cloth acquires a negative charge. Charging by rubbing
also explains why we experience a weak electric shock when we suddenly touch a metallic object after
walking on a carpeted floor.
The charge acquired by rubbed materials can be determined using the triboelectric series. The
triboelectric series is a list of common materials that were experimented on and found to behave in a
predictable way. When these materials are rubbed together, those that appear first in the list tend to lose
their electrons, making the negative. In other words, if you rub any two of the materials in the series,
[Grade 12 – General Physics 2]
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
the material in the upper part of the list will be positive, and the other material in the lower part will be
negative.
COULOMB’S LAW
Through experimentation, Coulomb discovered that the magnitude of the electrical force
between two charged particles is directly proportional to the product of the magnitudes of the charges
and inversely proportional to the square of the distance between them. This statement is called
Coulomb’s law. Mathematically, this can be written as
|q1q2|
Eq. (4.1)
FE = k
r2
where FE is the magnitude of the electric force in newtons, q1 and q2 are charges in coulombs, r is the
distance in meters, and k is the proportionality constant called Coulomb’s constant with an approximate
value of 9×109 N·m2/C2. Note that only the absolute value of the product of q1 and q2 is determined
because only the magnitude of the force is to be computed. The force magnitude F is always positive.
However, the electric force F is a vector quantity, which may be positive or negative depending
on its direction. Its direction is along the line joining q1 and q2. The vectors representing the force that
q1 exerts on q2 and vice versa are pointing toward each other for attraction and away from each other
for repulsion. These forces are equal but opposite in direction in accordance with Newton’s third law
of motion.
Coulomb’s law bears a strong resemblance to Newton’s law of universal gravitation. Both
equations have the same form. Recall that the gravitational force FG between two masses m1 and m2 is
given by
m1m2
Eq. (4.2)
FG = G
r2
where G is the universal gravitational constant approximately equal to 6.674×10 –11 N·m2/kg2.
The electric force and the gravitational force have similarities and differences. Both forces
follow the inverse square law for distance. Both forces are also proportional to the product of the
quantity that causes the force, that is, mass for gravitational force and charge for electric force. Lastly,
both forces are conservative and noncontact. However, the gravitational force is only an attractive force,
while the electric force may be attractive or repulsive. The gravitational force, in general, is much
weaker than the electric force.
Example: Two protons (q = 1.602×10–19 C) are separated by 3.8×10–10 m in the air. (a) Find the
magnitude of the electric force one proton exerts on the other. Is this force attractive or
repulsive? (b) Find the magnitude of the gravitational force one proton exerts on the other. Is
this force attractive or repulsive?
Solution: (a) Using Eq. 4.1 to solve for the electric force and substituting values,
|q1q2|
|(1.602×10–19 C)(1.602×10–19 C)|
9
2
2
=
(9×10
N·m
/C
)
r2
(3.8×10–10 m)2
–9
–9
FE = 1.596×10 N ≈ 2×10 N
This is a repulsive force because the charges are both protons.
(b) Using Eq. 4.2 to solve for the gravitational force and substituting values,
FE = k
(1.673×10–27 kg)2
m1m2
–11
2
2
=
(6.674×10
N·m
/kg
)
r2
(3.8×10–10 m)2
–45
FG = 1.29×10 N
Gravitational force is an attractive force.
FG = G
Example: Two small conducting and identical spheres A and B have charges –25 nC and +15 nC,
respectively. They are separated by 0.02 m distance. (a) What is the magnitude of the electric
force between the two spheres? Is this force attractive or repulsive? (b) The spheres are then
allowed to touch each other and then separated. What is the magnitude of the force between
the two spheres? Is this a repulsive force or an attractive force?
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
Solution: (a) Using Eq. 4.1 to solve for the electric force and substituting values,
|q1q2|
|(–25×10–9 C)(+15×10–9 C)|
= (9×109 N·m2/C2)
FE = k
2
r
(0.02 m)2
–3
FE = 8.44×10 N
Since the two charged spheres have unlike charges, the electric force is attractive.
