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Motion in Electric Fields
SACE Stage 2 Physics
1
Energy Changes in Electric
Fields
Consider the movement of a small charge in a uniform electric field
To lift a charge towards the top (positive)
plate we exert an external force
B
A
Fext= FE in size (opposite direction). Increase
10 m
q = +2 C
q = +4 C
in electrical Potential Energy (EP)
= Work Done by external Force
= Fext x S
= qE h (since F = E.q)
E = 10 N C-1
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Energy Changes in Electric
Fields
The work done on each charge is,
A
B
E = 10 N C-1
q = +2 C
w  qEh
 2 10 10
 200 J
10 m
q = +4 C
w  qEh
 4 10 10
 400 J
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Energy Changes in Electric
Fields
We define the Work done (in moving a small positive test charge from
one position to another) per unit positive charge as the change in
electric potential, V.
W
V 
q
Work done in moving each charge,
+2C Charge,
200 J
V
2C
 100 JC 1
+4C Charge,
400 J
V
4C
 100 JC 1
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Energy Changes in Electric
Fields
A
B
E = 10 N C-1
q = +2 C
10 m
q = +4 C
We say the top plate is 100J/C higher in potential than the bottom
plate.
A potential difference of 1J/C is also referred to as 1 volt.
ie. we define the unit of potential difference as, one volt
(V) equals one joule per coulomb (J.C-1)
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Energy Changes in Electric
Fields
Both charges are at the same potential compared to the bottom plate.
They are on an equipotential line.
The electric field is affecting the potential of the charges in a similar way.
Each charge is at the same potential relative to the bottom plate. The larger
charge has the higher potential energy.
Eg, consider the 4 C charge
100V
W
V 
q
W  qV
 4C 100 JC 1
 400 J
+4 C
0V
6
Electron Volt
Work done when a charge of one electron moves through a
potential difference of 1 V is one electron volt (e.V). (Is a unit of
energy)
The equivalent energy is:
q = e = 1.6 x 10-19C,
hence
1 eV
and 1 V = 1 J C-1
= 1.6 x 10-19C x 1 J C-1
ie.
1 eV = 1.6 x 10-19 J
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Relationship Between E and ΔV
Calculate the work done by
considering the force that needs to
be exerted to move the charge
against the field.
V2

s
V
W = Fs
=qEs
FE = qE
Separation
between the
plates = d
V1
Calculate the work done using
voltage
W = qV
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Relationship Between E and ΔV
These expressions for the Work Done should be the same. Therefore
qEs = qV
ie.
V
E
s
Electric field strength is equal to the number of volts potential difference per
metre of distance in the field
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Electric Field Strength Between
Parallel Plates
For parallel plates separated by a distance d, with a potential difference V,
the uniform electric field within the plates has strength:
V
E
d
We sometimes use an alternate unit for E.
Voltage
distance
Volts

metre
 Vm 1
E
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Electric Field Strength Between
Parallel Plates
Example – 2 parallel plates separated by 0.1m have a potential difference
ΔV = 100V. What is the Electric Field strength between the plates?
V
E
d
100V

0.1m
 1000Vm1 or 1000 NC 1
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Motion in an Electric Field
Charged particles that move through electric fields behave the same way
as mass does in a gravitational field. The corolation is as follows,
Mass  Charge
Gravitational Field  Electric Field
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Motion in an Electric Field
Consider a positive charge placed in a uniform electric field, as shown in
the diagram below. (Note the direction of the Electric field is the direction
that a positive charge would move in that field)
Electric Field
+ + + + + + + + + + + +
0.1m
q=10μC
+
FE
- - - - - - - - - - - -
1000V
M=0.1g
0V
Find the velocity of the charge after it has travelled a distance of 5 cm. Use the
following information:
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Motion in an Electric Field
Electric Field
+ + + + + + + + + + + +
V
E
d
1000

 1 10 4 Vm 1
0 .1
 1 10 4 NC 1
F  qE
 10  10 6  1 10 4
 10 1 N
q=10μC
+
0.1m
FE
- - - - - - - - - - - -
1000V
M=0.1g
0V
F
m
10 1
  4  103 ms  2
10
a
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Motion in an Electric Field
Can use the equations of motion to determine the speed of particle after
travelling for 5cm.
v1  0ms 1
s  0.05m
a  103 ms 2
v2  ?
v 22  v12  2as
v22  2as (v1  0ms 1 )
v2  2as
v2  2 103  0.05
v2  10ms 1 towards the -' ve plate
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Motion in an Electric Field
Can also determine the velocity by using the change in kinetic energy of
the particle.
Electric Field
+ + + + + + + + + + + +
1000 V
V
E
d
 10000Vm1
A
0.05 m
0.1 m
B
0V
- - - - - - - - - - - -
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Motion in an Electric Field
To find the potential difference between A
and B, rearrange the equation,
Electric Field
+ + + + + + + + + +
+ +
A
1000 V
0.05 m
0.1 m
B
V
s
 V  Es
E
- - - - - - - - - - -
0V
 V  10000Vm 1  0.05m
 V  500V
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Motion in an Electric Field
Now calculate the kinetic energy at point B. If the charge is released at rest
K at B (Gain in K) = electric Ep lost
1
2
mv 2  qV
2qV
v 
m


