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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.
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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.
11 Advancing Physics
.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.
12 Advancing Physics
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:
13 Advancing Physics
.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
14 Advancing Physics
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.
15 Advancing Physics
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.
16 Advancing Physics
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.
17 Advancing Physics
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.
18 Advancing Physics
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.
19 Advancing Physics
.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.
20 Advancing Physics
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.
21 Advancing Physics
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
22 Advancing Physics
.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:
23 Advancing Physics
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
24 Advancing Physics
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
25 Advancing Physics
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
26 Advancing Physics
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.
27 Advancing Physics
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.
28 Advancing Physics
.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
29 Advancing Physics
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
30 Advancing Physics
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
31 Advancing Physics
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:
32 Advancing Physics
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 ):
33 Advancing Physics
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
34 Advancing Physics
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.
35 Advancing Physics
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.
36 Advancing Physics
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
37 Advancing Physics
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.