(b) Since spheres A and B are identical, they will share the net charge equally.
q0A + q0B
(–25×10–9 C)+(15×10–9 C)
qA = qB =
= –5×10–9 C
=
2
2
Using Eq. 4.1 and substituting values,
|qAqB|
|(–5×10–9 C)( –5×10–9 C)|
9
2
2
=
(9×10
N·m
/C
)
FE = k
r2
(0.02 m)2
FE = 5.63×10–3 N
Since the two charged particles have like charges, the electric force between them is repulsive.
SUPERPOSITION PRINCIPLE
So far, only the force between two-point charges has been
discussed. What if there are more than two charges? For instance, if
there are three charges q1, q2, and q3, what will be the net force on q1
due to q2 and q3?
Electric forces obey the superposition principle. The
superposition principle states that each charge will exert a force on
another charge as if no other charges are present. The total force that
a particular charge experience due to a collection of charges is the
vector sum of all the individual forces.
Fig. 4.2 The net force F on q1 due to
Example: Three-point charges are located along the x–axis. Point q2 and q3 is the vector sum of F3 on 1
charge q1 = +3.5×10–6 C is at x = 0, point charge q2 = and F2 on 1.
+8.5×10–6 C is at x = 2 m, and point charge q3 = –5×10–6
C is at x = 3m. Find the resultant electric force acting on
q1.
Solution: We have the following given:
q1 = +3.5×10–6 C; q2 = +8.5×10–6 C; q3 = –5×10–6 C
r2 on 1 = 2m – 0 = 2m; and r3 on 1 = 3m – 0 = 3m
Solve first for the individual forces F2 on 1 (forces exerted by q2 on q1) and F3 on 1 (forces exerted
by q3 on q1). Using Eq. 4.1,
F2 on 1 = k
|q1q2|
= (9×109 N·m2/C2)
|(+3.5×10–6 C)(+8.5×10–6 C)|
(r2 on 1) 2
(2 m)2
F2 on 1 = 0.0669 N ≈ 0.07
Point charge q2 repels q1. Thus,
N F2 on 1 is directed to the left.
|q1q3|
|(+3.5×10–6 C)(–5×10–6 C)|
= (9×109 N·m2/C2)
F3 on 1 = k
2
(r3 on 1)
(3 m)2
F3 on 1 = 0.0175 N ≈ 0.02 N
Point charge q3 attracts q1. Therefore, F3 on 1 is directed to the right. Setting the left direction to
be positive and the right direction to be negative, the magnitude F of the resultant force is F = 0.0669
N – 0.0175 N = 0.0494 N, directed to the left.
Example: Three identical point charges with charge q = +3×10–6 C are placed at each vertex of an
equilateral triangle ABC as shown. If the side of the equilateral triangle is 0.01 m, find the
resultant electric force on the charge at vertex A.
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
Solution: We have the following given: qA = qB = qC = +3×10–6 C; r = 0.01 m
|qAqB|
|(+3×10–6 C)(+3×10–6 C)|
= (9×109 N·m2/C2)
FB on A = k
(rB on A)
(0.01 m)2
2
FB on A = 810 N
Point charge qB repels qA. Thus, FB on A is directed to the right up at
60° (since the triangle is equilateral) with the positive x–axis. Force
FC on A is computed using the same equation.
|qAqC|
|(+3×10–6 C)(+3×10–6 C)|
= (9×109 N·m2/C2)
FC on A = k
(rC on A)
(0.01 m)2
2
FB on A = 810 N
Point charge qC repels qA. Thus, FC on A is directed upward to the left
at an angle of 60° with the negative x–axis.
To visualize FB on A and FC on A, qA is placed at the origin of the
Cartesian coordinate system as shown. The component method is
used to determine the resultant electric force F.
Force
Horizontal Component
Vertical Component
FB on A = 810 N
+810 N (cos 60°) = 405 N
+810 N (sin 60°) = 702 N
FC on A = 810 N
–810 N (cos 60°) = –405 N
+810 N (sin 60°) = 702 N
∑Fx = 0
∑Fy = 1404 N
Therefore, the resultant electric force is equal to 1404 N acting vertically up.
ELECTRIC FIELD
The electric force is a noncontact force. An electric charge q can exert force on other charged
objects even though they are at some distance away. The space surrounding a charged body is called an
electric field. An electric field causes any charged particle placed in it to experience an electric force.
