- The document describes objectives and concepts related to electrostatics, including charging by rubbing and induction, conservation of charge, and Coulomb's law. - Coulomb's law states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. - Examples show how to calculate the electric force between two or more point charges using Coulomb's law and calculate unknown values like charge, distance, or force.

1. ELECTRO
STATICS

2. OBJECTIVES
At the end of this lesson, you will be able to:
• Describe using a diagram charging by rubbing and charging by induction
• Explain the role of electron transfer in electrostatic charging by rubbing
• Describe experiments to show electrostatic charging by induction
• State that there are positive and negative charges, and that charge is measured in
coulombs
• Predict charge distributions, and the resulting attraction or repulsion, in a system of charged
insulators and conductors
• Calculate the net electric force on a point charge exerted by a system of point charges

3. Electricity and magnetism
Electricity and magnetism play a profound role in almost all aspects of
life. Applications of electricity and magnetism provide tools, light, warmth,
transportation, communication, and even entertainment.

4. ELECTROSTATICS
- science that deals with phenomena due to attractions
or repulsions of electric charges but not dependent
upon their motion (the science that deals with the study
of the effects of charges at rest.)
https://www.youtube.com/watch?v=ViZNgU-Yt-Y

5. ELECTRIC CHARGE
- the basic element of electricity. It is a fundamental property of
matter.
PROTON POSITIVE CHARGE + 1.6021765 x 10^19 C
NEUTRON NO CHARGE 0 C
ELECTRON NEGATIVE CHARGE − 1.6021765 x 10^19 C

6. LAW OF CONSERVATION OF CHARGES
• A charge is neither created nor destroyed during
this charging process; it is simply transferred from
one object to the other object in the form of
electrons.
• The overall charge on the system of two objects is
the same after the charging process as it was before
the charging process.

7. LAW OF CHARGES
“Like charges repel, unlike charges
attract each other.”

8. METHODS
OF
CHARGING

9. - means gaining or losing electron.
CHARGING
Note: Electron (Negative charges) can move freely from one object to another.
Protons and neutrons never move from object to object.

10. • charging process that involves
rubbing of one particle on
another resulting in electrons
moving from one surface to
another.
• This method is useful for
charging insulators.
CHARGING BY FRICTION

11. • charging process that involves
touching a charged particle to a
conductive material. This way,
the charges are transferred
from the charged material to the
conductor.
• This method is useful for
charging conductors.
CHARGING BY CONDUCTION

12. • charging process that charges an object
without actually touching the object to any
another charged object.
• The charging by induction process is
where the charged particle is held near an
uncharged conductive material that is
grounded on a neutrally charged material.
• The charge flows between two objects
and the uncharged conductive material
develop a charge with opposite polarity.
CHARGING BY INDUCTION

13. COULOMB’S
LAW

14. COULUMB’S LAW
Charles-Augustin de Coulomb
“The magnitude of the
electric force between two
point charges is directly
proportional to the product
of their charges and
inversely proportional to the
square of the distance
between them. ”

15. COULUMB’S LAW
SI unit for charge: Coulombs(C)
Coulomb's Law constant for air is
9.0x109 (Nm2/C2)

16. EXAMPLE 1
A negative charge of –2.0 x 10–4 C and a positive charge of 8.0 x 10–4 C are separated by
0.30 m. What is the force between the two charges?
Formula:
Given:
q1 = –2.0 x 10–4 C
q2 = 8.0 x 10–4 C
d = 0.30m
k = 9.0x109
Unknown:
F = ?
Solution:
𝐹 = −16000 𝑁

17. EXAMPLE 2
A negative charge of –2.0 x 10–4 C exerts a repulsive force of 65 N on a second charge
0.050 m away. What is the magnitude of the second charge?
Formula:
Given:
q1 = –2.0 x 10–4 C
d = 0.050m
F = 65N
k = 9.0x109
Unknown:
q2 = ?
Solution:

18. EXAMPLE 3
Two positive charges of 6.0 µC are separated by 0.50 m. What force exists between the
charges?
Formula:
Given:
q1 & q2 = 6.0 µC/6.0 x 10–6 C
note: 1C = 1000000μC.
d = 0.50m
k = 9.0x10
Unknown:
F = ?
Solution:
"The force is repulsive since the two charges are
both positive charges."

