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The uniform electric fieldQuestion 10W: Warm-up Exercise

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Instructions and informationWrite your answers in the spaces provided.

The following data will be needed:

magnitude of the electronic charge = 1.6 10–19 C

mass of electron = 9.1 10–31 kg

1 eV = 1.6 10–19 J

QuestionsA table-tennis ball covered in conducting paint is suspended between two metal plates, A and B,which are connected via a sensitive ammeter to a high-voltage power supply. If the ball is initiallygiven a positive charge it oscillates between plates A and B and the ammeter shows a small reading.

+e.h.t.

–

A B

ping pong ball paintedwith liquid graphite

1. After being given its initial charge, will the ball move to the right or to the left?

2. State two ways in which the reading on the ammeter can be increased.

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Here are two closely spaced metal plates connected to a 500 V supply.

+ 500 V

0 V

5 cm

3. Draw solid lines to represent the electric field both between the plates and just outside the plates.Add arrows to indicate the direction of the field.

4. Add, and label, dotted lines to represent lines of equipotential at 100 V intervals.

In an experiment to measure the charge on an oil drop the potential difference between two parallelmetal plates 5 mm apart was 300 V.

5. Calculate the electric field strength between the plates.

6. Calculate the electrical force on a small oil drop carrying a charge of 32 10–19 C.

7. Calculate the energy, in joules, gained by an electron accelerated through a potential differenceof 50 kV in an x-ray machine.

8. Calculate the speed of an electron with a kinetic energy of 100 eV.

Getting F = q v BQuestion 80W: Warm-up Exercise

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BackgroundIn chapter 15, the equation F = I L B is introduced, giving the magnitude of the force F on a wire oflength L in which a current I runs at right angles to a magnetic field. These questions are aboutfinding a similar expression for the force on an electron having a charge e and a velocity v. The sameargument can be applied to any charge carriers, for example ions in a mass spectrometer.

QuestionsSuppose that, in a time t, N electrons pass any place in the beam, such as Y.

v v v vwire

X Y

1. What is the charge passing Y in time t?

2. What is the current I (charge per second) at Y?

3. If the speed is v, what length, L, of beam passes Y in time t?

4. Use F = I L B and substitute the value of I from question 2 and of L from question 3. What do youget for the force on N electrons?

5. What is the force on a single electron?

6. What would be the force on a particle with charge Q?

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Speed and energy of particles – Newtonian calculationQuestion 10S: Short Answer

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In these questions you will calculate the speed of charged particles accelerated by different potentialdifferences. You will find where relativistic effects start to be important.

Relationships to useThe kinetic energy given to a particle with charge q accelerated through a potential difference V is:

Ekinetic

= qV

The speed of the particle, given its mass m can be found from the Newtonian relationship for kineticenergy:

221

kinetic mvE

This calculation of speed is only valid for speeds much smaller than the speed of light.

Data to useCharge on electron = 1.6 10–19 C

Mass of electron = 9.1 10–31 kg

Speed of light = 3.0 108 m s–1

Speed from kinetic energy

1. Given the equation for kinetic energy Ekinetic

= ½ mv 2, show that the speed v can be found from

m

Ev kinetic2

2. Calculate the kinetic energy of a 500 kg car travelling at 20 m s–1.

3. Calculate the speed of the same car, if its kinetic energy is double the previous value.

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4. Calculate the kinetic energy of an electron accelerated through a p.d. of 1000 V, typical of aschool laboratory electron tube. Use the value of the energy to calculate the speed of theelectron. What percentage of the speed of light is the speed you have calculated?

5. Repeat the calculations of kinetic energy and speed for electrons accelerated through a p.d. of 64000 V in a dental x-ray tube. What fraction of the speed of light does this speed represent?

6. In fact the calculation in question 5 is not quite correct. The speed cannot increase indefinitelybecause no material particle can reach the speed of light. But, if the calculation were correct, atwhat potential difference does the calculation say the electron would reach the speed of light?

0v / c

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

the two equationsgive the same resultfor all speeds smallcompared to thespeed of light

relativistickineticenergy

Newtoniankinetic energy

1.0

0

The graph above compares the Newtonian and relativistic calculations of kinetic energy. You can see

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that the two calculations are extremely close up to speeds of about 20% of the speed of light.

7. Find the potential difference through which an electron has to be accelerated to reach 20% of thespeed of light.

Speed and energy of particles – relativistic calculationQuestion 20S: Short Answer


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In these questions you will make calculation about charged particles accelerated by different potentialdifferences, at energies where relativistic effects are important.

Relationships to useThe kinetic energy given to a particle with charge q accelerated through a potential difference V is:

qVE kinetic

At any speed up to close to the speed of light the total energy is given by:

2total mcE

where

22 /1

1

cv

At rest, with v = 0, = 1 and the energy is rest energy Erest = mc2. The relativistic factor can be found

from the ratio of total energy to rest energy:

rest

total

E

E

The kinetic energy is the difference between total energy and rest energy, Ekinetic = Etotal – Erest ,

which can be written as

2kinetic 1 mcE

Data to useCharge on proton = 1.6 10–19 C

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Mass of proton = 1.7 10–27 kg, equivalent to rest energy nearly 1 GeV


Advice: work with energy unitsFor making relativistic calculations it is best to express quantities in energy units. The electron volt(eV) is often convenient.

Kinetic energy and factor1. The table below shows the energies achieved by various early particle accelerators. Taking Erest

= 1 GeV for a proton, add to the table giving the relativistic factor . Use = Etotal / Erest with Etotal =

Ekinetic + Erest.

Accelerator Date Particleaccelerated

Kineticenergy

factor Ratio v / c

Cockcroft and Walton 1932 Proton 770 keV

Van de Graaff 1932 Proton 1.5 MeV

Cyclotron 1937 Proton 3 MeV

Synchrocyclotron 1947 Proton 500 MeV

Synchrotron 1954 Proton 6 GeV

2. Add to the table the ratio v / c for the protons in each accelerator. Use

22 /1

1

cv

so that

2

11/

cv

3. By 1960, an accelerator giving protons a kinetic energy of 30 GeV had started operating atBrookhaven, USA. Explain why you can at once write down the value 30 as an approximation tothe factor for these protons.

4. As a rule of thumb, relativistic effects only need to be taken into account for speeds exceeding20% of the speed of light. For which of these accelerators are relativistic effects certainlyimportant?

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Comparing relativistic and Newtonian kinetic energyQuestion 30S: Short Answer


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In these questions you will show numerically that the relativistic calculation of kinetic energy isequivalent to the Newtonian value E

kinetic = ½ mv 2 at speeds much smaller than the speed of light.

Relationships to useThe relativistic equation for the total energy of a particle of mass m is:

2total mcE

with the relativistic factor given by

22 /1

1

cv

At zero speed, the factor = 1. Thus there is still energy at speed v = 0, at rest. It is given by

2rest mcE

The kinetic energy of a particle moving at relative speed v is the difference between total energy andrest energy:

resttotalkinetic EEE

Subtracting the two equations above gives, for the kinetic energy at any speed

222kinetic )1( mcmcmcE

You will show that this expression is equivalent to the classical Newtonian expression

221

kinetic mvE

Graphical comparison

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0v / c

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9




1.0

0

The graph above compares the Newtonian and relativistic calculations of kinetic energy.

1. Describe how the two quantities compare in the range v / c from 0.5 to 0.9.

2. Describe how the two quantities compare in the range v / c from 0 to 0.3.

You will now compare the two equations for kinetic energy using a calculator. The equations are:

2kinetic )1(energykineticicrelativist mcE

where

22 /1

1

cv

and

221

kineticenergykineticNewtonian mvE

They look very different. They would become the same if

2212)1( vc

This can be written more simply as:

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221 /)1( cv

3. Copy the table below, and use your calculator (or a spreadsheet) to calculate values for (v / c)2,and then for , where

22 /1

1

cv

Finally, compare the values of – 1 and of ½ (v / c)2. The first row has been done for you.

v / c (v / c)2 – 1 0.5 (v / c)2

0 0 1 0 0

0.1

0.2

0.3

0.4

0.5

4. Show that the discrepancy between the two calculations is less than 5% up to v / c = 0.2.

Particles at extremely high energyQuestion 40S: Short Answer


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This question is about extremely high energy cosmic ray particles, for which relativistic effects areall-important.

Relationships to useAt any speed up to close to the speed of light the total energy of a particle is given by

2total mcE

where


22 /1

1

cv

At rest, with v = 0, = 1 and the energy is rest energy Erest = mc 2. The relativistic factor can be

found from the ratio of total energy to rest energy:

rest

total

E

E

The kinetic energy is the difference between total energy and rest energy:


which can be written as

2kinetic 1 mcE

Data to useCharge on proton = 1.6 10–19 C

Mass of proton = 1.7 10–27 kg, equivalent to rest energy nearly 1 GeV


Advice: work with energy unitsFor making relativistic calculations it is best to express quantities in energy units. The electron volt(eV) is often convenient.

Cosmic ray protonsCosmic rays contain the highest energy particles ever detected. Incoming high-energy protons createhuge showers of secondary particles as they hit the atmosphere. Their energy can be estimated fromthe size of the shower produced. One such detector is the extensive air shower array at HaverahPark, near Leeds.

Less than once a year, a particle of extraordinary energy is detected. These extremely energeticcosmic ray protons can have a total energy of 1020 eV. That’s more than 10 million times the energyof a proton in the Large Hadron Collider at CERN.

1. The rest energy of the proton is about 1 GeV. Calculate the ratio of total energy to rest energy.State the relativistic factor for a proton having energy 1020 eV.

2. Explain why negligible error will be made in treating this proton as travelling at the speed of light.


3. Suppose this proton came from the other side of the Milky Way galaxy, a distance of 100 000 lightyears. What time t in seconds does it take to travel this distance, as seen from Earth? (1 year =3 107 s).

4. The relativistic factor found in question 1 also measures the time dilation for the proton, with t =. Show that, as measured in proton wristwatch time , the proton only took less than 1minute to cross the galaxy.

5. Now look at the problem another way. The proton must ‘know’ that it is travelling close to thespeed of light. If the trip across the galaxy takes the time found in question 4, what distance doesthe photon experience as it travels across the galaxy?

Two uses for uniform electric fieldsQuestion 60S: Short Answer

Teaching Notes | Key Terms | Hints | Answers | KeySkills

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Instructions Here are two sets of questions. Answer all the questions in the spaces provided. The following datawill be needed:


gravitational field strength = 9.8 N kg–1

Questions Questions 1–5 are about an experiment designed to measure the charge on an electron. In thisexperiment, the ‘Millikan oil drop experiment’, two parallel metal plates, 3.2 10–2 m apart, areconnected to a 600 V power supply.


600 V+

–

3.2 × 10–2 m

1. Draw four arrowed lines on the diagram to show the electric field between the plates, well awayfrom the edges of the plates.

2. Add dotted lines to the diagram to represent the 200 V and 400 V equipotentials between theplates. Indicate which is which.

3. Calculate the electric field strength between the two plates.

The electric field between the plates just supports the weight of an oil drop of mass 1.8 10–15 kg,which has acquired a charge due to a few excess electrons.

600 V+

–

3.2 × 10–2 m charged oil drop

electrical force

weight of oil drop

4. Calculate the charge on the oil drop.

5. What is the most likely number of excess electrons acquired by the oil drop?

Questions 6–12 are about an idea for separating minerals using an electric field. Two differentminerals acquire opposite charges through crushing and friction.

In an experimental arrangement, two such minerals fall from a conveyor belt through the electric fieldbetween two vertical conducting plates. A 54 kV potential difference is applied between the 2.0 mlong plates which are 0.30 m apart.


+ 54 kV0 V

0.3 m

collectors

crushed minerals

conveyor belt

In a feasibility study, a 1.5 mg particle was found to take 1.1 s to fall 2.0 m in air. This is longer thanwould be expected from a calculation of the time taken in free fall.

6. How would you account for the longer time?

7. On the diagram, draw field lines to represent the electric field in the central region between theplates.

8. Calculate the magnitude of the electric field between the plates.

9. The charge on one 1.5 mg particle is 106 electron charges. Calculate the magnitude of theelectric force on this particle when it is between the plates.

10. Use your answer to question 9 to calculate the sideways displacement after the 2.0 m fall throughthe electric field.

11. Suggest why this method of separation will not work if the mass of the particles is too small.


12. Suggest why this method of separation will not work if the mass of the particles is too large.

Deflection with electric and magnetic fieldsQuestion 90S: Short Answer


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Instructions and informationWrite your answers in the spaces provided. Use these data:



The force on an electric current in a magnetic field is perpendicular to both the current and the field.This diagram may help you.

magnetic field

current

force

QuestionsA beam of particles passes in a straight line from a source X, to a spot, E, on a fluorescent screen.When the electromagnet shown is switched on, the beam hits the screen at one of the spots A, B, C,D or E.


N

SX A

B

CD

E

1. Which spot does the beam hit if the particles are positively charged?

2. Where does it go if the particles are neutral?

3. What happens to the positively charged particles if the magnetic field is reversed?

In one form of mass spectrometer, charged ions in the beam fan out, moving in the paths shown inthe diagram. Parts of the paths include a magnetic field whose direction is perpendicular to the planeof the paper.

source

detector

4. Indicate places where there must be no magnetic field in the direction suggested. Do not useshading.

5. Shade the area where there must be a magnetic field.

6. What is the shape of the path followed by a proton that is projected into a uniform magnetic fieldat right angles to its velocity? Justify your answer.


