Transcript PPT

Coulomb’s Law
Physics 12
Electroscope
► Draw
a diagram showing the forces that
exist on the leaves of the electroscope when
it is placed close to the Van de Graff
generator
Coulomb’s Pendulum
► Consider
the following
pendulum
► The mass initially shares
charge with an identical
sphere
► The two spheres then have
identical charge and (in this
case) the second sphere is
moved toward the pendulum
bob from the left, causing the
pendulum to deflect
Coulomb’s Pendulum
► Draw
a free body diagram for the pendulum
bob; you will know the mass of the bob and
the separation of the two spheres and the
angle the pendulum support makes with the
vertical
► Develop an equation that will allow you to
solve for the electrostatic force experienced
by the pendulum (and the second sphere)
Pith Ball
Fe  Tx  0
Fe  Tx
Fe  T cos 
mg
T
sin 
cos 
Fe  mg
sin 
mg
Fe 
tan 
Fg  Ty  0
Fg  Tg
Fg  T sin 
Coulomb’s Pendulum
Separation (m)
Angle
(with vertical)
0.150
11.52319
0.130
15.18589
0.110
20.76193
0.090
29.52358
0.070
43.11136
Coulomb’s Pendulum
► What
type of relationship does this display
as the distance between the point charges
is increased or decreased?
Coulomb’s Law
► Coulomb’s
Law is given as:
kq1q2
Fe  2
r
Electric Force: Coulomb’s Law
► Electric
Force is the force felt by separated
(positive or negative) charges.
 Opposite charges attract
 Like charges repel
kq1q2
Fe  2
r
+
-
+
+
k  9.0 10
 
9 Nm 2
C2
Coulomb’s Law
► Positive
force will result when:
 Two positively charged particles
 Two negatively charged particles
► Negative
kq1q2
Fe  2
r
force will result when:
 One positively and negatively charged particle
► Positive
force indicates repulsion
► Negative force indicates attraction
Practice
► What
attractive force does an electron in a
hydrogen atom experience?
► What attractive force does an electron in
helium experience?
Applications of
Coulomb’s Law
Physics 12
Coulomb Sample Problem
► Three
charges are arranged in a line; if the
three charges are 15μC, -12μC and 18μC
respectively. The distance between the first
two charges is 0.20m and the second and
third charges is 0.30m. What is the force
experienced by the first charge?
Coulomb Sample Problem
► What
force is experienced by the remaining
two charges?
Force (Vector) Addition
► To
add forces, resolve each force into its
components and treat the forces in the xdirection and y-direction independently
► Once you sum the x and y components, use
Pythagorean Theorem and Trigonometry to
resolve into a resultant force
Example
►A
point P has forces of 12.0N at 24.3°,
17.6N at 112°, 6.78N at 241° and 10.2N at
74.4°.
 Determine the resultant vector
► 25.5N,
81.4°
Coulomb’s Law and Vector Addition
► When
we consider an electrostatic system,
we need to use Coulomb’s Law to determine
the magnitude and direction of each force
► Once the magnitude and direction of each
force has been determined, then the vector
sum can be completed
Coulomb’s Law in 2D
► Three
charges are arranged as follows; a 2.0μC is placed 4.0m due north of a 3.0μC
charge and 3.0m due west of a 5.0μC
charge. What is the force experienced by
the -2.0μC charge?
Coulomb’s Law in 2D
1 -2.0μC
4.0m
2 3.0 μ C
3.0m
5.0 μ C 3
kq1q2
Fe12  2
r
Fe12  3.4 x10 3 N
kq3q2
Fe 23  2
r
3
Fe 23  5.4 x10 N
Coulomb’s Law in 2D
  53o  180o
2 3.0 μ C
  233
o
Fe12 x  0 N
Fe12 y  3.4 x10 3 N
Fe 23x  5.4 x10 3 N cos 233
Fe 23x  3.2 x10 3 N
Fe 23 y  5.4 x10 3 N sin 233
Fe 23 y  4.3 x10 3 N
Coulomb’s Law in 2D
► Use
the x and y component data to
determine the resultant force vector
Fex  0 N  3.2 x10 N
Fe  3.3x10 3 N
Fex  3.2 x10 3 N
  16
3
3
3
Fey  3.4 x10 N  4.3 x10 N
Fey  9.0 x10  4 N
o

