Electric fields and field lines — Paper 2 · A Level Physics 9702

6 Two horizontal metal plates X and Y are at a distance 0.75 cm apart. A positively charged For particle of mass 9.6 ×10–15kg is situated in a vacuum between the plates, as illustrated in Examiner’s Fig. 6.1. Use plate X + 0.75 cm plate Y Fig. 6.1 The potential difference between the plates is adjusted until the particle remains stationary. (a) State, with a reason, which plate, X or Y, is positively charged. … … … [2] (b) The potential difference required for the particle to be stationary between the plates is found to be 630 V. Calculate (i) the electric field strength between the plates, field strength = …………………………….. N C–1 [2] (ii) the charge on the particle. For Examiner’s Use charge = …………………………….. C [3]

7 marks

Mark scheme: 6 (a) force must be upwards (on positive charge) M1 so plate Y is positive A1 [2] (b) (i) E = V / d C1 = 630/(0.75 × 10-2) = 8.4 × 104 N C-1 A1 [2] (ii) qE = mg C1 q = (9.6 × 10-15 × 9.8) / (8.4 × 104) C1 = 1.12 × 10-18 C A1 [3]

This question in 9702/21 Oct/Nov 2005

2 (a) Define electric field strength. … … [1] (b) Two flat parallel metal plates, each of length 12.0 cm, are separated by a distance of 1.5 cm, as shown in Fig. 2.1. +210 V electron 1.5 cm speed 5.0 x 107 m s–1 12.0 cm Fig. 2.1 The space between the plates is a vacuum. The potential difference between the plates is 210 V. The electric field may be assumed to be uniform in the region between the plates and zero outside this region. Calculate the magnitude of the electric field strength between the plates. field strength = … N C–1 [1] Examiner’s Use (c) An electron initially travels parallel to the plates along a line mid-way between the plates, as shown in Fig. 2.1. The speed of the electron is 5.0 × 107 m s–1. For the electron between the plates, (i) determine the magnitude and direction of its acceleration, acceleration = … m s–2 direction … [4] (ii) calculate the time for the electron to travel a horizontal distance equal to the length of the plates. time = … s [1] (d) Use your answers in (c) to determine whether the electron will hit one of the plates or emerge from between the plates. [3]

10 marks

Mark scheme: 2 (a) force per unit positive charge (on a small test charge) B1 [1] (b) field strength = (210/{1.5 × 10-2} =) 1.4 ×104 N C-1 A1 [1] (c) (i) acceleration = Eq / m C1 = (1.4 × 104 × 1.6 × 10-19) / (9.1 × 10-31) C1 = 2.5 × 1015 m s-2 (2.46 × 1015) A1 towards positive plate / upwards (and normal to plate) B1 [4] (ii) time = 2.4 × 10-9 s A1 [1] (d) either vertical displacement after acceleration for 2.4 × 10-9 s = ½ × 2.46 × 1015 × (2.4 × 10-9)2 C1 = 7.1 × 10-3 m A1 (0.71 cm < 0.75 cm and) so will pass between plates A1 [3] i.e. valid conclusion based on a numerical value or 0.75 × 10-2 = ½ × 2.46 × 1015 × t2 (C1) t is time to travel ‘half-way across’ plates = 2.47 × 10-9 s (A1) (2.4 ns < 2.47 ns) so will pass between plates (A1) i.e. valid conclusion based on a numerical value

This question in 9702/21 May/June 2007

4 Two parallel plates P and Q are separated by a distance of 7.6 mm in a vacuum. There is a For potential difference of 250 V between the plates, as illustrated in Fig. 4.1. Examiner’s Use 7.6 mm plate P plate Q X 250 V Fig. 4.1 Electrons are produced at X on plate P. These electrons accelerate from rest and travel to plate Q. The electric field between the plates may be assumed to be uniform. (a) (i) Determine the force on an electron due to the electric field. force = …………………….. N [3] (ii) Show that the change in kinetic energy of an electron as it moves from plate P to plate Q is 4.0 × 10–17 J. [2] (iii) Determine the speed of an electron as it reaches plate Q. For Examiner’s Use speed = … m s–1 [2] (b) The positions of the plates are adjusted so that the electric field between them is not uniform. The potential difference remains unchanged. State and explain the effect, if any, of this adjustment on the speed of an electron as it reaches plate Q. … … … … [3]

10 marks

Mark scheme: 4 (a) (i) either force = e × (V / d) or E = V/d C1 = 1.6 × 10–19 × (250 / 7.6 × 10–3) C1 = 5.3 × 10–15 N A1 [3] (ii) either ∆EK = eV or ∆EK = Fd C1 = 1.6 × 10–19 × 250 = 5.3 × 10–15 × 7.6 × 10–3 M1 = 4.0 × 10–17 J A0 [2] (allow full credit for correct working via calculation of a and v) GCE A/AS LEVEL – October/November 2008 9702 02 (iii) either ∆EK = ½mv2 4.0 × 10–17 = ½ × 9.1 × 10–31 × v2 C1 v = 9.4 × 106 m s–1 A1 [2] or v2 = 2as and a = F/m v2 = (2 × 5.3 × 10–15 × 7.6 × 10–3)/(9.11 × 10–31) (C1) v = 9.4 × 106 m s–1 (A1) (b) speed depends on (electric) potential difference M2 (If states ∆EK does not depend on uniformity of field, then award 1 mark, treated as an M mark) so speed always the same A1 [3]

