The uniform electric field
Question 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 increase
in the electric field strength between the plates, and hence the force on the ball. The frequency at
which the ball moves between the plates increases so the current increases also. If the voltage is
increased the electric field strength increases and the ball picks up more charge on contact with
the plates. Both factors contribute to an increased current.
3.
+ 500 V
+ 400 V
+ 300 V
+ 200 V
0V
+ 100 V
4. See the solution to question 3.
5.
E
300 V
V

d 5  10 3 m
 6.0  10 4 V m 1 or N C 1.
6.
F  qE
 (32  10 19 C)  (6.0  10 4 N C 1 )
 1.9  10 13 N.
7.
W  qV
 (1.6  10 19 C)  (50  10 3 V )
 8.0  10 15 J.
8. The energy gained will have a positive value. Both the charge and the potential difference are
negative:
100 eV  100  (1.6  10 19 J)
 1.6  10 17 J
1
2
mv 2  1.6  10 17 J
so
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v  {[2  (1.6  10 17 J)] /(9.1 10 31 kg)}1/ 2
 5.9  10 6 m s 1.
Getting F = q v B
Question 80W: Short Answer
1. N e
2. N e / t
3. v t
4.
F  ILB 
Ne
 vt  B
t
F  NevB
5. e v B
6. Q v B
Speed and energy of particles – Newtonian calculation
Question 10S: Short Answer
1. Starting from Ekinetic = ½ mv 2, multiply both sides by 2, getting 2 Ekinetic = mv 2. Then divide both
sides by m to obtain
2Ekinetic
 v2
m
Finally, take the square root of both sides:
2Ekinetic
v
m
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
v
2
2  1.6  10 16 J
9.1 10 31 kg
 1.9  10 7 m s 1
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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
2  102  10 16 J
v
9.1 10 31 kg
 1.5  10 8 m s 1
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 difference
needed = 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 calculation
Question 20S: Short Answer
Accelerator
Date
Particle
accelerated
Kinetic
energy
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

E total 30 GeV + 1 GeV

 30
Erest
1 GeV
4. At 3 MeV the ratio v / c is less than 10%. But at 500 MeV it is 75%. Relativistic effects are
certainly important for the synchrocyclotron and synchrotron.
Comparing relativistic and Newtonian kinetic energy
Question 30S: Short Answer
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1. The relativistic kinetic energy increases from a little more than to several times the Newtonian
kinetic 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 energy
Question 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
v / c  1  1/  2
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 fields
Question 60S: Short Answer
1.
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400 V
3.2 × 10–2 m
200 V
+
600 V
–
2. See solution for question 1
3.
E V /d
 600 V /(3.2  10 2 m)
 1.9  10 4 V m 1 or N C 1.
4.
E
F mg

Q
Q
Q
mg 1.8  10 15 kg  9.8 N kg –1

E
1.9  10 4 N C –1
Q  9.3  10 19 C
5.
(9.3  10 19 C) /(1.6  10 19 C)  6.
6. The time is increased due to air resistance which leads to a lower acceleration.
7.
crushed minerals
0.3 m
conveyor belt
0V
+ 54 kV
collectors
8.
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E V /d
 54 000 V / 0.30 m
 1.8  10 5 V m 1 or N C 1.
9.
F  QE
 (10 6  1.6  10 19 C)  1.8  10 5 N C 1
 2.9  10 8 N
10.
s
1
F 2
t
m
2.9  10 8 N
2 at
2

1
2
1
2

 1.1 s
2 1.5  10 -6 kg
s  0.012 m
s
(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 fields
Question 90S: Short Answer
1. C
2. E
3. They hit A.
4.
B field
source
no field
detector
5. See the solution for question 4.
6. Circular since the proton experiences a force of constant magnitude at right angles to its path
regardless of its direction.
7.
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force
8. See the solution for question 7.
9.
0V
+ 1000 V
10 cm
10. Parabola since the force is constant and remains in the same direction.
The cyclotron
Question 100S: Short Answer
Solutions
1.
v

