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1 Advancing Physics

The uniform electric fieldQuestion 10W: Warm-up Exercise

1. The positively charged ball experiences a force to the left towards the negatively charged plate.

2. The current can be increased by decreasing the distance between the plates causing an increasein the electric field strength between the plates, and hence the force on the ball. The frequency atwhich the ball moves between the plates increases so the current increases also. If the voltage isincreased the electric field strength increases and the ball picks up more charge on contact withthe plates. Both factors contribute to an increased current.

3. + 500 V

0 V

+ 400 V

+ 300 V

+ 200 V

+ 100 V

4. See the solution to question 3.

5.

.CNormV100.6

m105

V300

114

3

d

VE

6.

.N109.1

)CN100.6()C1032(13

1419

qEF

7.

.J100.8

)V1050()C106.1(15

319

qVW

8. The energy gained will have a positive value. Both the charge and the potential difference arenegative:

J106.1

J106.1

)J106.1(100eV100

17221

17

19

mv

so

2 Advancing Physics

.sm109.5

)}kg101.9/()]J106.1(2{[16

2/13117

v

Getting F = q v BQuestion 80W: Short Answer

1. N e

2. N e / t

3. v t

4.

NevBF

Bvtt

NeLBF

I

5. e v B

6. Q v B

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

1. Starting from Ekinetic = ½ mv 2, multiply both sides by 2, getting 2 Ekinetic = mv 2. Then divide bothsides by m to obtain

2kinetic2v

m

E

Finally, take the square root of both sides:

vm

Ekinetic2

2. Ekinetic = ½ mv 2 so Ekinetic = ½ 500 kg 202 m2 s-2 = 105 J

3. v is proportional to (Ekinetic ), so doubling the energy increases the speed by a factor 2 = 1.4.

Thus the speed is 20 m s–1 1.4 = 28 m s–1

4. Ekinetic = qV = 1.6 10–19 C 1000 V = 1.6 10–16 J. The speed v is given by

1731

16

sm109.1kg 109.1

J 106.12

v

3 Advancing Physics

This is about 6% of the speed of light.

5. Ekinetic = qV = 1.6 10–19 C 64 000 V = 102 10–16 J. The speed v is given by

1831

16

sm105.1kg 109.1

J 101022

v

This is about 50% of the speed of light.

6. Speed has to be multiplied by 2, so energy needed is multiplied by 22 = 4. Potential differenceneeded = 4 64 000 V = 256 000 V. Note that the National Grid system works at potential

differences larger than this.

7. Ekinetic = ½ mv 2 = 0.5 9.1 10-31 kg (0.6 108 m s-1)2 = 1.6 10-15 J. The potential difference

V = E / q = 1.6 10-15 J / 1.6 10-19 C = 10 000 V.

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

Accelerator Date Particleaccelerated

Kineticenergy

factor Ratio v / c

Cockcroft and Walton 1932 Proton 770 keV 1.00077 0.039

Van de Graaff 1932 Proton 1.5 MeV 1.0015 0.055

Cyclotron 1937 Proton 3 MeV 1.003 0.077

Synchrocyclotron 1947 Proton 500 MeV 1.5 0.75

Synchrotron 1954 Proton 6 GeV 7 0.99

1. See table above.

2. See table above.

3. The proton rest energy is only 1 GeV so

30GeV 1

GeV 1 +GeV 30

rest

total E

E

4. At 3 MeV the ratio v / c is less than 10%. But at 500 MeV it is 75%. Relativistic effects arecertainly important for the synchrocyclotron and synchrotron.

Comparing relativistic and Newtonian kinetic energyQuestion 30S: Short Answer

4 Advancing Physics

1. The relativistic kinetic energy increases from a little more than to several times the Newtoniankinetic energy over this range energy.

2. The relativistic and Newtonian calculations are very similar over this range.

v / c (v / c)2 – 1 0.5 (v / c)2

0 0 1 0 0

0.1 0.01 1.005 0.005 0.005

0.2 0.04 1.021 0.021 0.020

0.3 0.09 1.048 0.048 0.045

0.4 0.16 1.091 0.091 0.080

0.5 0.25 1.155 0.155 0.125

3. See table above.

4. The discrepancy at v / c = 0.2 is 5%; at lower speeds it is smaller.

Particles at extremely high energyQuestion 40S: Short Answer

1. Proton rest energy = 1 GeV = 109 eV. Ratio of total energy to rest energy = 1020 eV / 109 eV =

1011. This ratio is the relativistic factor = 1011.

2. The ratio

2/11/ cv

differs negligibly from 1 because is so large.

3. Time t = 105 years = 105 year 3 107 s year-1 = 3 1012 s

4. Wristwatch time t / 3 1012 s / 1011 = 30 s

5. Distance = 1010 m = 107 km

Two uses for uniform electric fieldsQuestion 60S: Short Answer

1.

5 Advancing Physics

600 V+

–

3.2 × 10–2 m400 V

200 V

2. See solution for question 1

3.

