Transcript Electric field lines

Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 16: Electric Forces and
Fields
•Electric Charge
•Conductors & Insulators
•Coulomb’s Law
•Electric Field
•Motion of a Point Charge in a Uniform E-field
•Conductors in Electrostatic Equilibrium
•Gauss’s Law
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§16.1 Electric Charge
There are two kinds of electric charge: positive and negative.
A body is electrically neutral if the sum of all the charges in a
body is zero.
Charge is a conserved quantity.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The elementary unit of charge is e = 1.60210-19 C.
The charge on the electron is -1e.
The charge on the proton is +1e.
The charge on the neutron is 0e.
Experiments show that likes charges will repel each other
and unlike charges will attract each other and that the force
decreases with increasing distance between charges.
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- - +
+ +
+
+ -
This body is electrically neutral.
An object can become polarized if the charges within it can
be separated.
By holding a
charged rod near
the body, it can
be polarized.
-
+ + + + +
-
+
+
+
+
-
+
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§16.2 Conductors and Insulators
A conductor is made of material that allows electric charge to
move through it easily.
An insulator is made of material that does not allow electric
charge to move through it easily.
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§16.3 Coulomb’s Law
The magnitude of the force
between two point charges is:
F
k q1 q2
r2
where q1 and q2 are the charges and r is the separation
between the two charges.
k  8.99 109 Nm2 /C 2
where k 
1
40
and  0  8.85 10 12 C 2 /Nm 2
and 0 is called the permittivity of free space.
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The electric force is directed between the centers of the two
point charges.
q1
F12
F21
q2
Attractive force
between q1 and q2.
r
Repulsive force
between q1 and q2.
F12
q1
q2
F21
r
The electric force is an example of a long-range or field
force, just like the force of gravity.
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Example: What is the net force on the charge q1 due to the
other two charges? q1 = +1.2 C, q2 = -0.60 C, and q3 =
+0.20 C.
F21

F31
The net force on q1 is Fnet = F21 + F31
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Example continued:
The magnitudes of the forces are:
F21 
k q1 q2
r212

9 10


9
Nm2 /C 2 (1.2 10 6 C)(0.60 10 6 C)
(1.2 m) 2  (0.5 m) 2
9
Nm2 /C 2 (1.2 10 6 C)(0.20 10 6 C)
(1.2 m) 2
 3.8 10 3 N
F31 
k q1 q3
r312

9 10


 1.5 10 3 N
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Example continued:
The components of the net force are:
Fnet, x  F31, x  F21, x  F31  F21 cos  5.0 103 N
Fnet, y  F31, y  F21, y  0  F21 sin   1.4 103 N
1. 2 m
 0.92
1.3 m
0.5 m
sin  
 0.38
1.3 m
cos  
Where from the figure
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Example continued:
The magnitude of the net force is:
2
2
3
Fnet  Fnet

F

5
.
2

10
N
,x
net , y
The direction of the net force is:
tan  
Fnet , y
Fnet , x
 0.28
  16
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 16.13): What is the ratio of the electric
force and gravitational force between a proton and an
electron separated by 5.310-11 m (the radius of a Hydrogen
atom)?
Fe 
k q1 q2
r2
Gm1m2
Fg 
r2
q1  q2  e
m1  m p  1.67  10  27 kg
m2  me  9.11 10 31 kg
Fe k q1 q2
ke2
The ratio is:


 2.3  1039
Fg Gm1m2 Gme m p
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§16.4 The Electric Field
Recall :
Fg  mg
Fe  qE
Where g is the strength
of the gravitational field.
Similarly for electric forces
we can define the strength
of the electric field E.
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For a point charge of charge Q, the
magnitude of the force per unit charge
at a distance r (the electric field) is:
Fe k Q
E
 2
q
r
The electric field at a point in space is found by adding all of
the electric fields present.
E net   Ei
i
Be careful! The electric
field is a vector!
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: Find the electric field at the point P.
P
x
q1 = +e
q2 = -2e
x = 0m
x = 1m
x = 2m
E is a vector. What is its direction?
Place a positive test charge at the point of interest. The
direction of the electric field at the location of the test
charge is the same as the direction of the force on the
test charge.
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Example continued:
q1 = +e
P
x
q2 = -2e
Locate the
positive test
charge here.
P
x
q1 = +e
q2 = -2e
Direction of E due
to charge 2
Direction of E due
to charge 1
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The net electric field at point P is:
Enet  E1  E2
The magnitude of the electric field is:
Enet  E1  E2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
E1 
k q1
E2 
k q2
r2
r2

