Download 16-5 and 16-6 Coulomb`s Law

Chapter 16
Electric Charge and
Electric Field
Objectives: The students will be able to:
Apply Coulomb's law to determine the magnitude of the electrical
force between point charges separated by a distance r and state
whether the force will be one of attraction or repulsion.
Two types of charge:
Positive Charge: A shortage of electrons.
Negative Charge: An excess of electrons.
Conservation of charge – The net charge of a
closed system remains constant.
Nucleus
-
-
n + n
+ +
n
+
n
n
+
+ n
-
-
-
-
Negative
Neutral Atom
Atom
Positive
Atom
Number
Numberof
ofelectrons
electrons><=Number
Numberof
ofprotons
protons
Number
of
electrons
Number
of
protons
-19
-2e
=
-3.2
x
10
CC
+2e = +3.2 x 10-19
Inquiry Activity
Electric Field Hockey
phET
Electric Forces
Like Charges - Repel
F
+
+
Unlike Charges - Attract
-
F
F
+
F
16.5 Coulomb’s Law
Experiment shows that the electric force
between two charges is proportional to the
product of the charges and inversely
proportional to the distance between them.
Coulomb’s Law – Gives the electric force
between two point charges.
q1q2
F k 2
r
Inverse Square
Law
k = Coulomb’s Constant = 8.988x109 Nm2/C2
q1 = charge on mass 1
q2 = charge on mass 2
r = the distance between the two charges
The electric force is much stronger than the
gravitational force.
16.5 Coulomb’s Law
Coulomb’s law:
(16-1)
This equation gives the magnitude of
the force.
16.5 Coulomb’s Law
The force is along the line connecting the
charges, and is attractive if the charges are
opposite, and repulsive if they are the same.
16.5 Coulomb’s Law
Unit of charge: coulomb, C
The proportionality constant in Coulomb’s
law is then:
Charges produced by rubbing are
typically around a microcoulomb:
Coulomb's Law
• The force between
two charges gets
stronger as the
charges move closer
together.
• The force also gets
stronger if the
amount of charge
becomes larger.
Coulomb's Law
• The force between
two charges is
directed along the
line connecting their
centers.
• Electric forces
always occur in pairs
according to
Newton’s third law,
like all forces.
Coulomb's Law
• The force between
charges is directly
proportional to the
magnitude, or amount,
of each charge.
• Doubling one charge
doubles the force.
• Doubling both charges
quadruples the force.
Coulomb's Law
• The force between charges is
inversely proportional to the
square of the distance between
them.
• Doubling the distance reduces
the force by a factor of 22 = (4),
decreasing the force to onefourth its original value (1/4).
• This relationship is called an
inverse square law because
force and distance follow an
inverse square relationship.
16.5 Coulomb’s Law
Charge on the electron:
Electric charge is quantized in units
of the electron charge.
Unit of charge is a Coulomb (C)
Example 1
Two charges are separated by a distance r and have a force
F on each other.
qq
F k
F
1 2
2
r
q2
q1
F
r
If r is doubled then F is :
¼ of F
If q1 is doubled then F is :
2F
If q1 and q2 are doubled and r is halved then F is : 16F
Example 2
Two 40 gram masses each with a charge of 3μC are
placed 50cm apart. Compare the gravitational force
between the two masses to the electric force
between the two masses. (Ignore the force of the
earth on the two masses)
3μC
40g
3μC
40g
50cm
m1m2
Fg  G 2
r
 6.67 10
11
(.04)(.04)
2
(0.5)
 4.27 10
q1q2
FE  k 2
r
6
6
(
3