2E
m
2  80  10 3  (1.6  10 19 J)
1.7  10 27 kg
2.
F  BQv 
mv 2
r
so
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 3.9  10 6 m s 1.
B

mv
Qr
(1.7  10 27 kg)  (3.88  10 6 m s 1 )
(1.6  10 19 C)  (50  10 3 m)
 0.82 T.
3. Since
v  E, v 
1
2
 (3.9  10 6 m s 1 )
and
r 

mv
QB
(1.7  10 27 kg)  (3.9  10 6 m s 1 )
(1.6  10 19 C)  0.82 T  2
 0.036 m
 36 mm.
4.
T 

2r
v
2  50  10 3 m
3.88  10 6 m s 1
 8.1 10 8 s
since
mv
r 
, r v
QB
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 gets
faster, the particle spirals out into orbits with larger and larger radii. So all particles can be
accelerated across a ‘dee’ gap at the same time.
The Hall effect
Question 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 direction
to 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 electric
field E = B v.
5. Front edge negative, rear edge positive.
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6. V = E d = B v d
7. bd
8. n = B I / V b Q
Charged particles moving in a magnetic field
Question 150S: Short Answer
1.
Ek 
1
2
mv 2
so
v 
2  (2.9  10 16 J)
(9.1  10
 31
kg)
 2.5  10 7 m s 1.
The kinetic energy is considerably less than the electron's rest energy, so the Newtonian
approximation can be used.
2.
F  Bqv
 (2.2  10 3 T)  (1.6  10 19 C)  (2.5  10 7 m s 1 )
 8.8  10 15 N.
3.
F  mv 2 / r
so
r
(9.1 1031 kg)  (2.5  107 m s 1 )2
8.8  1015 N
 6.5  102 m.
4. Electrons lose speed through collisions with other particles. Since r = mv/Be, r decreases as v
decreases.
5.
proton
motion
uniform B field
acts into plane of
screen / paper
over shaded area
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.
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9.
B
2fm 2  10 9 Hz  (9.1  10 31 kg)

q
1.6  10 19 C
 3.6  10  2 T.
10. There will be no effect since the cyclotron frequency depends only upon B, q and m which do not
change under normal conditions.
Fields in nature and in particle accelerators
Question 160S: Short Answer
1.
V  Ed
 10 6 V m 1  200 m
 2  10 8 V.
2.
Q   0 AE
 (8.9  10 12 F m 1 )  (1000 m) 2  10 6 V m 1
 8.9 C.
3.
1
QV
2

1
2
 8.9 C  (2  10 8 V )
 8.9  10 8 J.
4.
eV 
1
2
mv 2
so
v 
2 eV
m
 8 . 39  10
 8 . 4  10
5.
2  (1 . 6  10

6
9 . 1  10
6
m s
kg
1
m s 1 .
(9.1  10 31 kg )  (8.39  10 6 m s 1 )
mv

Be
0.001 T  (1.6  10 19 C )
 0.048 m.
eE  Bev
so
v  E / B.
7.
10
C )  200 V
 31
Bev  mv 2 / r
so
r 
6.
 19
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E 
3200 V
V

d
0.024 m
 1 .33  10 5 V m 1.
v 
E 1.33  10 5 V m 1

B
8 .2  10  3 T
 1 .63  10 7 m s 1
 1 .6  10 7 m s 1.
8.
eV 
1
2
mv 2
so
(1 . 63  10 7 m s  1 ) 2
e
v2


m
2V
2  750 V
 1 . 8  10 11 C kg 1 .
9.
qV 
1
2
mv 2
so
V 
1
2
mv 2
q

(3.3  10 27 kg )  (9  10 6 m s 1 ) 2
2  (1.6  10 19 C)
 8.4  10 5 V.
10.
Bqv  mv 2 / r
so
B 
( 3 . 3  10  27 kg )  ( 9  10 6 m s  1 )
mv

qr
(1 . 6  10 19 C )  0 . 5 m
 0 . 37 T .
Non-uniform electric fields
Question 180S: Short Answer
Solutions
1.
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2.
+
–
3. See solution to question 2.
4.
F
Q1Q2
4 0 r 2