This question in 9702/21 Oct/Nov 2008

6 Two vertical parallel metal plates are situated 2.50 cm apart in a vacuum. The potential For difference between the plates is 350 V, as shown in Fig. 6.1. Examiner’s Use 350 V electron + – 2.50 cm Fig. 6.1 An electron is initially at rest close to the negative plate and in the uniform electric field between the plates. (a) (i) Calculate the magnitude of the electric field between the plates. electric field strength = … N C–1 [2] (ii) Show that the force on the electron due to the electric field is 2.24 × 10–15 N. [2] (b) The electron accelerates horizontally across the space between the plates. Determine For Examiner’s (i) the horizontal acceleration of the electron, Use acceleration = … m s–2 [2] (ii) the time to travel the horizontal distance of 2.50 cm between the plates. time = … s [2] (c) Explain why gravitational effects on the electron need not be taken into consideration in your calculation in (b). … … … [2]

10 marks

Mark scheme: 6 (a) (i) E = V / d … C1 = 350 / (2.5 × 10-2) = 1.4 × 104 N C-1 … A1 [2] (ii) force = Eq … C1 = 1.4 × 104 × 1.6 × 10-19 … M1 = 2.24 × 10-15 … A0 [2] (b) (i) F = ma … C1 a = (2.24 × 10-15) / (9.1 × 10-31) = 2.46 × 1015 m s-2 …(allow 2.5 × 105) … A1 [2] (ii) s = ½at2 … C1 2.5 × 10-2 = ½ × 2.46 × 1015 × t2 t = 4.5 × 10-9 s … A1 [2] (c) either gravitational force is normal to electric force or electric force horizontal, gravitational force vertical … B2 [2] special case: force/acceleration due to electric field >> force/acceleration due to gravitational field, allow 1 mark GCE A/AS LEVEL – May/June 2009 9702 21

This question in 9702/21 May/June 2009

5 (a) State what is meant by an electric field. For Examiner’s … Use … [1] (b) The electric field between an earthed metal plate and two charged metal spheres is illustrated in Fig. 5.1. earthed metal plate charged charged sphere sphere Fig. 5.1 (i) On Fig. 5.1, label each sphere with (+) or (–) to show its charge. [1] (ii) On Fig. 5.1, mark a region where the magnitude of the electric field is 1. constant (label this region C), [1] 2. decreasing (label this region D). [1] (c) A molecule has its centre P of positive charge situated a distance of 2.8 × 10–10 m from For its centre N of negative charge, as illustrated in Fig. 5.2. Examiner’s Use 2.8 × 10–10 m P applied electric field 30° 5.0 × 104 V m–1 N molecule Fig. 5.2 The molecule is situated in a uniform electric field of field strength 5.0 × 104 V m–1. The axis NP of the molecule is at an angle of 30° to this uniform applied electric field. The magnitude of the charge at P and at N is 1.6 × 10–19 C. (i) On Fig. 5.2, draw an arrow at P and an arrow at N to show the directions of the forces due to the applied electric field at each of these points. [1] (ii) Calculate the torque on the molecule produced by the forces in (i). torque = … N m [2]

7 marks

Mark scheme: 5 (a) region/area where a charge experiences a force ……………….………………….. B1 [1] (b) (i) left-hand sphere (+), right-hand sphere (–) ……………………..………………. B1 [1] (ii) 1 correct region labelled C within 10 mm of central part of plate otherwise within 5 mm of plate ………….…………………………………….. B1 [1] 2 correct region labelled D area of field not included for (b)(ii)1 …….………. B1 [1] (c) (i) arrows through P and N in correct directions …………………………………… B1 [1] (ii) torque = force × perpendicular distance (between forces) ….………………… C1 = 1.6 × 10–19 × 5.0 × 104 × 2.8 × 10 –10 × sin 30 = 1.1 × 10–24 N m …….…………….……………………………………… A1 [2]

This question in 9702/21 May/June 2010

7 Two oppositely-charged parallel metal plates are situated in a vacuum, as shown in Fig. 7.1. For Examiner’s Use negatively-charged metal plate – particle, mass m charge + q speed v positively-charged metal plate + L Fig. 7.1 The plates have length L. The uniform electric field between the plates has magnitude E. The electric field outside the plates is zero. A positively-charged particle has mass m and charge +q. Before the particle reaches the region between the plates, it is travelling with speed v parallel to the plates. The particle passes between the plates and into the region beyond them. (a) (i) On Fig. 7.1, draw the path of the particle between the plates and beyond them. [2] (ii) For the particle in the region between the plates, state expressions, in terms of E, m, q, v and L, as appropriate, for 1. the force F on the particle, … [1] 2. the time t for the particle to cross the region between the plates. … [1] (b) (i) State the law of conservation of linear momentum. For Examiner’s … Use … … [2] (ii) Use your answers in (a)(ii) to state an expression for the change in momentum of the particle. … [1] (iii) Suggest and explain whether the law of conservation of linear momentum applies to the particle moving between the plates. … … … [2]