.CNormV109.1

)m102.3/(V600

/

114

2

dVE

4.

C 103.9

C N 109.1

kg N 9.8kg 108.1

19

–14

–115

Q

E

mgQ

Q

mg

Q

FE

5. .6)C106.1/()C103.9( 1919

6. The time is increased due to air resistance which leads to a lower acceleration.

7.

+ 54 kV0 V

0.3 m

collectors

crushed minerals

conveyor belt

8.

6 Advancing Physics

.CNormV108.1

m30.0/V00054

/

115

dVE

9.

N 109.2

C N 101.8C) 106.110(8

15196

QEF

10.

m 012.0

s 1.1kg 101.5

N 109.2

2

1 2

6-

8

22

122

1

s

s

tm

Fats

(if air resistance is neglected).

11. The particles may stick to the plates.

12. There will be inadequate separation of the particles.

Deflection with electric and magnetic fieldsQuestion 90S: Short Answer

1. C

2. E

3. They hit A.

4.

source

detector

no field

B field

5. See the solution for question 4.

6. Circular since the proton experiences a force of constant magnitude at right angles to its pathregardless of its direction.

7.

7 Advancing Physics

force

8. See the solution for question 7.

9. 0 V

+ 1000 V

10 cm

10. Parabola since the force is constant and remains in the same direction.

The cyclotronQuestion 100S: Short Answer

Solutions1.

.sm109.3kg107.1

J)106.1(10802

2

1627

193

m

Ev

2.

r

mvBQvF

2

so

8 Advancing Physics

.T82.0

m)1050(C)106.1(

)sm1088.3(kg)107.1(319

1627

Qr

mvB

3. Since

)sm109.3(, 1621 vEv

and

mm.36

m036.0

2T82.0)C106.1(

)sm109.3()kg107.1(19

1627

QB

mvr

4.

s101.8

sm1088.3

m10502

2

8

16

3

v

rT

since

vrQB

mvr ,

and T is the same for all paths.

5. The time it takes for a particle to go around is independent of its speed (and energy). As it getsfaster, the particle spirals out into orbits with larger and larger radii. So all particles can beaccelerated across a ‘dee’ gap at the same time.

The Hall effectQuestion 140S: Short Answer

1. Q v B

2. Towards the front edge.

3. The moving charge carriers are pushed towards the front edge, so the density there will build up.It will continue to increase until there is an equally strong electrical force in the opposite directionto the magnetic field, i.e. from the front to the back.

4. The electric force (E Q) must be equal in magnitude to the magnetic force (B Q v). So the electricfield E = B v.

5. Front edge negative, rear edge positive.

9 Advancing Physics

6. V = E d = B v d

7. bd

8. n = B I / V b Q

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

1. 2

21 mvEk

so

.sm105.2)kg101.9(

)J109.2(2 1731

16

v

The kinetic energy is considerably less than the electron's rest energy, so the Newtonianapproximation can be used.

2.

.N108.8

)sm105.2()C106.1()T102.2(15

17193

BqvF

3. rmvF /2so

.m105.6N108.8

)sm105.2()kg101.9( 215

21731

r

4. Electrons lose speed through collisions with other particles. Since r = mv/Be, r decreases as vdecreases.

5.

uniform B fieldacts into plane ofscreen / paperover shaded area

protonmotion

6. The force on the particle is always perpendicular to v.

7. Equate q v B to m v2 / r .

8. Frequency f = v / 2 r . Substituting r from question 7 gives f = q B / 2 m.

10 Advancing Physics

9.

.T106.3

C106.1

)kg101.9(Hz1022

2

19

319

q

fmB

10. There will be no effect since the cyclotron frequency depends only upon B, q and m which do notchange under normal conditions.

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

1.

.V102

m200mV108

16

EdV

2.

.C9.8

mV10)m1000()mF109.8( 162112

0

AEQ

3.

.J109.8

)V102(C9.8

8

821

21

QV

4. 2

21 mveV

so

.sm104.8

sm1039.8

kg101.9

200)C106.1(22

16

16

31

19

V

m

eVv

5. rmvBev /2so

.m048.0

)C106.1(T001.0

)sm1039.8()kg101.9(19

1631

Be

mvr

6. BeveE so

./ BEv

7.


.sm106.1

sm1063.1

T102.8

mV1033.1

.mV1033.1

m024.0

3200

17

17

3

15

15

B

Ev

V

d

VE

8. 2

21 mveV

so

.kgC108.1

V7502

)sm1063.1(

2111

2172

V

v

m

e

9. 2

21 mvqV

so

.V104.8

)C106.1(2

)sm109()kg103.3(

5

19

21627221

q

mvV

10.

rmvBqv /2so

.T37.0

m5.0)C106.1(

)sm109()kg103.3(19

1627

qr

mvB

Non-uniform electric fieldsQuestion 180S: Short Answer

Solutions1.


2.