(9.0  10 9 N m 2 C 2 )  (1.0  10 9 C) 2
(0.060 m) 2
 2.5  10 6 N.
5. E is inversely proportional to r 2 and the distance r is doubled, so:
E  (1.44  1011 ) / 4
 3.6  1010 N C 1 or V m 1.
6. V is inversely proportional to r and the distance r is doubled, so:
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V
14.4
 7.2 V.
2
7.
( 9 . 0  10 9 N m 2 C  2 )  (1 . 6  10  19 C )
Q

4  0 r
0 . 50  10 10 m
 29 V .
V 
8.
Ep 
Q 1Q 2
( 9 .0  10 9 N m 2 C 2 )  (  1 .6  10 19 C )  (  1 .6  10 19 C )

4  0 r
0 .50  10 10 m
  4 .6  10 18 J.
9.
E 
Q
4  0 r 2

(9 .0  10 9 N m 2 C 2 )  (1 .4  10 17 C)
( 2 .0  10 14 m ) 2
 3 .15  10 20
 3 .2  10 20 V m 1 or N C 1 .
10.
F  qE
 (3.2  10 19 C)  (3.15  10 20 N C 1 )
 101
 100 N.
11.
a  F /m
 101 N /(6.6  10 27 kg)
 1.5  10 28 m s 2 .
12. Measuring the gradient at r = 0.3 m involves errors so answers may differ from the calculated
value. The calculation can be done simply by using the relationship E = V / r which only applies
outside a spherical charge distribution:
E V /r
 50 000 V / 0.3 m
 1.7  10 5 V m 1 or N C 1.
Charged spheres:
Force and potential
Question 190S: Short Answer
Solutions
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1.
F  mg
 44  10 6 kg  9.8 N kg 1
 4.3  10  4 N
2.
F 
kQ1Q 2
r2
Q1  Q 2
so
Q

Fr 2
k

4.3  10  4 N  12  10 -3 m

2
9.0  10 9 N m 2 C – 2
 2.6  10 – 9 C
You assume that the charge on the spheres acts as if it is concentrated at the centre.
3.
V 
V 
kQ
r
9.0  10 9 N m 2 C – 2  2.6  10 -9 C
5  10 -3 m
 4.7 kV
4.
++
+ ++ +
+ + +
5.
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0
0
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/(40r)
so
Q  Vr 4  0 
 1 . 5  10
8
450 V  0 . 300 m
9  10
9
N m 2 C 2
C.
9. Less than 900 V. Mutual repulsion between the charges causes them to be displaced towards the
outer sides of the spheres. This increases the effective separation of the charges and so reduces
the potential at O due to each of the spheres.
Using the 1 / r 2 and 1 / r laws for point charges
Question 200S: Short Answer
1.
F  (12  10 3 kg)  9.8 N kg 1
 1.18  10 4 N
 1.2  10 4 N.
2.
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F 
kQ1Q 2
r2
Q1  Q 2
so
Q

Fr 2
k

1.18  10  4 N  24  10 3 m

2
9.0  10 9 N m 2 C – 2
 2.8  10 9 C
3.
V 
V 
kQ
r
9.0  10 9 N m 2 C – 2  2.8  10 9 C
5  10 -3 m
3
 4.99  10 V
 5.0  10 3 V
4.
E
E
kQ
r2
9.0  10 9 N m 2 C – 2  3.0  10 6 C
0.15 m2
 1.2  10 6 N C 1
5.
kQ
r
9.0  10 9 N m 2 C – 2  3.0  10 6 C
V 
0.15 m
V 
 1.8  10 5 V
6. The potential difference between adjacent equipotentials (V) is 5 kV. The distance between
adjacent equipotentials (r ) increases as the distance from the cable increases. Since E is
proportional 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 ions
have more energy for the next collision.
9. The air becomes a conductor so energy will be wasted in warming the air.
10.
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EP 
EP 
kQ1Q 2
r
9.0  10 9 N m 2 C – 2  37  1.6  10 19 C  55  1.6  10 19 C