9 marks

Mark scheme: 7 (a) (i) path: reasonable curve upwards between plates B1 straight and at a tangent to the curve beyond the plates B1 [2] (ii) 1. (F =) E.g B1 [1] 2. (t =) L / v B1 [1] (b) (i) total momentum of a system remains constant or total momentum of a system before a collision equals total momentum after collision M1 provided no external force acts on the system A1 [2] (do not accept ‘conserved’ but otherwise correct statement gets 1/2) (ii) (∆p =) EqL / v allow ecf from (a)(ii) B1 [1] (iii) either charged particle is not an isolated system M1 so law does not apply A1 [2] or system is particle and ‘plates’ (M1) equal and opposite ∆p on plates / so law applies (A1) GCE A LEVEL – October/November 2010 9702 23 2

This question in 9702/23 Oct/Nov 2010

4 (a) Define electric field strength. For Examiner’s … Use … [1] (b) Two horizontal metal plates are 20 mm apart in a vacuum. A potential difference of 1.5 kV is applied across the plates, as shown in Fig. 4.1. metal plate +1.5 kV oil drop 20 mm 0 V metal plate Fig. 4.1 A charged oil drop of mass 5.0 × 10–15 kg is held stationary by the electric field. (i) On Fig. 4.1, draw lines to represent the electric field between the plates. [2] (ii) Calculate the electric field strength between the plates. electric field strength = … V m–1 [1] (iii) Calculate the charge on the drop. charge = … C [4] (iv) The potential of the upper plate is increased. Describe and explain the subsequent motion of the drop. … … … [2]

10 marks

Mark scheme: 4 (a) electric field strength = force / positive charge B1 [1] (b) (i) at least three equally spaced parallel vertical lines B1 direction down B1 [2] (ii) E = 1500 / 20 × 10–3 = 75000 V m–1 A1 [1] (iii) F = qE C1 (W = mg and) qE = mg C1 q = mg / E = 5 × 10–15 × 9.81 / 75000 = 6.5 × 10–19 C A1 negative charge A1 [4] (iv) F > mg or F now greater B1 drop will move upwards B1 [2]

This question in 9702/22 Oct/Nov 2011

6 Two horizontal metal plates are separated by distance d in a vacuum. A potential difference V For is applied across the plates, as shown in Fig. 6.1. Examiner’s Use +V metal plate radioactive d source metal plate beam of α-particles 0V Fig. 6.1 A horizontal beam of α-particles from a radioactive source is made to pass between the plates. (a) State and explain the effect on the deflection of the α-particles for each of the following changes: (i) The magnitude of V is increased. … … [1] (ii) The separation d of the plates is decreased. … … [1] (b) The source of α-particles is replaced with a source of β-particles. For Compare, with a reason in each case, the effect of each of the following properties on Examiner’s the deflections of α- and β-particles in a uniform electric field: Use (i) charge … … … [2] (ii) mass … … … [2] (iii) speed … … … [1] (c) The electric field gives rise to an acceleration of the α-particles and the β-particles. Determine the ratio acceleration of the α-particles . acceleration of the β-particles ratio = … [3]

10 marks

Mark scheme: 6 (a) (i) greater deflection M0 greater electric field / force on α-particle A1 [1] (ii) greater deflection M0 greater electric field / force on α-particle A1 [1] (b) (i) either deflections in opposite directions M1 because oppositely charged A1 or β less deflection (M1) β has smaller charge (A1) [2] (ii) α smaller deflection M1 because larger mass A1 [2] (iii) β less deflection because higher speed B1 [1] (c) either F = ma and F = Eq or a = Eq / m C1 ratio = either (2 × 1.6 × 10–19) × (9.11 × 10–31) (1.6 × 10–19) × 4 × (1.67 × 10–27) or [2e × 1 / 2000 u] / [e × 4u] C1 ratio = 1 /4000 or 2.5 × 10–4 or 2.7 × 10–4 A1 [3]

This question in 9702/23 Oct/Nov 2011

4 (a) Define electric field strength. For Examiner’s … Use … [1] (b) A uniform electric field is produced by applying a potential difference of 1200 V across two parallel metal plates in a vacuum, as shown in Fig. 4.1. 1200 V 14 mm metal plates P Fig. 4.1 The separation of the plates is 14 mm. A particle P with charge 3.2 × 10–19 C and mass 6.6 × 10–27 kg starts from rest at the lower plate and is moved vertically to the top plate by the electric field. Calculate (i) the electric field strength between the plates, electric field strength = … V m–1 [2] (ii) the work done on P by the electric field, work done = … J [2] (iii) the gain in gravitational potential energy of P, gain in potential energy = … J [2] (iv) the gain in kinetic energy of P, For Examiner’s Use gain in kinetic energy = … J [1] (v) the speed of P when it reaches the top plate. speed = … m s–1 [2]

