Download + + + + + + + v

Uniform Fields
Fields
Gravity Field g
Field strength
Also the acceleration of gravity
Only 9.8 m/s2 on Earth’s surface.
Leave the surface and g changes
Electric Field E
Field strength
Does not give acceleration directly
Different in every problem (there is no 9.8 equivalent)
Fields of Flat Objects
Charged plates usually consist of two equal sized plates separated by a distance d .
The electric field between two charged plates most closely resembles the gravity field
close to the earth, where the flat earth appears most similar to the flat plates.
+ + + + + + + +
+ sky
m
g
m
m
− ground
m
E+
+
+
+
− − − − − − − −
To find the direction of each field test it with a test mass or test charge.
Position an imaginary mass or charge at the location where you want the field
direction. Then determine the direction that the test mass or charge will move.
There is only one kind of mass, but two kinds of charge. Use a positive test
charge, since positive charges move in the same direction as the electric field.
Force on Objects Placed in the Field
+ + + + + + + +
+ sky
m
+
E
g
Fg
− ground
When a mass is placed in the gravity
field of the earth, the earth’s gravity
field puts a force on the mass.
Fg = mg
The direction of force is the same as
the direction of the field.
FE
FE
−
− − − − − − − −
When charges are placed in an
electric field, the electric field puts
a force on the charges.
FE = qE
FE = qE
The direction of force on positive
charges matches field direction.
However, negative charges move
opposite the field (minus = opposite).
Force on Objects Placed in the Field
+ + + + + + + +
+ sky
m
+
E
g
Fg
FE
FE
− ground
− −
How do you solve for acceleration of each particle ?
−
− − − − − −
SF = Fg
ma = mg
SFE = FE
ma = qE
SFE = FE
ma = qE
a= g
qE
a=
m
qE
a=
m
Now how would you solve for the distance
moved, the time traveled, and the final speed?
Kinematics
Two Types of Force and Field Problems
Static
Charges placed inside an electric field remain stationary.
This results when forces are in equilibrium
Strategy: Use balanced forces. Set opposite forces equal to each
other. The sum of forces and acceleration is zero.
Dynamic
Charges placed inside an electric field accelerate.
This occurs when forces are unbalanced.
Strategy: Sum forces to find acceleration, and then use kinematics to
find speed, distance, and time.
Example 1
Milikan’s Oil Drop
Milikan suspended a charged oil drop between two
charged plates as shown in the diagram at the right.
Solve for the charge on the oil drop.
FE = Fg
qE = mg
mg
q=
E
+ + + + +
FE
−
Fg
− − − − −
Milikan ran several trials. His results consisted of charge values such as
6.4×10-19 C , 3.2×10-19 C, 1.6×10-19 C, and 4.8×10-19 C.
All of these values are multiples of 1.6×10-19 , and he never recorded a
result smaller than 1.6×10-19 C.
Milikan concluded that on trials with a charge of 1.6×10-19 C the oil drop
held one extra electron, and that 1.6×10-19 C was the charge of an electron.
The other trials contained more than one electron. This means that all charge
is a multiple of the charge on one electron, and charge is conserved.
Example 2
A spherical conductor has a charge of −8.0 μC
and a mass of 5.0×10−2 kg. It is tied to a string
attached to the ceiling. The room has an electric
field passing through it, which causes the
charged spherical mass to hang at an angle of 30o
from the vertical as shown in the diagram.
a.
30o
−0.8 μC
Draw a FBD showing the forces acting on the sphere.
When an object is at rest forces are often the
key to the problem. Start by drawing a FBD.
Then remember that the force vectors have to
cancel when drawn tail to tail, or add to zero if
drawn tip to tail.
This time we include gravity. If you can tie a
string to the mass, then it is certainly large
enough that gravity matters.