 
10
14

m
 4.7  10 11 J
11. Electrical potential energy to kinetic energy.
12. 4.7  10–11 J
Controlling charged particles
Question 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 the
beam does not get absorbed.
3.
– 500 V
0V
4.
Beam current / magnitude of electronic charge  (20  10 3 A ) /(1.6  10 19 C)
 1.25  1017 s 1
 1.3  1017 s 1.
5.
Force  rate of change of momentum
 number of electrons per second  mass of electron  speed of electron
 (1.25  1017 s 1)  (9.1  10 31 kg)  (3.0  107 m s 1)
 3.4  10 6 N.
6. There is a constant magnitude of force due to the magnetic field at right angles to the field and to
the motion of the ion.
7. Radius of path depends on m and B. By varying B ions of different mass can be given the same
radius path.
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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:
B  mv / qR
and
v  (2qV / m)1/ 2 .
When v is eliminated
B  (2mV / q )1/ 2 / R.
Since V, q and R do not change, B is proportional to m1/2.
12.
most abundant isotope
heaviest isotope
0
B-field strength
13. See solution to question 12.
14. The right-hand peak corresponds to the heaviest ion since B is proportional to m1/2. The highest
peak corresponds to the most abundant ion since more charge arrives at the collector per
second.
The uniform electric field and its effect on charges
Question 20M: Multiple Choice
Solutions
1.
E V /d
 (100  10 6 V ) / 400 m
 2.5  10 5 N C 1.
The answer is therefore D.
2.
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W  QV
 (1 10 6 C)  (100  10 6 V )
 100 J.
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 the
force 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 is
A.
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 time
for 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 the
answer is D.
8.
F  qE
 qV / d
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
E V /d
 3 V / 0.002 m
 1500 N C 1
gives the value for the electric field strength. So the answer is D.
Charged particles in electric and magnetic fields
Question 110M: Comprehension
1. Reducing the potential difference between the filament and the anode will reduce the speed of
the electrons so they will spend more time within the vertical electric field and will experience
more deflection. The answer is B.
2. The electrons only experience a force due to the electric field between the plates so the answer is
D. (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 this
force. 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 so
response C is also wrong as are D and E. The greatest deflection possible in a uniform electric
field is 90 so response A is the correct answer.
5. Since r = m v / B q the answer must be C.
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6. Since r = m v / B q the ion with the largest charge and the smallest mass will have the smallest
radius. The answer is B.
7. The formulae give t =  m / B q which is independent of velocity. The answer is C. This illustrates
the 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 energy
Question 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 in
addition 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 the
point 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 twice
the magnitude of the charge at X. The field strength at any point is the sum of the field strength
contributions 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 are
correct 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 T
because S is closer to both P and R so the field contributions from each of the charges are
greater than they would be at T.
7. Electric field is a vector whereas potential is a scalar. The magnitude of the potential at any point
will depend on the sign of the charges contributing to that potential. The answer is D because the
fields 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 be
finite. 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 distance
from 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 particles
Question 240M: Multiple Choice
1. The distance between the electron gun and the deflecting plates has no effect on the force acting
on the electron or on the time taken to pass through the field so the answer is C.
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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 to
another. 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 would
accelerate 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 so
the 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 according
to the inverse square law. The answer is D.
Estimating with fields
Question 260E: Estimate
These solutions are for guidance only. Students will tackle the problems in different ways and will use
different values for the estimated data.
1. The potential difference between the dome and the earth is given by:
V  IR
 (0.5  10 6 A )  (3  1011 )
 1.5  10 5 V.
Since the earth is at zero potential the sphere will be at a potential of + 1.5  105 V. Assume that
the dome behaves like a sphere with an estimated radius of 0.15 m. For a spherical charge:
E  Q / 4  0 r 2
and
V  Q / 4 0 r
so the electric field strength at the surface is given by:
E 
V
1 . 5  10 5 V

r
0 . 15 m
 1  10 6 V m  1.
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 and
using
E  Q / 4  0 r 2 .
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 electric
field 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 the
dome is uniform so E = V / d . Sparking will occur when
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1 . 5  10 5 V
V