10 marks

Mark scheme: 4 (a) electric field strength is the force per unit positive charge (acting on a stationary charge) B1 [1] (b) (i) E = V / d C1 = 1200 / 14 × 10–3 = 8.57 × 104 V m–1 A1 [2] (ii) W = QV or W = F × d and therefore W = E × Q × d C1 = 3.2 × 10–19 × 1200 = 3.84 × 10–16 J A1 [2] (iii) ∆U = mgh C1 = 6.6 × 10–27 × 9.8 × 14 × 10–3 = 9.06 × 10–28 J A1 [2] (iv) ∆K = 3.84 × 10–16 – ∆U = 3.84 × 10–16 J A1 [1] (v) K = ½mv2 C1 v = [(2 × 3.8 × 10–16) / 6.6 × 10–27]1/2 = 3.4 × 105 m s–1 A1 [2]

This question in 9702/23 May/June 2012

7 (a) An electric field is set up between two parallel metal plates in a vacuum. The deflection For of α-particles as they pass between the plates is shown in Fig. 7.1. Examiner’s Use metal plate path of _-particles electric field metal plate Fig. 7.1 The electric field strength between the plates is reduced. The α-particles are replaced by β-particles. The deflection of β-particles is shown in Fig. 7.2. metal plate path of `-particles electric field metal plate Fig. 7.2 (i) State one similarity of the electric fields shown in Fig. 7.1 and Fig. 7.2. … … [1] (ii) The electric field strength in Fig. 7.2 is less than that in Fig. 7.1. State two methods of reducing this electric field strength. 1. … 2. … [2] (iii) By reference to the properties of α-particles and β-particles, suggest three reasons For for the differences in the deflections shown in Fig. 7.1 and Fig. 7.2. Examiner’s Use 1. … … 2. … … 3. … … [3] (b) A source of α-particles is uranium-238. The nuclear reaction for the emission of α-particles is represented by 23892U WXQ + YZ α. State the values of W … X … Y … Z … [2] (c) A source of β-particles is phosphorus-32. The nuclear reaction for the emission of β-particles is represented by 3215P ABR + CDβ. State the values of A … B … C … D … [1]

9 marks

Mark scheme: 7 (a) (i) the direction of the fields is the same OR fields are uniform OR constant electric field strength OR E = V / d with symbols explained B1 [1] (ii) reduce p.d. across plates B1 increase separation of plates B1 [2] (iii) α opposite charge to β (as deflection in opposite direction) B1 β has a range of velocities OR energies (as different deflections) and α all have same velocity OR energy (as constant deflection) B1 α are more massive (as deflection is less for greater field strength) B1 [3] (b) W = 234 and X = 90 B1 Y = 4 and Z = 2 B1 [2] (c) A = 32 and B = 16 and C = 0 and D = –1 B1 [1]

This question in 9702/21 Oct/Nov 2013

7 (a) An electric field is set up between two parallel metal plates in a vacuum. The deflection For of α-particles as they pass between the plates is shown in Fig. 7.1. Examiner’s Use metal plate path of _-particles electric field metal plate Fig. 7.1 The electric field strength between the plates is reduced. The α-particles are replaced by β-particles. The deflection of β-particles is shown in Fig. 7.2. metal plate path of `-particles electric field metal plate Fig. 7.2 (i) State one similarity of the electric fields shown in Fig. 7.1 and Fig. 7.2. … … [1] (ii) The electric field strength in Fig. 7.2 is less than that in Fig. 7.1. State two methods of reducing this electric field strength. 1. … 2. … [2] (iii) By reference to the properties of α-particles and β-particles, suggest three reasons For for the differences in the deflections shown in Fig. 7.1 and Fig. 7.2. Examiner’s Use 1. … … 2. … … 3. … … [3] (b) A source of α-particles is uranium-238. The nuclear reaction for the emission of α-particles is represented by 23892U WXQ + YZ α. State the values of W … X … Y … Z … [2] (c) A source of β-particles is phosphorus-32. The nuclear reaction for the emission of β-particles is represented by 3215P ABR + CDβ. State the values of A … B … C … D … [1]

9 marks

Mark scheme: 7 (a) (i) the direction of the fields is the same OR fields are uniform OR constant electric field strength OR E = V / d with symbols explained B1 [1] (ii) reduce p.d. across plates B1 increase separation of plates B1 [2] (iii) α opposite charge to β (as deflection in opposite direction) B1 β has a range of velocities OR energies (as different deflections) and α all have same velocity OR energy (as constant deflection) B1 α are more massive (as deflection is less for greater field strength) B1 [3] (b) W = 234 and X = 90 B1 Y = 4 and Z = 2 B1 [2] (c) A = 32 and B = 16 and C = 0 and D = –1 B1 [1]