2  10 10 6  500

10  4
 10ms 1 towards the -' ve plate
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Motion in an Electric Field
The work done by the field on the charge can be calculated easily because
it is equal to the gain in kinetic energy by the charge.
EK  W  qV
 10 5  500
 5 10 3
 5mJ
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Motion Perpendicular to the
Electric Field
Assumptions
- ignore fringe effects (ie. assume that the field is completely uniform)
- ignore gravity (it is quite easy to show that the acceleration due to
gravity is insignificant compared with the acceleration caused
by the electric field).
-Before entering electric field, the charge follows a straight line path
(no net force)
-As soon as it enters the field, the charge begins to follow a parabolic
path (constant force always in the same direction)
- As soon as it leaves the field, the charge follows a straight line
path (no net force)
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Motion Perpendicular to the
Electric Field
Straight line
parabola
+q
Straight line
Horizontal Component of the velocity (H component)
vh  v1h  v2 h horizontal velocity is constant
L
vh 
t
a0
L
so t 
vh
as
v  0
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Motion Perpendicular to the
Electric Field
Vertical Component of the
velocity (v component)
As it is initially travelling
horizontally, vy1 = 0 m/s
F qE
Now av  
m m
So, v y2  at

qE L
m vx
L
So, s v  2 av  
 vx 
2
L
 sv  1 2 av 2
vx
2
1
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Motion Perpendicular to the
Electric Field
Electric Field
+ + + + + + + + + + + +
v
Direction of F on +
+q
-q
v
Direction of F on -
- - - - - - - - - - - - - - -
The direction of the force depends on the charge on the particle.
However, the force at all times is parallel (or "anti-parallel') to the
field.
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Motion Perpendicular to the
Electric Field
Example: Consider a negative
charge entering a uniform electric
field initially perpendicular to the
field. The acceleration will always
be in the opposite direction to
the electric field lines.
Find: (a) The time taken for an electron to
pass through the field
(b) The sideways deflection of the
electron beam in the field.
(c) The final velocity of the electrons
when they leave the field.
(d) The trajectory (path) of the electron
beam on exit from the field.
2000 V
Electric
Field
+
+
+
+
+
+
+
+
+
+
+
d = 10
cm
0V
-
L = 10 cm
v = 5.9 x 107 m/s
Electron Beam
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Motion Perpendicular to the
Electric Field
(a) The time taken for an electron to pass
through the field.
v (perpendicular) does not change as the force
only acts parallel to the field.
The time taken for the beam to pass through the
field is given by
s L
t

v
v
0.1

5.9 10 7
 1.7 10 9 s or 1.7ns
2000
Elect
V
ric
Field
+
+
+
d=
10
cm
0V
L = 10
cm
v = 5.9- x
107 m/s
Electron
Beam
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Motion Perpendicular to the
Electric Field
(b) The sideways deflection of the
electron beam in the field.
To find the sideways deflection, we
need to consider the component
of the velocity || to the field.
9
t  1.7 10 s
vi parallel  0ms 1
Need to find the acceleration,
V
F qE
a 
 s
m m
m
qV
1.6 10 19 C  2000V 


ms 9.1110 31 Kg  0.1m 
q




 a  3.5 1015 ms  2 towards ' ve plate
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Motion Perpendicular to the
Electric Field
Can now use the value for acceleration to substitute in to the following
equation to determine the deflection.
s  1 2 at 2

1
2

 3.5 10
15
 1.7 10 
9 2
 s  5.110 3 m
The deflection of the electron beam is 5.1 x 10-3m towards the +’ve
plate.
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Motion Perpendicular to the
Electric Field
(c) The final velocity of the electrons when they leave the field.
Need to use a vector diagram to calculate
Final velocity = final  velocity + final || velocity
Need to calculate vparallel,
v parallel  at

 
 3.5  1015  1.7 10 9

 5.95  106 ms 1
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Motion Perpendicular to the
Electric Field
By Pythagoras’ Theorem,
v
v 2  v 2parallel  v 2perpendicular