Every charge has an electric field associated with it.
ELECTRIC FIELD DUE TO A POINT CHARGE
An electric field exists in the region of space around a charged object or a source charge. When
another charged object enters this electric field, it will experience an electric force. The strength of the
electric field at a point due to the source charge is called electric field intensity. (In this text, the electric
field will simply refer to electric field intensity.) The 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 electric field are usually represented by q0 and E,
respectively.
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
Eq. (4.3)
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 distance 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
|qq0|
FE = k
r2
Thus, the magnitude of the electric field due to the point charge is
FE
|q|
E= q =k 2
Eq. (4.4)
0
r
It follows that E has the unit of Newton/coulomb (N/C). Like electric force, the 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 principle.
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
Example: Calculate the magnitude and direction of the electric field 0.45 m from +7.85×10–9 C point
charge
Solution: We have the following given: q = 7.85×10–9 C; r = 0.45 m
The magnitude of the electric field is calculated using Eq. 4.4.
|q|
|(+7.85×10–9 C)|
E = k 2 = (9×109 N·m2/C2)
= 348.89 N/C
r
(0.45 m)2
The charge q will repel the positive test charge q0. Hence, the electric field will be directed
away from charge q.
Example: An electric dipole consists of two equals but unlike charges separated by a distance. Twopoint charges, q1 = +4.5×10–6 C and q2 = –4.5×10–6 C, are separated by 6.4×10–2 m, forming
an electric dipole as shown in the figure. Find the electric field halfway between the dipoles.
Solution: We have the following given: q1 = +4.5×10–6 C; q2 = –4.5×10–6 C; r = 6.4×10–2 m.
Imagine a unit positive test charge q0 placed halfway between the two charges. Eq. 4.4 will
be used to calculate the magnitude of the electric field E1 due to q1 and E2 due to q2.
|q1|
|(+4.5×10–6 C)|
E1 = k 2 = (9×109 N·m2/C2)
= 3.96×107 N/C
r
(3.2×10–2 m)2
E2 = k
|q2|
= (9×109 N·m2/C2)
r2
|(–4.5×10–6 C)|
(3.2×10–2 m)2
= 3.96×107 N/C
Charge q1 will repel q0. Hence, E1 will be directed to the right. Charge q2 will attract q0. Hence,
E2 will be directed to the right. In accordance with the superposition principle,
E = E1 + E2 = = 3.96×107 N/C + 3.96×107 N/C
= 7.92×107 N/C, directed to the right.
If the electric field at a point is known, then the force on any other charge placed at that point
is determined by multiplying the charge by the electric field.
Eq. (4.5)
FE = E|q|
If the charge happens to be negative, the direction of the force on the negative charge is
opposite the direction of the field.
Example: An electron enters a uniform electric field that is directed downward and has a magnitude of
5 N/C. (a) Find the magnitude and direction of the force experienced by the electron. (b) Also,
find its acceleration.
Solution: We have the following given: E = 5 N/C; q = –1.602×10–19 C; m = 9.109×10–31 kg
a. The direction of the force is upward because the electron is negatively charged. To solve
for the magnitude of the force experienced by the electron, Eq. 4.5 is used. Only the
absolute value of the charge of the electron is considered.
FE = E|q|
= (5 N/C)|–1.602×10–19 C|
= 8.01×10–19 N
b. Newton’s second law of motion is used to solve for acceleration. The acceleration is also
directed upward.
F
8.01×10–19 N
= 8.78×1011 m/s2
a=
=
–31
m
9.109×10 kg
[Grade 12 – General Physics 2]
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
Example: A tiny ball weighs 0.0055 kg and carries a charge of +3.25×10–6 C. What electric field
(magnitude and direction) is needed for the ball to remain suspended in the air?