The diagram shows the initial path of an electron fired into a uniform magnetic field. The magneticfield is at right angles to the direction of the electron and is directed away from the reader.

electron

7. Add a labelled arrow to the show the direction of the force on the electron.

8. Draw the subsequent path of the electron.

A uniform electric field is produced by maintaining a potential difference of 1000 V across a pair ofparallel plates 5 cm apart. An electron enters the field at right angles as shown with a velocity of 4.0 107 m s–1 and emerges from the plates without hitting them.

0 V

+ 1000 V

electron

9. Draw in the path of the electron on the diagram.

10. What name is given to this type of path?

The cyclotronQuestion 100S: Short Answer


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BackgroundIn a cyclotron, protons are kept moving in a circular path by a uniform magnetic field at right angles tothe plane of the path.

uniformB-field

protons incircular path

The American physicist E O Lawrence designed and built the first cyclotron in 1930. For thisachievement, he was awarded the Nobel prize for physics in 1939. Only a few centimetres indiameter, the Lawrence cyclotron accelerated protons to energies of 80 keV.

QuestionsYou will need the following data:

mass of proton = 1.7 10–27 kg

charge on proton = 1.6 10–19 C

1. What is the velocity of a proton with an energy of 80 keV?

2. The largest possible path had a radius of about 50 mm. What strength of magnetic field musthave been used?

3. What would be the radius of the path followed by a proton with half this maximum energy in thesame field?


4. How long would it take 80 keV protons to travel once round their path? How long would it take forthose with half this energy?

Constructed of two D-shaped chambers (later called ‘dees’ because of their shape), the cyclotronworked by giving particles of any energy, on their respective paths, a push every time they hadcompleted half an orbit.

beam of protons

push

push

5. How was it possible for particles of differing energy all to be accelerated together?

The Hall effectQuestion 140S: Short Answer


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BackgroundIn 1879 the American physicist E H Hall found that when a current was flowing in a conductor, amagnetic field at right angles to the current caused a very small potential difference across theconductor. If semiconductors are used instead of ordinary conductors, there is a much larger p.d. TheHall effect is widely used in industry for measuring magnetic fields.

QuestionsIf I is the current when there are n charge carriers per unit volume, each with electric charge Qpassing through an area A with a speed v, then I = n A Q v.


1. When the magnetic field B is switched on, what force will be exerted on one moving chargecarrier?

2. What is the direction of the force? Mark it on the diagram.

negative charge carriers going from left to right;there are n of them per cubic metre

area A

IIQ

v

B

3. After a time there will be higher density of negative charge near the front edge than near the rearedge. Why?

II

4. A charge carrier in the middle will still experience a force due to the magnetic field, but it will alsobe repelled by the extra charge carriers near the ‘front edge’.


II

v

magnetic force

electrical force

How big must the electrical force be so that there is no further change in the number of negativelycharged particles near the edge?

5. Which edge of the strip will be positively, and which negatively, charged?

If, in another experiment, there were two parallel plates separated by a distance, d , and one platewas kept positively charged and the other negatively charged, then there would be an electric field, E,present between the plates, given by E = V / d where V is the p.d. between the plates.

d

V d.c. supply

A charge, Q, placed between the plates would experience a force due to this electric field: force = EQ.

The electric force acting on a charge moving in the conductor has been given by your answer toquestion 4.

6. What is the potential difference that has been produced between the front and rear edges of thestrip of conductor?

7. But I = n A Q v. What is A if the strip has thickness b?


You need an relationship for the number of charge carriers which does not contain the speed of thecharge carriers.

II

A

b

d

8. Use I = n A Q v to eliminate v from your equation for the potential difference and obtain anequation relating the number of charge carriers per unit volume to the potential difference. (Checkthat your expression has the units m–3.)

Charged particles moving in a magnetic fieldQuestion 150S: Short Answer


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Instructions This question set contains two groups of questions. Answer all the questions in the spaces provided.The following data will be needed:



Questions 1–5 Questions 1–5 are about the motion of charged particles in a bubble chamber.

1. An electron gun in a vacuum accelerates electrons up to a kinetic energy of 2.9 10–16 J. Showthat the speed acquired by each electron is 2.5 107 m s–1.


The electron beam enters a region of uniform magnetic field of strength, B, perpendicular to thebeam. The magnetic field causes the beam to follow a circular path as in the diagram below.

uniform B fieldacts into plane ofscreen / paperover shaded area

electronmotion

2. Show that the force experienced by the electron is about 8.8 10–15 N, when B is 2.2 10–3 T.

3. Use your answer to question 2 to find the radius of this circular path.

4. Evidence for the motion of electrons in magnetic fields can be observed from the trail of bubblesthey leave as they pass through liquid hydrogen. In these bubble-chamber experiments a singleelectron tends to produce a track which is a spiral rather than a circle. Explain why.


electronmotion


5. On the diagram below draw the likely path of a proton travelling at the same speed in an identicalfield.


protonmotion

Questions 6–10 are about the motion of charged particles in magnetic fields and the cyclotronfrequency.

6. Explain why a charged particle, moving with a constant speed v perpendicular to a uniformmagnetic field B, will follow a circular path.

7. Show that for a particle of mass m and charge q the radius of the circular path is given by theexpression r = m v / B q .

8. Using your answer to question 7, show that the frequency of this circular motion, known as thecyclotron frequency, is given by the expression f = q B / 2 m.

9. Some astrophysicists believe that the radio signals of 109 Hz reaching us from Jupiter are emittedby electrons orbiting in Jupiter’s magnetic field. Assuming the frequency of the radio emission isidentical to the cyclotron frequency, find the strength of the magnetic field around Jupiter.


10. The electrons lose energy as they emit radiation. What effect, if any, will this have on thefrequency of the radio signals detected? Explain your answer.

Fields in nature and in particle acceleratorsQuestion 160S: Short Answer


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Instructions and information Write your answers in the spaces provided. The following data will be needed when answering thesequestions:

electronic charge = – 1.6 10–19 C


0 = 8.9 10–12 F m–1

Questions Suppose that a thundercloud has a flat horizontal base of area 1 km2. The base of the cloud is 200 mabove the Earth’s surface so that a uniform electric field is formed between the cloud and the Earth. Afield of 106 V m–1 between the base of the cloud and the Earth is sufficient to cause a lightning flash.

1. Calculate the potential difference between the cloud and the ground at the moment the lightningflash begins.

2. The charge on the base of the cloud is given by the equation Q = 0 A E where A is the area of

the base of the cloud and E is the electric field strength between the base and the Earth.Calculate the charge on the base of the cloud.

3. The cloud and the Earth can be thought of as a parallel plate capacitor which stores energy whencharged. Assuming that the potential difference immediately after the flash is very smallcompared with the potential difference at the beginning of the flash, calculate a value for theenergy released during the flash.


Electrons accelerated through a potential difference of 200 V enter a uniform magnetic field of 0.001T perpendicular to the direction of motion.

4. Calculate the speed of the electrons when they enter the magnetic field.

5. Use your answer to question 4 to calculate the radius of the orbit in the magnetic field.

In an electron tube, electrons were passed through a region containing a vertical electric field E and ahorizontal magnetic field B. When the forces on the electron were balanced the electrons passedthrough the tube undeflected.

horizontal magnetic field

undeflected pathof electrons

vertical electric field

6. Show that the electrons of charge e pass undeflected when they have a speed v = E / B.

The separation of the deflector plates was 24 mm and no deflection was observed when the voltageacross the plates was 3.2 kV and the magnetic field was 8.2 10–3 T.

7. Calculate the velocity of the electrons.

The potential difference used to accelerate the electrons to this velocity was 750 V.

8. Use your answer to question 7 to calculate the ratio e/m for electrons where m is the mass of anelectron.


A proton joined to a neutron is known as a deuteron or deuterium ion and is used in nuclearscattering experiments. A deuteron has a mass of 3.3 10–27 kg and a charge of + 1.6 10–19 C.

9. Calculate the potential difference required to accelerate a deuteron from rest in a vacuum to avelocity of 9 106 m s–1 (3% of the speed of light).

In an early form of particle accelerator, deuterons were made to move in a circular path within atoroidal tube of diameter 1 m. A toroidal tube is like a hollow ring.

10. Calculate the magnetic field required to constrain a deuteron within the tube at the velocity of 9 106 m s–1.

Non-uniform electric fieldsQuestion 180S: Short Answer


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Instructions and informationWrite your answers in the spaces provided. The following data will be needed:


1 / (40) = 9.0 109 N m2 C–2

Questions1. Part of the electric field pattern has been drawn in each of the diagrams below. Use your

knowledge of symmetry to complete the electric field pattern in each case.


Each of the diagrams below shows a pair of electrodes connected to a potential difference supply.

+ – + –

2. For each of the diagrams, add solid lines to illustrate the shape of the electric field between theelectrodes. Draw arrows to indicate the direction of the field.

3. Add dotted lines to show the shape of equipotential lines associated with the field. The spacing ofthe equipotential lines should give an indication of the relative strength of the electric field.

4. In an experiment to test Coulomb’s law, two expanded polystyrene spheres, each with a chargeof 1.0 nC were 0.060 m apart (measured from their centres). Calculate the force acting on eachsphere.

The electric field strength field at a distance of 1.0 10–10 m from an isolated proton is 1.44 1011 NC–1 and the electrical potential is 14.4 V.

5. Calculate the electric field strength at a distance of 2.0 10–10 m from the proton.


6. Calculate the electrical potential at a distance of 2.0 10–10 m from the proton.

A simple model of a hydrogen atom can be thought of as an electron 0.50 10–10 m from a proton.

7. Calculate the electrical potential 0.50 10–10 m from a proton.

8. The electron is in the electric field of the proton. Calculate the electrical potential energy of theelectron and proton in joules.

When a uranium nucleus containing 92 protons and rather more neutrons emits an alpha particle ofcharge + 3.2 10–19 C the remaining nucleus then behaves like a sphere of charge of magnitude +1.4 10–17 C.

9. Assuming that the alpha particle is 2.0 10–14 m from the centre of the nucleus on release,calculate the electric field experienced by the alpha particle.

10. Calculate the force on the alpha particle when at this distance.

11. Calculate the maximum acceleration of the alpha particle given that its mass is 6.6 10–27 kg.


160

140

120

100

80

60

40

20

00 0.1 0.2 0.3 0.4 0.5

r / m

The graph shows the variation of potential with distance from the charged dome of a van de Graaffgenerator.

12. Use the graph, together with the equation E = – V / r , to find the electric field strength at adistance of 0.3 m from the dome. (You may like to check your answer with an alternativecalculation.)

Charged spheres:Force and potentialQuestion 190S: Short Answer


Quick Help


gravitational field strength g = 9.8 N kg–1


k = 1/(40) = 9.0 109 N m2 C–2

Questions Questions 1–5 are about the electrical force between charged spheres. A light conducting sphere A,20 mm in diameter and coated in conducting paint, is mounted on an insulating rod fixed to the pan ofa sensitive top-pan balance. The balance is adjusted to read zero when it is supporting the weight ofthe sphere A and its rod. A second identical sphere B can be moved up or down to adjust theseparation between the spheres.

10 mm diameter

2 mm gap

10 mm diameter

sphere B

sphere A

top pan balance

insulating rod

insulating rod

The two spheres are given identical positive charges using an EHT power supply. Sphere B is placedsuch that the gap between them is 2 mm. The reading on the top-pan balance is 44 mg.

1. Calculate the force between the two spheres.

2. Using your answer to question 1, show that the magnitude of the charge on each sphere is about2.5 10–9 C. What assumptions are you making?


3. Calculate the magnitude of the potential to which each of the spheres must be raised in order tocharge it to the value in question 2.

4. When charged spheres are close, charge is not distributed evenly on the surfaces. Draw a few ‘+’signs to indicate those regions of each sphere where the charge density is likely to be greatest.

In an experiment, the force, F, between the spheres is measured for a range of values of separation, r, between the centres of the spheres. The charges on the spheres remain constant throughout theexperiment. The graph shows part of the line obtained when F is plotted against 1 / r 2. It is suggestedthat, when r is small, the new distribution of charge will result in each sphere’s ‘centre of charge’being slightly displaced. This will cause the graph to curve for small values of r.

5. Draw a continuation of one end of the line to show this deviation from a straight line.

00

1 / r2

Questions 6–9 are about electrical potential near a sphere. A metal sphere A is connected by meansof a long fine wire to the output terminal of an EHT power supply so that it is at a potential of 1000 V.


An insulated stand supports the sphere sufficiently far above the bench for the effect of the bench tobe negligible. The potential 300 mm from the centre of the sphere is 450 V.

6. Calculate the potential 500 mm from the centre of the sphere.

7. Calculate the radius of the sphere.

8. Calculate the charge on the sphere.

Another identical sphere (B), also connected to the same output terminal of the EHT power supply, issimilarly supported with its centre 600 mm from that of the first sphere (A):

A BO

600 mm

9. Would you expect the potential at O, a point midway between A and B, to be equal to, greaterthan or less than 900 V? Explain your answer.

Using the 1 / r 2 and 1 / r laws for point chargesQuestion 200S: Short Answer


Quick Help

Instructions and information Write your answers in the spaces provided. The following data will be needed:


1 eV = 1.6 10–19 J


k = 1/(40)= 9.0 109 N m2 C–2

g = 9.8 N kg–1

Questions These questions are about the electrical forces between charged spheres.

A light conducting sphere A, 10 mm in diameter and coated in conducting paint, is mounted on aninsulating rod fixed to the pan of a sensitive top-pan balance. The balance is adjusted to read zerowhen it is supporting the weight of the sphere A and its rod. A second identical sphere B can bemoved up or down to adjust the separation between the spheres.

