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 1 Advancing Physics 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 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 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 3 Advancing Physics 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. 4 Advancing Physics 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. 5 Advancing Physics 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. 6 Advancing Physics 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 7 Advancing Physics 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 2r 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. 8 Advancing Physics 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 1031 kg) (2.5 107 m s 1 )2 8.8 1015 N 6.5 102 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. 9 Advancing Physics 9. B 2fm 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 Advancing Physics 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. 11 Advancing Physics 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: 12 Advancing Physics 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 13 Advancing Physics 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. 14 Advancing Physics 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/(40r) 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. 15 Advancing Physics 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 m2 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. 16 Advancing Physics 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. 17 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: 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. 18 Advancing Physics 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. 19 Advancing Physics 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. 20 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 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 21 Advancing Physics 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: 22 Advancing Physics 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 23 Advancing Physics 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 24 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 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 26 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 25 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 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. 26 Advancing Physics 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. 27 Advancing Physics 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 28 Advancing Physics 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 29 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 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 30 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 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: 31 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 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 ): 32 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 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 33 Advancing Physics 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. 34 Advancing Physics 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. 35 Advancing Physics 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 36 Advancing Physics 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. 37 Advancing Physics