+ –

3. See solution to question 2.

4.

.N105.2

)m060.0(

)C100.1()CmN100.9(

46

2

29229

20

21

r

QQF

5. E is inversely proportional to r 2 and the distance r is doubled, so:

.mVorCN106.3

4/)1044.1(1110

11

E

6. V is inversely proportional to r and the distance r is doubled, so:


.V2.72

4.14V

7.

.V29

m1050.0

)C106.1()CmN100.9(

4 10

19229

0

r

QV

8.

.J106.4

m1050.0

)C106.1()C106.1()CmN100.9(

4

18

10

1919229

0

21

r

QQE p

9.

.CNormV102.3

1015.3

)m100.2(

C)104.1()CmN100.9(

4

1120

20

214

17229

20

r

QE

10.

.N100

101

)CN1015.3()C102.3( 12019

qEF

11.

.sm105.1

)kg106.6/(N101

/

228

27

mFa

12. Measuring the gradient at r = 0.3 m involves errors so answers may differ from the calculatedvalue. The calculation can be done simply by using the relationship E = V / r which only appliesoutside a spherical charge distribution:

.CNormV107.1

m3.0/V00050

/

115

rVE

Charged spheres:Force and potentialQuestion 190S: Short Answer

Solutions


1.

N 103.4

kg N 9.8kg 10444

16

mgF

2.

C 10 6.2

C m N 109.0

m 1012N 103.4

so

9–

2–29

23-4

2

21

221

k

FrQ

QQr

QkQF

You assume that the charge on the spheres acts as if it is concentrated at the centre.

3.

kV 7.4m 105

C 10 6.2 C m N 109.03-

9-2–29

V

r

kQV

4. +++ ++ +++ +

5.


00

1 / r2

6. Since V is inversely proportional to r , V r is constant so V = (300 mm / 500 mm) 450 V = 270 V.

7. Since V is inversely proportional to r , V r is constant so r = (450 V / 1000 V) 300 mm = 135 mm.

8. V = Q/(40r)so

.C105.1

CmN109

m300.0V4504

8

2290

VrQ

9. Less than 900 V. Mutual repulsion between the charges causes them to be displaced towards theouter sides of the spheres. This increases the effective separation of the charges and so reducesthe potential at O due to each of the spheres.

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

1.

.N102.1

N1018.1

kgN8.9)kg1012(

4

4

13

F

2.


C 10 8.2

C m N 109.0

m 1042N 1018.1

so

9

2–29

234

2

21

221

k

FrQ

QQr

QkQF

3.

V10 0.5

V10 99.4

m 105

C 10 8.2 C m N 109.0

3

3

3-

92–29

V

r

kQV

4.

16

2

62–29

2

C N10 2.1

m 0.15

C 10 0.3 C m N 109.0

E

r

kQE

5.

V10 8.1

m 0.15

C 10 0.3 C m N 109.0

5

62–29

V

r

kQV

6. The potential difference between adjacent equipotentials (V) is 5 kV. The distance betweenadjacent equipotentials (r ) increases as the distance from the cable increases. Since E isproportional to V / r , E must decrease with distance form the cable.

7. The magnitude of E is V/r which equals (5 103 V)/(5 10–3 m) = 1 106 V m–1 or N C–1.

8. The electric field, and hence the acceleration of the ions, is largest near the cable so the ionshave more energy for the next collision.

9. The air becomes a conductor so energy will be wasted in warming the air.

10.


J10 7.4

m 10

C 10 6.155C 10 6.137 C m N 109.0

11

14

19192–29

P

21P

E

r

QkQE

11. Electrical potential energy to kinetic energy.

12. 4.7 10–11 J

Controlling charged particlesQuestion 250S: Short Answer

1. Electrical potential energy to kinetic energy.

2. This prevents electrons from being scattered by collisions with gas molecules in the tube so thebeam does not get absorbed.

3. 0 V– 500 V

4.

.s103.1

s1025.1

)C106.1/()A1020(chargeelectronicofmagnitude/currentBeam

117

117

193

5.

.N104.3

)sm100.3()kg101.9()s10(1.25

electronofspeedelectronofmasssecondperelectronsofnumber

momentumofchangeofrateForce

6

1731117

6. There is a constant magnitude of force due to the magnetic field at right angles to the field and tothe motion of the ion.

7. Radius of path depends on m and B. By varying B ions of different mass can be given the sameradius path.


8. The kinetic energy of the ion.

9. The loss in electrical potential energy of the ion.

10. The force on the ion.

11. Using the equations given:qRmvB /

and

.)/2( 2/1mqVv When v is eliminated

./)/2( 2/1 RqmVB

Since V, q and R do not change, B is proportional to m1/2.

12.

B-field strength

0

most abundant isotope

heaviest isotope

13. See solution to question 12.

14. The right-hand peak corresponds to the heaviest ion since B is proportional to m1/2. The highestpeak corresponds to the most abundant ion since more charge arrives at the collector persecond.

