Download Electric Charges

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Electric Charges,
Forces, and
Fields
Physics 231
Lecture 1-1
Fall 2008
Electric Charges
Electric charge is a basic property of matter
Two basic charges
Positive and Negative
Each having an absolute value of
1.6 x 10-19 Coulombs
Experiments have shown that
Like signed charges repel each other
Unlike signed charges attract each other
For an isolated system, the net charge of the
system remains constant
Charge Conservation
Physics 231
Lecture 1-2
Fall 2008
Two basics type of materials
Conductors
Materials, such as metals, that allow the free
movement of charges
Insulators
Materials, such as rubber and glass, that don’t
allow the free movement of charges
Physics 231
Lecture 1-3
Fall 2008
Coulomb’s Law
Coulomb found that the electric force between
two charged objects is
Proportional to the product of the charges on
the objects, and
Inversely proportional to the separation of the
objects squared
F k
q1q2
r
2
k being a proportionality constant, having a value
of 8.988 x 109 Nm2/c2
Physics 231
Lecture 1-4
Fall 2008
Electric Force
As with all forces, the electric force is a Vector
So we rewrite Coulomb’s Law as

q1q2
F12  k 2 r̂12
r
This gives the force on charged object 2 due to charged
q1
object 1
q2
r̂12 is a unit vector pointing from object 1 to object 2
The direction of the force is either parallel or
antiparallel to this unit vector depending upon the
relative signs of the charges
Physics 231
Lecture 1-5
Fall 2008
Electric Force
The force acting on each charged object has the
same magnitude - but acting in opposite directions


F12  F21
Physics 231
(Newton’s Third Law)
Lecture 1-6
Fall 2008
Example 1
A charged ball Q1 is fixed to a horizontal surface
as shown. When another massive charged
ball Q2 is brought near, it achieves an
equilibrium position at a distance d12 directly
above Q1.
When Q1 is replaced by a different charged ball
Q3, Q2 achieves an equilibrium position at a
distance d23 (< d12) directly above Q3.
Q2
Q2
d12
d23
g
Q1
Q3
For 1a and 1b which is the correct answer
1a: A) The charge of Q3 has the same sign of the charge of Q1
B) The charge of Q3 has the opposite sign as the charge of Q1
C) Cannot determine the relative signs of the charges of Q3 & Q1
1b: A) The magnitude of charge Q3 < the magnitude of charge Q1
B) The magnitude of charge Q3 > the magnitude of charge Q1
C) Cannot determine relative magnitudes of charges of Q3 & Q1
Physics 231
Lecture 1-7
Fall 2008
Example 1
A charged ball Q1 is fixed to a horizontal surface
as shown. When another massive charged
ball Q2 is brought near, it achieves an
equilibrium position at a distance d12 directly
above Q1.
When Q1 is replaced by a different charged ball
Q3, Q2 achieves an equilibrium position at a
distance d23 (< d12) directly above Q3.
Q2
Q2
d12
d23
g
Q1
Q3
1a: A) The charge of Q3 has the same sign of the charge of Q1
B) The charge of Q3 has the opposite sign as the charge of Q1
C) Cannot determine the relative signs of the charges of Q3 & Q1
• To be in equilibrium, the total force on Q2 must be zero.
• The only other known force acting on Q2 is its weight.
• Therefore, in both cases, the electrical force on Q2 must be directed upward
to cancel its weight.
• Therefore, the sign of Q3 must be the SAME as the sign of Q1
Physics 231
Lecture 1-8
Fall 2008
Example 1
A charged ball Q1 is fixed to a horizontal surface
as shown. When another massive charged ball
Q2 is brought near, it achieves an equilibrium
position at a distance d12 directly above Q1.
When Q1 is replaced by a different charged ball
Q3, Q2 achieves an equilibrium position at a
distance d23 (< d12) directly above Q3.
Q2
Q2
d12
d23
g
Q1
Q3
1b: A) The magnitude of charge Q3 < the magnitude of charge Q1
B) The magnitude of charge Q3 > the magnitude of charge Q1
C) Cannot determine relative magnitudes of charges of Q3 & Q1
• The electrical force on Q2 must be the same in both cases … it just cancels
the weight of Q2
• Since d23 < d12 , the charge of Q3 must be SMALLER than the charge of Q1
so that the total electrical force can be the same!!
Physics 231
Lecture 1-9
Fall 2008
More Than Two Charges
Given charges q, q1, and q2
If q1 were the only other charge,
we would know
the force on q

due to q1 - Fq q
1
If q2 were the only other charge,
we would know the force on q

due to q2 - Fq q

Fq
q1
1q

Fnet
q

Fq
2q
q2
2
What is the net force if both charges are present?
The net force is given by the Superposition Principle