This question in 9702/22 Oct/Nov 2013

7 (a) Two horizontal metal plates are connected to a power supply, as shown in Fig. 7.1. For Examiner’s metal plate Use S + 1.2 kV 40 mm − metal plate Fig. 7.1 The separation of the plates is 40 mm. The switch S is then closed so that a potential difference of 1.2 kV is applied across the plates. (i) On Fig. 7.1, draw six field lines to represent the electric field between the metal plates. [2] (ii) Calculate the electric field strength E between the plates. E = … V m–1 [2] (b) The switch S is opened and the plates lose their charge. Two very small metal spheres A and B joined by an insulating rod are placed between the metal plates as shown in Fig. 7.2. metal plate S 15 mm + A C B 1.2 kV 40 mm − −e +e insulating rod metal plate Fig. 7.2 Sphere A has charge –e and sphere B has charge +e, where e is the charge of a proton. For The length AB is 15 mm. The rod is supported at its centre C so that the rod is horizontal Examiner’s and in equilibrium. Use The switch S is then closed so that the potential difference of 1.2 kV is applied across the plates. (i) There is a force acting on A due to the electric field between the plates. Show that this force is 4.8 × 10–15 N. [2] (ii) The insulating rod joining A and B is fixed in the position shown in Fig. 7.2. Calculate the torque of the couple acting on the rod. torque = … unit … [3] (iii) The insulating rod is now released so that it is free to rotate about C. State and explain the position of the rod when it comes to rest. … … … … [2]

11 marks

Mark scheme: 7 (a) (i) six vertical lines from plate to plate equally spaced across plates B1 [only allow if greatest to least spacing is < 1.3, condone slight curving on the two edges. There must be no area between the plates where an additional line(s) could be added.] arrow downwards on at least one line B1 [2] (ii) E = V / d C1 E = 1200 / 40 × 10–3 = 3.0 × 104 V m–1 (allow 1 s.f.) A1 [2] (b) (i) F = Ee C1 E = 3 × 104 × 1.6 × 10–19 = 4.8 × 10–15 N A1 [2] (ii) couple = F × separation of charges C1 couple = 4.8 × 10–15 × 15 × 10–3 = 7.2 × 10–17 A1 unit: N m or unit consistent with unit used for the separation B1 [3] (iii) A at top/next to +ve plate B at bottom/next to −ve plate vertically aligned M1 [could be shown on the diagram] forces are equal and opposite in same line / no resultant force and no resultant torque A1 [2]

This question in 9702/23 Oct/Nov 2013

3 (a) Define electric field strength. … … [1] (b) A sphere S has radius 1.2 × 10–6 m and density 930 kg m−3. Show that the weight of S is 6.6 × 10−14 N. [2] (c) Two horizontal metal plates are 14 mm apart in a vacuum. A potential difference (p.d.) of 1.9 kV is applied across the plates, as shown in Fig. 3.1. metal plate +1.9 kV 14 mm metal plate sphere S Fig. 3.1 A uniform electric field is produced between the plates. The sphere S in (b) is charged and is held stationary between the plates by the electric field. (i) Calculate the electric field strength between the plates. electric field strength = … V m−1 [2] (ii) Calculate the magnitude of the charge on S. charge = … C [2] (iii) The magnitude of the p.d. applied to the plates is increased. Explain why S accelerates towards the top plate. … … [2]

9 marks

Mark scheme: 3 (a) electric field strength is force per unit positive charge B1 [1] (b) mass = volume × density (any subject, allow usual symbols or defined symbols) C1 = 4/3 × π × (1.2 × 10–6)3 × 930 (= 6.73 × 10–15) weight = 4/3 × π × (1.2 × 10–6)3 × 930 × 9.81 = 6.6 × 10–14 N M1 [2] (c) (i) E = 1.9 × 103 / 14 × 10–3 C1 = 1.4 (1.36) × 105 V m–1 A1 [2] (ii) F = QE Q = 6.6 × 10–14 / 1.36 × 105 C1 = 4.9 (4.86) × 10–19 C [allow 4.7 × 10–19 C if 1.4 × 105 used] A1 [2] (iii) electric force increases / is greater (than weight) B1 charge (on S) is negative to give resultant / net / sum / total force up B1 [2]

This question in 9702/22 Oct/Nov 2014

7 (a) Explain what is meant by an electric field. … … [1] (b) A uniform electric field is produced between two vertical metal plates AB and CD, as shown in Fig. 7.1. A C _-particle 16 mm B D 450 V + – Fig. 7.1 The potential difference between the plates is 450 V and the separation of the plates is 16 mm. An α-particle is accelerated from plate AB to plate CD. (i) On Fig. 7.1, draw lines to represent the electric field between the plates. [2] (ii) Calculate the electric field strength between the plates. electric field strength = … V m–1 [2] (iii) Calculate the work done by the electric field on the α-particle as it moves from AB to CD. work done = … J [3] Question 7 continues on page 16. (iv) A β-particle moves from AB to CD. Calculate the ratio work done by the electric field on the α-particle work done by the electric field on the β-particle. Show your working. ratio = … [1]