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

Solutions 1.

.CN105.2

m400/)V10100(

/

15

6

dVE

The answer is therefore D.

2.


.J100

)V10100()C101( 66

QVW

The answer is therefore C.

3. F = Q E so the force changes when E and Q change. E is constant between the plates so theforce does not depend on position. So the answer is C.

4. a = F / m and F = e V / d , so a is proportional to V since e and d are constant. So the answer isA.

5. The electric field direction is from positive to negative charge, so the answer is B.

6. Since a = F / m and F = q V / d , a depends on V, m and q so all three factors will affect the timefor the ion to go from G to P. The answer is E.

7. The acceleration is proportional to the charge and inversely proportional to the mass so theanswer is D.

8.

dqV

qEF

/

so the answer is C.

9. The charge q transfers energy from the supply at a voltage V so using W = q V, the answer is B.This energy is not the same as the energy stored in a capacitor of course.

10. Having checked the accuracy of statement B using V = W / q the equation

1CN1500

m002.0/V3

/

dVE

gives the value for the electric field strength. So the answer is D.

Charged particles in electric and magnetic fieldsQuestion 110M: Comprehension

1. Reducing the potential difference between the filament and the anode will reduce the speed ofthe electrons so they will spend more time within the vertical electric field and will experiencemore deflection. The answer is B.

2. The electrons only experience a force due to the electric field between the plates so the answer isD. (The linear motion beyond the field region is a nice example of Newton’s first law of motion.)

3. The force acts at right angles to the direction of travel so no work is done on the electron by thisforce. The answer is C.

4. The speed of the electron does not change when moving in a circle so response B is wrong.There will always be a force on the electron in the electric field regardless of its speed soresponse C is also wrong as are D and E. The greatest deflection possible in a uniform electricfield is 90 so response A is the correct answer.

5. Since r = m v / B q the answer must be C.


6. Since r = m v / B q the ion with the largest charge and the smallest mass will have the smallestradius. The answer is B.

7. The formulae give t = m / B q which is independent of velocity. The answer is C. This illustratesthe principle of the cyclotron.

8. Using t = m / B q the time would halve. The answer is B.

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

1. Since the force obeys an inverse square law the answer is A.

2. The uniform electric field acting on the positively charged sphere causes a constant force inaddition to that provided by the charged sphere on the fixed rod. The answer is B.

3. Using the inverse square law the field at X will be 43.2 / 22. Y is the same distance from Q as thepoint where the field is 8.7 so the answer must be B.

4. It is important to remember the vector nature of the electric field and that the charge at Y is twicethe magnitude of the charge at X. The field strength at any point is the sum of the field strengthcontributions from each of the charges and will depend on the distance from each of the charges,the direction of the electric field and the magnitude of the charge producing it. All statements arecorrect so the answer is E.

5. The graphs show a proportional relationship. Field is proportional to 1 / r 2 and potential against 1/ r so the answer is D.

6. All statements are correct so the answer is E. The field at S will be greater than the field at Tbecause S is closer to both P and R so the field contributions from each of the charges aregreater than they would be at T.

7. Electric field is a vector whereas potential is a scalar. The magnitude of the potential at any pointwill depend on the sign of the charges contributing to that potential. The answer is D because thefields cancel along the x- and y-axes and there is as much positive as negative potential.

8. The potential a long way from the three charges will be zero whereas the potential at O will befinite. Statements 2 and 3 are correct so the answer is D.

9. The force will change direction as the charge moves from P to Q but will decrease with distancefrom each charge most rapidly when it is close to each charge. The answer must be C.

10. When the distance doubles the potential energy must halve so the answer is D.

Fields and charged particlesQuestion 240M: Multiple Choice

1. The distance between the electron gun and the deflecting plates has no effect on the force actingon the electron or on the time taken to pass through the field so the answer is C.


2. Since e V = ½ m v2, v is proportional to the square root of V so the answer is C.

3. Equipotentials run perpendicular to the field lines and the particle will move from one potential toanother. The answer is D.

4. Using r = m v / B e it can be seen that 1 is wrong and 2 is correct. An electric field wouldaccelerate the electron along line of the field. The answer is B.

5. B

6. C

7. Within a uniform field the potential increases linearly with the distance from the earthed plate sothe answer is A.

8. Graph E shows the expected 1 / x variation.

9. Gravitational forces are weaker than electrical forces and both decrease with distance accordingto the inverse square law. The answer is D.

Estimating with fieldsQuestion 260E: Estimate

These solutions are for guidance only. Students will tackle the problems in different ways and will usedifferent values for the estimated data.

