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General Physics 2 module 1 Part 1
college of education (Catanduanes State University)
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Christian Polytechnic Institute of Catanduanes, Inc.
Francia Virac, Catanduanes
GENERAL PHYSICS 2
SENIOR HIGH SCHOOL
MODULE 1
ELECTRIC CHARGE,
COULOMB’S LAW,
ELECTRIC FIELDS
AND
ELECTRIC FLUX
(Part 1)
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Lesson 1: ELECTRIC CHARGE
We can trace all electrical effects to electrons
and protons inside every atom. This is because these
particles have a property called electric charge.
The electrons are negative and surround a
dense, positive nucleus. Protons, which are positive,
and neutrons reside in this nucleus. Neutrons are
neutral and do not participate in electrostatic
interactions.
Just like mass, charge is fundamental property of subatomic particles. The
smallest amount of charge is called the elementary charge, indicated universally
by the symbol the element charge has magnitude of
1.60 x 10-19 C
Particles
proton
Electron
Neutron
Mass (in kg)
9.1093897 x 10-31 kg
1.6726231 x 10-27 kg
1.6749286 x 10-27 kg
Charge
(in Coulomb (C))
+1.60217733 x 10-19 C
-1.60217733 x 10-19 C
None
Location in atom
Nucleus
Outside nucleus
nuclues
The unit C stands for coulomb, named after French physicist Charles Agustin
Coulomb.
Example 1.1
How many electrons must an object lose so that it has a net positive charge of
+1C?
SOLUTION
If the object loses one electron, its charge is +1.60 x 10 -19 C; if two
electrons are lost, the object gains has a net charge of
2 x 1.60 x 10 -19 = 3.20 x 10 -19C
And so forth. If we let n be the number of electrons lost and e the charge of each
electron, then the total charge Q is
Q = ne
Thus, the number of electrons the object must lose so that it has a net positivity
charge of +1C is
or more than six quintillion electrons!
This problem simply shows us that a charge of 1C is enormous.
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Example 1.2
Aluminum has atomic number 13. This means that it has 13 electrons and 13
protons. What is the total charge of all electrons in an aluminum atom.
SOLUTION
We are given the number of electrons n. the charge of one electron is
-1.60 x 10 -19 C. So if we have 13 electrons, each of which has a charge of
-1.60 x 10 -19 C, then the total charge is:
Q = ne = 13 (-1.60 x 10
-19
C) = -2.08 x 10
-18
C
Thirteen electrons altogether still have very small charge
Conductor and Insulator
In certain materials such as aluminum, copper, and other metlas, the outermost or
valence electrons are free to move around the entire materials, such materials are
classified as conductor. In other materials, such as glass and wood, the electrons are
more lightly bound to atom that they do not easily move around. These materials are
insulators. good conductor are poor insulators poor conductors are good insulators. most
metals are conductors while most nonmetals are insulators.
In a conductor, the outermost electrons are farther away from the nucleus that
they are more weakly bound to it. When atoms of a conductor form a bulk material, their
outer electrons are no longer bound to the atoms but are free to float over the material.
These are called free electrons. When energy is applied to these electrons, they can
move in a more organized way, producing electric current.
Free electrons can also be moved from one materials to another in process called
charging. Removing an electron from an atom creates a positive ion.
Charging
In general, a material can be give a net charge by adding or removing electrons.
There are several ways of doing this:
1. Using friction. By rubbing thighs together, electrons can peel off one
material and remain in the other.
2. By touching. When a charged object comes into contact with another
object, electrons are transferred, thereby charging the second object.
3. By induction. In this method, there is no actual contact between the
charged object and that which is being charged. In this way, the charged
object does not lose its charge to the object that gets charged in the process.
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Using Friction
By touching
By induction
An easy way to see charging by induction is by running a comb through your
hair a few times (or rubbing it repeatedly with dry hands) and then holding it up over
some small bits of paper.
Lesson 2: COULOMB’S LAW
In the previous lesson, we saw that like charges repel while unlike charges
attract. This suggest the presence of the electrostatic force. This force is either a
force of attraction between positive and a negative charges or a force of repulsion
between two like charges. Electrostatic force holds the atom together. The positive
nucleus attracts the negative electrons around it in a similar way that the Earth
and other planets are held in orbit around the Sun by gravity.
And this is not where the similarity ends. Recall that the force of gravity
between two masses m1 and m2 is given by the equation:
In which G is the gravitational constant and r is the distance between the
centers of the masses.