E
3  10 6 V m  1
d 
 0 . 05 m .
2. The electrons are travelling from the electron gun at a speed
2eV

m
v 
2  (1 .6  10 19 C )  10 4 V
9 . 1  10  31 kg
 5 .93  10 7 m s 1 .
If the distance from the electron gun to the front of the screen is estimated as 0.30 m the
electrons will travel through the gravitational and magnetic field for a time given by:
0.30 m /(5.93  10 7 m s 1 )  5.06  10 9 s.
The deflection due to the gravitational field is given by:
s
1
2
at 2

1
2
 9.8 m s 2  (5.06  10 9 s) 2
 10 16 m.
Clearly this deflection is negligible.
The electrons will experience a force due to the Earth’s magnetic field that has a horizontal and
vertical component. Students might estimate the Earth’s magnetic field to be between 10–6 T and
10–3 T. Measured values are about 10–5 T for both the horizontal and vertical components. The
electrons will experience maximum force when directed at right angles to the horizontal
component and no force when directed along it. The force will cause a vertical deflection. The
electrons will always be at right angles to the vertical component and so will experience a force
that will be horizontal.
The maximum force will be given by:
F  Bev
 10 5 T  (1.6  10 19 C)  (5.93  10 7 m s 1 )
 9.49  10 17 N
 10 16 N.
Whilst the force will be changing direction due to the circular path of the electron, the radius of the
orbit will be very large so we can assume that the force will remain in one direction only. This will
cause an acceleration
a  F /m
 (9.49  10 17 N) /(9.1 10 31 kg)
 1.04  1014 m s 2
and so a deflection of
s
1
2
at 2

1
2
 (1.04  1014 m s 2 )  (5.06  10 9 s) 2
 1.33  10 3 m
 1 mm.
As we can ignore the effect of gravity, and the force due to the Earth’s magnetic field is very
small, we can assume that the effect will not cause any significant problems.
3. The force on the alpha particle is given by:
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F 
F 
kQ1Q 2
r2
9.0  10 9 N m 2 C – 2  90  1.6  10 19 C  2  1.6  10 19 C

10
14
m
 


2
F  415 N
As the alpha particle moves from the nucleus the force will decrease, as will the acceleration. The
acceleration would be greatest at the moment of release from the nucleus and is given by:
a

F
m
415 N
6.6  10 27 kg
 6.28  10 28 m s 2
Energy from the electrical potential will be carried away as kinetic energy of the alpha particle and
nucleus. The mass of the nucleus is much greater than that of alpha particle so the alpha particle
will have the greatest share of the kinetic energy. If the nucleus is assumed to have no kinetic
energy, the final kinetic energy of the alpha particle will be equal to the electrical potential energy
of the nucleus and the alpha particle at the point of release.
Using
kQ1Q2
r
the kinetic energy of the alpha particle is:
1 mv 2
2

EP 

 
9.0  10 9 N m 2 C –2  90  1.6  10 19 C  2  1.6  10 19 C
10
14

m
 4.15  10 12 J
so
1
2 mv
2
 4.15  10 12 J
v 
2  4.15  10 12 J
6.6  10 27 kg
v  4  10 7 m s 1
This velocity is reached when the distance between the nucleus and the alpha particle is very
large. 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:
E


hc
λ
6.6  10 34 J s  3  10 8 m s 1
2500  10
15
7.9  10 14 J
1.6  10 19 J eV 1
 4.9  10 5 eV
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m
 7.9  10 14 J
The gamma ray is emitted when a nucleus drops from a more excited state to a less excited
state. The nucleus remains intact so the potential energy of the nucleus must be greater than this
value.
5.
If the size of the atom is taken as 10–10 m the electrical potential energy will be given by:
kQ1Q 2
EP 
r
9.0  10 9 N m 2 C – 2   1.6  10 19 C   1.6  10 19 C
EP 
10 10 m