1. The potential difference between the dome and the earth is given by:

.V105.1

)103()A105.0(5

116

RV I

Since the earth is at zero potential the sphere will be at a potential of + 1.5 105 V. Assume thatthe dome behaves like a sphere with an estimated radius of 0.15 m. For a spherical charge:

204/ rQE

andrQV 04/

so the electric field strength at the surface is given by:

.mV101

m15.0

V105.1

16

5

r

VE

This field strength is below the critical value for sparking to occur.Students might arrive at the surface field strength by calculating the charge from the potential andusing

.4/ 20 rQE

The dome is at a fixed potential so the potential gradient, and hence the electric field strength,increases when the plate is brought close to the dome. Sparking will then occur when the electricfield strength reaches and exceeds 3 106 V m–1.When the plate is very close we may assume that the electric field between the plate and thedome is uniform so E = V / d . Sparking will occur when


.m05.0

mV103

V105.116

5

E

Vd

2. The electrons are travelling from the electron gun at a speed

.sm1093.5

kg101.9

V10)C106.1(22

17

31

419

m

eVv

If the distance from the electron gun to the front of the screen is estimated as 0.30 m theelectrons will travel through the gravitational and magnetic field for a time given by:

.s1006.5)sm1093.5/(m30.0 917 The deflection due to the gravitational field is given by:

.m10

)s1006.5(sm8.9

16

29221

221

ats

Clearly this deflection is negligible.The electrons will experience a force due to the Earth’s magnetic field that has a horizontal andvertical component. Students might estimate the Earth’s magnetic field to be between 10–6 T and10–3 T. Measured values are about 10–5 T for both the horizontal and vertical components. Theelectrons will experience maximum force when directed at right angles to the horizontalcomponent and no force when directed along it. The force will cause a vertical deflection. Theelectrons will always be at right angles to the vertical component and so will experience a forcethat will be horizontal.The maximum force will be given by:

.N10

N1049.9

)sm1093.5()C106.1(T10

16

17

17195

BevF

Whilst the force will be changing direction due to the circular path of the electron, the radius of theorbit will be very large so we can assume that the force will remain in one direction only. This willcause an acceleration

214

3117

sm1004.1

)kg101.9/()N1049.9(

/

mFa

and so a deflection of

.mm1

m1033.1

)s1006.5()sm1004.1(

3

2921421

221

ats

As we can ignore the effect of gravity, and the force due to the Earth’s magnetic field is verysmall, we can assume that the effect will not cause any significant problems.

3. The force on the alpha particle is given by:


N 415

m 10

C 10 6.12C 10 6.190 C m N 109.0214

19192–29

221

F

F

r

QkQF

As the alpha particle moves from the nucleus the force will decrease, as will the acceleration. Theacceleration would be greatest at the moment of release from the nucleus and is given by:

228

27

s m 1028.6

kg 106.6

N 415

m

Fa

Energy from the electrical potential will be carried away as kinetic energy of the alpha particle andnucleus. The mass of the nucleus is much greater than that of alpha particle so the alpha particlewill have the greatest share of the kinetic energy. If the nucleus is assumed to have no kineticenergy, the final kinetic energy of the alpha particle will be equal to the electrical potential energyof the nucleus and the alpha particle at the point of release.

Using

r

QkQmv 212

21

the kinetic energy of the alpha particle is:

17

27

12

1222

1

12

14

1919–229

P

s m 104

kg 106.6

J10 15.42

J10 15.4

so

J10 15.4

m 10

C 10 6.12C 10 6.190 C m N 109.0

v

v

mv

E

This velocity is reached when the distance between the nucleus and the alpha particle is verylarge. It assumes that no other interactions occur after release.

4. If the size of a nucleon is taken as 10–15 m the photon energy will be given by:

eV 109.4

eV J 101.6

J 109.7

J 109.7m 102500

s m 103s J 106.6

λ

5

119

14

1415

1834

hcE


The gamma ray is emitted when a nucleus drops from a more excited state to a less excitedstate. The nucleus remains intact so the potential energy of the nucleus must be greater than thisvalue.

5.

If the size of the atom is taken as 10–10 m the electrical potential energy will be given by:

eV 14eV J 101.6

J10 3.2

J10 3.2

m 10

C 10 6.1C 10 6.1 C m N 109.0

–1–19

18

18

10

19192–29

P

21P

E

r

QkQE

The most energetic photon that can be emitted will have energy of 14 eV. The wavelength of theelectromagnetic radiation emitted is given by:

m 106.8

J 103.2

s m 103s J 106.6

Eλ

8

–18

–1834

hc

This is in the ultraviolet region.

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

1. LHC diameter = 8.5 km. Length of semicircle = (d / 2). Thus the beam has to travel

dddd

57.0122

further than the control signal = 0.57 8.5 103 m

At speed c this takes

s 16s m 103.0

m 105.857.018

3

2. Speed ~ c . Time for one circuit = distance / speed = 26.7 103 m / 3.0 108 m s-1 = 8.9 10-5 s,so number of circuits s-1 = 1 / 8.9 10-5 s = 11 240 s–1.

3. Average bunch spacing is equal to

m 5.92808

m 107.26 3


9.5 m at speed c takes 9.5 m / 3.0 108 m s-1 = 31.7 ns ~ 30 ns.

