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1 Electrostatics Coulombs Law

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ECE

Engr. Marcelo V. Rivera


Electrostatics – electricity at rest
Electrostatic Field - is an electric field
produced by static electric charges.

Two fundamental laws of electrostatic fields:

1) Coulomb's law
2) Gauss's law
Atomic Number – The number of protons
or electrons in a neutral atom of an element.
±𝟏. 𝟔𝟎𝟐 × 𝟏𝟎−𝟏𝟗 =charge per electron or
proton
Excess electrons are placed on a small lead sphere with mass
8.00 g so that its net charge is −3.20 × 10−9 𝐶. (a) Find the
number of excess electrons on the sphere. (b) How many
excess electrons are there per lead atom? The atomic number
of lead is 82, and its atomic mass is 207 g/mol.
a) 2.00 × 1010 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
−13 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
b) 8.59 × 10
𝑎𝑡𝑜𝑚
Lightning occurs when there is a flow of electric charge
(principally electrons) between the ground and a
thundercloud. The maximum rate of charge flow in a lightning
bolt is about 20,000C/s; this lasts for 100𝜇𝑠 or less. How
much charge flows between the ground and the cloud in this
time? How many electrons flow during this time?
2C
1.25 × 1019
Assume a body has a mass of 70kg. Estimate how many
electrons there are in the body. What is the combined charge
of all these electrons?
2.1× 1028 electrons
−3.35 × 109 C
- states that the force 𝑭 between two point charges 𝑸𝟏 and
𝑸𝟐 is:

1. Along the line joining them


2. Directly proportional to the product 𝑸𝟏 𝑸𝟐 of the charges
3. Inversely proportional to the square of the distance 𝑹
between them.
𝑸𝟏 𝑸𝟐
𝑭=𝒌
𝑹𝟐
𝑸𝟏 𝑸𝟐
𝑭=𝒌 𝟐
𝑹
Where:
𝑘 = 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑡𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝟏
𝒌= ≅ 𝟗 × 𝟏𝟎𝟗 𝒎/𝑭
𝟒𝝅𝜺𝟎
𝜀0 = 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑓𝑟𝑒𝑒 𝑠𝑝𝑎𝑐𝑒 𝑖𝑛 𝐹/𝑚 (ability of a region to accept
electricity)
𝟏𝟎 −𝟗
𝜺𝟎 = 𝟖. 𝟖𝟓𝟒 × 𝟏𝟎−𝟏𝟐 ≅ 𝑭/𝒎
𝟑𝟔𝝅
𝑄1 𝑎𝑛𝑑 𝑄2 = 𝑐ℎ𝑎𝑟𝑔𝑒𝑠 𝑖𝑛 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠
𝑅 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑚𝑒𝑡𝑒𝑟𝑠
𝑸𝟏 𝑸𝟐
𝑭=
𝟒𝝅𝜺𝟎 𝑹𝟐
𝑸𝟏 𝑸𝟐
𝑭=
𝟒𝝅𝜺𝟎 𝑹𝟐
If point charges 𝑸𝟏 and 𝑸𝟐 are located at points having position
vectors 𝒓𝟏 and 𝒓𝟐 , then the force 𝑭𝟏𝟐 on 𝑸𝟐 due to 𝑸𝟏 .

𝑸 𝟏 𝑸𝟐
𝑭𝟏𝟐 = 𝟐
𝐚𝑹𝟏𝟐
𝟒𝝅𝜺𝟎 𝑹
𝑸 𝟏 𝑸𝟐
𝑭𝟏𝟐 = 𝟐
𝐚𝑹𝟏𝟐
𝟒𝝅𝜺𝟎 𝑹
𝑅12 = 𝑟2 − 𝑟1
𝑅 = 𝑅12 = 𝑟2 − 𝑟1
𝑅12
a𝑅12 =
𝑅
𝑄1 𝑄2
𝐹12 = 𝑅
4𝜋𝜀0 𝑅 3 12
𝑸𝟏 𝑸𝟐 𝒓𝟐 − 𝒓𝟏
𝑭𝟏𝟐 =
𝟒𝝅𝜺𝟎 𝒓𝟐 − 𝒓𝟏 𝟑
1. 𝑭𝟐𝟏 = −𝑭𝟏𝟐
.

2. Like charges (charges of the same sign) repel each other while
unlike charges attract.
3. The distance 𝑅 between the charged bodies 𝑄1 and 𝑄2 must be large
compared with the linear dimensions of the bodies; that is, 𝑄1 and 𝑄2
must be point charges.

4. 𝑄1 and 𝑄2 must be static (at rest).

5. The signs of 𝑄1 and 𝑄2 must be taken into account in eq.


𝑸 𝟏 𝑸𝟐
𝑭𝟏𝟐 = 𝟐
𝐚𝑹𝟏𝟐
𝟒𝝅𝜺𝟎 𝑹
What is the distance between two charges 𝑞1 = 26𝜇𝐶 and
𝑞2 = −47𝜇𝐶 if the electrostatic force between them is 5.7𝑁?
1.4m
Charges 𝑞1 and 𝑞2 lie on the x-axis at 𝑥 = −𝑎 and 𝑥 = +𝑎.
How must 𝑞1 and 𝑞2 be related for the net electrostatic force
𝑎
on a third charge +𝑄, placed at 𝑥 = + , to be zero?
2
𝑞1 = 9𝑞2
Two small spheres have positive charge, and the combined
charge is 5.0 × 10−5 𝐶. If each sphere is repelled from the
other by a force of 1.0𝑁 when the spheres are 2.0m apart,
what is the charge on each sphere?
𝑞2 = 3.8 × 10−5
𝑞1 = 1.2 × 10−5
It states that if there are N charges 𝑄1 , 𝑄2 , …𝑄𝑁 located, respectively, at
points with position vectors 𝑟1 , 𝑟2 , … 𝑟𝑁 , the resultant force F on a
charge Q located at point r is the vector sum of the forces exerted on Q
by each of the charges 𝑄1 , 𝑄2 , …𝑄𝑁 .
𝑄𝑄1 𝑟 − 𝑟1 𝑄𝑄2 𝑟 − 𝑟2 𝑄𝑄𝑁 𝑟 − 𝑟𝑁
𝐹= 3
+ 3
+ ⋯+
4𝜋𝜀0 𝑟 − 𝑟1 4𝜋𝜀0 𝑟 − 𝑟2 4𝜋𝜀0 𝑟 − 𝑟𝑁 3
𝑵
𝑸 𝑸𝒌 𝒓 − 𝒓𝒌
𝑭= ෍
𝟒𝝅𝜺𝟎 𝒓 − 𝒓𝒌 𝟑
𝒌=𝟏
Point charges 1𝑚𝐶 and −2𝑚𝐶 are located at (3, 2, −1) and
−1, −1,4 , respectively. Calculate the electric force on a 10nC
charge located at 0,3,1 .
𝐹 = −6.507𝐚𝑥 − 3.817𝐚𝑦 + 7.506𝐚𝑧 mN
Five identical 15𝜇𝐶 point charges are located at the center
and corners of a square defined by −1 < 𝑥, 𝑦 < 1, 𝑧 = 0. Find
the force on the 10𝜇𝐶 point charge at 0,0,2 .
𝐹 = 1.072𝐚𝑧 N
Thank You!

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