10 mm diameter

14 mm gap

10 mm diametersphere B

sphere A

top pan balance

insulating rod

insulating rod

The two spheres are given identical positive charges using an EHT power supply. Sphere B is placedsuch that the gap between them is 14 mm. The reading on the top-pan balance is 12 mg.

1. Calculate the force between the two spheres.


2. Use your answer to question 1 to show that the magnitude of the charge on each sphere is about3 10–9 C.

3. Calculate the magnitude of the potential to which each of the spheres must be raised in order tocharge it to the value obtained in question 2.

Assume that the dome of a van de Graaff generator can be treated as a sphere of radius 0.15 m.

4. Calculate the electric field at the surface of the sphere when the charge on the sphere is 3.0 10–6 C.

5. Calculate the electrical potential at the surface of the sphere when the charge on it is 3.0 10–6

C.

These questions are about the transmission of electricity by overhead power cables. The diagramshows a transmission cable (assumed to be cylindrical) at a potential of 200 kV and threeequipotentials near the cable.


cable

25 mm

185 kV

190 kV

195 kV

6. Explain how these equipotentials indicate that the electric field strength decreases with distancefrom the centre of the cable.

7. Use information from the diagram to calculate the average value of the electric field strength overthe first 5 mm from the surface of the cable.

There are always a few stray ions in the air.

8. Why is there likely to be far more ionisation near the surface of the cable?

9. Suggest a problem this may cause given that the cable is designed to transmit electricity.

The energy available from the fission of a single nucleus of uranium can be estimated by consideringthe change in electrical potential energy as the fission fragments move apart. A uranium nucleus maysplit into fission fragments of rubidium with a charge of + 37 e and caesium with a charge of + 55 e,where e is 1.6 10–19 C.


10. Calculate the electrical potential energy of the fission fragments at the moment fission occurs,assuming that the centres of the fragments are 10–14 m apart.

11. What energy changes will occur to two isolated fission fragments immediately after fission?

12. Write down an estimate for the maximum total kinetic energy of two isolated fission fragments.

Controlling charged particlesQuestion 250S: Short Answer


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QuestionsQuestions 1–5 are about the production of an electron beam in a cathode ray tube. Electrons areejected from a hot metal filament and are accelerated towards the screen by a potential differencebetween the filament and anodes.


screenelectronbeam

0 V– 500 V– 2500 V

filamentanodes

1. Describe the changes of energy that occur as the electrons are accelerated.

To get a clear picture on the screen the tube has to be highly evacuated and the beam has to befocused.

2. Why must the tube be highly evacuated?

Focusing can be achieved by means of the electrostatic field between two cylindrical anodes. Theshape of the electric field between these anodes is shown below.

0 V– 500 V

3. Use the electric field lines to draw three equipotentials in the region where the electron beam isbeing focused.

4. The electrons carry charge away from the gun at a rate of 20 mC s–1. Calculate how manyelectrons leave the gun each second.

5. Electrons emerge from the gun with a mean speed of 3 107 m s–1 and are brought to a halt bythe screen. Calculate the force exerted by the electrons on the screen.


Questions 6–14 are about the separation of isotopes of xenon by a magnetic field.

ion source

electromagnet

collector

sensitiveammeter A

This is a schematic view of the basic components of a simple mass spectrometer. Ions, of charge q ,are accelerated through a potential difference and leave the source in a narrow beam. Theelectromagnet deflects the beam to a collector and the collector current is measured with a sensitiveammeter.

6. Why is the path of an ion in the magnetic field an arc of a circle?

When xenon ions are used, this graph shows how the collector current varies with magnetic field.

B-field strength

0

7. How is it possible to collect ions of different mass without moving the source or collector?

The following is an extract from a student’s notes on mass spectrometers:

‘When an ion (mass m) is accelerated through a potential difference V,


½ m v2 = q V.

When the ion (speed v) moves through the B-field in a circular path (radius R),

q v B = m v2 / R ''

8. Name the physical quantity represented by ½ m v 2.

9. Name the physical quantity represented by q V.

10. Name the physical quantity represented by q v B.

11. Assuming that all ions have the same charge q , show that the value of B needed to bring eachisotope to the collector is proportional to m1/2.

12. Add a label to the graph to show which of the peaks corresponds to the heaviest isotope ofxenon.

13. Add a label to the graph to show which of the peaks corresponds to the most abundant isotope ofxenon.

14. Explain how you arrived at your answers to questions 12 and 13.

The uniform electric field and its effect on chargesQuestion 20M: Multiple Choice


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Instructions and information Each of the questions has five possible answers, A, B, C, D or E. Only one answer is correct for eachquestion.

QuestionsA thundercloud has a flat lower surface which is parallel to the ground and 400 m above it. There is a


potential difference of 100 MV between the base of the cloud and the ground.

ground

cloud

400 m

1. Which one of A to E below is the magnitude of the electric field (in N C–1) between the cloud andthe ground?

2. Which one of A to E below is the work done (in J) when 1 C of charge moves between the cloudand the ground?

(You may assume that moving the charge does not alter the potential difference between thecloud and the ground.)

A: 0.25

B: 4.0

C: 100

D: 2.5 105

E: 4.0 1010

3. The diagram shows a charged drop between two plates connected to a constant high voltage V.

+

–

Vchargeddrop

Which of the following correctly describes(s) changes to the electrical force on the drop?

(a) If the plates were closer together, the electrical force on the drop would be larger (withthe same voltage V, and the same charge on the drop).

(b) If the drop moves towards the top plate, the electrical force on it decreases.

(c) If the drop acquires more charge, the electrical force on it increases (with the samevoltage and plate spacing)

A: (a) only

B: (b) only


C: (a) and (c) only

D: (b) and (c) only

E: (a), (b) and (c)

4. Electrons are accelerated from rest in a uniform electric field from one electrode to another.

Which sketch graph correctly displays the relationship between the acceleration, a, of an electronand the voltage, V, applied across the electrodes as the electron passes between them?

0

0

A

0

0

B

0

0

C

0

0

D

0

0

E

V / V V / V V / V

V / VV / V

5. A gas-filled tube containing low-pressure xenon is used to investigate the excitation of xenon. Thecircuit is shown below.

2 V

20 V

collecting plate

grid (positive)

cathode

heater

–

+

+

–


If the three electrodes of the tube are connected as shown above, which one of the followingdiagrams best represents the direction(s) of the electric field within the tube?

–

+

–

–

+

–

–

+

–

–

+

–

A B

C D

6. The diagram shows a source of ions S, and a grid G beyond which is a collecting plate P. Thereis a voltage V between P and G so that ions drifting into the space between P and G areaccelerated to P. The whole apparatus is in an evacuated chamber.


S

G

P

+++

+

V+ –

Which of the following will affect the time an ion takes to go from G to P?

(a) The voltage V.

(b) The mass of the ion.

(c) The charge on the ion.

A: (a) only

B: (b) only

C: (a) and (c) only

D: (b) and (c) only

E: (a), (b) and (c)

7. Differently charged ions of several isotopes are all being accelerated by the same uniform electricfield.

Which of the ions A to E below has the greatest acceleration?

A 147N2+

B 147N+

C 126C2+

D 63Li2+

E 63Li+

A very small conducting ball on a very long insulating thread shuttles to and fro between two largeclosely space metal plates which are connected to a high-voltage supply.


+V 0

q

d

Here are five expressions involving: the supply voltage V, the charge q on the ball as it moves, thedistance d between the plates, the time t the ball takes to go from one plate to the other and thecapacitance C of the plates:

A: q / t

B: V q

C: V q / d

D: C V / t

E: C V / d

8. Which one of these expressions gives the force on the ball at the mid-point between the plates?

9. Which one of these expressions gives the energy taken from the supply as the ball moves fromone plate to the other?

10. An electron with charge – 1.6 10–19 C moves between two points R and S, 2 mm apart, in anelectric field. The graph shows how the potential energy (E) of the electron varies with its distance(x) along RS from R (R being arbitrarily taken as origin for both E and x). There is no componentof electric field at right angles to the line RS.

Which one of the following deductions from the data presented in the graph is not correct?


+ 4.8 × 10–19 J

0

R P S

1 mm 1 mm

distance

A: The electric field strength between R and S is uniform.

B: The magnitude of the potential difference between R and S is 3 V.

C: The potential difference between R and P is the same as that between P and S.

D: The electric field strength at P is 3000 N C–1.

E: The force on the electron at R is the same as the force on it at P.

Charged particles in electric and magnetic fieldsQuestion 110M: Multiple Choice


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Instructions and information Each of the questions has five possible answers, A, B, C, D or E. Only one answer is correct for eachquestion.

1. In a cathode ray tube, the beam of electrons from the electron gun is deflected vertically when apotential difference is applied between the deflection plates, PQ, of length l .


filament

electron gun

anode P

Q

screen+

+

l

Which one of the following changes to the tube will increase the deflection of the beam for a givenpotential difference across the plates?

A: Moving the deflection plates closer to the screen.

B: Reducing the potential difference between the filament and the anode in the electron gun.

C: Moving the deflection plates further apart.

D: Reducing the length l of the deflection plates.

E: Increasing the heating current in the filament of the electron gun.

2. An electron beam is travelling in a vacuum and is deflected by the electric field between twoplates, F and G. The electron beam strikes a screen at S. Which of the diagrams below bestrepresents a possible path for the electron?


F

G

R

SF

G

R

S

F

G

R

S

F

G

R

S

F

G

R

S

A B

C D

E

3. An electron moves in a circular orbit in a uniform magnetic field.

Which of the following statements about the force acting on the electron is / are correct?

(a) ‘The force on the electron depends on its speed.’

(b) ‘The force acts in the direction of travel.’

(c) ‘The work done by the force when the electron makes one complete revolution is zero.’

A: (a) only

B: (b) only

C: (a) and (c) only

D: (b) and (c) only

E: (a), (b) and (c)

4. Which of the following statements is a correct comparison of deflections of electrons in uniformelectric and magnetic fields?

A: Only in a magnetic field can an electron be deflected by more than 180.


B: In both electric and magnetic fields the speed as well as the direction of the electron mustchange.

C: In both electric and magnetic fields there is no force on the electron unless it is moving.

D: The force is always perpendicular to the direction of motion in an electric field, but not in amagnetic field.

E: Only in an electric field does the force depend on the magnitude of the charge on the electron.

5. A particle of mass m, charge q , and speed v, enters a uniform magnetic field B and describes acircular path of radius R.

Which of the following statements is correct?The radius R of the circular path is:

A: unaffected by m.

B: directly proportional to q .

C: inversely proportional to B.

D: inversely proportional to m.

E: inversely proportional to v.

6. The differently charged ions of several isotopes are all fired at the same speed into a regionwhere there is a uniform magnetic field at right angles to their initial path.

Which of the ions A to E will travel in the circular path of smallest radius?

A 63Li+

B 63Li2+

C 126C2+

D 147N+

E 147N2+

A particle P with charge q , mass m and constant speed v crosses line QR into a region where auniform magnetic field B


Q

P

R

B field inthis region

Within the magnetic field it moves in a circular path of radius r , determined by the equation q v B = mv2

/ r , so that r / v = m / B q .

The particle P takes a time t = r / v to travel in a semicircle from Q to R.

7. A second particle, identical to P but with a speed 2v enters the field at Q. What time will thisparticle take to go around from Q so as to recross the straight line QR?

8. A third particle, identical to P but with a charge 2q enters the field at Q. What time will this particletake to go around from Q so as to recross the straight line QR?

A: t / 4

B: t / 2

C: t

D: 2 t

E: 4 t

Relationships for force and field, potential and potential energyQuestion 220M: Multiple Choice


Quick Help

Instructions and information These questions are to review the inverse square law for force and field and the inverse relationshipfor potential and potential energy. Each of the questions has five possible answers, A, B, C, D or E.Only one answer is correct for each question.

Questions The diagram shows apparatus used to investigate the electrostatic force between two identical smallconducting spheres whose centres are a distance r apart. For small angles of deflection, isproportional to the horizontal force acting on the hanging sphere. The spheres have equal positivecharge.


r

fixedinsulatingrod

chargedsphere

chargedsphere

light, insulatingsuspension

not to scale

1. Which one of the graphs below, A to E, would you expect from the experiment?

2. When an additional uniform horizontal electric field is introduced, acting from left to right, whichone of the graphs below, A to E, would you now expect from the experiment?

0

0

1 / r2

0

0

1 / r2

0

0

1 / r2

0

0

1 / r

0

0

1 / r

A B C

D E


3. The diagram gives the magnitude of the electric field strength (in arbitrary units) at various pointsnear an isolated point charge Q.

Y

21.6 8.7

X 4.843.2Q

Which of the following correctly describes the electric field (in these units) both at X and at Y?

Field at X Field at Y

A 10.8 10.8

B 10.8 8.7

C 21.6 10.8

D 21.6 8.7

E 10.8 14.4

4. In the isolated system of point charges shown below, there is a charge q at X and a charge – 2q at Y.

X Y

+ q – 2q

Which of the following statements about the electric field along the line through the two charges is/ are correct?

(a) ‘Between X and Y the components of the electric field due to each charge are in thesame direction.’

(b) ‘To the left of X there is a point where the electric field is zero.’

(c) ‘To the right of Y the electric field is always directed to the left.’

A: (a) only

B: (b) only

C: (a) and (c) only


D: (b) and (c) only

E: (a), (b) and (c)

5. Gravitational and electrical forces both obey an inverse square law. For both, field and potentialvary with distance, r .

Which one of the following graphs correctly indicates a relation between the magnitude of eitherfield or potential, and distance r ?

