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Phys 2180 Lecture (1)
Electrical Charges and Coulomb’s Law
1
Structure of Matter
• Fundamental building blocks of the
matter are atoms.
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-
-
+ +
+
+ +
+ +
-
-
2
Structure of Matter
• Neutral atom – electron = Positive ion
-
-
-1.602  10
1 electron charge

C
+ +
-
-
+
+ +
+ +
-
-19
-
3
Structure of Matter
• Neutral atom + electron = negative ion.
-
-
-
+ +
+
+ +
+ +
-
-
4
Electrical Charges
• Electrostatics is the study of electric charge at rest.
• Electric charge is a fundamental property of matter.
– Two types of electric charges
• Positive charge - every proton has a single positive charge.
• Negative charge - every electron has a single negative
charge.
5
Fundamental Charges
Note that the electron and proton both have the samemagnetud
charge, with the electron being negative and the proton being
positive. This amount of charge is often called the
electronic charge, e. This electronic charge is generally considered
a positive value (just like g in gravity). We add the negative sign
when we need to:
qe = -e; qp = +e.
6
•Electrostatics is the study of electric charge at rest.
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Electric charge
Two different kinds of charge, positive and negative.
Electricity, with attractive and repulsive forces, needs two kinds of charge.
•
Electric charge is measured in coulombs (C)
Properties of electric charge
1.Two kinds of charges occur in nature,
o like charges repel one another,
o and unlike charges attract one another.
2. Charge is conserved.
3. Charge is quantized.
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• An object becomes electrostatically
charged by:
1. Friction, which transfers electrons
between two objects in contact,
1. Contact with a charged body which
results in the transfer of electrons,
2. Induction which produces a charge
redistribution of electrons in a material.
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Types Of Forces
• There are only four fundamental forces of nature.
• Gravitational Force
• Electromagnetic Force
• Strong Nuclear Force
• Weak Nuclear Force
10
Electric Field - Definition
• the electric field vector E at a point in space is defined as the electric force
Fe acting on a positive test charge q placed at that point divided by the
test charge:
F = K |q||q|/r2 = magnitude of the electric force
k = Coulomb's constant = 8.9875 x 109Nm2/c2
E = K |q| / r2
Note that since F is a vector and q is a scalar,
E must be a vector.
the units of Electric Field in SI units of newton's per coulomb (N/C)
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Electric Field Lines
• The electric field lines for a point charge.
• (a) For a positive point charge, the lines are directed radially outward.
• (b) For a negative point charge, the lines are directed radially inward.
• Or the electric field lines extend away from positive charge (where they
originate) and towards negative charge (where they terminate)
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Electric Force: Coulomb’s Law
13
Electric Force: Coulomb’s Law
• we can express Coulomb’s law as an equation giving the
magnitude of
• the electric force (sometimes called the Coulomb force)
between two point charges:
where ke is a constant called the Coulomb constant, q1 and q2 the
charges, r is the distance between the charges.
We also found that the force decreases with distance between the
charges just like gravity.
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Electric Force: Coulomb’s Law


The value of the Coulomb constant depends on the choice of units. The SI unit of charge is the
coulomb (C).
The Coulomb constant ke in SI units has the value
where the constant epsilon is known as the permittivity of free space
ε0 ≅ (8.854 187 817)*10-12 C2/(N m2)
1 Coulomb = 106 microCoulomb
1 Coulomb = 109 nanoCoulomb
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Problem (1)
• The average distance r between the electron and the proton in the hydrogen
atom is 5.3x10-11 m. (a) What is the magnitude of the average electrostatic
force that acts between these two particles? (b) What is the magnitude of
the average gravitational force that acts between these particles?
• Solution
• the electrostatic force,
𝒒𝟏 𝒒𝟐
𝑭𝒆 = 𝑲 𝟐
𝒓
2
(8.99 X 109N.m /C2)(1.60X l0−19C)2
=
(5.3 X 10−11 m)2
= 8.2 X 10-8 N
𝒎𝟏 𝒎𝟐
• the gravitational force,
𝑭𝑮 = 𝑮
𝒓𝟐
=
(6.67 X 10−11N.m2/kg2)(9.11X l0−31kgX1.67X l0−27kg)
(5.3 X 10−11 m)2
16
-57
= 1.91 X 10 N
Problem (2)
• The nucleus of an iron atom has a radius of about 4x10-15 m and
contains 26 protons. What repulsive electrostatic force acts between two
protons in such a nucleus if a distance of one radius separates them?
• Solution
𝒒𝟏 𝒒𝟐
𝑭𝒆 = 𝑲 𝟐
𝒓
2
(8.99 X 109N.m /C2)(1.60X l0−19C)2
=
(4 X 10−15 m)2
= 14 N
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Problem (3)
Two balloons with charges of +3.37 µC and -8.21 µC attract each other with a force of
0.0626 N. Determine the separation distance between the two balloons.
Solution:
Given:
q1 = +3.37 µC = +3.37 x 10-6 C , k= 8.99 X 109N.m2/C2
q2 = -8.21 µC = -8.21 x 10-6 C
Fe = -0.0626 N (use a - force value since it is attractive)
𝑞1 𝑞2
𝐹𝑒 = 𝐾 2
𝑟
𝑞1 𝑞2
2
𝑟 =𝐾
𝐹𝑒
r=
=
𝐾
Find:
r = ???
𝑞1 𝑞2
𝐹𝑒
(8.99 x 109 Nm2/C2) (−8.21 x 10−6 C)(+3.37 x 10−6 C)
(−0.0626 N)
r = +3.98 m2
=1.99 m
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Problem (4)
Three Charges on a Line
Determine the magnitude and direction of the net force on q1.
Problem (4)
F12  k
F13  k
q1 q2
r
2
q1 q3
r
2
8.99 10

9
8.99 10







N  m 2 C 2 3.0 10 6 C 4.0 10 6 C
0.20m 2
9
N  m 2 C2 3.0 106 C 7.0 106 C
0.15m2
 

F  F12  F13  2.7 N  8.4N  5.7N
 2 .7 N
 8.4 N
18.5 Coulomb’s Law
Find the net force on q1
F  F12 sin   F13 cos 
18.5 Coulomb’s Law
F12  k
q1 q2
F13  k
q1 q3
r
r
2
2

8.99 10

8.99 10






9
N  m 2 C 2 4.0 10 6 C 2 6.0 10 6 C 2
0.15m 2
9
N  m 2 C 2 4.0 10 6 C 2  5.0 10 6 C 2
0.10 m 2
  9.59 N

 17.98 N
F  F12 sin   F13 cos    9.59 sin 73  17.98 cos 73  14.43 N