19. EXAMPLE 4
A force of –4.4 x 103N exists between a positive charge of 8.0 x 10-4C and a negative charge
of –3.0 x 10-4C. What distance separates the charges?
Formula:
Given:
𝐹=−4.4×10−3𝑁
𝑞2=−3.0×10−4𝐶
𝑞1=8.0×10−4𝐶
𝑘=9×109
Unknown:
r = ?
Solution:

20. EXAMPLE 5
Two identical positive charges exert a repulsive force of 6.4 x 10–9N when separated by a
distance of 3.8 x10–10m. Calculate the charge of each.
Formula:
Given:
𝐹=6.4×10−9𝑁
𝑘=9×109
𝑟=3.8×10−10𝑚
Unknown:
q= ?
Solution:

21. EXAMPLE 6
A –2.0 x 10–4 C charge is placed at the origin, A -7.0 x 10–5 C charge is place at x = 3m and a
2.0 x 10–4 C charge is place at x= -5m
a. What is the net electric force on the –2.0 x 10–4 charge?
b. What is the net electric force acting on the -7.0 x 10–5 C charge?
-7.0 x 10–5 C
2.0 x 10–4 C –2.0 x 10–4 C
3m
5m

22. EXAMPLE 6
A –2.0 x 10–4 C charge is placed at the origin, A -7.0 x 10–5 C charge is place at x = 3m and a
2.0 x 10–4 C charge is place at x= -5m
a. What is the net electric force on the –2.0 x 10–4 charge?
-7.0 x 10–5 C
2.0 x 10–4 C –2.0 x 10–4 C
3m
5m
𝐹2,1 = 9𝑥109
2𝑥10−4
−2𝑥10−4
|
52
𝐹2,3 = 9𝑥109
| −7𝑥10−5
−2𝑥10−4
|
32
𝐹2,1 = −14.4𝑁 𝐹2,3 = −14𝑁
𝐹𝑛𝑒𝑡 = 𝐹2,1 + 𝐹2,3 𝐹𝑛𝑒𝑡 = −14.4N +(−14𝑁) = −28.4𝑁

23. EXAMPLE 6
A –2.0 x 10–4 C charge is placed at the origin, A -7.0 x 10–5 C charge is place at x = 3m and a
2.0 x 10–4 C charge is place at x= -5m
b. What is the net electric force acting on the -7.0 x 10–5 C charge?
-7.0 x 10–5 C
2.0 x 10–4 C –2.0 x 10–4 C
3m
5m
𝐹3,2 = 9𝑥109
| −2𝑥10−4
−7𝑥10−5
|
32
𝐹3,1 = 9𝑥109
| 2𝑥10−4
−7𝑥10−5
|
82
𝐹3,2 = 14𝑁 𝐹3,1 = −1.97
𝐹𝑛𝑒𝑡 = 𝐹3,2 + 𝐹3,1 𝐹𝑛𝑒𝑡 = 14N -1.97𝑁 = 12.03𝑁

24. EXAMPLE 7
Consider the three point charges located at the corners of a right triangle as shown in
the figure, where q1 = q3 = 5.00 µC, q2 = - 2.00 µC and a = 0.100 m. Find the resultant
force exerted on q3.

25. EXAMPLE 7
Consider the three point charges located at the corners of a right triangle as shown in
the figure, where q1 = q3 = 5.00 µC, q2 = - 2.00 µC and a = 0.100 m. Find the resultant
force exerted on q3.
Given:
𝑞1=𝑞3= 5.00𝑥10−6
𝐶
𝑞2=−2.00𝑥10−6
𝐶
a=0.100𝑚
𝑘=9×109
Formula:

26. EXAMPLE 7
Consider the three point charges located at the corners of a right triangle as shown in
the figure, where q1 = q3 = 5.00 µC, q2 = - 2.00 µC and a = 0.100 m. Find the resultant
force exerted on q3.
Force Fcosθ(xcom) Fsinθ(ycom)
𝐹1,3 11.25cos45 = 7.95 11.25sin45 = 7.95
𝐹2,3 9cos180 = -9.0 9sin0 = 0
෍ 𝑥/𝑦 ෍ 𝑥 = −1.05 ෍ 𝑦 = 7.95

27. EXAMPLE 8
Solve for the force exerted on charge +13 µC

28. EXAMPLE 8
Solve for the force exerted on charge +13 µC
𝐹1,2 = 9𝑥109
(13𝑥10−6
)(15𝑥10−6
)
0.0752
𝐹1,2 = 312𝑁
𝐹1,3 = 9𝑥109
(13𝑥10−6
)(11𝑥10−6
)
0.102
𝐹1,2 = 128.7𝑁

29. Force Fcosθ(xcom) Fsinθ(ycom)
𝐹1,2 0 312sin270
𝐹1,3 128.7cos0 0
෍ 𝑥/𝑦 ෍ 𝑥 = 128.7 ෍ 𝑦 = −312
𝐹𝑛𝑒𝑡 = 128.72 + −3122
𝐹𝑛𝑒𝑡 = 337.50𝑁
θ = tan−1(
−312
128.7
)
θ = −67.58deg
EXAMPLE 8
Solve for the force exerted on charge +13 µC