Transcript PHy 184 lecture 3 - MSU Department of Physics and Astronomy

PHY 184
Spring 2007
Lecture 3
Title: The Coulomb Force
1/10/2006
184 Lecture 3
1
Announcements
 Don’t forget to register your clicker!
 Homework Set 1 is due
Tuesday, January 16, at 8:00 am
 We will soon post the complete SLC schedule.
Strosacker Learning Center
Room 1248 B. P. S.
1/10/2006
184 Lecture 3
2
Outline
1 – Review
2 – Electrostatic charging
3 – Coulomb’s Law
1/10/2006
184 Lecture 3
3
Review
 There are two types of charge: negative and positive.
 Most objects are electrically neutral; they have equal numbers of
negative and positive charges (net charge is 0).
 An object becomes charged by adding or removing electrons.
 An electron carries negative charge of magnitude e = 1.602×10-19 C.
 Law of Charges
• Like charges repel and opposite charges attract.
 Law of charge conservation
• The total charge of an isolated system is strictly conserved.
 Conductors are materials where some of the electrons can move
freely.
 Insulators are materials where none of the charges can move freely.
1/10/2006
184 Lecture 3
4
Electrostatic Charging
 There are two ways to charge an object
• Conduction
• Induction
 Charging by conduction
• We can charge an object by connecting a source of
charge directly to the object and then disconnecting the
source of charge
• The object will remain charged
– Conservation of charge
1/10/2006
184 Lecture 3
5
Charging by Conduction
+++++++++
Electroscope
We brought charge onto the
electrode by contact.
1/10/2006
184 Lecture 3
6
Induction
+++++++++
---------
Induction
The presence of the
positively charged rod
leads to a redistribution
of charge (a kind of
polarization).
It pulls electrons up to
the electrode.
1/10/2006
184 Lecture 3
7
Charging by Induction
 We can also charge an object without physically
connecting to it
• First we charge a rod positively
• Then we ground the object to be charged
• Connecting the object to the Earth provides an effectively
infinite sink for charge
• We bring the charged rod close to the object but do not
touch it
• We remove the ground connection and move the rod away
• The object will be charged by induction
1/10/2006
184 Lecture 3
8
Charging by induction
+++++++++
---------
Induction
The presence of the
positively charged rod
leads to a redistribution
of charge.
Grounding pushes
positive charge to Earth
(or rather pulls electrons
from Earth!) leaving the
electroscope negative.
ground
1/10/2006
184 Lecture 3
9
Coulomb’s Law
1/10/2006
184 Lecture 3
10
Electric Force - Coulomb’s Law
 Consider two electric
charges: q1 and q2
 The electric force F
between these two charges
separated by a distance r is
given by Coulomb’s Law
 The constant k is called
Coulomb’s constant and is
given by
1/10/2006
kq1q2
F 2
r
k  8.99  10 Nm /C
9
184 Lecture 3
2
2
11
Coulomb’s Law (2)
 The coulomb constant is also written as
k
1
4  0
where  0  8.85  10
12
2
C
Nm2
 0 is the “electric permittivity of
vacuum”
• A fundamental constant of nature
1 q1q 2
F
2
4 0 r
1/10/2006
184 Lecture 3
12
Example:
What is the force between two charges of 1 C separated by 1 meter?
Answer: 8.99 x 109 N,
i.e., huge!
1/10/2006
184 Lecture 3
13
Electric Force
 The electric force is given by
 The electric force, unlike the
gravitational force, can be positive
or negative
• If the charges have opposite
signs, the force is negative
• Attractive
• If the charges have the same
sign, the force is positive
• Repulsive
1/10/2006
184 Lecture 3
q1q2
F k 2
r
14
Electric Force Vector
Electric force in vector form
q1
y
r
r1
q2
x
r2

q1q2
F2  k 2 rˆ
r
1/10/2006
  
r  r2  r1
  
r
r

r
2
1
ˆr  
r
r

q1q2
F1  k 2  rˆ 
r
184 Lecture 3
15
Superposition Principle
 The net force acting on any charge is the vector
sum of the forces due to the remaining charges in
the distribution.
F1,net  F1, 2  F1,3    F1,n
F  F
F
 F
1n, x
 1x 12, x 13, x