9 marks

Mark scheme: 7 (a) a region/space/area where a (stationary) charge experiences an (electric) force B1 [1] (b) (i) at least four parallel equally spaced straight lines perpendicular to plates B1 consistent direction of an arrow on line(s) from left to right B1 [2] (ii) electric field strength E = V / d C1 E = (450 / 16 × 10–3) = 28 × 103 (28 125) V m–1 A1 [2] (iii) W = Eqd or Vq C1 q = 3.2 × 10–19 (C) C1 W = 28 125 × 3.2 × 10–19 × 16 × 10–3 or 450 × 3.2 × 10–19 = 1.4(4) × 10–16 J A1 [3] 450 × 3.2 × 10 −19 (iv) ratio = −19 (evidence of working required) 450 × − 1.6 × 10 = (–) 2 A1 [1]

This question in 9702/21 May/June 2015

7 Two parallel, vertical metal plates in a vacuum are connected to a power supply and a switch, as shown in Fig. 7.1. path of _-particles metal metal radioactive source + – power supply Fig. 7.1 A radioactive source emitting α-particles is placed below the plates. The path of the α-particles is shown on Fig. 7.1. The switch is closed producing a potential difference (p.d.) across the plates. This gives rise to a uniform electric field between the plates. The separation of the plates is 12 mm. (a) (i) On Fig. 7.1, draw the path of the α-particles. [1] (ii) Explain why the metal plates are placed in a vacuum. … … [1] (iii) Calculate the p.d. required to produce an electric field of 140 MV m–1. p.d. = … MV [2] (b) The α-particle source is replaced by a β-particle source. By reference to the properties of α-radiation and β-radiation, suggest three possible differences in the deflection observed with β-particles. 1. … … 2. … … 3. … … [3] (c) Complete Fig. 7.2 to show the changes in the proton number Z and the nucleon number A of different radioactive nuclei when either an α-particle or a β-particle is emitted. emitted particle change in Z change in A α-particle β-particle Fig. 7.2 [1]

8 marks

Mark scheme: 7 (a) (i) curved path towards negative (–) plate (right-hand side) B1 [1] (ii) range of α-particle is only few cm in air/loss of energy of the α-particles due to collision with air molecules/ionisation of the air molecules B1 [1] (iii) V = E × d C1 = 140 × 106 × 12 × 10–3 = 1.7 (1.68) MV A1 [2] (b) β have opposite charge to α therefore deflection in opposite direction B1 β has a range of velocities/energies hence number of different deflections B1 β have less mass or q / m is larger hence deflection is greater or β with (very) high speed (may) have less deflection B1 [3] (c) emitted particle change in Z change in A α-particle –2 –4 β-particle +1 0 A1 [1]

This question in 9702/21 Oct/Nov 2015

6 Two parallel vertical metal plates are connected to a power supply, as shown in Fig. 6.1. metal plate metal plate 16 mm + – Fig. 6.1 The separation of the plates is 16 mm. (a) On Fig. 6.1, draw at least six field lines to represent the electric field between the plates. [1] (b) An α-particle travels in a vacuum between the two plates. The electric field does work on the α-particle. The gain in kinetic energy of the α-particle is 15 keV. Calculate the electric field strength between the plates. electric field strength = … V m–1 [4] [Total: 5]

5 marks

Mark scheme: 6 (a) at least six horizontal lines equally spaced and arrow to the right B1 [1] (b) charge used 2e C1 gain in KE = 15 × 1.6 × 10–19 × 103 = 2 × 1.6 × 10–19 × V (p.d.across plates) or F (= W / d) = 15 × 1.6 × 10–19 × 103 / 16 × 10–3 C1 (hence V = 7500 V or F = 1.5 × 10–13 N) E = V / d or E = F / Q C1 E = (7500 / 16 × 10–3) or E = (1.5 × 10–13 / 3.2 × 10–19) E = 4.7 × 105 (468 750) V m–1 A1 [4] or KE (= ½mv2) = 15 × 103 × 1.6 × 10–19 v = [(2 × 15 × 103 × 1.6 × 10–19) / (6.68 × 10–27)]1/2 = 8.5 × 105 m s–1 (C1) a (= v2 / 2s) = (8.5 × 105)2 / 2 × 16 × 10–3 = 2.25 × 1013 m s–2 F (= 6.68 × 10–27 × 2.25 × 10–13) = 1.5 × 10–13 N E = F / Q (C1) Q = 2e (C1) E = 4.7 × 105 V m–1 (A1)