 

 2.3  10 18 J

 2.3  10 18 J
1.6  10 –19 J eV –1
 14 eV
The most energetic photon that can be emitted will have energy of 14 eV. The wavelength of the
electromagnetic radiation emitted is given by:
hc
λ
E
6.6  10 34 J s  3  10 8 m s –1

2.3  10 –18 J
 8.6  10 8 m
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
d
 
 d  d   1  0.57d
2
2 
further than the control signal = 0.57  8.5 103 m
At speed c this takes
0.57  8.5  10 3 m
3.0  10 8 m s 1
 16 s
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
26.7  10 3 m
 9 .5 m
2808
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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 per
proton = 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 ~ 350
MJ.
Accelerator
Maximum kinetic energy
Relativistic factor 
Linac
50 MeV
1.05
Proton synchrotron booster
1.4 GeV
2.4
Proton synchrotron
25 GeV
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Super proton synchrotron
450 GeV
451
LHC
7 TeV
7001
7. See table above.

E total Ekinetic  Erest
E

 1  kinetic
Erest
Erest
Erest
At the end of the second stage
v / c  1  1/  2  1  1/ 2.14 2  0.88
8.
p
E total 7.0  1012 eV  1.6  10 19 J eV 1

 3.7  10 15 kg m s 1
c
3  10 8 m s 1

E total 570  1000 GeV  207 GeV

 2750
Erest
207 GeV
B
E
p
7.0  1012 eV  1.6  10 19 J eV 1
 total 
 5 .5 T
qr
qrc
1.6  10 19 C x 4.25  10 3 m x 3.0  10 8 m s 1
9.
10.
The field required is larger because of the straight sections of the path, so that the curved
sections have radius less than the average radius of the accelerator.
Thunderclouds and lightning conductors
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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 produces
one flash every 20 seconds:
Total energy  energy per discharge  number of discharges
 1010 J  90
 9  1011 J.
4. The discharge heats the air directly. The sound increases the random motion of molecules and
light 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.
E  20 C /[(1.6  10 7 m 2 )  (8.9  10 12 F m 1 )]
 1.4  10 5 V m 1 or N C 1
V  Ed
 (1.4  10 6 V m 1 )  1200 m
 1.7  10 8 V.
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 upper
positive charge in the cloud has been ignored; this would tend to reduce E. The equation
assumes a vacuum as the medium that would give an answer similar to that for dry air, but the
medium is actually wet air.
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9.
F  qE
 (1.6  10 19 C)  (3  10 4 N C 1 )
 4.8  10 15 N
a  F /m
 ( 4.8  10 15 N) /( 4.7  10 26 kg)
 1.0  1011 m s 2 .
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 when
water drops are present.
+
–
+
–
+
–
deformed rain drop
14. Because the greatest concentration of deformed raindrops is just below the base of the cloud.
15.
I  1.16  10 4 (4)(30 000  295)
 14 A
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 some
of the ions removed the current would increase as the rate of ionisation is increased. The faster
the 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 speed
would make no difference to the current.
19. With a wind speed of 8 m s–1 the discharge current is 42 A.
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Time  Q / I
 20 C /(42  10 6 A )
 4.8  10 5 s
 5.5 days.
This is rather a long time and establishes that lightning conductors do not work by discharging the
cloud.
20. They provide a low-resistance path to earth and act as a source for the positive streamer. The
lighting conductor actually triggers the lightning discharge.
Electrical breakdown in a vacuum
Question 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 be
sustained.
3. The plasma provides charge carriers.
4. The second figure shows that the breakdown voltage is 37 kV. From the first figure the resistance
of the circuit is 5  108 :
I V /R
 (37  10 3 V ) /(5  10 8 )
 7.4  10 5 A
 75 A.
5.
cathode
6.
emission here
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7. A point which emits electrons at a constant rate.
8. Energy transfer to low grade thermal energy
9.
F  qE
 (1.6  10 19 C)  (10 9 V m 1 )
 (1.6  10 19 C)  (10 9 N C 1 )
 1.6  10 10 N.
10. Noise is fluctuations in the current (the spikes shown in the second figure). These can be caused
by 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:
mass  density  volume
 (9000 kg m 3 )  34    (10 6 m) 3
 3.8  10 14 kg.
13. q V = kinetic energy gained by microparticle where V = 50 kV so:
q  mv 2 / 2V

