2. motion in electric fields

7/29/2019 2. Motion in Electric Fields

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Motion in Electric Fields

SACE Stage 2 Physics


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Energy Changes in Electric

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

B A

q = +4 Cq = +2 C

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

F ext = F E in size (opposite direction). Increasein electrical Potential Energy (EP)

= Work Done by external Force= F ext x S = qE h ( since F = E.q)

10 m

E = 10 N C -1


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Fields

J

hqE w

200

10102

The work done on each charge is,

J

hqE w

400

10104

B A

q = +4 Cq = +2 C

10 mE = 10 N C -1


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Fields

1

100

2200

JC

C J

V

We define the Work done (in moving a small positive test charge fromone position to another) per unit positive charge as the change inelectric potential, V.

Work done in moving each charge,

+2C Charge,

1100 4

400

JC C J

V

+4C Charge,

V W q


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FieldsB A

q = +4 Cq = +2 C

10 mE = 10 N C -1

We say the top plate is 100J/C higher in potential than the bottomplate.

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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FieldsBoth 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 largercharge has the higher potential energy.

Eg, consider the 4 C charge

+4 C

100V

0V

J

JC C

V qW q

W V

400

1004 1


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Electron VoltWork done when a charge of one electron moves through apotential 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 -19 C, 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


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Relationship Between E and V

F E = qE

s Separationbetween theplates = d

V1

V2

V

Calculate the work done byconsidering the force that needs tobe exerted to move the chargeagainst the field.

W = F s = qE s

Calculate the work done usingvoltage

W = q V


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Relationship Between E and V

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

qE s = q V

ie. E V s

Electric field strength is equal to the number of volts potential difference permetre 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:

E V d

We sometimes use an alternate unit for E.

1

distance

Vm

metreVolts

Voltage E


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Electric Field Strength Between

Parallel PlatesExample 2 parallel plates separated by 0.1m have a potential difference V = 100V . What is the Electric Field strength between the plates?

11 1000 1000 1.0

100

NC or Vmm

V d V

E


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Motion in an Electric FieldCharged particles that move through electric fields behave the same wayas 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 FieldConsider a positive charge placed in a uniform electric field, as shown inthe diagram below. ( Note the direction of the Electric field is the direction that a positive charge would move in that field )

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

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

F E

+

0V

1000V

0.1mq=10 C

M=0.1g

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


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Motion in an Electric Field

N

qE F

NC

Vm

d

V

E

1

46

14

14

10

1011010

101

1011.0

1000

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

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

F E

+

0V

1000V

0.1mq=10 C

M=0.1g

234

1

101010

ms

m F a


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Motion in an Electric FieldCan use the equations of motion to determine the speed of particle aftertravelling for 5cm.

plateve-' thetowards10

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


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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 Vmd V

E


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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,

V V mVmV

s E V sV

E

50005.010000

1


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

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

plateve-' thetowards10

1050010102

2

1

4

6

22

1

ms

mV q

v

V qmv


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The work done by the field on the charge can be calculated easily becauseit is equal to the gain in kinetic energy by the charge.

mJ

V qW E K

5 105

50010 3

5


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Motion Perpendicular to theElectric Field

+ q Straight line

parabolaStraight line

Horizontal Component of the velocity (H component)

0 as 0

so

constantisvelocityhorizontal 21

va

v L

t t

Lv

vvv

hh

hhh


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Vertical Component of the velocity (v component)

As it is initially travellinghorizontally , v y1 = 0 m/s

2

2

21

2

21

v

y 2

s ,

v,

Now

xvv

xv

x

v

v L

a s

v LaSo

v L

mqE

t aSomqE

m F

a


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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 thefield.


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Example: Consider a negativecharge entering a uniform electricfield initially perpendicular to the

field. The acceleration will alwaysbe in the opposite direction tothe electric field lines.

L = 10 cm

ElectricField

+++++++++++

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

v = 5.9 x 10 7 m/s

0 V2000 V d = 10cm

Electron Beam

Find: (a) The time taken for an electron topass through the field(b) The sideways deflection of theelectron beam in the field.(c) The final velocity of the electronswhen they leave the field.(d) The trajectory (path) of the electronbeam on exit from the field.