This question in 9702/22 May/June 2016

2 (a) Define electric field strength. … … [1] (b) A potential difference of 2.5 kV is applied across a pair of horizontal metal plates in a vacuum, as shown in Fig. 2.1. metal plate \ 2.0 cm electron B + 2.5 kV velocity A – 2.0 cm 3.7 × 107 m s–1 metal 5.9 cm plate Fig. 2.1 (not to scale) Each plate has a length of 5.9 cm. The separation of the plates is 4.0 cm. The arrangement produces a uniform electric field between the plates. Assume the field does not extend beyond the edges of the plates. An electron enters the field at point A with horizontal velocity 3.7 × 107 m s–1 along a line mid-way between the plates. The electron leaves the field at point B. (i) Calculate the time taken for the electron to move from A to B. time taken = … s [1] (ii) Calculate the magnitude of the electric field strength. field strength = … N C–1 [2] (iii) Show that the acceleration of the electron in the field is 1.1 × 1016 m s–2. [2] (iv) Use the acceleration given in (iii) and your answer in (i) to determine the vertical distance y between point B and the upper plate. y = … cm [3] (v) Explain why the calculation in (iv) does not need to include the gravitational effects on the electron. … … [1] (vi) The electron enters the field at time t = 0. On Fig. 2.2, sketch graphs to show the variation with time t of 1. the horizontal component vX of the velocity of the electron, 2. the vertical component v of the velocity of the electron. Y Numerical values are not required. vX vY 0 0 0 0 t t Fig. 2.2 [2] [Total: 12]

12 marks

Mark scheme: 2 (a) force per unit positive charge B1 [1] (b) (i) time = 5.9 × 10–2 / 3.7 × 107 = 1.6 × 10–9 s (1.59 × 10–9 s) A1 [1] (ii) E = V / d C1 = 2500 / 4.0 × 10–2 = 6.3 × 104 N C–1 (6.25 × 104 or 62500 N C–1) A1 [2] (iii) a = Eq / m or F = ma and F = Eq C1 = (6.3 × 104 × 1.60 × 10–19) / 9.11 × 10–31 = 1.1 × 1016 m s–2 A1 [2] (iv) s = ut + ½at 2 = ½ × 1.1 × 1016 × (1.6 × 10–9)2 C1 = 1.4 × 10–2 (m) C1 distance from plate = 2.0 – 1.4 = 0.6 cm (allow 1 or more s.f.) A1 [3] (v) electric force ≫ gravitational force (on electron)/weight or acceleration due to electric field ≫ acceleration due to gravitational field B1 [1] (vi) vX–t graph: horizontal line at a non-zero value of vX B1 vY–t graph: straight line through the origin with positive gradient B1 [2]

This question in 9702/21 Oct/Nov 2016

2 (a) Define electric field strength. … … [1] (b) A potential difference of 2.5 kV is applied across a pair of horizontal metal plates in a vacuum, as shown in Fig. 2.1. metal plate y 2.0 cm electron B + 2.5 kV velocity A – 2.0 cm 3.7 × 107 m s–1 metal 5.9 cm plate Fig. 2.1 (not to scale) Each plate has a length of 5.9 cm. The separation of the plates is 4.0 cm. The arrangement produces a uniform electric field between the plates. Assume the field does not extend beyond the edges of the plates. An electron enters the field at point A with horizontal velocity 3.7 × 107 m s–1 along a line mid-way between the plates. The electron leaves the field at point B. (i) Calculate the time taken for the electron to move from A to B. time taken = … s [1] (ii) Calculate the magnitude of the electric field strength. field strength = … N C–1 [2] (iii) Show that the acceleration of the electron in the field is 1.1 × 1016 m s–2. [2] (iv) Use the acceleration given in (iii) and your answer in (i) to determine the vertical distance y between point B and the upper plate. y = … cm [3] (v) Explain why the calculation in (iv) does not need to include the gravitational effects on the electron. … … [1] (vi) The electron enters the field at time t = 0. On Fig. 2.2, sketch graphs to show the variation with time t of 1. the horizontal component vX of the velocity of the electron, 2. the vertical component v of the velocity of the electron. Y Numerical values are not required. vX vY 0 0 0 0 t t Fig. 2.2 [2] [Total: 12]

12 marks

Mark scheme: 2 (a) force per unit positive charge B1 [1] (b) (i) time = 5.9 × 10–2 / 3.7 × 107 = 1.6 × 10–9 s (1.59 × 10–9 s) A1 [1] (ii) E = V / d C1 = 2500 / 4.0 × 10–2 = 6.3 × 104 N C–1 (6.25 × 104 or 62500 N C–1) A1 [2] (iii) a = Eq / m or F = ma and F = Eq C1 = (6.3 × 104 × 1.60 × 10–19) / 9.11 × 10–31 = 1.1 × 1016 m s–2 A1 [2] (iv) s = ut + ½at 2 = ½ × 1.1 × 1016 × (1.6 × 10–9)2 C1 = 1.4 × 10–2 (m) C1 distance from plate = 2.0 – 1.4 = 0.6 cm (allow 1 or more s.f.) A1 [3] (v) electric force ≫ gravitational force (on electron)/weight or acceleration due to electric field ≫ acceleration due to gravitational field B1 [1] (vi) vX–t graph: horizontal line at a non-zero value of vX B1 vY–t graph: straight line through the origin with positive gradient B1 [2]