4. Bunches pass a collision point 2808 11 240 s-1 = 31.6 MHz.

5. Current I = 2808 bunches 1.1 1011 protons per bunch 11 240 rev s-1 1.6 10-19 C perproton = 0.55 A ~ 0.5 A.

6. 7 1012 eV 1.6 10-19 J eV-1 1.1 1011 protons per bunch 2808 bunches = 346 MJ ~ 350MJ.

Accelerator Maximum kinetic energy Relativistic factor

Linac 50 MeV 1.05

Proton synchrotron booster 1.4 GeV 2.4

Proton synchrotron 25 GeV 26

Super proton synchrotron 450 GeV 451

LHC 7 TeV 7001

7. See table above.

rest

kinetic

rest

restkinetic

rest

total 1E

E

E

EE

E

E

At the end of the second stage

88.014.2/11/11/ 22 cv

8.

11518

11912total smkg107.3

s m 103

eV J 101.6 eV 100.7

c

Ep

9.

2750GeV 207

GeV 207 GeV 1000570

rest

total

E

E

10.

T5.5s m 103.0 x m 104.25 x C 101.6

eV J 101.6 eV 100.718319

11912total

qrc

E

qr

pB

The field required is larger because of the straight sections of the path, so that the curvedsections have radius less than the average radius of the accelerator.

Thunderclouds and lightning conductors


Question 40C: Comprehension

Solutions 1. The cloud is neutral initially because atoms are neutral. If charge separates, there must be equal

amounts of positive and negative charge.

2. The Sun warms the Earth. Heated air is less dense so it rises (convection).

3. There will be 90 discharges from a thundercloud if the cloud exists for 30 minutes and producesone flash every 20 seconds:

.J109

90J10

dischargesofnumberdischargeperenergyenergyTotal

11

10

4. The discharge heats the air directly. The sound increases the random motion of molecules andlight is absorbed by molecules. All the energy eventually becomes thermal energy.

5.

–––– –– – – –

cloud

Earth+ + + ++ + + + + + +

6. The diameter of the base of the cloud is about 4.5 km so the area is ((4.5 103)/2)2 1.6 107

m2. The distance from the ground about 1200 m.

7.

115

11227

CNormV104.1

)]mF109.8()m106.1/[(C20

E

.V107.1

m1200)mV104.1(8

16

EdV

The answers will depend on the estimates made in question 6.

8. Any sensible answer such as the field is not uniform as assumed by the equation. The upperpositive charge in the cloud has been ignored; this would tend to reduce E. The equationassumes a vacuum as the medium that would give an answer similar to that for dry air, but themedium is actually wet air.


9.

N108.4

)CN103()C106.1(15

1419

qEF

.sm100.1

)kg107.4/()N108.4(

/

211

2615

mFa

10. The ions do not travel far before they collide with other molecules and lose their energy. So,whilst the acceleration may be very large, the ions have insufficient time to reach a high speed.

11. A stream of positive and negative ions travelling in opposite directions.

12. The electric field is greater near the points.

13. The electric field is greater near the top and bottom of the drop so ionisation is more likely whenwater drops are present.

+

–

++

––

deformed rain drop

14. Because the greatest concentration of deformed raindrops is just below the base of the cloud.

15.

A14

)29500030)(4(1016.1 4

I

16. With W = 8 m s–1, I becomes (8 + 4) / 4 times greater which is an increase by a factor of 3.

17. The wind blows away ions that would otherwise reduce the effective field at the point. With someof the ions removed the current would increase as the rate of ionisation is increased. The fasterthe wind blows the more quickly the ions are removed so the greater the current.

18. When the ions are being removed as fast as they are formed, a further increase in wind speedwould make no difference to the current.

19. With a wind speed of 8 m s–1 the discharge current is 42 A.


.days5.5

s108.4

)A1042/(C20

/

5

6

IQTime

This is rather a long time and establishes that lightning conductors do not work by discharging thecloud.

20. They provide a low-resistance path to earth and act as a source for the positive streamer. Thelighting conductor actually triggers the lightning discharge.

Electrical breakdown in a vacuumQuestion 50C: Comprehension

Solutions 1. There are no particles to carry charge in a vacuum.

2. Breakdown now occurs at lower voltages so the working voltages inside the apparatus cannot besustained.

3. The plasma provides charge carriers.

4. The second figure shows that the breakdown voltage is 37 kV. From the first figure the resistanceof the circuit is 5 108 :

.A75

A104.7

)105/()V1037(

/

5

83

RVI

5.

cathode

6.

emission here


7. A point which emits electrons at a constant rate.

8. Energy transfer to low grade thermal energy

9.

.N106.1

)CN10()C106.1(

)mV10()C106.1(

10

1919

1919

qEF

10. Noise is fluctuations in the current (the spikes shown in the second figure). These can be causedby a protrusion vaporising and creating a temporary increase in the number of ions.