3
o
Fe  3.3x10 N ,196
Electric Fields
Physics 12
Field Theory
► When
forces exist without contact, it can be
useful to use field theory to describe the
force experienced by a particle at any point
in space
► We live in a gravitational field where the
separation between massive objects results
in attractive forces
► In a similar way, we can think of an electric
field
Electric Field Mapping
► To
map an electric field, a small positive
test charge is placed in the field and the
magnitude and direction of the force is
recorded
► The test charge is then moved throughout
the electric field and a map of the field is
created
► If the force experienced by the test charge
can be measured, then we can map the
field
Test Charge
► The
test charge that is used
must be small compared to the
charge creating the field
► If not, the test charge’s field will
change the field that is being
investigated
► The electric field should be the
same regardless of the test
charge used
Field Lines – Two Positive
Charges
Field Lines – Two Opposing
Charges
Problem
► What
are the relative
magnitudes of the
charges in the
diagram?
► What is the polarity of
each of the charges?
Multiple Charges
► It
is also possible to
consider what happens
with multiple charges:
Electric Field Intensity
► The
electric field can be determined using
the force experienced by a particle and the
charge on the particle
 Fonq'
E 
q'
Electric Field
► The
electric field also has a direction; since
the field is the superposition of all the
electric field vectors at a given point in
space where:

E1 

Fonq'1
q'


E  Ei
Electric Field
►A
charge of 2.0mC is placed at the origin
and a charge of -5.0mC is placed at the
point (3,0); what electric field exists at:
 (1,0)
 (4,0)
 (-1,0)
► Where
is the electric field equal to zero?
Electric Potential
Physics 12
Electric Potential Energy
► Gravitational
potential
energy is due to
mass, gravitational
field intensity and
separation
► Electric potential
energy is due to
charge, electric field
intensity and
separation
E p  Eqr
F
E p  qr
q
kqQ
Ep  2 r
r
kqQ
Ep 
r
Potential Difference
► Similar
to gravitational potential
difference, electric potential difference is
measured with respect to a reference
point (usually the ground) which we call
zero
► This concept is not as useful for
gravitational difference as objects have
different masses, but since each charge
carrier has the same charge, this concept
has value for electric potential difference
Voltage or Potential Difference
► Electric
potential
difference is known as
voltage
► One volt is defined as
one joule per coulomb
► Electric Potential is
NOT electric potential
energy
V
Ep
q
kQ
V
r
These plots show the
potential due to (a)
positive and (b) negative
charge.
What minimum work is required by an external
force to bring a charge q = 3.00 μC from a great
distance away (r = infinity) to a point 0.500 m
from a charge Q = 20.0 μC ?
Analogy between gravitational and electrical
potential energy:
Both rocks have the same
gravitational potential, but
the bigger rock has more Ep.
Both charges have the
same electric potential,
but the 2Q charge has
more Ep.
What is the change in
potential energy of the
electron in going from a
to b?
What is the speed of the
electron as a result of
this acceleration?
Repeat both calculations
for a proton.
Uniform Electric Field
► As
previously seen, it
is possible to produce
a uniform electric field
► The intensity of the
field is a function of
the voltage and
separation of the
plates
V  Ed
Equipotential Lines
•An equipotential
(represented by the green
dashed lines) is a line or
surface over which the
potential is constant.
•Electric field lines are
perpendicular to
equipotentials.
•The surface of a conductor
is an equipotential.
Equipotential Lines
The Electron Volt, a Unit of Energy
A Joule is too large when dealing with
electrons or atoms, so electron volts are
used. One electron volt (eV) is the energy
gained by an electron moving through a
potential difference of one volt.
Elementary Charge
► Robert
Millikan investigated the charge on
an electron in his famous oil-drop
experiment
► He was awarded the Nobel Prize in 1923 for
his 1917 research that led to the elementary
charge of 1.60x10-19C
► Today, the accepted value of the elementary
charge is 1.60217733x10-19C
Elementary Charge
► Since
we know the
value of the
elementary charge, we
can determine the
number of charge
carriers or the total
charge with the
following equation
q  Ne