10
)(
3

10
)
9
 9.0 10
(0.5) 2
13
N
 0.324 N
The electric force is much greater than the
gravitational force
Homework
Problems page 465
#s 1, 3, 5, 7, 10
Objectives: The students will be able to:
• Apply Coulomb's law to determine the magnitude of the
electrical force between point charges separated by a
distance r and state whether the force will be one of
attraction or repulsion.
• For practical reasons, the coulomb is defined using
current and magnetism giving
k = 8.988 x 109 Nm2/C2
• Permittivity of free space
e0 =
Then
1
4p k
= 8.85´10
1 q1q2
F=
2
4pe0 r
-12
2
C / Nm
2
16.5 Coulomb’s Law
The proportionality constant k can also be
written in terms of
, the permittivity of free
space:
(16-2)
Coulombs Law
Lab Experiment
In 1785 Charles Augustin Coulomb reported in the
Royal Academy Memoires using a torsion balance
two charged mulberry pithballs repelled each other
with a force that is inversely proportional to the
distance.
kq1q 2
F= 2
r
q1
r
where k=8.99*109 Nm2/C2 in SI unit
k ~ 1010 Nm2/C2
+
q2
Point charges
Spheres same
as points
+
Repulsion
+
-
--
Attraction
Repulsion
Summer July 06
PHYS632 E&M
24
Superposition of electric forces
Net force is the vector sum of forces from each
charge
F3
q1
q2
q
q3
F2
F1
Net force on q:
F = F1 + F2 + F3
F
Principle of Superposition
Three charges In a line
•
In the previous example we tacitly assumed that the forces between nuclei
simply added and did not interfere with each other. That is the force between
two nuclei in each penny is the same as if all the others were not there. This
idea is correct and is referred to as the Principle of Superposition.
•
Example of charges in a line
x
2
1
–
3
Three charges lie on the x axis: q1=+25 nC at the origin, q2= -12 nC at x =2m, q3=+18
nC at x=3 m. What is the net force on q1? We simply add the two forces keeping track
of their directions. Let a positive force be one in the + x direction.
æ q2
q3 ö
F=-kq1 ç
+
è (2m)2 (3m)2 ÷ø
(
= - 10
10 Nm 2
C2
= 2.5 ´ 10 -7 N
)
æ -12 ´ 10 -9 C 18 ´ 10 -9 C ö
(25 ´ 10 C) ç
+
2
(3m)2 ÷ø
è (2m)
-9
Force from many charges
Q2
F41
Q1
-
F21
+
F31
Principle of
superposition
-
+
Q4
Q3
Force on charge is
vector sum of forces
from all charges
F1  F21  F31  F41
Example 16-3 p.448
Three charges in a line. Three charged particles
are arranged in a line, as shown below. Calculate the
net electrostatic force on particle 3 (the -4.0μC on the
right) due to the other two charges.
Example 16-3 p.448
Three charges in a line. Three charged particles are arranged in a line, as shown
below. Calculate the net electrostatic force on particle 3 (the -4.0μC on the
right) due to the other two charges.
16.5 Coulomb’s Law
Coulomb’s law strictly applies only to point charges.
Superposition: for multiple point charges, the forces
on each charge from every other charge can be
calculated and then added as vectors.
16.6 Solving Problems Involving
Coulomb’s Law and Vectors
The net force on a charge is the vector
sum of all the forces acting on it.
16.6 Solving Problems Involving
Coulomb’s Law and Vectors
Vector addition review:
Example 3
Three charged objects are placed as shown. Find the net
force on the object with the charge of -4μC.
F k
- 5μC
45º
20cm
202  202  28cm
q1q2
r2
(5 106 )(4 106 )
F1  9 10
 4.5N
2
(0.20)
9
(5 106 )(4 106 )
F2  9 10
 2.30 N
2
(0.28)
9
F1 45º
- 4μC
5μC
20cm
F2
F1 and F2 must be added together as vectors.
F1
2.3cos45≈1.6
45º
F2
2.3sin45≈1.6
F1 = < - 4.5 , 0.0 >
+ F2 = < 1.6 , - 1.6 >
Fnet = < - 2.9 , - 1.6 >
- 1.6
- 2.9
29º
θ
3.31
Fnet  2.9 2  1.6 2  3.31N
  1.6 

  tan 

29

  2.9 
1
3.31N at 209º
Homework
Problems on pages 465 and 465
#s 12, 13, 18
Closure
Kahoot