Solution: We have the following given: m = 0.0055 kg; and q = 3.25×10–6 C
For the ball to remain suspended in the air, the electric force must be directed
vertically up and numerically equal to the weight of the object. Since the charge of
the ball is positive, the direction of the electric field is also vertically up. Using the
first condition for equilibrium,
∑F = 0
FE – mg = 0
FE = (0.0055 kg)(9.8 m/s2)
= 0.0539 N
Using Eq. 4.5 to solve for the electric field,
FE = E |q|
0.0539 N = E |+3.25×10–6 C|
= 1.66×104 N/C, directed upward
ELECTRIC FLUX
The electric field can be quantitatively described using the concept of electric flux. The word
flux comes from the Latin word Fluxus meaning “flow.” Electric flux (Φ) is a measure of the number
of field lines passing through a surface.
Fig. 4.3 Electric flux is the number of
field lines passing perpendicularly
through a surface.
Fig. 4.4 Angle θ between the area
vector and the electric field.
Mathematically, electric flux is the dot product of the electric field and area vector. The
direction of the area vector is the same as that of a vector perpendicular to the area.
Φ = E · A = EA cos θ
Eq. (4.6)
where θ is the angle between the electric field and the area vector. Note that electric flux is scalar and
has the unit N·m2/C.
Example: A flat surface of the area of 1.25 m2 is rotated through a uniform horizontal electric field of 5
N/C. What is the electric flux if the surface is (a) parallel and (b) perpendicular to the electric
field?
Solution: We have the following given: A = 1.25 m2; E = 5 N/C
a. If the surface is parallel to the flux, the area vector is perpendicular to the electric field. The
angle between the area vector and the electric field is 90°. Using 4.6,
Φ = EA cos θ
= (5 N/C)( 1.25 m2)cos 90° = 0
b. If the surface is perpendicular to the flux, the area vector is parallel to the electric field. The
angle between the area vector and the electric field is 0°. Using Eq. 4.6,
Φ = EA cos θ
= (5 N/C)( 1.25 m2)cos 0° = 6.25 N·m2/C
GAUSS’S LAW
Carl Friedrich Gauss, a German scientist, formulated a law, which relates electric field, electric
flux, and electric charge. This is known as Gauss’s law which states that the total electric flux through
a surface is the total electric charge qtotal inside the surface divided by ∈0. The constant ∈0 is called the
permittivity of free space and has an approximate value of 8.8542×10–12 C2/N·m2. In symbols
qtotal
Eq. (4.7)
Φtotal = EA cosθ =
∈0
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
or in integral form:
Eq. (4.8)
Φ = ∫E·dA
The surface mentioned in Gauss’s law is called the Gaussian surface. Since a line of force starts
from a positive charge to a negative charge, the lines going out of the surface are considered positive,
while the lines going into the surface is negative.
Example: Calculate the total electric flux Φ for each of the closed surface’s a, b, c, and d as shown.
Note that q1 = +3C, q2 = +1C, q3 = –5C, and q4 = –9C.
Solution: Surface a encloses all the four given charges. Therefore,
+3C + 1C – 5C – 9C
8.8542×10–12 C2/N·m2
= –1.1294×1012 N·m2/C
Surface b encloses q2 only. Therefore,
1C
Φb =
8.8542×10–12 C2/N·m2
= +1.1294×1011 N·m2/C
Φa =
Surface c encloses q1 and q2 only. Therefore,
+3 C + 1 C
Φc =
8.8542×10–12 C2/N·m2
= +4.5176×1011 N·m2/C
Surface d encloses all four given charges. Therefore,
+3C + 1C – 5C – 9C
Φd =
8.8542×10–12 C2/N·m2
= –1.1294×1012 N·m2/C
CHARGE DISTRIBUTION AND GAUSS’S LAW
Gauss’s law can be used to compute the electric field due to a system point charge as well as for
the continuation of the charge distribution. In practice, the charge distribution must be uniform and
symmetrical. The charge distribution may be expressed in terms of linear charge density, surface charge
density, or volume charge density. Linear charge density λ is a charge per unit length, surface charge
density σ is a charge per surface area, and volume charge density ρ is a charge per unit volume of a
body.
Example: If a solid insulating sphere of radius 50 cm carries a total charge of
150 nC uniformly distributed throughout its volume, what is its (a)
volume charge density? What is the magnitude of the electric field
at (b) 10 cm and (c) 65 cm from the center of the sphere?