1
2
 (9  10 15 kg)  (1500 m s 1 ) 2 /(50  10 3 )
 2  10 13 C
 10 13 C.
14. A surge in current increases the potential difference across the resistor at the expense of the
potential difference across the gap. The reduction in the potential difference across the gap acts
to reduce the current surge.
15. Conditioning removes or blunts microprotrusions by evaporation and removes loose
microparticles.
Millikan's oil drop experiment
Question 70D: Data Handling
1.
+ + + + +
+ + + + + + + +
electrical force
– –
– –
– –
–
weight
– –
2. See solution for question 1
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–
– – –
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 formula
[(measured charge  ne ) / ne ]  100
which gives values of 0, 0, 2%, 6%, 0, 3%. The greatest uncertainty is likely to be as a result of
deducing the weight from the graph since V and d can be measured accurately.
The proton synchrotron
Question 130D: Data Handling
Solutions
1.
Ek 
1
2
mv 2
so
v  [(2  8  10 12 J) /(1.66  10 27 kg)]1/ 2
 9.82  10 7 m s 1
 10 8 m s 1.
2.
Time  distance/speed
 628 m / 10 8 m s 1
 6.28  10 6 s
 6.3  10 6 s.
3.
mv  (1.66  10 27 kg)  (9.8  10 7 m s 1 )
 1.63  10 19 kg m s 1
 1.6  10 19 kg m s 1.
4. The total charge is I t = N e where N is the number of protons injected in a time t, so:
N  It / e
 [(100  10 3 A )  (6.28  10 6 s)] /(1.60  10 19 C)
 3.9  1012.
5. At very low pressures the protons are less likely to collide with air molecules. Collisions would
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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 the
opposite 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 per
revolution, i.e.
(28  10 9 J) /(56  10 3 J)  5  10 5.
9. To give a proton the same energy the length of the linear accelerator would have to be equal to
the number of revolutions of the synchrotron multiplied by the circumference of synchrotron. So:
length  (5  10 5 )  628 m
 3.1 10 8 m.
This is excessively long given that the radius of the Earth is about 6  106 m.
10.
F  Bev
 mv 2 / r
so
B  mv / er .
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 m
s–1. So:
B  [(1.66  10 27 kg)  (9.82  10 7 m s 1 )] /[(1.60  10 19 C)  100 m]
 0.010 T.
12. As the speed of the protons increases each bunch of protons takes less time to travel between
the accelerating points. To maintain the acceleration the potential difference must change in less
time.
13. We have already seen that B is directly proportional to the momentum, p, if e and r remain
constant. So
Bf  ( pf / pi )  Bi
where the subscripts i and f refer to the initial and final states. Hence:
Bf / Bi  (1.6  10 17 kg m s 1 ) /(1.63  10 19 kg m s 1 )
 98.2
 98.
14.  = Etotal / Erest = (28 GeV + 1 GeV)/1 GeV = 29.
The electric dipole
Question 170D: Data Handling
The completed spreadsheet table is shown below:
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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 further
analysis:
6. The graph looks non-linear. It is difficult to tell if the graph has a definite linear region so further
analysis is recommended:
7. There is a clearly defined range over which E d 3 has an approximately constant value so E must
be proportional to 1 / d 3 over this range (i.e. for large values of d ):
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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 single
charge falls off as 1 / d 2. At very large distances from a dipole the field will be approximately that
of two superimposed equal and opposite charges. This means that the field will be very small
indeed 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. The
field outside the ionic crystal is the field due to a large number of dipoles at a relatively large
distance. This field is very weak.
Testing Coulomb’s law
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Question 210D: Data Handling
1. A typical set of values is as follows:
d/m
r/m
1/r2
/ 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 become
non-linear when r is very small, as might be expected if the charges on the spheres become
redistributed when the balls are too close.
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0.14
0.12
0.10
0.08
0.06
0.04
0.02
0
0
200
400
(
1/r2
600
)/
800
1000
m–2
2.
Q  CV
 (0.01 10 6 F)  0.5 V
 5  10 9 C
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 W
d since tan  = d / 1. The gradient can be measured using the point (780 m–2, 0.14 m) which
becomes (780 m–2, 1.4  10–4 N) when d is converted into newtons. The gradient of the graph is
1.8  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 will
range from 5  109 to 11  109 N m2 C–2.
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Using uniform electric fields
Question 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. The
volume decreases by a factor of 14 so the pressure increases by similar factor. The temperature
is not constant so an accurate calculation cannot be made. Air molecules are closer together at a
higher pressure so the molecules have a shorter distance to move before colliding with another
molecule. Before each collision an ion must gain sufficient energy (about 30 eV) to ionise the
molecule it hits. If the distance between the molecules is decreased the voltage must be
increased 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 potential
energy 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 acceleration
the ions will have a range of velocities as seen on the graph. For a given ion, v is proportional to
V 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 electrode
will be in the ratio 21/2:1. The pulses reaching the collecting electrode will have a lower amplitude
than the pulse leaving the accelerating electrode and both will show more broadening than this
pulse as the ions are travelling over a finite distance with a range of velocities. The pulse for ions
of mass 2 m should have a larger amplitude than the pulse for ions of mass m.
pulse leaving acceleration electrode
mass 2m
time 2 t
mass m
time t
0
0
time
Deflecting charged particles in a magnetic field
Question 120X: Explanation–Exposition
Solutions
1. The accelerating voltage is given by
V  Ed
 (5.0  10 3 N C 1 )  0.1 m
 500 V.
Using
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eV 
1
2
mv 2
the velocity v of electrons leaving the electron gun is
{[2  (1.6  10 19 C)  500 V] /(9.1 10 31 kg)}1/ 2  1.3  10 7 m s 1.
The force on the electrons during acceleration due to the electric field is given by
F  eE
 (1.6  10 19 C)  5000 N C 1
 8.0  10 16 N.
The force on the electrons due to the magnetic field is given by
F  evB