The electrostatic force between two charges Q1 and Q2 is given by a very
similar equation. The magnitude of the electrostatic force Fe is:
Where Fe is the electric force
k is the Coulomb constant
k = 8.9875 x 109 N · m2 / C2 or simplified by 9 x 109 N · m2 / C2
Q is the charge on object
and r is the distance between the charges
Take Note: If you encounter the nC (nanocoulomb) will show 10-9 when you convert
the 109 coulomb.
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This equation is called Coulomb’s law. It tells us that the force in newton’s
between two charges is directly proportional to the product of the magnitude of the
charges and inversely proportional to the square of the distance between them.
Note:
Two things in Coulomb’s law. First, the absolute value sign means that the result is
always positive. This is, after all, the magnitude of the force, so it has to be positive.
Then, we see a constant k. This is the electrostatic constant. It is related to some
fundamental constant of nature as follows:
k=
the constant ε0 is called the permittivity of free space and is equal to 8.85 x 10-12
coulomb squared per newton meter squared (C2/N · m2). The permittivity of a medium is
a measure of its ability to store electrical charge.
Substituting the constant needed to get the value of k, we get an approximately
value of:
k==k=
9
= 9 x 10 N · m2 / C2
Example 2.1
Two charge spheres Q1 = -2.00 x 10-10 C and Q2 = -5.0 x 10-8 C are held fixed
at positions 5.00 cm apart.
1. Calculate the magnitude of each electrostatic force between the two
spheres. Are the force attractive or repulsive?
2. If the first sphere is set loose, what will be its resulting acceleration? It
has a mass of 5.00 x 10-3 kg.
Solution
1. To solve for the magnitude of the electrostatic force Fe, we substitute
the values into Equation 1:
= 9 x 109 N · m2/ C2
= 3.6 x 10-5 N
The electrostatic forces are repulsive because both charges are
negative.
2. If the first sphere were to be released, then it will accelerate at a rate
given by Newton’s second law of motion:
Direction of Electrostatic Force
The calculation we made ( Example 2.1) does not tell whether the force is
directed toward the north, south, west, and east. The only way to find out is to sketch the
problem out.
Let us assume that given positive charge Q1 and negative charge Q2,
separately by distance r on a horizontal line. We already know that the force between
them is attractive. In addition, the attraction is mutual. This means that Q1 pulls on Q2
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and at the same time, Q2 also pulls on Q1 with the same amount (or magnitude) of force.
This implies that there are two forces present.
Let us design the force exerted by Q1 and Q2 as F12 because this force is
attractive, the said force is directed toward Q1 (to the left), acting on Q2. At the same
time, Q2 is attracting Q1 with the Force F21. This force is directed toward Q2 and is acting
on Q1. Therefore to the right
Q
1
+
F1
2
F2
1
Q
2
-
F2
1
Q
1
Q
2
+
+
r
F1
2
r
Picture 1:Attractive forces between two unlike charge
Picture 2:Repulsive forces between two like charges
On the other hand, if two like charges are involved, they push each other away. So
in each case, the force is directed away from whichever force is exerting it. For example,
that charges Q1 and Q2 are both positive (Picture 2), then Q1 push away. So the force F12
acts on Q2 and is directed to the right. And at the same time, Q2 pushes Q1 away. So the
force F12 acts on Q1 and is directed to the left.
Compare the attractive force (Picture 1) with the repulsive forces in (Picture 2), and
remember that electrostatic forces always come in pairs that are equal in magnitude and
opposite direction.
Example 2.2
Two charge spheres Q1 = -2.00 x 10-5 C and Q1 = -1.50 x 10-5 C are 10.0 cm
apart. Calculate the force that each charge exerts on a third charge Q 3 = 5.00
x 10-6 C as shown below.
Q
1
Q
2
Q
3
-
-
+
10.0cm
6.00cm
Figure 1.1
Solution
There are two things we need to do: calculate the magnitude of the
force exerted by Q1 on Q3 (we call it F13) and the magnitude of the force
exerted by Q2 on Q3 (we call it F23).
While we solve for the magnitude of F13, we will ignore F23. So leaving
out Q2, our problem now looks like these:
Q
1
F1
3
-
Q
3
+
16.0 cm
Figure 1.2
We are not connected about the force exerted by Q3 on Q1 so we will
leave that out too. Using r13 = 16.0 cm = 0.16 m, we can get the
magnitude of F13:
= 9 x 109 N · m2/C2
= 35.1 N
This force is directed to the left
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To find F23, we ignore Q1 (Figure 1.3). using r23 = 6 cm = 0.06 m, we
can get the magnitude of F23:
F2
3
Q
2
Q
3
-
+
Figure 1.3
6.00cm
= 9 x 109 N · m2/C2
= 187.5 N
This force is also to the left.