11. The microparticles are electrically charged and in an electric field.

12. Consider each microparticle as a sphere of radius 1 m:

.kg108.3

)m10()mkg9000(

volumedensitymass

14

36343

13. q V = kinetic energy gained by microparticle where V = 50 kV so:

.C10

C102

)1050/()sm1500()kg109(

2/

13

13

3211521

2

Vmvq

14. A surge in current increases the potential difference across the resistor at the expense of thepotential difference across the gap. The reduction in the potential difference across the gap actsto reduce the current surge.

15. Conditioning removes or blunts microprotrusions by evaporation and removes loosemicroparticles.

Millikan's oil drop experimentQuestion 70D: Data Handling

1.

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–weight

electrical force

2. See solution for question 1


3. When balanced, Q E = W where E = V / d . Hence, Q V / d = W and so Q = W d / V.

4. The weight was balanced by upward force due to air resistance.

5. The weights of drops 1–6 are as follows: 5.1 10–14 N, 5.4 10–14 N, 3.4 10–14 N, 2.9 10–14 N,3.7 10–14 N, 6.4 10–14 N.

6. Using Q = W d / V, the charges of drops 1–6 are as follows: 4.8 10–19 C, 2.9 10–19 C, 6.5 10–19 C, 1.7 10–19 C, 1.6 10–19 C, 7.7 10–19 C.

7. Using n = Q / e, the values of n for drops 1–6 are as follows: 3, 2, 4, 1, 1, 5.

8. For each drop the percentage uncertainty can be calculated using the formula100]/)charge[(measured nene

which gives values of 0, 0, 2%, 6%, 0, 3%. The greatest uncertainty is likely to be as a result ofdeducing the weight from the graph since V and d can be measured accurately.

The proton synchrotronQuestion 130D: Data Handling

Solutions 1.

221 mvEk

so

.18

17

2/12712

sm10

sm1082.9

)]kg1066.1/()J1082[(

v

2.

.s103.6

s1028.6

sm10/m628

peeddistance/sTime

6

6

18

3.

.smkg106.1

smkg1063.1

)sm108.9()kg1066.1(

119

119

1727

mv

4. The total charge is I t = N e where N is the number of protons injected in a time t, so:

.109.3

)C1060.1/()]s1028.6()A10100[(

/

12

1963

etN I

5. At very low pressures the protons are less likely to collide with air molecules. Collisions would


reduce the energy of the protons and would reduce the number reaching the target.

6. Protons have a positive charge so they must approach a negative charge on the electrode on theopposite side of the gap if they are to be accelerated.

7. The proton gains 4 keV at each accelerating point and it passes 14 such points in one revolution.The energy gained in one revolution is 4 keV 14 = 56 keV.

8. The number of revolutions is equal to the total energy gained divided by the energy gained perrevolution, i.e.

.105)J1056/()J1028( 539

9. To give a proton the same energy the length of the linear accelerator would have to be equal tothe number of revolutions of the synchrotron multiplied by the circumference of synchrotron. So:

.m101.3

m628)105(length8

5

This is excessively long given that the radius of the Earth is about 6 106 m.

10.

rmv

BevF

/2

so./ ermvB

Since e and r are constant, the B field is directly proportional to the momentum, m v.

11. On injection the energy of the protons is 50 MeV and the speed (see question 1) is 9.82 107 ms–1. So:

.T010.0

]m100)C1060.1/[()]sm1082.9()kg1066.1[( 191727

B

12. As the speed of the protons increases each bunch of protons takes less time to travel betweenthe accelerating points. To maintain the acceleration the potential difference must change in lesstime.

13. We have already seen that B is directly proportional to the momentum, p, if e and r remainconstant. So

iiff BppB )/(

where the subscripts i and f refer to the initial and final states. Hence:

.98

2.98

)smkg1063.1/()smkg106.1(/ 119117

if BB

14. = Etotal / Erest = (28 GeV + 1 GeV)/1 GeV = 29.

The electric dipoleQuestion 170D: Data Handling

The completed spreadsheet table is shown below:


Open the Excel Worksheet

1. See spreadsheet.

2. See spreadsheet.

3. See spreadsheet.

4. See spreadsheet.

5. The graph shows a decrease of E with d , but no relationship can be established without furtheranalysis:

6. The graph looks non-linear. It is difficult to tell if the graph has a definite linear region so furtheranalysis is recommended:

7. There is a clearly defined range over which E d 3 has an approximately constant value so E mustbe proportional to 1 / d 3 over this range (i.e. for large values of d ):


8. When E d 3 is 1% larger it has a value of 5.81 10–19 N C–1 m3 which corresponds to d = 1.5 10–9 m or 1.6 10–9 m, so the range is 1.5 10–9 m to 1 m or 1.6 10–9 m to 1 m.

9. The graph shows that E is inversely proportional to d 3 over the range of values considered.

10. The electric field strength falls off more rapidly with distance for a dipole. The field of a singlecharge falls off as 1 / d 2. At very large distances from a dipole the field will be approximately thatof two superimposed equal and opposite charges. This means that the field will be very smallindeed and certainly much smaller than the field of a single charge.