 5.95  10
  5.9 10 
6 6
 v  5.9  10 ms
7
1
7 2
v
v

5.95  10 6
tan  
5.9  10 7
  5.80
The electron beam leaves the field at 5.9 x 107ms-1 at 5.8o towards the +’ve plate.
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Motion Perpendicular to the
Electric Field
(d)
The trajectory (path) of the electron beam on exit from the field.
As soon as an electron leaves the field there is no force on it and hence
the path of the beam is a straight line. (Newton's First Law).
ie. Velocity is in the same direction as the final velocity in part (c)
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Use of Electric Fields in
Cyclotrons
The cyclotron is a device for accelerating particles to high velocities, generally
for the purpose of allowing them to collide with atomic nuclei in a target to
cause a nuclear reaction. The results of these reactions are used in research
about the nucleus, but can be used to make short lived radioactive isotopes
used in nuclear medicine.
31
Use of Electric Fields in
Cyclotrons
Alternating potential
difference
Main components of a cyclotron
There are 4 main parts to a cyclotron
1.
2.
3.
4.
evacuated outer
chamber
The ion source
Two semicircular metal
containers called 'dees'
because of their shape
An evacuated outer container
An electromagnet.
plan view
dee
target
ion
source
Magnet
dee
side view
Magnet
32
Use of Electric Fields in
Cyclotrons
The ion source
Produces the charged particles for the cyclotron.
Done by passing the gas over a hot filament where electrons being
emitted can ionise the gas. More modern cyclotrons pass the gas to be
tested through an electric arc.
33
Use of Electric Fields in
Cyclotrons
Semicircular Metal Containers, or
'dees‘
The dees are two hollow dee shaped
metal electrodes with their straight edges
facing. A large alternating voltage is
applied between the dees (~ 1KV and 10
MHz). The high voltage establishes an
electric field between the dees which
reverses every time the alternating
current reverses. Note that there is no
electric field within the dees - there is no
electric field inside a hollow conductor.
Alternating potential
difference
evacuated outer
chamber
plan view
dee
target
ion
source
Magnet
dee
side view
Magnet
34
Use of Electric Fields in
Cyclotrons
Evacuated outer container
The dees are placed within an outer evacuated container so that the ions
being accelerated do not suffer energy loses due to collisions with air
molecules, or that they are not scattered away from their intended paths by
the collisions.
In addition to reducing the beams intensity, scattered ions may collide with
the walls at high energies. This can cause the walls to become radioactive.
35
Use of Electric Fields in
Cyclotrons
The electromagnets.
The electromagnets produce a strong magnetic field , but unlike the electric
field, this field can pass through the dees. This magnetic field causes the
ions to move in a circular pathway.
36
Use of Electric Fields in
Cyclotrons
Principles of Operation
Ions in the cyclotron are accelerated when they cross the electric field
between the dees.
The magnetic field in the apparatus ensures the ionic particles travel a
circular path. The particles will the repeatedly keep crossing the Electric
field which only exists between the dees.
37
Use of Electric Fields in
Cyclotrons
At position A, the ionic particle is
accelerated to the right dur to the electric
field. It then enters the dee where there
is no electric field (no electric field inside
of a conductor). The magnetic field
causes the ionic particle to traverse a
circular path. While doing so the Electric
field changes direction due to the
alternating current across the dees.
38
Use of Electric Fields in
Cyclotrons
At position B, the ionic particle is
accelerated across the gap and
then enters the other side of the
dee where there is only a
magnetic field to keep the ionic
particle travelling in a circular
path. The Electric field changes
again to the opposite direction.
39
Use of Electric Fields in
Cyclotrons
The particle is now at C where
the whole process is repeated.
The radius of the circle is only
depended upon the velocity of
the ionic particle and as it speeds
up the radius becomes bigger.
40
Use of Electric Fields in
Cyclotrons
The accelerated ionic particles
can be evacuated by placing
deflecting electrodes that will
cancel the magnetic field in
that region so the path of the
particles becomes a straight
line.
41
Use of Electric Fields in
Cyclotrons
The Energy Transfer to the ions by the Electric Field.
When the ions pass the gap, their speed increases, hence they have a gain
in kinetic energy.
Work done on the ion is given by,
W  qV
The work done is equal to the increase in Kinetic Energy. As the particles
pass the gap many times, they end up with a large increase of Kinetic
Energy.
42
Use of Electric Fields in
Cyclotrons
The energetic protons are bombarded with stable atoms of carbon, nitrogen,
oxygen, or fluorine to produce certain radioactive forms of these elements.
The radioactive forms are combine with elements commonly found n the
body such as glucose.
After small amounts are administered into a patient, it can then be traced to
determine the functions of the body. It can also be used in the treatment of
certain kinds of cancer in certain organs where the chemical tends to
concentrate.
43