0

0

r2

0

0

1 / r

0

0

r

0

01 / r

0

0

1 / r2

C D E

A B

6. The diagram shows two charges P and R, of opposite sign but equal magnitude. Points S and Tlie on an equipotential surface.


+

P

–

R

S

T

Which of the following statements is / are true?

(a) ‘A positive charge placed at S can be moved to T without doing any work on it.’

(b) ‘The field at S is in the direction shown by the arrow in the diagram.’

(c) ‘The magnitude of the electric field at S is bigger than at T.’

A: (a) only

B: (b) only

C: (a) and (c) only

D: (b) and (c) only

E: (a), (b) and (c)

7. Diagrams A, B, C, D and E show five different arrangements of positive and negative charges ofequal magnitude.

A B C D E

X+ +

X– –

X+ +

X– –

X+ –

+

+

+

+

–

+

In which one of these arrangements are the electric field and the electric potential at the centralpoint X both equal to zero?

8. The map shows computer predictions for some equipotentials near to three equal positivecharges.


++

+

O

Z

Y

X

Which of the following statements is / are true?

(a) ‘The potential at O is the same as the potential a long way from the three charges.’

(b) ‘The electric field at X is in the direction shown by the arrow in the diagram.’

(c) ‘The magnitude of the electric field at Z is bigger than that at Y.’

A: (a) only

B: (b) only

C: (a) and (c) only

D: (b) and (c) only

E: (a), (b) and (c)

9. The graph below shows the way the electrical potential varies with distance on the line betweenthe centres of two spheres P and Q which have equal positive charges.


0

0

distance

P Q

A small positively charged sphere moves from the surface of P to the surface of Q on the linebetween their centres. Which one of the graphs below correctly represents the force acting onthis small sphere as it travels across the space between P and Q on this line? In each graph theforce is plotted on the y-axis and the distance from P on the x-axis. A force towards Q is reckonedas positive.


P Q

A

xP Q

B

P Q

C

D E

P Q P Q

10. A student is attempting to plot the potential energy of an electron near a proton against thedistance r between them, as part of a calculation about a hydrogen atom.

distance r

E

D

C

X

B

A

–21.8 × 10–19

00 r0

The student has marked in one correct point X at the energy – 21.8 10–19 J of the lowest energystate of the hydrogen atom, at a distance r 0.


Which one of the points A to E is another correctly plotted value of the potential energy of the twoparticles?

Fields and charged particlesQuestion 240M: Multiple Choice


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Each of the questions has five possible answers, A, B, C, D or E. Only one answer is correct for eachquestion.

QuestionsThe diagram shows part of an evacuated tube in which a stream of electrons from an electron gunpasses between a pair of parallel deflecting plates.

The vertical displacement of the electron beam as it leaves the parallel plates is x.

acceleratingvoltage, V– +

electronpath

+deflecting voltage

–

x

1. Which one of A to E below will not change the displacement, x, of the beam as it leaves theparallel plates?

A: Increasing the accelerating voltage.

B: Increasing the deflecting voltage.

C: Increasing the distance between the electron gun and the deflecting plates.

D: Increasing the distance between the two deflecting plates.

E: Increasing the length of the deflecting plates.

2. When the accelerating voltage is V the speed of the electrons emerging from the gun is v. Whichone of A to E is the speed of the electrons when the accelerating voltage is doubled?


A: v

B: v

C: v 2

D: 2 v

E: 4 v

3. The diagram shows a region of space in which an electric field exists. The horizontal lines witharrows are the field lines.

X Y

A positively charged particle moves from X to Y parallel to the uniform electric field. Which of thestatements below is / are correct?

(a) ‘The electric potential is constant whilst moving from X to Y.’

(b) ‘The force on the particle is constant whilst moving from X to Y.’

(c) ‘Work is done by the field on the particle as it moves from X to Y’.

A: (a) only

B: (b) only

C: (a) and (c) only

D: (b) and (c) only

E: (a), (b) and (c)

4. The spiral track shown is that of an electron, moving in a magnetic field which is perpendicular tothe plane of the track.

P

Q


Which of the following suggestions, on its own, could explain the fact that the radius of the track islarger at P than at Q?

(a) ‘The magnetic field at P is larger than the magnetic field at Q.’

(b) ‘The electron slows down in travelling from P to Q.’

(c) ‘There is also an electric field, parallel to the plane of the track.’

A: (a) only

B: (b) only

C: (a) and (c) only

D: (b) and (c) only

E: (a), (b) and (c)

The diagrams below are all taken from a physics textbook. Lines representing equipotentials aredrawn with equal potential intervals.

BA C

D E

5. Which diagram would best represent the equipotential lines of a uniform field?

6. Which diagram could represent field lines for a point electric charge?


0

0

y

x

A

0

0

y

x

B

0

0

y

x

C

0

0

y

x

D

0

0

y

x

E

Which one of the graphs A to E represents the variation of y with x when:

7. y is the electric potential at a point in the uniform field between two charged parallel plates and xis the distance of that point from the earthed negative plate?

8. y is the electric potential at a distance x from an isolated point positive charge?

9. The table lists pairs of particles, a force between them, and gives a distance between them.

Particles Force Distance betweenparticles / m

1 2 protons Electrical 1.0

2 2 protons Gravitational 1.0

3 2 protons Gravitational 0.1

4 2 electrons Electrical 0.1

Which sequence, A to E below, puts the forces between the particles in order of increasingmagnitude?

A: 2 1 3 4

B: 1 4 2 3

C: 3 2 1 4

D: 2 3 1 4


E: 2 3 4 1

Estimating with fieldsQuestion 260E: Estimate


Quick Help

Instructions and informationEach of the questions below involves making estimates. Estimating is not guessing. Some data aresupplied for use in calculations but you may need to estimate other values. You will need to makesimplifications so that you can use the physics you have learned. You should always make it clearwhat simplifications you have made.

The following data will be needed:


k = 1/(40) = 9.0 109 N m2 C–2

rest mass of electron = 9.1 10–31 kg

gravitational field strength at surface of Earth = 9.8 N kg–1

speed of electromagnetic radiation = 3.0 108 m s–1

Planck constant = 6.6 10–34 J s

1 eV = 1.6 10–19J

1. The dome of a Van de Graaff generator is maintained at a high potential by a moving belt whichcarries charge to the sphere at a rate of 0.5 A. The Perspex column which supports the domehas a resistance of 3 1011 . Sparking will occur between the dome and a small, earthed metal

plate when the plate is brought close.


column of resistance R

sphere at potential V

current I

Explain why the plate must be close to the dome for sparking to occur and estimate how close theplate needs to be. Use the fact that air conducts when the electric field strength reaches 3 106 Vm–1.

2. Electrons in an old-fashioned vacuum-filled television tube are accelerated through a potentialdifference of 10 kV before travelling horizontally to the screen at the front. A physics studentrealises that moving electrons will also be affected by the Earth’s magnetic and gravitationalfields.

Explain why the beam will be deflected. Estimate the maximum deflection of the television imageand say whether this is likely to cause problems.


3. A simplified model of a uranium nucleus is a sphere of radius of order 10–14 m containing 92protons and many more neutrons. An unstable uranium nucleus can release an alpha particle ofcharge + 3.2 10–19 C and mass 6.6 10–27 kg from its surface.

Estimate the maximum velocity and acceleration of the alpha particle after release from thenucleus.

4. A gamma ray has a wavelength approximately 2500 times the size of a nucleon (10–15 m).

Estimate the energy of the gamma ray in electron volts and suggest how your answer is related tothe total potential energy of the nucleus.

5. Use a simple potential energy argument to estimate the maximum energy of a photon that can beemitted from a hydrogen atom.

Calculate the wavelength of the electromagnetic radiation emitted and indicate which part of thespectrum it comes from.

The Large Hadron Collider (LHC)Question 30C: Comprehension


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The Large Hadron Collider (LHC) is the world’s most powerful particle accelerator, situated at CERNnear Geneva. Its name comes from:


Large – the accelerator is a ring nearly 27 km in circumference

Hadron – it accelerates two kinds of massive hadrons, either protons or Pb-207 nuclei (one of theheaviest stable nuclides)

Collider – because its two counter-rotating beams are arranged on a collision course inside thedetectors at four points around the circumference.

The protons are accelerated to an energy of 7 TeV, so the energy available when they collide is 14TeV. Although this sounds a lot it is about the same as the kinetic energy of a grain of sand beingblown by the wind at a few metres per second.

Even as its predecessor the LEP (Large Electron–Positron Collider) was being designed in the early1980s the question of what to do next was being already being considered, culminating in 1994 withthe decision to eventually convert the LEP into the LHC.

The LHC involves more than 5000 scientists from more than 30 countries. Its purpose is threefold:

1. to find the Higgs boson that will explain why subatomic particles have the mass they do

2. to help answer the question why visible matter makes up only 4% of total mass in the Universe

3. to understand why there seems to be much more matter than antimatter in the Universe.

The LHC is approximately circular, made up of a series of straight sections where the energy of thehadrons is boosted, and arcs where the bending magnets are situated to steer the beam into a closedcircuit.

The maximum energy the hadrons can achieve is related to the radius of the curved sections: a largerradius means a smaller centripetal acceleration which in turns means less energy loss due tosynchrotron radiation – that is, the electromagnetic waves emitted whenever charged particlesaccelerate. The radius of the curve along which the beam can be deflected depends on the fieldstrength of the bending magnets.

The LHC is in fact the last in a cascade of 5 accelerators, each of which increases the energy of theprotons.

Accelerator Maximum kinetic energy

Linac 50 MeV

Proton synchrotron booster 1.4 GeV

Proton synchrotron 25 GeV

Super proton synchrotron 450 GeV

LHC 7 TeV

The LHC beamsThe protons are assembled and accelerated in 2808 bunches, with 1.1 1011 protons in each bunch.The energy stored in each beam is equivalent to a 200 tonne train travelling at 200 km h–1. This beamtakes some controlling. Here the essentially circular design helps. The beam is monitored at variouspoints and the information fed across a diameter to activate the desired corrections while the beamitself has to travel round a semicircle. Both the beam and the correction signal travel at virtually thespeed of light, but as the diameter is less than the length of a semicircle, the electronics has just


enough time to respond.

The total volume of the tubes in which the hadrons circulate is 6500 m3 which is equivalent to thevolume of over 20 average houses. To keep the beam as intact as possible the tubes are evacuatedto a pressure of 10–13 atmospheres (i.e. 10–8 Pa).

You may be surprised to learn that the energy of the beam is influenced by the position of the Moon.Tidal effects on Lake Geneva change the loading on the underground tunnels, and there is also adirect effect of the Moon attracting the ground itself (ground tides). Between a new and full Moon theEarth’s crust rises by about 25 cm in the Geneva area – this translates into variation of 1 mm in theLHC circumference. This in turn leads to changes in beam energy of 1 in 20 000 (0.005%). As thebeam energy is measured to a precision ten times better than this, this varying systematic errorneeds to be taken into account.

Data:1 eV = 1.6 10–19 J

c = 3.0 108 m s–1

mp = 1.67 10–27 kg

LHC data: Circumference = 26.7 km

Diameter = 8.5 km

2808 proton bunches in beam

1.1 1011 protons per bunch

Bending magnet strength = 8.3 T

Proton energy 7.0 TeV

Questions1. Show that beam control electronics has a maximum of 16 s to activate the control systems.

2. Show that a proton makes over 11 000 circuits of the LHC every second.

3. Show that the average spacing between the bunches of protons is about 10 m or 30 ns traveltime.


4. Show that the bunches of protons pass a collision point about 30 106 times each second (i.e. atabout 30 MHz)

5. Show that the circulating beam of bunches of protons is equivalent to a current of about 0.5 A.

6. Show that the energy stored in each proton beam is about 350 MJ.

7. The LHC is in fact the last in a cascade of five accelerators each increasing the kinetic energy ofthe protons. Calculate the relativistic factor for protons after each stage of acceleration. Takethe value of Erest for protons to be 1 GeV.

Accelerator Maximum kinetic energy Relativistic factor

Linac 50 MeV

Proton synchrotron booster 1.4 GeV

Proton synchrotron 25 GeV

Super proton synchrotron 450 GeV

LHC 7 TeV

Show that at the end of the second stage (1.4 GeV) the protons have reached nearly 90% of thespeed of light.

8. The LHC provides protons with a total energy of 7.0 TeV travelling at almost the speed of light.What is the momentum of each colliding proton? At this energy the momentum is given to a goodapproximation by


c

Ep total

9. The LHC is also designed to accelerate Pb nuclei (atomic mass number 207) at a kinetic energyof 570 TeV. What will the value of their factor be? Take the value of Erest

for both protons and

neutrons to be 1 GeV.

10. For particles deflected by a magnetic field B into a curved path of radius r, momentum p = q r B.Given that the average radius of the LHC ring is 4.25 km and the maximum total energy ofprotons is 7.0 TeV, show that this suggests that the deflecting magnets need a field of nearly 6 T.Suggest why that actual maximum magnetic field is rather larger than this, at 8.33 T.

Thunderclouds and lightning conductorsQuestion 40C: Comprehension


Quick Help

Instructions The following passage is taken from an article by P F Martin in the School Science Review (volume54, number 189, June 1973). Read the passage carefully and then answer the questions.

The lightning conductor The lightning conductor was one of the first useful applications of the science of electricity. Itsinvention by Franklin occurred 30 years before Galvani and Volta began the experiments that led tothe first electric batteries. While today there is little doubt of the effectiveness of well-designedlightning conductors, for a considerable time the mechanism by which they protected buildings waspoorly understood. In fact a satisfactory explanation was not found until the twentieth century whenthunderclouds and lightning flashes were studied by electronic and photographic methods.