9 10


9 10


9
Nm2 / C2 (1.6 1019 C)
10

3
.
6

10
N/C
2
(2 m)
9
Nm2 / C2 (2 *1.6 1019 C)
9

2
.
9

10
N/C
2
(1 m)

9
Enet  E1  E2  2.5 10 N/C
The net E-field is
directed to the left>
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Electric field lines
Electric field lines are a useful way to indicate what the
magnitude and direction of an electric field is in space.
Rules:
1. The direction of the E-field is tangent to the field lines at
every point in space.
2. The field is strong where there are many field lines and
weak where there are few lines.
3. The field lines start on + charges and end on – charges.
4. Field lines do not cross.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Pictorial representation of the rules on the previous slide:
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§16.5 Motion of a Point Charge in a
Uniform E-Field
A region of space with a uniform
electric field containing a particle
of charge q (q>0) and mass m.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
FBD for the
charge q
y
Fe
x
Apply Newton’s 2nd Law and
solve for the acceleration.
F
x
 Fe  ma
Fe  qE  ma
q
a E
m
One could now use the kinematic equations to solve for
distance traveled in a time interval, the velocity at the end of
a time interval, etc.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: What electric field strength is needed to keep an
electron suspended in the air?
y
FBD for the
electron:
Fe
x
w
To get an upward force on the electron, the electric field
must be directed toward the Earth.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
Apply Newton’s 2nd Law:
F
y
 Fe  w  0
Fe  w
qE  eE  mg
mg
E
 5.6  10 11 N/C
e
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 16.44): A horizontal beam of electrons
moving 4.0107 m/s is deflected vertically by the vertical
electric field between two oppositely charged parallel plates.
The magnitude of the field is 2.00104 N/C.
(a) What is the direction of the field between the plates?
From the top plate to the bottom plate
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) What is the charge per unit area on the plates?
Q

E

0 A 0
This is the electric field
between two charged plates.
Note that E here is independent of the distance from
the plates!
  E 0  2.00 104 N/C8.85 1012 C 2 /Nm 2 
 1.77 10 7 C/m 2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(c) What is the vertical deflection d of the electrons as they
leave the plates?
y
FBD for an electron
in the beam:
Fe
x
w
Apply Newton’s 2nd Law and solve for the acceleration:
F
y
 Fe  w  ma y
Fe  w Fe
qE
ay 

g 
 g  3.52 1015  9.8 m/s
m
m
m


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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
What is the vertical position of the electron after it travels a
horizontal distance of 2.0 cm?
1 02
x  x0  voxt  a x t
2
x  x0
0.02 m
10
t


5
.
0

10
sec
7
v0 x
4.0 10 m/s
Time interval to
travel 2.00 cm
horizontally
0
1 2
y  y0  voyt  a y t
2
y  y0  d 
1 2
a y t  4.4 10  4 m
2
Deflection of an
electron in the
beam
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§16.6 Conductors in Electrostatic
Equilibrium
Conductors are easily polarized. These materials have free
electrons that are free to move around inside the material.
Any charges that are placed on a conductor will arrange
themselves in a stable distribution. This stable situation is
called electrostatic equilibrium.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
When a conductor is in electrostatic equilibrium, the E-field
inside it is zero.
Any net charge must reside on the surface of a conductor
in electrostatic equilibrium.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Just outside the surface of a conductor in electrostatic
equilibrium the electric field must be perpendicular to the
surface.
If this were not true, then any surface
charge would have a net force acting
on it, and the conductor would not be
in electrostatic equilibrium.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Any excess charge on the
surface of a conductor will
accumulate where the
surface is highly curved
(i.e. a sharp point).
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§16.7 Gauss’s Law
Enclose a point
charge +Q with an
imaginary sphere.
+Q
Here, E-Field lines exit the sphere.
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Look at a small patch
of the surface of the
imaginary sphere.
With a positive charge
inside the sphere you
would see electric field
lines leaving the surface.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
number of field lines
Recall that E 
A
so that the
number of field lines  EA
It is only the component of the electric field that is
perpendicular to the surface that exits the surface.
E

Surface
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Define a quantity called flux, which is related to the number
of field lines that cross a surface:
flux   e  E A  E cos  A
E
This picture defines
the value of .

Flux > 0 when field lines exit the surface and flux < 0
when field lines enter the surface.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 16.58): Find the electric flux through
each side of a cube of edge length a in a uniform electric field
of magnitude E.
A cube has six sides: The field lines enter one face and exit
through another. What is the flux through each of the other
four faces?
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
There is zero electric flux though the other four faces.
The electric field lines never enter/exit any of them.
The flux through the left face is –EA.
The flux through the right face is +EA.
The net flux through the cube is zero.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Since Eq, the flux through a surface can also be written as
e 
Qinside
0
This is Gauss’s Law.
The flux through a surface depends on the amount of
charge inside the surface. Based on this, the cube in
the previous example contained no net charge.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Summary
•Properties of Conductors/Insulators
•Charge Polarization
•Coulomb’s Law
•The Electric Field
•Motion of a Point Charge in an Electric Field
•Gauss’s Law
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