Q
3
F2
3
The Superposition Principle
+
Figure 1.4
F1
3
Electrostatic force, being a vector, ca be added just like other forces. In the
previous example, Q3 is simultaneously attracted by two charges. The end result is that
Q3 experiences the combined effect of the two attractive forces, which we can calculate
using vector addition. This is the superposition principle. We can state the superposition
principle as:
The net electrostatic force on a charged particle is equal to the vector sum of the
electrostatic force exerted by each point charge on that particle.
Example 2.3
Going back to Example 2.2, find the magnitude and direction of the net
force acting on Q3.
Solution
As shown in Figure 1.4, find the magnitude and direction of the net
force acting on Q3. Since both forces are to the left, we just add them
algebraically (i.e., use negative signs for forces directed to the left) to
get the net force on Q3.
Fnet = -35.1N – 187.5 N = -222.6 N
Thus, the magnitude of the force is 222.6N going to the left.
Example 2.4
Figure 1.5 shows three charges arranged on the x- and y- axis where Q 1 =
3µC, and Q2 = 7µC, and Q3 = 4µC. what is the net force acting on Q1?
Q1
+
0.3 m
Figure 1.5
Q3
Q2
+
0.4m
Solution
First, we draw the forces acting only on Q1. There is an upward force F21
on it. This is the repulsive force exerted by Q2. There is a force along a
straight line drawn from Q1 and Q3. This is F31 as shown in Figure 1.6.
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To calculate the force F21, we need to determine that r21 = 0.3 m.
We are given Q1 = 3 µC = 3 x 10-6C and Q2 =7µC = 7 x 10-6C.
To calculate F31, we need to determine r13, the distance between Q1
and Q2, which is also the hypotenuse of the right triangle. Using
the Pythagorean theorem:
Now we can get the force F31.
We must just calculate the magnitude of the two forces on Q1. To
get the magnitude of the resultant, we can use any convenient
vector addition method. Let us use the cosine law. First, we have
to draw the triangle defined by the three forces.
The angle opposite Fnet is 53.1°. We get this by doing back to
Figure 1.6 (it is a 3-4-5 triangle). Now we can apply the cosine law:
What is 3-4-5 triangle and
where is 53.1° came from?
Look for “Asking why” below
=1.87 N
Finally, we need to get another internal angle in the triangle
defined by the three forces. The angle opposite F21, call it , will
give us the direction Fnet.
Using the sine law:
the angle is:
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Thus, the net force acting on Q1 has a magnitude of 1.87 N
and is directed 26.1° above the x-axis.
Asking why?
What is 3-4-5 triangle?
 it is to determine the absolutely certainly of an angle 90 degrees.
 This rule says that if one side of a triangle measures 3 and the adjacent
side measures 4, then the diagonal between those two points must
measure.
 It has a ratio of the sides in whole numbers called Pythagorean Triples.
 It is classified as a scalene triangle because all its three side lengths and
internal angles are different.
That’s why there is 53.13°
53.13°
3
came
5
90°
36.87°
4

The ratio is
 Side 1: 3n
 Side 2: 4n
 Hypotenuse: 5n
We can prove this by using the Pythagorean Theorem as
follows:
a2 + b2 = c2
32 + 42 = 52
or in right triangle 3 + 4 = 5
9 + 16 = 25
53.13° + 36.87° = 90°
25 = 25
90° = 90°
ACTIVITY 1
a rubber balloon has a charge of 1.4 x 10-16 C. how many electrons has it
lost?
ACTIVITY 2
Charge Q1 = 2.00 nanocoulombs (nC) is placed 15.00 cm away from the
origin to the left. Charge Q2 = -3.00 nC is placed 15.00 cm from the
origin to the right. A third charge Q3 = 1.00 nC is placed 5.00 cm to the
right of the origin.
1. Sketch the position of the three charges. Show the distance (in
meters) between Q1 and Q3, and between Q2 and Q3. Show also
the force vectors acting on Q3; label them F13 and F23.
2. What is the magnitude of F13?
3. What is the magnitude of F23?
4. What is the magnitude and direction of the net force on Q3?
5. Calculate the net force on Q1 due to the forces exerted on it by
Q2 and Q3.
6. Calculate the net force on Q2 due to the forces exerted on it by
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Q1 and Q3.
Write you’re Answer on a Clean Sheet of paper with your Name, Strand and
name of Teacher.
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