Franklin first used a kite as a lightning conductor and the strands of the kite string were seen to standon end, so showing that the thundercloud was electrified. From his other experimental work, Franklin


knew that a charged object could be discharged by holding an earthed pointed conductor nearby. Sohe suggested that a long, earthed rod might discharge the cloud before it produced a lightning stroke:he had invented the lightning conductor. The first ones were installed in the American colonies and itwas not until 1760 that the first conductor was installed in England, on the Eddystone lighthouse. Itsoon became apparent that sometimes the conductor was struck and the lightning current carried toearth without damaging the building. But sometimes the building was damaged and this was used asevidence that lightning conductors in fact increased the chance of a building being struck becausethey attracted lightning. King George III ordered the lightning conductors to be removed from hispalace.

Eventually lightning conductors became accepted as the best way of protecting buildings, although itwas not clear whether they discharged the cloud or merely conducted the lightning strokes safely toearth. To decide what really happens, both the charge present in the cloud and the rate at which thecharge flows to earth through the conductor must be known. It is only during the last 50 years thatthese quantities have been determined.

The thundercloud Thunderclouds usually have two main centres of charge, with the lower charge being negative. Bothcharges are of the order of 20 coulombs and are separated by a vertical distance of about 3 km. Thewhole cloud is between 5 and 10 km high and has a base diameter of several kilometres.

+

–

4.5 km

Each individual thundercloud usually exists for about 30 minutes, during which time it may producelightning flashes as often as once every 20 seconds. As each flash will remove most of the 20coulombs of charge, the thundercloud must be a generator of static electricity. 20 coulombs removedevery 20 seconds is equivalent to a current of 1 ampere, on average. About 66 per cent of all lightningflashes are between charges in the cloud and not to earth. So the average current carried to earth bylightning in a single thunderstorm is about 0.3 amperes.

The lightning flash An electric discharge will occur in dry air at atmospheric pressure in electric fields of above 3 106 V


m–1; however, the average field below the thundercloud is only of the order of 3 104 V m–1. Severalfactors assist the development of a lightning stroke. In small regions within the cloud the electric fieldmay be locally very high due to localised patches of charge. Water drops tend to become deformed inthe high electric fields and discharge will start from the pointed ends. Larger drops are more easilydeformed, and if drops of several millimetres in diameter are present, electrical breakdown may occurin fields as low as 5 105 V m–1. Once the lightning stroke has started in one part of the cloud it willcontinue in the smaller fields below.

The lightning flash is quite complex. The first visible sign of the discharge is the stepped leader inwhich negative charge from the base of the cloud moves downwards in a series of steps of 10 to 200metres in length, as the figure below. The leader follows a zig-zag path with the tip advancing at aspeed of about 105 m s–1. When the leader nears the ground it is met by an upward-moving streamerof positive charge. A conducting path has now been established and the main or return stroke flowsalong this ionised path. The return stroke carries the main current of the discharge – about 104 A –and lasts for about 100 s. If the cloud has not been fully discharged further strokes along the samechannel are likely to occur.

+

++

++

positive streamer

– –––– ––

––

––

––

––

–––

–

–

–

––

–

––

––

–

–

–

––

––

––

––

––

––

––

––

–– –

–

–

–

stepped leader

The estimated potential difference between cloud and earth just before a lightning flash is typicallybetween 108 and 109 volts. As the quantity of charge flowing to earth is about 20 coulombs, theenergy released is of the order of 1010 joules. Seventy five per cent of this energy is dissipated inheating up the air column surrounding the discharge path, the consequent rapid expansion producingthe thunder.

Point discharge currents If the electric field due to a charged cloud is great enough then an exposed earthed metal rod willpass a current into the atmosphere. This current is called the point discharge current and is apossible way in which a conductor can discharge a thundercloud. Point discharge occurs when theelectric field lines near a point are sufficiently concentrated to accelerate electrons to such a speedthat they will ionise neutral air molecules by collision. Clearly the greater the electric field the greaterthe point discharge current. The current will also depend on how well the pointed rod is able to


concentrate the electric field lines and so will be greater if the rod stands out from its surroundings.

–––– –– – – –

––

–

ground

cloud

Because the positive ions move at only a few metres per second under the influence of the field, thecloud of ions produced by the point can reduce the effective field at the point and so reduce the pointdischarge current. If a strong wind is blowing, the positive ions will be swept away and the dischargecurrent will be higher.

Research on point discharge currents suggests that they may be calculated using the followingformula, deduced for a 5 metre rod placed on a 22 metre mast:

)295)(4(1016.1 4 FWI

where I is the discharge current in A, W the wind speed in m s–1 and F the electric field strength in Vm–1.

Point discharge can occur from any pointed object, especially the branches of trees, and thepresence of a large number of discharging points under a thundercloud will tend to reduce itselectrical activity.

The attraction theory of the lightning conductor Few buildings are tall enough to trigger a lightning discharge. However, very tall ones, such as theEmpire State Building, are sometimes close enough to the lower charge in the cloud to cause thestepped leader to form.

Many lightning stokes are inclined to the vertical and the actual point of strike is not usually decideduntil the leader is about 50 m from the ground, when the positive streamer leaves the ground tocomplete the ionised path. If the lightning conductor is to be effective then this streamer must startfrom the conductor and not from the building to be protected. For a vertical rod an acceptable degreeof protection is produced within a cone of angle 45, as shown. Horizontal conductors can also beused as in the protection of long low buildings.

So the major factor in the ability of a lightning conductor to attract lightning is that it is the mostexposed object in the vicinity of the protected building. It is also well-earthed and so will provide alower resistance path to earth than the building. It is interesting to note that the conductor does nothave to be pointed.


45° 45°

vertical conductor horizontal conductor

Questions 1. People might think, wrongly, that thunderclouds are either positively or negatively charged. In

fact, as the first diagram shows, the clouds, which are initially neutral, end up with ‘two maincentres of charge’. Give a simple physics explanation of why it would be reasonable to expectthere to be two regions of charge.

2. Observations suggest that strong vertical winds are involved in the charging of thunderclouds.Suggest where this kinetic energy comes from.

3. In the section entitled ‘The lightning flash’ the author estimates the energy released in a lightningdischarge to be of the order of 1010 J. Use this, and any other information given in the passage, toestimate the total electrical energy generated during a typical thunderstorm.

4. Suggest and explain what eventually happens to this energy generated during a typicalthunderstorm.

The passage quotes various values of field strengths, potential differences, etc, involved inthunderclouds. Questions 5–10 ask you to check some of these values using a very simple model ofa thundercloud. In this model the electric field strength E under the cloud is given by the expression E= Q / A 0 where A is the area of the cloud base and Q is the charge on the cloud base. The constant


0 is 8.9 10–12 F m–1.

Suppose a negative charge is spread evenly over the base of a cloud, in an even simplerarrangement than that shown.

5. Sketch the likely pattern of the electric field between the cloud and the ground.

6. Use the scale given in the diagram in the text to estimate the area of the cloud base and itsdistance from the ground.

7. Use your estimated value from question 6 to calculate the strength of the electric field under thecloud and the potential difference between it and the ground, assuming that the charge on thecloud base is 20 coulombs.

8. Actually, the ‘average field below the thundercloud is only of the order of 3 104 V m–1’. Suggestone reason why the result of your calculation in question 7 is different from this value.

9. Use the value of the field strength under a thundercloud as 3 104 V m–1 to calculate theacceleration of a typical gas ion in air. Take the mass of a typical gas ion as 4.7 10–26 kg andthe charge on the ion as 1.6 10–19 C.

10. The author states that the ‘positive ions move at only a few metres per second under theinfluence of the field’. How can you reconcile your answer to question 9 with this low resultantspeed?

11. The main current of the discharge is of the order of 104 amperes. What does this current actuallyconsist of?


12. A diagram provided shows that electric field lines tend to concentrate near the sharp end ofpointed objects. What does this suggest to you about the electric field near such points,compared with a rounded object at the same potential?

13. The author says ‘Water drops tend to become deformed in the high electric fields’. Draw adiagram showing the kind of deformation you would expect and use it to explain why this may bea factor which may ‘assist the development of a lightning stroke’.

14. Why should the ‘stepped leader’ of a lightning discharge start from the base of the cloud ratherthan from the ground?

One theory about how lightning conductors work is that they gradually and safely discharge the baseof the thundercloud. The formula given in the section entitled ‘Point discharge currents’ makes itpossible to calculate the discharge current in certain circumstances.

15. What discharge current would flow with the 27 metre tall conductor referred to with a field of 3 104 V m–1, in the absence of wind?

16. By how many times would this current change in a wind of speed 8 m s–1?

17. Explain why the discharge current depends on the speed of the wind.

18. Suggest a reason why the relationship might break down at very high wind speeds.


19. Estimate how long it would take such a lightning conductor to discharge the base of athundercloud in a wind of 8 m s–1, and comment on your answer.

20. Explain in what way lightning conductors actually help to provide a protection for tall buildingsagainst being damaged by lightning.

Electrical breakdown in a vacuumQuestion 50C: Comprehension


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Instructions The following passage is adapted from an article by R V Latham ‘Initiation of electrical breakdown invacuum’ (Phys. Technol. 9, 1978). Read the passage carefully and then answer the questions.

Introduction Surprisingly, even a vacuum does not provide a perfect insulating barrier. The current–voltage graphfor the pair of electrodes in the graph below shows that there is a considerable range of voltages forwhich the vacuum gap will conduct a small current.


+e.h.t.

–

A

current-limitingresistor (~ 500 M)

vacuumpumpingsystem

cathode anode

Beyond a certain critical voltage the current becomes unstable and will shoot up to a valuedetermined solely by the resistance of the external circuit. When this happens a glowing arc is seenbetween the electrodes and electrical breakdown is said to have occurred.

Electrical breakdown is an irreversible and catastrophic process, since a high-voltage gap willsubsequently break down at a much lower voltage. This has profound implications for the design ofequipment which uses high-voltage vacuum insulation (e.g. x-ray tubes, electron microscopes etc),since breakdown will impair performance and be expensive to remedy.

Two physical mechanisms which can start the electrical breakdown of a vacuum gap are describedbelow. In both cases electrode material is vaporised and then ionised to form a plasma (a mixture ofelectrons and ions). Usually these plasmas are short lived, their charges rapidly separate in thestrong field causing microdischarges.


10-6

10-7

10-8

10-9

10-10

10-11

10-5

0 10 20 30 40

electricalbreakdown

voltage across electrodes / kV

However, if the plasma current is sufficiently large, then heating of the anode or cathode surface cancause further vaporisation, so sustaining the plasma and creating an arc.

Electron emission initiated breakdown The tiny currents which occur before breakdown are caused by electrons being emitted from isolatedsites on the cathode surface. Since this type of electron emission requires an electric field of over 109

V m–1, which is some 20 times higher than the typical fields found between high-voltage electrodes, itis necessary to assume that the gap field is locally enhanced at the emission sites. An examination ofthe cathode using an electron microscope shows that the electrode surface contains a number of verysmall isolated protrusions or whiskers. The electric field in the gap is distorted by the protrusion,leading to very strong fields in the region surrounding the tip of the whisker. For low emissioncurrents, these protrusions act as stable electron sources. However, as emission current levelsincrease, as they would following an increase in the gap voltage, a critical condition is reached wherethe heating effect of the current in the protrusion is greater than the dissipation of energy byconduction and radiation. The protrusion is vaporised leading to plasma formation.

Microparticle initiated breakdown To reduce the above effect, the final stage of manufacture of high-voltage electrodes involvespolishing with a fine abrasive to remove surface irregularities. Unfortunately these freshly polishedelectrodes have numerous spherical particles of diameter 1–2 m loosely embedded in their surfaces.When a field is applied, these particles experience strong electrical forces which tend to tear themfrom the surface. Those that escape are accelerated towards the opposite electrode. If the impactvelocity of the microparticle is high enough it will cause sufficient vaporisation of the collidingmaterials for the creation of a plasma. It has been established that impact velocities of over 1500 ms–1 are needed to create the microcraters that are characteristic evidence of electrodes that havebroken down. Theoretical calculations based on estimates of the size and charge carried bymicroparticles suggest that microcratering should occur for gap voltages exceeding 50 kV.

Technological implications To minimise the risk of breakdown between a new pair of high-voltage electrodes they must be‘conditioned’ after installation without precipitating breakdown. To do this, the high-voltage electrodesare connected to a circuit such as shown in the first diagram. The large series resistor prevents


breakdown occurring by reducing the gap voltage whenever there is a surge in the pre-breakdowncurrent. The conditioning procedure involves gradually increasing the applied voltage in steps, suchthat the pre-breakdown current is allowed to stabilise at each stage before progressing. This processis continued until the operating voltage is reached.

Questions Questions 1–4 are based on the Introduction.

1. Why might it be thought likely that a vacuum would provide perfect electrical insulation?

2. How does the author suggest that electrical breakdown between high-voltage electrodes willimpair performance in apparatus such as electron microscopes and x-ray tubes?

3. Why does the current between the electrodes increase when a plasma is formed?

4. With reference to the graph in the second diagram and the value of the protective resistor in thefirst, show that the breakdown current in the circuit is limited to a maximum value of about 75 A.

Questions 5–10 are based on the section entitled ‘Electron emission initiated breakdown’.

5. Draw a diagram representing part of the surface of a perfectly flat negative terminal (cathode) andthe pattern of the electric field lines in the uniform field above the surface.

6. Draw a second diagram, suitably enlarged, showing the smooth surface of the cathode having asingle protrusion. Draw the field pattern in this case and indicate on your diagram where youwould expect electron emission to occur first, as the voltage is increased.