 
Fnet  F1 F 2
Physics 231
Lecture 1-10
Fall 2008
Superposition of Forces
If there are more than two charged objects
interacting with each other
The net force on any one of the charged
objects is
The vector sum of the individual Coulomb
forces on that charged object

qi
Fj  q j  k 2 rˆij
i  j rij
Physics 231
Lecture 1-11
Fall 2008
Example Two
qo, q1, and q2 are all point charges
where qo = -1mC, q1 = 3mC, and
q2 = 4mC
y (cm)
4
3
2
1
What is the force acting on qo?
1
F20  k
q0q2
Physics 231
r102
2
r20
3
4
x (cm)
5

Decompose F20 into its x and y

F10  F10 yˆ

F20   F20 rˆ
2
q2
What are F0x and F0y ?


Needto calculate F10 and F20
F10  k

q1



We have that F0  F10  F20
q0q1
qo
components

ˆ  F20 sin  y
ˆ
F20  F20 cos   x
20
cos  
Lecture 1-12
x 2  x0
r20
sin  
y0  y 2
r20
Fall 2008
Example Two - Continued
Now add the components of


F10 and F20 to
X-direction: F0 x  F10 x  F20 x
find
F0 x
and
F0 y
y (cm)
F10 x  0
4
F0 x  F20 cos
3
qo
2

F10
1
Y-direction: F  F  F
0y
10 y
20 y

F20

F0
q1
q2
1
2
3
4
5
x (cm)
F0 y   F10  F20 sin 
Physics 231
Lecture 1-13
Fall 2008
Example Two - Continued
Putting in the numbers . . .
y (cm)
4
cos  0.8
r10  3cm
r20  5cm
3
qo
2

F10
1
F10  30 N F20  14.4N

F20

F0
q1
q2
1
2
3
4
5
x (cm)
We then get for the components
F0 x  11.52N F0 y  38.64N

The magnitude of F0 is
F0  F02x  F02y  40.32 N
At an angle given by
  tan 1 F0 y F0 x   tan 1 (38.64 / 11.52)  73.40
Physics 231
Lecture 1-14
Fall 2008
Note on constants
k is in reality defined in terms of a more
fundamental constant, known as the
permittivity of free space.
k
1
4 0
with  0  8.854 x10
Physics 231
12
Lecture 1-15
2
C
2
Nm
Fall 2008
Electric Field
The Electric Force is like the Gravitational
Force
Action at a Distance
The electric force can be thought of as
being mediated by an electric field.
Physics 231
Lecture 1-16
Fall 2008
What is a Field?
A Field is something that can be defined anywhere
in space
A field represents some physical quantity
(e.g., temperature, wind speed, force)
It can be a scalar field (e.g., Temperature field)
It can be a vector field (e.g., Electric field)
It can be a “tensor” field (e.g., Space-time curvature)
Physics 231
Lecture 1-17
Fall 2008
A Scalar Field
73
77
82
84
83
72
71
75
77
68
80
64 73
82
88
55
66
88
80
75
88
83 90 91
92
A scalar field is a map of a quantity that has
only a magnitude, such as temperature
Physics 231
Lecture 1-18
Fall 2008
A Vector Field
73
77
72
71
82
84
83
88
75
68 64
80
73
57 56 55
66
88
75 80
90
83
92
91
77
A vector field is a map of a quantity that is
a vector, a quantity having both magnitude
and direction, such as wind
Physics 231
Lecture 1-19
Fall 2008
Electric Field
We say that when a charged object is put at
a point in space,
The charged object sets up an Electric
Field throughout the space surrounding
the charged object
It is this field that then exerts a force on
another charged object
Physics 231
Lecture 1-20
Fall 2008
Electric Field
Like the electric force,
the electric field is also a vector
If there is an electric force acting on an
object having a charge qo, then the
electric field at that point is given by