F1y  F12, y  F13, y    F1n, y

F1z  F12, z  F13, z    F1n, z
1/10/2006
184 Lecture 3
16
Example - The Helium Nucleus
Part 1: The nucleus of a helium atom has two protons and
two neutrons. What is the magnitude of the electric force
between the two protons in the helium nucleus?
Answer: 58 N
Part 2: What if the distance is doubled; how will
the force change?
Answer: 14.5 N
Inverse square law: If the distance is
doubled then the force is reduced by a
factor of 4.
1/10/2006
184 Lecture 3
17
Example - Equilibrium Position
 Consider two charges located on the x axis
x2
x1
 The charges are described by
• q1 = 0.15 C
• q2 = 0.35 C
x1 = 0.0 m
x2 = 0.40 m
 Where do we need to put a third charge for
that charge to be at an equilibrium point?
• At the equilibrium point, the forces from the two
charges will cancel.
1/10/2006
184 Lecture 3
18
Example - Equilibrium Position (2)
x1
x2
 The equilibrium point must be along the xaxis.
 Three regions along the x-axis where we
might place our third charge
x3 < x 1
x1 < x 3 < x 2
x3 > x 2
1/10/2006
184 Lecture 3
19
Example - Equilibrium Position (3)
x1
x2
 x3<x1
• Here the forces from q1 and q2 will always point in the
same direction (to the left for a positive test charge)
• No equilibrium
 x2<x3
• Here the forces from q1 and q2 will always point in the
same direction (to the right for a positive test charge)
• No equilibrium
1/10/2006
184 Lecture 3
20
Example - Equilibrium Position (4)
x2
x1
q3
x 1 < x3 < x 2
Here the forces from q1 and q2 can balance.
q1q3
q 2 q3
k
k
0
2
2
( x3  x1 )
( x 2  x3 )
Check the signs!!
1/10/2006
Answer: x3 = 0.16 m
184 Lecture 3
21
Example - Charged Pendulums
 Consider two identical charged
balls hanging from the ceiling
by strings of equal length 1.5 m
(in equilibrium). Each ball has a y
charge of 25 C. The balls hang
at an angle  = 25 with respect
to the vertical. What is the
mass of the balls?
x
Ball on left :
Step 1: Three forces act on
each ball: Coulomb force,
gravity and the tension of the
string.
1/10/2006
184 Lecture 3
kq
Fx  T sin   2
d
Fy  T cos   mg
2
22
Example - Charged Pendulums (2)
Step 2: The balls are in equilibrium
positions. That means the sum of all
forces acting on the ball is zero!
T sin  kq2 / d 2

T cos
mg
kq
mg  2
d tan
2
d=2 l sin 
Answer: m = 0.76 kg
A similar analysis applies to the ball on the right.
1/10/2006
184 Lecture 3
23
Electric Force and Gravitational Force
 Coulomb’s Law that describes the electric force
and Newton’s gravitational law have a similar
functional form
Felectric
q1q2
k 2
r
Fgravity
m1m2
G 2
r
 Both forces vary as the inverse square of the
distance between the objects.
 Gravitation is always attractive.
 k and G give the strength of the force.
1/10/2006
184 Lecture 3
24
Example - Forces between Electrons
 What is relative strength of the electric force
compared with the force of gravity for two
electrons?
( e ) 2
Felectric  k 2
r
Felectric ke 2
42


(do
the
calculatio
n)

4
.
2

10
me2
Fgravity Gm2
Fgravity  G 2
r
• Gravity is irrelevant for atomic and subatomic
processes – the electric force is much much stronger.
• But sometimes gravity is most important; e.g, the
Why?
motion of the planets.
1/10/2006
184 Lecture 3
25
Example - Four Charges
Consider four charges
placed at the corners of
a square with sides of
length 1.25 m as shown
on the right. What is
the magnitude of the
electric force on q4
resulting from the
electric force from the
remaining three
charges?
Set up an xy-coordinate system with its origin at q2.
1/10/2006
184 Lecture 3
26
Example - Four Charges (2)
Answer:
F (on q4) = 0.0916 N
… and the direction?
1/10/2006
184 Lecture 3
27