7. Explain the meaning of the phrase ‘stable electron source’ used in the paragraph.


8. Explain the meaning of the phrase ‘dissipation of energy’ used in the paragraph.

9. Use the information in the paragraph to estimate the size of the electrical force required to pull anelectron from the surface of a metal. The magnitude of the electronic charge is 1.6 10–19

C.

10. How does the existence of the microprotrusions help explain the noisy nature of the current priorto breakdown?

Questions 11–13 are based on the section entitled ‘Microparticle initiated breakdown’.

11. Why do the particles experience strong electrical forces tending to tear them away from thesurface of the electrodes?

12. Use the data within this section of the passage to show that the mass of a microparticle istypically about 10–14 kg. The density of stainless steel is 9000 kg m–3.

13. Use the data within this section to show that the charge carried by each particle is about 10–13 C.

Questions 14 and 15 are about the conditioning procedure described in the section entitled‘Technological implications’.

14. Explain how the use of a large series resistor in the circuit in the first diagram prevents a currentsurge developing into electrical breakdown.


15. Suggest ways in which the conditioning procedure might physically change the electrodes.

Millikan's oil drop experimentQuestion 70D: Data Handling


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Introduction In 1910 the American physicist Robert Millikan carried out an experiment which allowed him tocalculate the charge on a single electron for the first time ever. He used oil drops that pick up anelectric charge by friction. These drops were introduced into a region of vertical uniform electric field.If the weight of the oil drop was measured, together with the balancing voltage, V, across the plates,and the distance, d , between them, the charge on the drop could be measured.

Data The data used in this question have been obtained from more modern apparatus. By adjusting thesize of the electric field, the experimenter was able to produce sufficient force on the oil drop toexactly balance its weight and so make it stand still. Study the data given and then answer thequestions below.

Here is a simplified arrangement for obtaining an electric field in the region of the oil drop:

voltagesupply

+

–

oil dropd

The distance, d , between plates = 4.42 mm.

The table shows some measurements for a number of different oil drops:

Dropnumber

Balancingvoltage, V / V

Time for oil drop to fall1 mm in air / s

1 470 6.8


Dropnumber

Balancingvoltage, V / V

Time for oil drop to fall1 mm in air / s

2 820 6.5

3 230 9.3

4 770 10.3

5 1030 8.8

6 395 5.8

Below is a graph of weight against time for oil drops falling 1 mm in air.

0

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

time in seconds, to fall 1 mm

Questions 1. Assuming that the charge on the oil drop is negative, draw the charges on the plates that would

set up the electric field required to hold the drop stationary and draw the electric field between theplates.

2. Add arrows to the oil drop to indicate the two forces on the drop when stationary. Label these


forces ‘weight’ and ‘electrical force’.

3. If the charge on the drop is Q, the potential difference across the plates V and the distancebetween the plates d , show that Q = W d / V when the oil drop is balanced where W is the weightof the oil drop.

In order to measure the weight of an oil drop Millikan had to allow the drop to fall freely under gravityand through air. The electric field was switched off for this part of the experiment and the drop wasfound to fall with a steady velocity. The data include a modern calibration graph in which the weight ofthe oil drop is plotted against the time for the drop to fall a distance of 1 mm in air at a steady velocity.

4. Explain why the drop fell with a steady velocity.

5. Use the calibration graph to deduce the weight of each of the drops numbered 1 to 6.

6. Use the formula derived in question 3 to calculate the charge Q for each of the drops 1 to 6 to anappropriate number of significant figures.

Millikan’s experiment was used to show that each charge Q should be some whole number multiple nof e so that Q = n e.

7. Given that other experiments have shown that the value of e is 1.6 10–19 C, deduce the multiplen for each of the drops 1 to 6.

8. Calculate the percentage uncertainty in each of the measurements 1 to 6 and suggest which islikely to be the largest uncertainty.


The proton synchrotronQuestion 130D: Data Handling


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This data analysis exercise is about the acceleration of protons in the proton synchrotron at theEuropean Centre for Nuclear Research (CERN) near Geneva. Study the data given before answeringthe questions.

Data Particle accelerators are used to increase the energy of charged particles such as protons andelectrons. The accelerated particles are made to collide with other particles in order to investigate thestructure of matter. Subatomic particles created in these collisions can be observed and studied.

One type of particle accelerator is the synchrotron. In this machine a magnetic field causes chargedparticles to travel in a circular path. The particles are accelerated by an electric field. As theirmomentum increases, the magnetic field is also increased to keep them travelling in a path ofconstant radius.

General data:

charge on proton = 1.60 10–19 C

mass of proton = 1.67 10–27 kg

1 MeV = 1 106 eV

1 GeV = 1 109 eV

injector

direction ofprotons

200 m

28 GeV proton synchrotron (PS)

particle beam

vacuum pipe

alternating E field in thisregion accelerates protons

acceleration cavity

acceleration point


For the proton synchrotron:

diameter of ring = 200 m

circumference of ring = 628 m

number of accelerating points = 14

average accelerating voltage = 4.00 kV

At injection:

energy of proton = 50 MeV

At ejection:

energy of proton = 28 GeV

momentum of proton = 1.60 10–17 kg m s–1

speed of proton almost 3.00 108 m s–1

For the Super Proton Synchrotron (SPS):

diameter of ring = 2.2 km

speed of proton almost 3.00 108 m s–1

Questions Protons are injected into the 28 GeV proton synchrotron ring at CERN with an energy of 50 MeV (8 10–12 J).

1. Ignoring relativistic effects, show that the speed of a proton at injection is about 108 m s–1.

2. Ignoring relativistic effects, show that a proton takes about 6 s to travel round the ring at thisspeed.

3. Ignoring relativistic effects, show that the momentum of a proton at injection is about 1.6 10–19

kg m s–1.

The accelerator ring is a vacuum pipe maintained at a very low pressure. It is ‘filled’ with protons byinjecting a proton current of 100 mA for the 6 s it takes for protons to make one revolution at theinjection energy of 50 MeV.


4. Calculate the number of protons injected.

5. Explain why the ring must be maintained at a very low pressure.

Before the protons are accelerated, an electric field is used to group the protons in the ring into anumber of bunches. The bunches of protons are then accelerated as they pass through each of 14acceleration points spaced equally round the ring. An acceleration point is essentially a pair ofelectrodes between which an alternating voltage is applied.

6. Suggest why an alternating voltage is applied between the electrodes in order to accelerate theprotons.

7. The proton bunches pass through the acceleration point when the voltage between its electrodesis about 4 kV. By how much does the energy of one proton increase in each revolution? (Giveyour answer in eV.)

The final energy of a proton is 28 GeV.

8. Estimate the number of times a proton travels round the ring in acquiring this energy.

9. Explain briefly why linear accelerators (as opposed to ring accelerators) are not used toaccelerate protons to these energies.

10. Show that the magnetic field required to maintain a proton of charge e in a circular path of radiusr is proportional to the momentum p of the proton.


11. Estimate the magnetic field required to maintain 50 MeV protons within the CERN protonsynchrotron.

12. Explain why the frequency of the accelerating voltage must be increased as the speed of theprotons increases.

When it reaches its maximum energy of 28 GeV, the momentum of a proton is 1.6 10–17 kg m s–1

and it is travelling at almost 3 108 m s–1. At this relativistic speed, the momentum p is givenapproximately by E / c where E is the energy.

13. The magnetic field needed is proportional to the particle momentum. Estimate by what factor themagnetic field must be increased during acceleration.

14. The mass of a proton corresponds to a rest energy of about 1 GeV. The relativistic time dilationfactor is equal to Etotal / Erest. Estimate .

The electric dipoleQuestion 170D: Data Handling


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Instructions This data analysis exercise is about the electric fields near an electrical dipole. The calculations andgraphs for this question should be done with the aid of a computer spreadsheet. Some numbers arevery large and some are very small so it is best to format the cells for ‘scientific notation’.

Data The electrical properties of many insulators are controlled by the fact that they contain an array of


dipoles. A dipole is a pair of charges equal in size and of opposite sign, separated by a shortdistance. The distribution of the electrons in many molecules often leads to the existence of smalldipoles. A and B below could be a pair of ions, one of which had given an electron to the other:

A

–1.6 × 10–19 C

B

+1.6 × 10–19 C

2 × 10–10 m

O P

d

The questions are about the way that the electric field strength varies with distance, d , from thecentre of the dipole. A has a charge of – 1.6 10–19 C and B has a charge of 1.6 10–19 C. Thedistance between the centres of charge is 2 10–10 m. Point O is midway between A and B and pointP is on the axis of the dipole at a distance d from point O.

Questions 1. Construct a table containing a column for values of d ranging from 2 10–10 m to 2.5 10–9 m. It

is best to increase d in 1 10–10 m intervals.

2. Use the next column to calculate the electric field strength at point P due to the charge A for eachof your values of d . The physics formula to use within your spreadsheet is E = kQ / r 2 where k =9.0 109 N m2 C–2.

3. Use the next column to calculate the electric field strength at point P due to the charge B for eachof your values of d .

4. Use the next column to calculate the resultant electric field strength, E, at point P due to bothcharges A and B for each of your values of d .

5. Use your spreadsheet to plot a graph of the resultant field strength, E, at P against distance d .Does your graph enable you to establish the relationship between the electric field strength anddistance d ?

6. Theory suggests that the resultant field at P is inversely proportional to d 3 for a range of values ofd . Use your spreadsheet to plot a graph of the resultant field against 1 / d 3 to establish whetherthis is the case. Is the suggestion correct?

7. Use another column in your spreadsheet to compute values of E d 3 and plot a graph of E d 3

against d . Comment on the graph you have obtained.

The value of E d 3 is 5.76 10–19 N C–1 m3 when d is 1 m (very large). This can be checked veryeasily with your spreadsheet.

8. Use your spreadsheet to find the value of d at which E d 3 is about 1% greater than the value at 1m and hence find the range of values of d over which E d 3 does not differ by more than 1%,starting from the value at 1 m.

9. Replot the graph of E against 1 / d 3 over the range for which you now think a 1 / d 3 relationshipapplies.

10. Does the electric field strength due to a dipole fall off more or less rapidly with distance than thatof a single charge? Support your answer by considering the likely field strength due to two closelyspaced, equal and opposite charges at a very large distance.


11. Ionic crystals such as sodium chloride can be thought of as consisting of many dipoles. Theelectric fields inside a crystal of sodium chloride are large enough to hold ions together but thereis very little field outside the crystal. Use your understanding of the field near a dipole to suggestwhy this might be the case.

Testing Coulomb’s lawQuestion 210D: Data Handling


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This data analysis exercise is about an experiment to establish the inverse square law for electriccharge, known as Coulomb’s law.

Data This is the apparatus for this experiment:

chargedball on rod

hangingcharged ball

double suspensionof nylon thread

insulatingrod

A fine nylon thread is used to suspend a small charged polystyrene ball. A second charged ball,attached to an insulating rod, is pushed towards the first ball. This causes a sideways deflection of


the suspended ball. The nylon thread is 1.00 m long.

F

repulsive forcefrom other sphere

r

d

1 m

The photographs show the positions that the suspended ball takes up as the distance of the other ballis changed. When the charged ball on the rod is a long way away, the suspended ball hangs vertically

Open the JPEG file

As the second ball is brought closer, the suspended ball is pushed to the right until there is a bigenough sideways force for the suspension to balance the repulsive force due to the second ball. Forsmall angles used in this experiment, the sideways repulsive force, F, on the suspended ball isdirectly proportional to the deflection, d .

Open the JPEG file

Questions1. Using the markings on the ruler shown, measure the deflection, d , and the separation, r ,


between the centres of the ball, for photographs.

Open the JPEG file

Open the JPEG file

Open the JPEG file

Open the JPEG file

Open the JPEG file

Plot a graph to test whether the sideways force, F, on the suspended balls varies as 1 / r 2.Remember, the value of F is proportional to d .

The balls were given equal charges, each about 5 10–9 C, from a high-voltage source. The chargingwas repeated for each picture and the charge was measured by sharing the charge on the ball with a0.01 F capacitor, which was then found, using a high-resistance voltmeter, to have a potentialdifference of 0.5 V across it.

2. Check that the charge stated above is correct.

3. The measurements of charge indicated that the charge varies by a few per cent on differentoccasions. Would such fluctuations explain any feature of your graph?

4. The sideways force, F, is W tan where W is the weight of the suspended ball and is the anglemade between the nylon thread and the vertical (as on the second diagram). The weight of theball was found to be 1.1 10–3 N.

5. Estimate the order of magnitude of the force constant in the equation:

2balls)between(distance

ball)othertheon(chargeball)oneon(chargeconstant)[(forceballsbetweenForce

The charge on each ball almost certainly lies between 4 10–9 C and 6 10–9 C. Between whatlimits does this suggest that the force constant lies? The usual value is 9 109 N m2 C–2.


Using uniform electric fieldsQuestion 30X: Explanation–Exposition


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How to answer these questions Write briefly for each of the questions below to show how the physical principles you have learnedapply to the situation described. You should make calculations, give equations and include diagramswhere relevant. You may like to use the ‘Hints’ to help you to get started.

Questions 1. A sparking plug of a petrol engine used in a motor car has two electrodes separated by a spacing

of 0.67 mm. Air at atmospheric pressure begins to ionise when the electric field is about 3 106 Vm–1 so the voltage that must be applied across the electrodes to cause a spark in a spark plugmust be very large. The volume of the gases in a petrol engine can decrease by a factor of 14before ignition occurs so the voltage needed will be significantly different from the valuecalculated at atmospheric pressure.