 
 
 1.6  10 –19 C  1.3  10 7 m s –1  5.0  10  4 T

 1.0  10 –15 N
This force causes electrons to move in circular path of radius
r  mv / Be
 [(9.1 10 31 kg)  (1.3  10 7 m s 1 )] /[(5.0  10 4 T)  (1.6  10 19 C)]
 0.15 m.
The force on the electrons due to the gravitational field is given by
F  mg
 (9.1 10 31 kg)  9.8 N kg 1
 8.9  10 30 N.
The electrons will take a time
t  s /v
 0.25 m /(1.3  10 7 m s 1 )
 1.9  10 8 s
to reach the end of the tube from the electron gun. The distance fallen under gravity during this
time is
s
1
2
at 2

1
2
 9.8 m s – 2  1.9  10 8 s


2
 1.8  10 15 m
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 to
the field and the direction of motion. This force remains at right angles to the path at any point in
its motion regardless of the direction of the particle. The force is q v B which causes a centripetal
force m v 2 / r leading to B q r = m v. The radius is directly proportional to m and v and inversely
proportional to B and q . When at an angle to a uniform field the component vx leads to circular
motion but the component vy results in movement along the field at a constant speed. This leads
to a helical path. In a non-uniform field, r is inversely proportional to B for a given v so r
decreases as the particle approaches the poles where B is greater. This leads to a spiral path.
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