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L = 10cm

ElectricField

+

+

+

-----

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

10 7 m/s

0 V2000 V

d =10cm

ElectronBeam

(a) The time taken for an electron to passthrough 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 thefield is given by

nsor s

v

L

v

st

7.1 107.1 109.51.0

9

7


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1

9

0

107.1

msv

st

paral le l i

(b) The sideways deflection of theelectron beam in the field.

To find the sideways deflection, weneed to consider the componentof the velocity || to the field.

Need to find the acceleration,

platevetowardsmsa

m Kg V C

smV q

m sV q

mqE

m F

a

' 105.3

1.01011.92000106.1

215

31

19


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

m s

at s

3

29152

1

22

1

101.5

107.1105.3

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


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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 v parallel ,

16

915

1095.5

107.1105.3

ms

t av parallel


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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 lar perpendicu parallel

The electron beam leaves the field at 5.9 x 107ms -1 at 5.8 o 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 hencethe 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 inCyclotrons

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 tocause a nuclear reaction. The results of these reactions are used in researchabout the nucleus, but can be used to make short lived radioactive isotopesused in nuclear medicine.


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Main components of a cyclotron

There are 4 main parts to a cyclotron

1. The ion source2. Two semicircular metal

containers called 'dees'because of their shape

3. An evacuated outer container 4. An electromagnet.

dee

plan view

side view

Magnet

Magnet

Alternating potentialdifference

ion

sourcetarget

dee

evacuated outer

chamber


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The ion source

Produces the charged particles for the cyclotron.Done by passing the gas over a hot filament where electrons beingemitted can ionise the gas. More modern cyclotrons pass the gas to betested through an electric arc.


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Semicircular Metal Containers, or'dees

The dees are two hollow dee shapedmetal electrodes with their straight edgesfacing. A large alternating voltage isapplied between the dees (~ 1KV and 10MHz). The high voltage establishes anelectric field between the dees whichreverses every time the alternatingcurrent reverses. Note that there is noelectric field within the dees - there is noelectric field inside a hollow conductor.

dee

plan view

side view

Magnet

Magnet

Alternating potentialdifference

ionsource

target

dee

evacuated outer chamber


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Evacuated outer container

The dees are placed within an outer evacuated container so that the ionsbeing accelerated do not suffer energy loses due to collisions with airmolecules, or that they are not scattered away from their intended paths bythe collisions.

In addition to reducing the beams intensity, scattered ions may collide withthe walls at high energies. This can cause the walls to become radioactive.


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The electromagnets .

The electromagnets produce a strong magnetic field , but unlike the electricfield, this field can pass through the dees. This magnetic field causes theions to move in a circular pathway.


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Principles of Operation

Ions in the cyclotron are accelerated when they cross the electric fieldbetween the dees.

The magnetic field in the apparatus ensures the ionic particles travel acircular path. The particles will the repeatedly keep crossing the Electricfield which only exists between the dees.


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At position A, the ionic particle isaccelerated to the right dur to the electricfield. It then enters the dee where thereis no electric field (no electric field insideof a conductor). The magnetic fieldcauses the ionic particle to traverse acircular path. While doing so the Electricfield changes direction due to thealternating current across the dees.


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At position B, the ionic particle isaccelerated across the gap and

then enters the other side of thedee where there is only amagnetic field to keep the ionicparticle travelling in a circularpath. The Electric field changesagain to the opposite direction.


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The particle is now at C wherethe whole process is repeated.

The radius of the circle is onlydepended upon the velocity of the ionic particle and as it speedsup the radius becomes bigger.


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The accelerated ionic particlescan be evacuated by placingdeflecting electrodes that will

cancel the magnetic field inthat region so the path of theparticles becomes a straightline.


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The Energy Transfer to the ions by the Electric Field .

When the ions pass the gap, their speed increases, hence they have a gainin kinetic energy.

Work done on the ion is given by,

V qW

The work done is equal to the increase in Kinetic Energy. As the particlespass the gap many times, they end up with a large increase of KineticEnergy.


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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 todetermine 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 toconcentrate.

2. motion in electric fields

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