T
FE
Fg
Example 2
A spherical conductor has a charge of −8.0 μC
and a mass of 5.0×10−2 kg. It is tied to a string
attached to the ceiling. The room has an electric
field passing through it, which causes the
charged spherical mass to hang at an angle of 30o
from the vertical as shown in the diagram.
30o
b. Calculate the force of electricity acting on the sphere.
Tip to tail the vectors form a right triangle.
Use right triangle trig to find FE .
FE
tan q =
Fg
Fg = mg
Fg = 0.05 10
Fg = 0.5 N
(
Fg is easy to calculate.
)( )
30o
T
FE
( )
FE = Fg tan q = 0.5 tan 30 = 0.29 N
o
Example 2
A spherical conductor has a charge of −8.0 μC
and a mass of 5.0×10−2 kg. It is tied to a string
attached to the ceiling. The room has an electric
field passing through it, which causes the
charged spherical mass to hang at an angle of 30o
from the vertical as shown in the diagram.
c.
30o
Determine the strength of the electricity field
generating the force on the sphere.
FE = qE
(0.29) = (8.0 ´ 10 ) E
-6
E = 36000 N C
Fg = 0.5 N
30o
T
FE = 0.29 N
d. State the direction of the electric field.
−x The negative charge is being forced to the right. Negative charges
move opposite the field, so the field must be to the left.
Possible Uniform Electric Field
− − − − −
− − − − −
+ + + + +
+ + + + +
− − − − −
− − − − −
+ + + + +
+ + + + +
Charged plates are the most similar to the uniform gravity field
found near the surface of a large mass, like Earth.
But, plates can be oriented in different ways, where at the Earth’s
surface we only consider an up/down arrangement.
And others, including
pointing into and out
of this screen. Always
think 3D.
Stationary Charge Released in a Uniform Electric Field
Positive charges follow the electric field lines.
Since positive charges move with the field they define the direction
the electric field direction. If the field is down positive charge
move down. If positive charges move down then the field in down.
However, negative charges (the opposite, wrong way, particle)
move opposite the electric field. They fall up toward positive.
+ + + + + +
+
−
− − − − − −
Object moving in direction of the Field
In gravity the particle is thrown downward and accelerates (forces
in direction of motion cause +a).
In electricity positive particles accelerate (forces in direction of
motion cause +a).
But, negative particles decelerate (forces opposing motion cause
−a). Negative sign means opposite.
v0
M
+ + + + +
+
v0
−
m
v0
− − − − −
Particle moving opposite the Field
In gravity the particle is thrown upward and decelerates (forces
opposing motion cause −a).
In electricity positive particle decelerate (forces opposing motion
cause −a).
But, negative particles accelerate (forces in direction of motion
cause +a). Negative sign means opposite.
m
v0
+
M
v0
−
v0
+ + + + +
− − − − −
Example 3
E
−
E
− − − − − − −
+ + + + + + +
An electron is located at a point in space where the electric
field has a magnitude 30 N/C in the positive x-direction.
a. Determine the magnitude of force on the electron..
Assume the simplest case: a uniform field
consisting of parallel field lines. While we were
never told what created this field, we know it
has to have been plates. If it helps to solve the
problem draw them or imagine them.
F
You have a field and a single charge try:
FE = qE
(
FE = 1.6 ´ 10
FE = 4.8 ´ 10
-19
) (30)
-18
N
Constants , like
the charge and
mass of an
electron will be
found on the
constant page.
b. Determine the direction of force on the electron..
1st method: Negative charges are opposite particles. Field +x , force is −x
2nd method: Objects released from rest accelerate in the direction of the
force acting on them. Which way will the electron go. −x
Example 1
FE
−
E
− − − − − − −
+ + + + + + +
An electron is located at a point in space where the electric
field has a magnitude 30 N/C in the positive x-direction.
c. Determine the acceleration of the electron.
SF = FE
ma = qE
(
) (
)( )
9.11´ 10-31 a = 1.6 ´ 10-19 30
a = 5.27 ´ 10 m s
12
2
d. If the electron is released from rest, how long will it take to travel 20 cm.
