2._Motion_in_Electric_Fields.ppt

1

Motion in Electric Fields

SACE Stage 2 Physics

2

Energy Changes in Electric Fields

Consider the movement of a small charge in a uniform electric field

BA

q = +4 Cq = +2 C

To lift a charge towards the top (positive) plate we exert an external force

Fext= FE in size (opposite direction). Increase in electrical Potential Energy (EP)

= Work Done by external Force= Fext x S= qE h (sinceF = E.q)

10 m

E = 10 N C-1

3


J

hqEw

200

10102

The work done on each charge is,

J

hqEw

400

10104

BA

q = +4 Cq = +2 C

10 mE = 10 N C-1

4


1100

2

200

JC

C

JV

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.

Work done in moving each charge,

+2C Charge,

1100

4

400

JC

C

JV

+4C Charge,

VW

q

5


BA

q = +4 Cq = +2 C

10 mE = 10 N C-1

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)

6


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

+4 C

100V

0V

J

JCC

VqW

q

WV

400

1004

1

7

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, and 1 V = 1 J C-1

hence 1 eV = 1.6 x 10-19C x 1 J C-1

ie. 1 eV = 1.6 x 10-19 J

8

Relationship Between E and ΔV

FE = qE

s

Separation between the plates = d

V1

V2

V

Calculate the work done by considering the force that needs to be exerted to move the charge against the field.

W = Fs =qEs

Calculate the work done using voltage

W = qV

9

Relationship Between E and ΔV

These expressions for the Work Done should be the same. Therefore

qEs = qV

ie. EV

s

Electric field strength is equal to the number of volts potential difference per metre of distance in the field

10

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:

EV

d

We sometimes use an alternate unit for E.

1

distance

Vm

metre

Volts

VoltageE

11

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?

11 1000 1000

1.0

100

NCorVm

m

Vd

VE

12

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

13

Motion in an Electric FieldConsider 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+ + + + + + + + + + + +

- - - - - - - - - - - -

FE

+

0V

1000V

0.1m

q=10μC

M=0.1g

Find the velocity of the charge after it has travelled a distance of 5 cm. Use the following information:

14


N

qEF

NC

Vm

d

VE

1

46

14

14

10

1011010

101

1011.0

1000

Electric Field+ + + + + + + + + + + +

- - - - - - - - - - - -

FE

+

0V

1000V

0.1m

q=10μC

M=0.1g

234

1

1010

10

ms

m

Fa

15


Can use the equations of motion to determine the speed of particle after travelling for 5cm.

plate ve-' the towards10

05.0102

2

)0(v 2

2v

? v10a 05.0s 0

12

32

2

11

22

21

22

2231

1

msv

v

asv

msasv

asv

msmmsv

16


Can also determine the velocity by using the change in kinetic energy of the particle.

Electric Field+ + + + + + + + + + + +

- - - - - - - - - - - -

1000 V

0 V

0.05 m0.1 m

A

B110000

Vm

d

VE

17

Motion in an Electric FieldElectric Field+ + + + + + + + + + + +

- - - - - - - - - - - -

1000 V

0 V

0.05 m0.1 m

A

B

To find the potential difference between A and B, rearrange the equation,

VV

mVmV

sEVs

VE

500

05.010000

1

18


Now calculate the kinetic energy at point B. If the charge is released at rest

K at B (Gain in K) = electric Ep lost

plate ve-' the towards10

10

50010102

2

1

4

6

22

1

ms

m

Vqv

Vqmv

19


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.

mJ

VqWEK

5

105

50010 3

5

20

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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+q

Straight line

parabolaStraight line

Horizontal Component of the velocity (H component)

0 as 0

so

constant is velocity horizontal 21

va

v

Lt

t

Lv

vvv

hh

hhh

22


Vertical Component of the velocity (v component)

As it is initially travelling horizontally, vy1 = 0 m/s

2

2

21

2

21

v

y2

s ,

v,

Now

xvv

xv

x

v

v

Las

v

LaSo

v

L

m

qE

taSom

qE

m

Fa

23


Electric Field+ + + + + + + + + + + +

- - - - - - - - - - - - - - -

+q

-q

v

v

Direction of F on +

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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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.

L = 10 cm

Electric Field + + + + + + + + + + +

-------------

v = 5.9 x 107 m/s

0 V2000 V

d = 10 cm

Electron Beam

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.

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L = 10 cm

Electric Field + + +

-------------v = 5.9 x

107 m/s

0 V2000 V

d = 10 cm

Electron Beam

(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

nsors

v

L

v

st

7.1 107.1

109.5

1.0

9

7

26


1

9

0

107.1

msv

st

paralleli

(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.

Need to find the acceleration,

platevetowardsmsa

mKg

VC

sm

Vq

msV

q

m

qE

m

Fa

' 105.3

1.01011.9

2000106.1

215

31

19

27


Can now use the value for acceleration to substitute in to the following equation to determine the deflection.

ms

ats

3

29152

1

22

1

101.5

107.1105.3

The deflection of the electron beam is 5.1 x 10-3m towards the +’ve plate.

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(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,

16

915

1095.5

107.1105.3

ms

tavparallel

29


v

v v

By Pythagoras’ Theorem,

0

7

6

17

2766

222

8.5

109.5

1095.5tan

109.5

109.51095.5

msv

vvv larperpendicuparallel

The electron beam leaves the field at 5.9 x 107ms-1 at 5.8o towards the +’ve plate.

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(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)

31

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.

32


Main components of a cyclotronThere are 4 main parts to a cyclotron

1. The ion source2. Two semicircular metal containers called 'dees' because of their shape3. An evacuated outer container 4. An electromagnet.

dee

plan view

side view

Magnet

Magnet

Alternating potential difference

ion sourcetarget

dee

evacuated outer chamber

33


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.

34


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.

dee

plan view

side view

Magnet

Magnet

Alternating potential difference

ion sourcetarget

dee

evacuated outer chamber

35


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.

36


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.

37


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.

38


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.

39


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.

40


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.

41


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.

42


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,

VqW

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.

43


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.

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2._Motion_in_Electric_Fields.ppt