0.67 mm


2. In a ‘time of flight’ mass spectrometer all the ions from the source are accelerated through thesame potential difference. A sharp burst of ions is then allowed to drift down a tube in which thereare no significant electric or magnetic fields. At the end of the tube the ions hit an electrode andan electrical pulse is produced on a recorder.

acceleration electrode sharp pulse of ions

collectingelectrode

no electricfield here

ion source

–+

time

00

pulse shape as it leavesthe acceleration electrode

Ions of the same kind have similar velocities when leaving the accelerating electrode. The shape ofthe pulse as it leaves the accelerating electrode is shown. Pulses reaching the collecting electrodearrive later. The speed of the ions leaving the accelerating electrode is related to the acceleratingvoltage and to the mass of the ions. For a given accelerating voltage, ions of different masses, say mand 2 m, would reach the collecting electrode at different times. If there were more ions of mass 2 mthe pulse for these ions would be different from the pulse for ions of mass m.


Deflecting charged particles in a magnetic fieldQuestion 120X: Explanation–Exposition


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How to answer these questions Write briefly for each of the questions below to show how the physical principles you have learnedapply to the situation described. You should make calculations, give equations and include diagramswhere relevant. You may like to use the ‘Hints’ to help you to get started.

electronic charge = 1.6 10–19 C

electron mass = 9.1 10–31 kg

gravitational field strength = 9.8 N kg–31

Questions 1. Electrons in an old fashioned television cathode ray tube are accelerated in the electron gun by

an electric field of 5.0 103 N C–1 over a distance of 10 cm. The electrons then pass betweensets of deflection coils which produce magnetic fields of up to 5.0 10–4 T. Although the electricand magnetic fields produce significant effects the designers of the tube can ignore the presenceof the gravitational field in their calculations.


deflection coils

electron gun

screen

35 cm

10 cm

2. Charged particles moving at right angles to a uniform magnetic field move in a circular path. Theradius of this circular path is related to the strength of the magnetic field and to properties of thecharged particle. A particle projected at an angle to a uniform magnetic field as shown will followa helical path.

v

vx

vy

B field B field

particle projectedat an angle to auniform magneticfield

helical path of acharged particlemoving in a uniformmagnetic field


When charged particles move in a non-uniform magnetic field the path is a helical (spiral) one. Anexample of this is when charged particles move in regions around the Earth known as the VanAllen radiation belts as shown. The radius of the spiral path varies with position in the Earth’smagnetic field.

spiral path of a charged particle moving in a non-uniform magnetic field

Atmospheric electricityReading 20T: Text to Read


Teaching Notes | Key Terms Quick Help

The Earth’s atmosphere is an insulator sandwiched between two good conductors, the surface of theEarth and the ionosphere (a charged layer that surrounds the Earth at an altitude above about 30km). Thunderstorms are electrical discharges in the atmosphere whose net effect is to maintain ahigh potential difference (about 3105 V) between these two conductors.

Each cloud-to-ground lightning strike carries a burst of current of several thousand amperes. Theseintermittent electrical discharges transfer positive charge upwards, and a moderate sized storm cloudmight transfer charge at an average rate of about 1 A. Charging mechanisms within the cloudsthemselves raise the potential at the top of a storm cloud to about 108 V. This is well above thepotential of the ionosphere, so there is a current from the top of the cloud to the ionosphere. If thischarging process continued in isolation it would eventually raise the potential of the ionosphere to thatof the cloud tops. No more current would flow. This does not happen.

There is a balancing process that transfers charge down from the ionosphere to the ground in regionsaway from storm clouds. This gradual leakage of charge is called the ‘fair-weather current’. On aglobal scale, the average current from the ground up to the ionosphere as a result of storms isbalanced by a ‘fair-weather current’ of about 2000 A in the other direction. This means that, onaverage, at any particular moment, there are the equivalent of about 2000 moderate sized electricalstorms taking place across the world. In practice, large tropical storms account for a disproportionateamount of the ‘storm current’, so 2000 is an overestimate. We can take this further and draw a simplecircuit diagram to represent the global flow of electricity. Storm clouds act as a kind of generator,‘pumping’ charge to higher altitude, so these can be represented by cells. Each ‘storm cell’ has itsnegative electrode connected to the ground and its positive electrode connected in series with aresistor and then the ionosphere. The resistor represents resistance in the cloud and atmosphereabove the cloud. The return circuit is represented by a single, smaller resistor, representing theresistance of the entire atmosphere excluding those parts that are above storm clouds.

Rstorm

108 V

RatmosRstormRstorm

+ + +

This simple circuit could be redrawn as a potential divider:


3 × 105 V

Rstorm

108 V

Ratmos

Here Rstorm is the parallel combination of all the individual Rstorm values from particular storm clouds.The potential divider circuit can now be used to find the ratio of Rstorm to Ratmos. If we assume that theresistance of the atmosphere above storm clouds is the same as elsewhere this can give a roughidea of the amount of the Earth’s surface that is affected by thunderstorms at any time. It is about0.3%.

Fair-weather electricity We have already said that the ionosphere and the Earth’s surface are good conductors. This makesthem, more or less, equipotential surfaces. The potential difference between them is about 3105 Vand they are separated by a distance of about 50 km. A simple calculation shows that we shouldexpect a fair-weather potential gradient of about 100 V m–1 vertically above the surface of the Earth.At first this seems a rather ludicrous idea. Adults are not much under 2 m tall, so does this mean thatthere is a 200 V p.d. between your scalp and the soles of your feet when you stand up? That wouldbe worrying to say the least, and fortunately it is not the case. Our bodies are pretty good conductors,so we too form part of the equipotential surface and simply distort the electric field where we stand.So do buildings, trees and other objects built on the surface.

The existence of a fair-weather current (about 2000 A across the Earth’s entire surface) implies thatthe atmosphere is not a perfect insulator. About 5 pA m–2 flow to Earth, so there must be a supply offree charges in the atmosphere. What creates them? At the beginning of the twentieth centuryionising radiation from radioactive isotopes in the Earth’s crust was thought the most likely cause. If


this hypothesis was correct the density of charge carriers in the air should decrease with height abovethe surface. In 1912 this prediction was tested by Hess who measured the conductivity of the air atdifferent heights from a balloon. To his surprise there are more ions per unit volume at high altitudethan at low altitude. This extra ionisation is created by extraterrestrial radiation, the cosmic rays thatbombard the Earth from space. A second effect also increases the conductivity at high altitude:high-altitude air has lower density, so the molecules have greater average separation and the ionscan move further before they hit something and lose kinetic energy (longer ‘mean free path’). Aboveabout 50 km the density of ions and their long mean free paths create an excellent conducting layer.Strictly speaking this is not actually the ionosphere (which is higher up where ionisation is dominatedby photoelectric interactions from solar radiation) but it is the layer from which the atmosphericcurrents flow to Earth.

The ultimate speedReading 30T: Text to Read

Teaching Notes | Key Terms

Quick Help

Drop a stone in a pond and the disturbance on the water surface sets up ripples that travel outwardsat a fixed speed. When Maxwell unified electricity and magnetism he realised that disturbances in theelectromagnetic field would move away from their source at a constant speed and that this speeddepends only on the medium through which they pass. In a vacuum the speed is given by theequation

.1

00c

This is 299 792 458 m s–1 (usually taken as 3.0 108 m s–1), the speed of light. Visible light is just asmall part of a much broader spectrum of electromagnetic waves all of which travel at this speed in avacuum.

After Maxwell, physicists made increasingly accurate measurements of the speed of light and beganto ask a simple question: speed relative to what? The speed of sound is measured relative to the air,the speed of water waves relative to the sea and seismic waves relative to the crust of the Earth, so itseemed reasonable to assume that the speed of light must also be measured relative to something,but what? They called this unknown medium the ‘ether’ and began to wonder how it might bedetected.

Is there an ‘ether’? If the ‘ether’ is anchored in space then the speed of light measured on Earth should increase ordecrease depending on the direction of the Earth’s own motion through the ether. The most famousexperiment designed to test this idea was carried out by Michelson and Morley in 1887. To theirsurprise they detected no change in the speed of light. This conclusion has been confirmed andextended by more recent experiments. In 1932 Kennedy and Thorndike carried out a more sensitiveversion of the Michelson–Morley experiment and confirmed that the Earth’s orbital motion affects themeasured speed of light by less than 1 part in 108. The constancy of the speed of light has also beentested using particle physics. This involves high-speed neutral pions, subatomic particles created


when proton beams crash into matter. Neutral pions are unstable and decay by emitting pairs ofphotons. The speed of these photons has been measured in the laboratory and is still c even whenthe pion source is moving at close to the speed of light before it decays.

The speed of light is invariant The above results confirm that:

The speed of light does not depend on the relative velocity of source and observer.

If you think about it, this is rather strange. Imagine a jet plane passing over your head at 100 m s–1. Ithas a blinking light underneath it. How fast does light leaving this lamp travel relative to the plane?How fast relative to you? The experiments above suggest that it has the same speed, c, in bothcases. But how can this be? Why doesn’t the light travelling ahead of the plane gain an extra 100 ms–1 from the plane’s motion?

Einstein made a start on the theory of relativity by asking, is it possible to run fast enough to catch upa light beam?

Accelerators according to Newton What if you do try to accelerate something up to the speed of light, which is what Einstein was talkingabout? Is it possible to catch a light beam? Particle accelerators take energy from electrical potentialenergy and give it to the particle as kinetic energy, by ‘dropping’ charges through a potentialdifference V. Taking the kinetic energy to be ½ m v2, using the classical Newtonian relationship, thevelocity v of a particle with mass m and charge q is:

.2

m

qVv

The accelerating potential difference V can be built up in stages by passing the charged particlethrough a sequence of accelerating electrodes (linear accelerator) or repeatedly past the sameaccelerating electrodes (circular accelerator or synchrotron). A quick calculation shows that about 260kV would be enough to accelerate an electron through the light barrier. So what happens if you try it?

Bertozzi’s experiment In 1962 William Bertozzi used a linear accelerator to fire electrons through a measured distance andtimed how long they took.


The diagram shows a highly simplified version of Bertozzi’s apparatus. Detectorsat the start and end of an 8.4 m section start and stop a timer. Speed is thencalculated by dividing distance by time.

electron gun

start timer

vacuum tube

constant velocity8.4 m

variable accelerating p.d.

stop timer

Bertozzi’s experiment.

He started with low accelerating voltages and gradually increased the potential difference. His resultsare shown below:

Potentialdifference /MV

Kineticenergy /MeV

Kineticenergy / J

Distancetravelled/ m

Time offlight /ns

Calculatedvelocity /

m s–1

Predictedvelocity / m s–1

0.50 0.50 8.0 10–14 8.40 32.3 2.60 108 2.96 108

1.0 1.0 1.6 10–13 8.40 30.8 2.73 108 4.19 108

1.5 1.5 2.4 10–13 8.40 29.2 2.88 108 5.13 108

0

0

2 4 6 8 10 12 14 16

accelerating p.d. / MV

2

4

6

8

10

12

14

16

18

Bertozzi’s experiment

experiment

prediction based on 1 mv22


Compare Bertozzi’s measured speeds with those predicted by the equation based on kinetic energy =½ m v2. Theory and experiment are in good agreement for low potential differences but disagree athigh accelerating voltages. However large the accelerating voltage the particle speed never quitereaches the speed of light. This seems to be generally true – there is no way to make a particle travelas fast as light.

A new start In 1905 Einstein approached the same problem from a theoretical point of view. He started from theprinciple that the laws of physics ought not to depend on mere relative velocity. Physics should be thesame in a galaxy moving relative to you as it is for you. He thought of the value of the speed of light,predicted by Maxwell’s theory, as one of these laws. So the invariance of the speed of light became aprinciple, not an awkward fact.

How, though, can a speed be the same for everybody? Einstein’s idea was to think of the speed oflight as just the fixed way, the same for everybody, that time and space must be related. Onelight-second of distance (300 million km) always takes light 1 second to cover. But to have this fixedrelation between space and time meant that measurements of space and time would have to dependon relative velocity.

Thus Einstein found that he could keep the speed of light constant if comparisons of space and timewere influenced by relative motion. How can two people moving at different speeds relative to oneanother measure the rate at which a light beam moves away from them and come up with the sameanswer? Einstein thought hard about what they would have to do to measure the speed of light. Theywould need to measure the time taken to cover a measured distance. This would need calibratedrulers and clocks that were moving with them. His solution was ingenious – two observers woulddisagree on distances and times but would agree on the speed of light! The laws of physics are thesame for everyone, and so is the relation between space and time, but measurements of spaces andtimes depend on how you move. So, it turns out, do values of energy and momentum. The detailswere worked out in Einstein’s Special Theory of Relativity.

New meaning for mass:new equations Einstein’s new thinking meant that the Newtonian equations momentum = m v and kinetic energy = ½m v2 are only approximations, valid only at speeds small compared with the speed of light. Inrelativity, mass takes on a new meaning. It is the total energy of a particle alongside which you aretravelling through space-time, that is, the rest energy. This is where the famous equation

2rest mcE

comes from. The mass has to be thought of as part of the total energy of a particle moving past you.The total energy E (including rest energy) and momentum p are now given by:

mvp

mcE

2

where is the same as the time dilation factor

.)/(1

122 cv

These equations are true for all speeds.


Classical equations still valid when v << c When v is much less than c, the quantity is approximately equal to 1. This gives the old relationshipp m v for the momentum. Because the total energy E includes the rest energy m c2, the kineticenergy is the total energy minus the rest energy:

).1(

energykinetic2

22

mc

mcmc

Some algebra shows that if v / c is small, the factor – 1 ½ v2 / c2, so that

.energykinetic 221 mv=

You can also see from the expression

)/(1

122 cv

what happens at higher speeds. As v approaches c, becomes larger and larger without limit. So itwould take infinite energy and momentum to reach the speed c.