Solution: We have the following given:
R = 50cm = 0.50m; q = 150 nC = 1.50×10–7 C
a. Volume charge density
q
q
ρ=
=
4 πR3
V
3
=
1.50×10–7 C
C
4 π(0.50m)
3
3
= 2.87×10–7 C3
b. Using Eq. 4.7 with θ = 0, the magnitude E1 of the electric field at r1 = 10 cm = 0.10 m
from the center of the sphere is
(
)
4
3
ρ 3 πr1
qtotal
ρV
E1 =
=
=
∈0A
∈04πr12
∈04πr12
[Grade 12 – General Physics 2]
=
(2.87×10–7 C/m3)(0.10 m)
ρr1
=
= 1080 N/C
3∈0
3(8.8542×10–12 C2/N·m2)
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
At distance r2 = 65 cm = 0.65 m, the total enclosed charge qtotal = 1.50×10–7 C. Using Eq. 4.7
with θ = 0 (since the angle between E and A is always zero in the Gaussian sphere),
qtotal
1.50×10–7 C
= 3191 N/C
E2 =
=
∈0A
(8.8542×10–12 C2/N·m2)[4π(0.65 m)2]
Example: Suppose the sphere in the previous example is conducting. Find (a) the charge density of the
sphere and the magnitude of the electric field at the following distances: (b) 10 cm and (c) 65
cm.
Solution:
q
1.50×10–7 C
a. charge density σ =
= 4.77×10–8 C/m2
=
A
4π(0.50 m)2
b. For a conductor, the charges reside at its surface. Thus, for r = 0.10 m, qtotal = 0.
Therefore, E = 0.
c. For r = 0.65 m, qtotal = 1.50×10–7 C. Using Eq. 4.7,
qtotal
1.50×10–7 C
= 3191 N/C
E=
=
∈0A
(8.8542×10–12 C2/N·m2)[4π(0.65 m)2]
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
APPLY YOUR UNDERSTANDING
Name: ______________________________________
Section: _____________________________
Score:
/20
x
Date: _____________
Directions: Read and analyze the given problems. Write your complete answer/solution on the space
provided and box your final answer. Write your interpretation and do not forget to affix your signature
in each solution. Erasures and alterations are not allowed. Unanswered questions/items will be marked
wrong.
1. As seen in the illustration, two identical metallic spheres are strung as bobs of a simple
pendulum. When a charge of +7.5×10 –9 C is applied to the spheres, each string forms a 5° angle
with the vertical. The length of the string is 0.5 m. Determine the electric force that exists
between the spheres.
2. As indicated, four-point charges (two with q = 2.50×10–6 C and two with q = –2.50×10–6 C) are
placed at the corners of a 1 m square. Calculate the resultant force experienced by the charge
at A as a result of the charges at the other corners of the square.
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
3. Two-point charges q1 and q2 are separated by a distance of 1 m. Find the neutral point if q1 = 3
nC and q2 = 4 nC.
4. When an electron encounters a homogeneous electric field of 5.6 N/C directed vertically down,
it is travelling horizontally at v0 = 3×10–6 m/s. After entering 2μs at the electric field, what are
the horizontal and vertical components of its velocity?
[Grade 12 – General Physics 2]
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Module 4 – Electric Charge, Coulomb’s Law, Electric Fields, and Electric Flux
LEARNING RESOURCES
References:
Cacanindin, Dan Agustin A (2016) General Physics 2. 1253 Gregorio Araneta Avenue,
corner Maria Clara St, Quezon City, 1114 Metro Manila, Philippines: Vibal
Group, Inc.
David, Oliver M (2017) DIWA Senior High School Series: General Physics 2. 120 Rada
Corner Legaspi Street, Legaspi Village, Makati, 1229 Metro Manila, Philippines:
Diwa Learning Systems Inc.
Silverio, Angelina A (2017) Exploring Life through Science Series: Senior High School
General Physics 2. Phoenix bldg., 927, Quezon Ave, Quezon City, 1121 Metro
Manila, Philippines: Phoenix Publishing House.
ABOUT MODULE
Module Author/Curator
Template & Layout Designer
[Grade 12 – General Physics 2]
: Mr. Jomari B. Montalbo
: Mr. Florence A. Somoria
49