 F
E
q0
Physics 231
(with the sign of q0 included)
Lecture 1-21
Fall 2008
Electric Field
The force on a positively
charged object is in the same
direction as the electric field at
that point,
While the force on a negative
test charge is in the opposite
direction as the electric field
at the point
Physics 231
Lecture 1-22
Fall 2008
Electric Field
A positive charge sets up
an electric field pointing
away from the charge
A negative charge sets up an
electric field pointing
towards the charge
Physics 231
Lecture 1-23
Fall 2008
Electric Field
Earlier we saw that the
force on a charged object
is given by




q

i
F j  q j   k 2 rˆij 


r
 i  j ij



The term in parentheses remains the same if we
change the charge on the object at the point in
question
The quantity in the parentheses can be thought of as the
electric field at the point where the test object is placed
The electric field of a point charge can then be
shown to be given by

q
E  k 2 rˆ
r
Physics 231
Lecture 1-24
Fall 2008
Electric Field
As with the electric force, if there are
several charged objects, the net electric
field at a given point is given by the
vector sum of the individual electric fields


E   Ei
i
Physics 231
Lecture 1-25
Fall 2008
Electric Field
If we have a continuous charge distribution
the summation becomes an integral

dq
E  k  2 rˆ
r
Physics 231
Lecture 1-26
Fall 2008
Hints
1) Look for and exploit symmetries in the
problem.
2) Choose variables for integration
carefully.
3) Check limiting conditions for
appropriate result
Physics 231
Lecture 1-27
Fall 2008
Electric Field
Ring of Charge
Physics 231
Lecture 1-28
Fall 2008
Electric Field
Line of Charge
Physics 231
Lecture 1-29
Fall 2008
Example 3
Two equal, but opposite charges are placed on the x axis. The
positive charge is placed at x = -5 m and the negative charge is
placed at x = +5m as shown in the figure above.
1) What is the direction of the electric field at point A?
a) up
b) down
c) left
d) right
e) zero
2) What is the direction of the electric field at point B?
a) up
b) down
c) left d) right
e) zero
Physics 231
Lecture 1-30
Fall 2008
Example 4
y
Two charges, Q1 and Q2, fixed along the x-axis as
shown produce an electric field, E, at a point
(x,y) = (0,d) which is directed along the negative
y-axis.
E
d
Which of the following is true?
(a) Both charges Q1 and Q2 are positive
Q1
Q2
x
(b) Both charges Q1 and Q2 are negative
(c) The charges Q1 and Q2 have opposite signs
E
E
(a)
Q1
Physics 231
(b)
Q2
(c)
Q1
Q2
Lecture 1-31
Q1
E
Q2
Fall 2008
Electric Field Lines
Possible to map out the electric field in a
region of space
An imaginary line that at any given point
has its tangent being in the direction of the
electric field at that point
The spacing, density, of lines is related to
the magnitude of the electric field at that
point
Physics 231
Lecture 1-32
Fall 2008
Electric Field Lines
At any given point, there can be only one
field line
The electric field has a unique direction at
any given point
Electric Field Lines
Begin on Positive Charges
End on Negative Charges
Physics 231
Lecture 1-33
Fall 2008
Electric Field Lines
Physics 231
Lecture 1-34
Fall 2008
Electric Dipole
An electric dipole is a pair of point charges
having equal magnitude but opposite sign that
are separated by a distance d.
Two questions concerning dipoles:
1) What are the forces and torques acting on a
dipole when placed in an external electric field?
2) What does the electric field of a dipole look
like?
Physics 231
Lecture 1-35
Fall 2008
Force on a Dipole
Given a uniform external field
Then since the charges are of
equal magnitude, the force on
each charge has the same
value
However the forces are in opposite directions!
Therefore the net force on the dipole is
Fnet = 0
Physics 231
Lecture 1-36
Fall 2008
Torque on a Dipole
The individual forces acting on the dipole
may not necessarily be acting along the
same line.
If this is the case, then there will be a
torque acting on the dipole, causing the
dipole to rotate.
Physics 231
Lecture 1-37
Fall 2008
Torque on a Dipole
The torque is then given by
 = qE dsinf
 
 q dE



d is a vector pointing from the negative charge to the
positive charge
Physics 231
Lecture 1-38
Fall 2008
Potential Energy of a Dipole
Given a dipole in an external field:
Dipole will rotate due to torque
Electric field will do work
The work done is the negative of the
change in potential energy of the dipole
The potential energy can be shown to be
 
U  q d  E

Physics 231
Lecture 1-39

Fall 2008
Electric Field of a Dipole
Physics 231
Lecture 1-40
Fall 2008