Transcript Coulomb’s Law

Forces and Fields Lesson 4
Charles Coulomb
 1736 -1806
 Discovered the inverse square law which states that as
you move charges apart the force between them
decreases exponentially
1
Fe
r2
 Also showed that the force between two charged
objects is proportional to the product of the charges
Fe  q1q2
 Unit of electrical charge is named after him
The Torsion Balance
 Coulomb used a
torsion balance
(Physics 20 – universal
gravitational
constant) to
determine “k” a
universal value
called Coulomb’s
Constant.
 http://www.youtube
.com/watch?v=_5Vp
Ije-R54
By measuring the amount of twist
when the charges repelled each other,
the force could be measured and the
constant could be determined.
Coulomb’s Law describes the electrostatic forces
between two charged objects.
Where:
kq1q2
Fe  2
r
 Fe is the electrostatic force
2
N m
 k is Coulomb’s Constant
(8.99  10
)
2
C
 q is the charge in Coulombs (C)
9
( the unit for charge is 1 C which equals the charge of
6.25 10 electrons)
18
 r is the distance separating the two charges (m)
 Notice the similarities to Newton’s Law of Universal
Gravitation
Gm1m2
Fg 
2
r
 This shows only attraction while Coulomb’s Law can be
used for both attractive and repulsive situations
 Gravity is a weak force requiring large masses
 Electric forces are very large in comparison and are
effective with small charges
Using Coulomb’s Law (1 D)
 To calculate the electric force between two charges,
always solve for the absolute value and ignore the sign
on the charge. Determine whether the charges attract
or repel at the end. ( A point charge is a small + or –
charge.)
 Eg) Calculate the magnitude of the electric force
between two point charges of +4.00 uC and +3.00 uC
when they are 4.50 cm apart.
r  0.045m
q1  4.00  C
q2  3.00  C
kq1 q2
Fe  2
r
N m2
6
6
(8.99  10
)(4.00

10
C
)(3.00

10
C)
2
C
Fe 
(0.045m) 2
 53.27 N
The magnitude of the repulsive force is 53.3N .
9
Eg 2) A comb with 2.0 uC of charge is 1.5 m to the left
from a hair with -3.0 uC of charge. Determine the
force the hair exerts on the comb.
r  1.5m
q1  2.0  C
kq1q2
q2  3.0  C
F 
e
r2
2
N
m
6
6
(8.99  109
)(2.0

10
C
)(3.0

10
C)
2
C
Fe 
(1.5m) 2
 2.4  102 N
The hair is pulling the comb with a force
2
of 2.4  10 N right.
6
6
1.8

10
C
2.4

10
C
 Two point charges of
and
produce a force of 2.2  103 N
on each other. How far apart are these two charges?
kq1q2
Fe  2
r
kq1q2
r
Fe
2
N
m
6
6
(8.99  109
)(1.8

10
C
)(2.4

10
C)
2
C
r
2.2  103 N
 4.2m
NOTE:
 Charges are usually in the range of microCoulombs.
Only lightning strikes have a charge of 1 or 2 coulombs.
The electron’s charge is so important it is
called an elementary charge. Its symbol is “e”
and is also on your formula sheet.
Net electrical force
Determine the net force on object B.
always interpret this to mean that the
object in question is free to move and the
other objects are stationary.
Forces on B
Free body diagram
F C on B
kq1q2
Fe  2
r
9
2
2
8.99 10 N m / C 1.00C 2.00C
Fe 
2
(1.00m)
Fe  1.798 10 N [left ]
10
F A on B
kq1q2
Fe  2
r
9
2
2
8.99 10 N m / C 3.00C 2.00C
Fe 
2
(2.00m)
Fe  1.3485 10 N [right ]
10
Fnet
Fnet  FC on B  FAon B


Fnet  1.798 10 N  1.3485 10 N
10
Fe  4.50 10 N [left ]
9
10
Example: Find the net electrical force on B.
A
2.0 x 10 -6
C
3.20 m
-1.5  C
B
-3.8  C
AB
4.1 m
C
k q1 q2
Fe 
r2
2

9 Nm 
6
6
8.99

10
2

10
C
3.8

10
C



2 
C 

Fe 
2
3.2
m


Fe  6.67 103 N
down on B
*
BC F  kq1q2
e
2
r
2


Nm
9
6
6
8.99

10
3.8

10
C
1.5

10
C



2 
C 

Fe 
2
 4.1 m 
3
Fe  3.05 10 N
left on B
*
3.05 x 10 -3
N
Draw Vector Diagram
6.67 x 10 -3
N

FR 
3.05  10   6.67  10 
3
2
3
FResultant
2

FR  7.3  10  3 N
O 6.67  10  3 N
tan  
A 3.05  10  3 N
   65 

3
FR  7.3  10 N , W 65  S
*
Example: Calculate the gravitational force between an electron and a
proton placed 1.0 cm apart.
G m1 m2
Fg 
R2
2

 11 Nm 
 31
 27
 1.0  10

9
.
11

10
kg
1
.
67

10
kg
2 
kg 

Fg 
1.0  10  2 m 2





Fg  1.0  10 63 N
*
Example: Calculate the electrical force between an electron and a proton
placed 1.0 cm apart.
kq1q2
Fe  2
r

Felectrical
 8.99  10 9




Nm 2 
 19
 19
1
.
6

10
C
1
.
6

10
C
2 
C 
1.0  10
2
m


2
Felectrical  2.3  10  24 N
2.3  10  24 N
electrical
force
 10

ratio :

63
gravitational force 1.0  10
N
39
times
*
 Three point charges, A, B, and C, are arranged in a line
as shown.
a.Determine the net electrostatic force acting on charge
B.
b.The charge on object C is negative. Identify the sign of
the charge (+ or -) on object
i. A.
ii. B.
2 D Example:
 Three point charged objects are arranged in a right
angle as shown in the diagram.
a. The charge on object A is positive. Identify the sign of
the charge (+ or -) on object
 i. B. (negative)
 ii. C. (negative)
b. Determine the net electrostatic force acting on charge
B.
Read WB p. 85-86 TB p. 532-6 Examples 10.3
to 10.6
Do WB p.87 2-5,8,10,14,16,18,19