11. We have seen from question 10 that the field of a dipole falls off very rapidly with distance. Thefield outside the ionic crystal is the field due to a large number of dipoles at a relatively largedistance. This field is very weak.

Testing Coulomb’s law


Question 210D: Data Handling

1. A typical set of values is as follows:

d / m r / m 1 / r 2

/ m–2

0.004 0.104 93

0.030 0.064 244

0.052 0.054 343

0.071 0.046 473

0.088 0.039 658

0.111 0.035 816

The graph shows that F is approximately proportional to 1 / r 2. The graph appears to becomenon-linear when r is very small, as might be expected if the charges on the spheres becomeredistributed when the balls are too close.


0.14

0.12

0.10

0.08

0.06

0.04

0.02

0

0 200 400 600 800 1000

( 1/r2 ) / m–2

2.

C105

V5.0)F1001.0(9

6

CVQ

as expected.

3. The points are scattered around a straight line. Changes in the charge could explain this.

4. The gradient of the d against 1 / r 2 graph can be used. The force F is W tan which becomes Wd since tan = d / 1. The gradient can be measured using the point (780 m–2, 0.14 m) whichbecomes (780 m–2, 1.4 10–4 N) when d is converted into newtons. The gradient of the graph is1.8 10–7 N m2 which equals k Q1 Q2. Since Q1 = Q2 = 5 10–9 C this yields a value for the

constant of 7.2 109 N m2 C–2.

5. Using the measured value for the gradient but with the new values of charge the constant willrange from 5 109 to 11 109 N m2 C–2.


Using uniform electric fieldsQuestion 30X: Explanation–Exposition

1. Assuming a uniform field, V = E d = (3 106 V m–1) (0.67 10–3 m) = 2 103 V in air. Thevolume decreases by a factor of 14 so the pressure increases by similar factor. The temperatureis not constant so an accurate calculation cannot be made. Air molecules are closer together at ahigher pressure so the molecules have a shorter distance to move before colliding with anothermolecule. Before each collision an ion must gain sufficient energy (about 30 eV) to ionise themolecule it hits. If the distance between the molecules is decreased the voltage must beincreased to ensure that the molecules reach the required energy before a collision.

2. Ions are accelerated by the electric field provided by the accelerating voltage. Electrical potentialenergy is transferred to kinetic energy of the ions. Assuming the ions are accelerated from rest,eV = ½ m v

2 where v is the velocity of the ions. Because of random motion before accelerationthe ions will have a range of velocities as seen on the graph. For a given ion, v is proportional toV

1/2. For a given accelerating voltage, v is proportional to 1 / m1/2 if the ions have the same

charge. Ions of mass ratio 2:1 will have a velocity ratio of 1:21/2. The times to reach the electrodewill be in the ratio 21/2:1. The pulses reaching the collecting electrode will have a lower amplitudethan the pulse leaving the accelerating electrode and both will show more broadening than thispulse as the ions are travelling over a finite distance with a range of velocities. The pulse for ionsof mass 2 m should have a larger amplitude than the pulse for ions of mass m.

pulse leaving acceleration electrode

mass m

mass 2m

time

time t

time 2 t

0

0

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

Solutions 1. The accelerating voltage is given by

.V500

m1.0)CN100.5( 13

EdV

Using


221 mveV

the velocity v of electrons leaving the electron gun is

.sm103.1)}kg101.9/(]V500)C106.1(2{[ 172/13119 The force on the electrons during acceleration due to the electric field is given by

.N100.8

CN5000)C106.1(16

119

eEF

The force on the electrons due to the magnetic field is given by

N 10 0.1

T 100.5s m 101.3C 10 6.1–15

4–17–19

evBF

This force causes electrons to move in circular path of radius

.m15.0

)]C106.1()T100.5/[()]sm103.1()kg101.9[(

/1941731

Bemvr

The force on the electrons due to the gravitational field is given by

.N109.8

kgN8.9)kg101.9(30

131

mgF

The electrons will take a time

s109.1

)sm103.1/(m25.0

/

8

17

vst

to reach the end of the tube from the electron gun. The distance fallen under gravity during thistime is

m 108.1

s109.1s m 8.9

15

282–21

221

ats

The effect of gravity can be ignored.

2. When a particle moves at right angles to the field there is a force on the particle at right angles tothe field and the direction of motion. This force remains at right angles to the path at any point inits motion regardless of the direction of the particle. The force is q v B which causes a centripetalforce m v 2 / r leading to B q r = m v. The radius is directly proportional to m and v and inverselyproportional to B and q . When at an angle to a uniform field the component vx leads to circularmotion but the component vy results in movement along the field at a constant speed. This leadsto a helical path. In a non-uniform field, r is inversely proportional to B for a given v so rdecreases as the particle approaches the poles where B is greater. This leads to a spiral path.

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