Accelerators add energy and momentum, not speed If an accelerating voltage of 10 MV is enough to make an electron travel at 99.5% of the speed oflight, what is the point of using accelerating voltages 10 000 times greater, as is actually done in thebiggest accelerators? The real point of a particle accelerator is not the final speed of the particles,which is always just a shade under the speed of light, but the energy and momentum they carry.Accelerators give particles enormous amounts of energy and momentum. When they collide, newparticles can be created with larger mass (rest energy) and lower kinetic energy than the collidingparticles got from the accelerator. This allows low-mass particles, like electrons and positrons, tocollide head-on and create new particles, such as the massive W+ and W– particles.

Accelerators according to relativity The Newtonian equation for velocity of a particle accelerated by potential difference V is:

.2

m

qVv

The relativistic equation can be put in a simple form if the rest energy Erest = m c2 is compared withthe total energy E, which is just Erest + q V:

2rest mc

E

E

E

where

.)(1

122 cv

Since the total energy E = m c2 + q V, then:


.12mc

qV

Having found you can calculate the ratio

).1(1/ 2cv

The way to decide whether the Newtonian expression is adequate is to compare the energy q Vprovided by the accelerator with the rest energy. If the energy q V is less than 10% of the rest energy,the Newtonian expression will be accurate to within about 5%. For electrons, with rest energy (mass)of 0.5 MeV, this allows energies up to about 50 keV. Many x-ray machines exceed potentialdifferences of 50 kV.

Summary of main ideas The speed of light is invariant, the same for all observers, independent of the motion of the

observer or the source of light observed.

It is impossible to accelerate any massive particle up to or beyond the speed of light.

Newtonian mechanics cannot be applied to objects whose velocity approaches the speed of light,but is an approximation valid at low velocities.

The mass of a particle must be thought of as part of its total energy, with Erest = m c 2.

At low velocities, the kinetic energy (approximately ½ m v2) is a small ‘add on’ to the rest energy.

At very large energies, with v c, the rest energy (mass) of particles is much smaller than theirtotal energy E. In this case the momentum p is given approximately by E p c.

Afterword: ‘increase in mass?’ Many books say that mass increases with velocity. They base this idea on the equation for themomentum:

.mvp

Suggesting that the momentum p should be equal to m v at all speeds, the idea is that the mass is m, equal to m only at rest. So mass m is said to be increased by the factor . The ‘effect’ is then called‘the relativistic increase of mass with velocity’. Some books are honest enough to say that theyhaven’t the slightest idea why it happens. The simplest answer is that the reason there isn’t anyknown mechanism for increasing the mass is that it doesn’t actually happen.

The trouble with the idea of ‘increase of mass’ is that, if mass is given the new meaning it has inrelativity, which is the energy of a body you are travelling with (i.e. ‘at rest’), there can’t be anyincrease in mass. The mass, that is the rest energy, must be the same for everybody. Otherwisephysics would be different in galaxies moving relative to us (for example atomic spectra would bedifferent).

The choice is between saying that the old Newtonian ideas still apply, with p = m v but now with massmodified to be m = m

rest or that the momentum is not actually given by p = m v but by a new rule

which reflects how momentum depends on velocity as compared to the speed c, namely p = m v.

Same difference? Well, the answers come out the same, but the ideas behind them are different.

That is why, in Advancing Physics, you will find that we refer to rest energy Erest, with Erest = mc2, and


do not refer to ‘rest mass’ or to increase of mass with velocity.

Why we believe in Erest = mc2

Reading 35T: Text to Read


Quick Help

Perhaps the most amazing prediction of Einstein's special theory of relativity is Erest = mc2. If it turnedout to be just slightly incorrect, there would be far-reaching consequences. This reading describessome of the experimental evidence that gives physicists empirical support for their belief in theequation.

First test: ‘Splitting the atom’Although Einstein’s mass–energy relationship was formulated in 1905, the first real experimental testof the equation came over a quarter of a century later in 1932 when Cockcroft and Walton first ‘splitthe atom’.

Cockcroft and Walton bombarded lithium nuclei with protons accelerated by a potential difference of125 kV, so that their energy was 125 keV (chapter 16). The result was that one in every 10 millionprotons combined with the lithium to form two alpha particles (helium nuclei).

Strikingly, the alpha particles flew apart with a much greater energy than that supplied by thebombarding protons. In fact, they flew apart with a total energy of 16 MeV. At first sight this lookedlike a violation of conservation of energy. Einstein’s equation shows that it is not. When the mass ofthe proton plus the lithium is compared with the mass of two alpha particles, the difference m,multiplied by c 2, is equal to the net kinetic energy of the particles. The equation appears to be correctto within the uncertainty of the measurements.

Second test: electron positron annihilationThe year 1932 also saw the discovery of the positron whose existence had been predicted by Dirac in1928. Electron positron annihilation (chapter 17) provides a better test of Erest = mc2. Now the m isthe sum of the identical masses of the electron and the positron, and the two gamma photons take upthe energy involved (chapter 17, Question 50C).

The best test so far (2007)The latest direct test exploits areas of physics drawn from five different chapters of the AdvancingPhysics course.

The nuclear reaction usedThe nuclear process used is called neutron capture. As its name suggests, neutron capture occurswhen a nucleus absorbs a neutron to increase its mass number A by one nucleon to become adifferent isotope of the nuclide concerned. In the reactions chosen this heavier isotope is formed in anexcited state and soon decays by emitting a gamma photon, while the extra neutron adds to the totalbinding energy of the new nucleus (chapter 18).


Two different neutron capture reactions were studied one starting with silicon-28 and the other withsulphur-32. Both were of the form AX + n A + 1X + .

Accurate measurements of the atomic mass differences and gamma ray energies following neutroncapture have been made.

If Erest = mc2, is valid, then the energy of the photon divided by c2 should be equal to the mass of the

original nucleus ( AX ) plus the mass of the neutron minus the mass of the heavier nucleus ( A + 1X ).

Measuring the massThe mass of the original free neutron is already known. The masses of the isotopes were found by'cyclotron resonance'. An ion moving at speed v in a magnetic field B will follow a circular orbit radiusr with a circulation (or cyclotron) frequency f C (chapter 16).

Equating the centripetal force (m v 2) / r and q v B, gives m = (r q B) / v .

The angular speed = v / r , and the frequency of revolution (the cyclotron frequency) f C

= / 2, so

m = (q B) / (2f C).

These frequencies are easy to measure with great precision, and the ions before and after neutroncapture were kept circulating in the magnetic field for several weeks to allow the precisemeasurement required.

Measuring the energyThe energy of the resulting gamma photon emitted after neutron capture was found via the Planckrelationship E = h f (chapter 7), where f was calculated from the wavelength using c = f(chapter 6).

The wavelength for the gamma photon was found by diffraction using a crystal and the Braggformula 2d sin = n(chapter 7). The crystal lattice spacing d was known from x-ray diffractionmeasurements. The gamma energies are typically 5 MeV, giving a wavelength for the gamma photon

of about 2.5 10–13 m. It is the measurement of the resulting diffraction angle of about 0.1 that sets thelimit to the overall precision of the experiment.

The result This latest direct test shows that Erest = mc2 is precise to 0.00004% or better than 1 part in a million.It is 55 times better than the previous best direct test comparing the electron and positron masses tothe gamma ray energies when they annihilate (chapter 17).

Reference

Rainville S et al 2005 World year of physics: a direct test of E = mc2 Nature 438 1096–7

Proving that at low speeds relativistic kinetic energy (1/2) mv2

Reading 40T: Text to Read


Quick Help


This reading shows you, using only ordinary algebra, that at low speeds with v << c, the relativisticequation for kinetic energy is equivalent to the familiar Newtonian result Ekinetic = ½ mv2.

The relativistic equationsThe relativistic equation for the total energy of a particle of mass m is

2total mcE

with the relativistic factor given by

22 /1

1

cv

At zero speed, the factor = 1. Thus there is still energy at speed v = 0, at rest. It is given by

2rest mcE

The kinetic energy of a particle moving at relative speed v is the difference between total energy andrest energy:


Subtracting the two equations above gives, for the kinetic energy at any speed

222kinetic )1( mcmcmcE

We will now show that at low speeds this expression is equivalent to the classical Newtonianexpression

221

kinetic mvE

The relationships shown graphicallyThe graphs below compare the relativistic kinetic energy

2kinetic )1( mcE

with the Newtonian kinetic energy

221

kinetic mvE

as the speed v varies.


0v / c

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9




1.0

0

You can see from the two graphs that the relativistic and Newtonian values of kinetic energy agreevery closely for speeds even up to 0.3c. For speeds such as those of cars, trains or aeroplanes, theagreement is very exact indeed.

Algebraic proofYou may like to see how the remarkable equality of the two very different looking expressions

2kinetic )1( mcE

and

221

kinetic mvE

for v << c can be demonstrated, knowing that

22 /1

1

cv

Step 1: Look at the term – 1At low speeds, is only just slightly bigger than 1. So we ask, how much bigger? We know that

22 /1

1

cv

So we know that

1/1

11

22

cv

Express this as a single fraction with a common denominator:


22

22

/1

/111

cv

cv

That does not look much better, but it does if you realise that the common denominator

22 /1 cv

is very close indeed to 1. Thus we can get as a very good approximation:

22 /111 cv

Step 2: A neat algebraic trick The next step is not obvious at all. We are going to show that

222122 /1/111 cvcv

The problem is to find the value of a difference 1 – x where

22 /1 cvx

Recall a well known algebraic equation which involves the quantity ‘1 minus something’. It is:

xxx 111 2

Rearranging to get an expression for 1 – x:

x

xx

1

11

2

You know that

22 /1 cvx

is very close to 1. Thus

21 x

and the expression for 1 – x approximates closely to:

2

11

2xx

Since

222 /1 cvx

the equation can be written, in terms of v and c as:

2221

2222 /

2

)/1(1/11 cv

cvcv

That is, we have shown that at low speeds, – 1 approximates closely to:

2221 /1 cv


Step 3: Put – 1 back into the relativistic energyThe relativistic kinetic energy is given by:

2kinetic )1( mcE

At speeds v << c the value of the – 1 is approximately:

2221 /1 cv

Thus, at these low speeds, the relativistic expression for kinetic energy becomes the same as theNewtonian expression:

221

kinetic mvE

AcknowledgementWe are grateful to Dr Jozef Hanc for suggesting the algebraic argument above.

Revision ChecklistI can show my understanding of effects, ideas and relationships bydescribing and explaining cases involving:a uniform electric field E = V /d (measured in volts per metre)

A–Z references: electric field

Summary diagrams: The electric field between parallel plates, Two ways of describing electricalforces, Field strength and potential gradient, Field lines and equipotential surfaces

the electric field of a charged body; the force on a small charged body in an electric field; theinverse square law for the field due to a small (point or spherical) charged object

A–Z references: electric field, inverse square laws

Summary diagrams: Inverse square law and flux, Radial fields in gravity and electricity, How anelectric field deflects an electron beam

electrical potential energy and electric potential due to a point charge and the 1 / r relationshipfor electric potential due to a point charge

A–Z references: electric potential, inverse square laws

Summary diagrams: Force, field, energy and potential, Radial fields in gravity and electricity

evidence for the discreteness of the charge on an electron

A–Z references: electron


Summary diagrams: Millikan's experiment

the force qvB on a moving charged particle due to a magnetic field

A–Z references: force on a moving charge

Summary diagrams: How a magnetic field deflects an electron beam, Force on current: force onmoving charge, Measuring the momentum of moving charged particles

Relativistic relationships between mass and energy

A–Z references: mass and energy, theory of special relativity

Summary diagrams: The ultimate speed – Bertozzi's demonstration, Relativistic momentum,Relativistic energy, Energy, momentum and mass

I can use the following words and phrases accurately whendescribing effects and observations:electric charge, electric field; electric potential (J C–1) and electrical potential energy (J);equipotential surface

A–Z references: electric field, inverse square laws, electric potential

Summary diagrams: Force, field, energy and potential

the electron volt used as a unit of energy

A–Z references: electron volt

I can sketch and interpret:graphs of electric force versus distance, knowing that the area under the curve between twopoints gives the electric potential difference between the points

graphs of electric potential and electrical potential energy versus distance, knowing thatthe tangent to the potential vs distance graph at a point gives the value of the electric field atthat point

Summary diagrams: Force, field, energy and potential, Radial fields in gravity and electricity

diagrams illustrating electric fields (e.g. uniform and radial ) and the corresponding equipotentialsurfaces


Summary diagrams: Field strength and potential gradient, Field lines and equipotential surfaces

I can make calculations and estimates making use of:the force qE on a moving charged particle in a uniform electric field



Summary diagrams: How an electric field deflects an electron beam

radial component of electric force due to a point charge

2electricr

kqQF

radial component of electric field due to a point charge

2electric

electricr

kQ

q

FE



electric field related to electric potential difference

x

VE

d

delectric

and

dVE electric

for a uniform field


Summary diagrams: Force, field, energy and potential, Field strength and potential gradient

electric potential at a point distance r from a point charge:

rkQ

V

A–Z references: electric potential


the force F on a charge q moving at a velocity v perpendicular to a magnetic field B:

F = q v B

A–Z references: force on a moving charge

Summary diagrams: How a magnetic field deflects an electron beam, Force on current: force onmoving charge, Measuring the momentum of a moving particle

the uniform electric field - as-a2-physics - home · pdf file · 2009-03-31the...

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