1 2
Dx = v 0 t + at
2
1 2
Dx = at
2
(
)
(
)
1
0.20 = 5.27 ´ 1012 t 2
2
t = 2.76 ´ 10-7 s
Motion Perpendicular to a Uniform Field
Close to a large mass the gravity is nearly parallel, and is uniform
in magnitude.
A mass that moves perpendicular to this field will experience
projectile motion.
Charges moving in electric fields also experience projectile motion.
m
v0
M
Motion Perpendicular to a Uniform Field
The electric plate configuration that looks most like gravity is the
configuration with a positive charge moving through plates where
the positive plate is on top.
+ + + + + + + + +
v0
− − − − − − − − −
Again electricity does not have a traditional up and down, and
charged plates can have many configurations.
Motion Perpendicular to a Uniform Field
The plates could be upside down
+ + + + + + + + +
− − − − − − − − −
+
v0
v0
− − − − − − − − −
+
• If the orientation is confusing just turn
the test problem upside down, or side
ways. Then it will look normal.
+ + + + + + + + +
or side ways
Motion Perpendicular to a Uniform Field
The charged particle could be negative.
Try to imagine all the possible scenarios. There are many.
Regardless, the motion perpendicular to the field is constant
velocity, while the motion parallel to the field is acceleration.
+ + + + + + + + +
−
v0
− − − − − − − − −
Motion Perpendicular to a Uniform Field
One other difference with gravity is the fact that the electric field is
limited only to the region between the plates.
Before entering the plates and after leaving the plates there is no
electric force, so the particle has only inertia.
+ + + + + +
+
v0
Inertia
Force field portion
Projectile motion
only in this region
− − − − − −
Inertia
Motion Perpendicular to a Uniform Field
Acceleration while in the electric field
is calculated as before.
This acceleration is directed along the
electric field lines for positive charges
It is opposite the field lines for negative charges.
+
+ + + + + +
v0
− − − − − −
qE
a=
m
In the diagram at the left the
acceleration is directed
downward, perpendicular to
the initial motion.
Motion Perpendicular to a Uniform Field
The projectile motion portion is analogous to a horizontal launch,
with initial velocity perpendicular to the field.
The projectile part only takes place during the time that the
projectile is between the plates.
+
Therefore, the Δx
displacement is equal to
the length of the plates.
+ + + + + +
v0
Δy
− − − − − −
Δx
The Δy displacement is
the vertical change in
motion (deflection) while
between the plates.
Motion Perpendicular to a Uniform Field
The formulas for x and y are the traditional horizontal launch
simplifications of the kinematic equations.
+
1 2
Dy = at
2
+ + + + + +
v0
Δy
− − − − − −
Δx
Dx = v0x t
The plates have an electric field E = 150 N/C and a length of
30 cm. An electron is fired between the plates, v = 8.0× 106 m/s
.
Draw the path of the electron and mark the vertical deflection.
Example 4
a.
+ + + + + + + +
Δy
−
– – – – – – – –
b. Determine the acceleration of the electron when it is between the plates.
SF = FE
ma = qE
(
) (
)( )
9.11´ 10-31 a = 1.6 ´ 10-19 150
a = 2.63 ´ 1013 m s2
The plates have an electric field E = 150 N/C and a length of
30 cm. An electron is fired between the plates, v = 8.0× 106 m/s
.
Determine the vertical deflection
Example 4
c.
+ + + + + + + +
a = 2.63 ´ 10 m s
13
2
Δy
−
– – – – – – – –
1 2
Dy = at
2
x = v0x t
(0.30) = (8.0 ´ 10 ) t
1
Dy = 2.63 ´ 1013 3.75 ´ 10-8
2
t = 3.75 ´ 10-8 s
Dy = 0.0185 m
6
(
)(
)
2