This question in 9702/23 Oct/Nov 2016

3 (a) Define electric field strength. … … [1] (b) An electron is accelerated from point A to point B by a uniform electric field, as illustrated in Fig. 3.1. electric field A electron B Fig. 3.1 The distance between A and B is 12 mm. The velocity of the electron at A is 2.5 km s–1 and at B is 18 Mm s–1. Calculate (i) the acceleration of the electron, acceleration = … m s–2 [2] (ii) the change in kinetic energy of the electron, change in kinetic energy = … J [3] (iii) the electric field strength. electric field strength = … V m–1 [3] (c) An α-particle moves from A to B in the electric field in (b). Describe and explain how the change in the kinetic energy of the α-particle compares with that of the electron. Numerical values are not required. … … … … … [3] [Total: 12]

12 marks

Mark scheme: 3(a) force per unit (positive) charge B1 3(b)(i) a = (v 2 − u 2) / 2s = [(18 × 106)2 − (2.5 × 103)2] / (2 × 12 × 10–3) B1 = 1.3 (1.35) × 1016 m s–2 A1 3(b)(ii) KE = ½ mv 2 or ½ m(v2 – u2) C1 change in KE = 0.5 × 9.11 × 10–31 × [(18 × 106)2 − (2.5 × 103)2] B1 = 1.5 (1.48) × 10–16J A1 3(b)(iii) E = F / e = ma / e or eV = ∆KE so E = ∆KE / (e × d) C1 E = (9.11 × 10–31 × 1.35 × 1016) / 1.60 × 10–19 or E = (1.48 × 10–16) / (12 × 10–3 × 1.60 × 10–19) C1 = 7.7 (7.69) × 104 V m–1 A1 3(c) charge on α opposite to electron/charge on α is positive B1 ∆KE is negative/KE reduced B1 charge of α greater/twice that of electron causes larger/twice ∆KE (in magnitude) B1

This question in 9702/23 May/June 2017

6 (a) Define electric field strength. … … [1] (b) Two parallel metal plates in a vacuum are separated by a distance of 15 mm, as shown in Fig. 6.1. + – particle mass 1.7 × 10–27 kg charge +1.6 × 10–19 C A B metal metal plate plate 15 mm Fig. 6.1 A uniform electric field is produced between the plates by applying a potential difference between them. A particle of mass 1.7 × 10–27 kg and charge +1.6 × 10–19 C is initially at rest at point A on one plate. The particle is moved by the electric field to point B on the other plate. The particle reaches point B with kinetic energy 2.4 × 10–16 J. (i) Calculate the speed of the particle at point B. speed = … m s–1 [2] (ii) State the work done by the electric field to move the particle from A to B. work done = … J [1] (iii) Use your answer in (ii) to determine the force on the particle. force = … N [2] (iv) Determine the potential difference between the plates. potential difference = … V [3] (v) On Fig. 6.2, sketch a graph to show the variation of the kinetic energy of the particle with the distance x from point A along the line AB. Numerical values for the kinetic energy are not required. kinetic energy 0 0 15 x / mm Fig. 6.2 [1] [Total: 10]

10 marks

Mark scheme: 6(a) force per unit positive charge B1 6(b)(i) EK = ½mv 2 C1 2.4 × 10–16 = ½ × 1.7 × 10–27 × v 2 v = 5.3 × 105 m s–1 A1 6(b)(ii) work done = 2.4 × 10–16 J A1 6(b)(iii) W = Fs C1 F = 2.4 × 10–16 / 15 × 10–3 = 1.6 × 10–14 N A1 6(b)(iv) V = Fd / Q or V = W / Q or E = V / d and E = F / Q C1 V = (1.6 × 10–14 × 15 × 10–3) / 1.6 × 10–19 or 2.4 × 10–16 / 1.6 × 10–19 C1 = 1500 V A1 6(b)(v) straight line with positive gradient starting at the origin and going as far as x = 15 mm B1

This question in 9702/21 Oct/Nov 2017

5 (a) Define electric field strength. … … [1] (b) Two parallel metal plates in a vacuum are separated by 0.045 m. A potential difference V is applied between the plates, as shown in Fig. 5.1. metal plate – 0.045 m V + metal plate proton Fig. 5.1 A proton is initially at rest on the surface of the positive plate. The proton in the uniform electric field takes a time of 1.5 × 10–7 s to reach the negative plate. (i) Show that the acceleration of the proton is 4.0 × 1012 m s–2. [2] (ii) Calculate the electric force on the proton. force = … N [1] (iii) Use your answer in (ii) to determine 1. the electric field strength, field strength = … N C–1 [2] 2. the potential difference V between the plates. V = … V [2] (c) An α particle is now accelerated between the two metal plates in (b) by the electric field. Calculate the ratio acceleration of α particle . acceleration of proton ratio = … [2] [Total: 10]

10 marks

Mark scheme: 5(a) force per unit positive charge B1 5(b)(i) s = ½at 2 C1 a = (2 × 0.045) / (1.5 × 10–7)2 = 4(.0) × 1012 m s–2 A1 5(b)(ii) F = 1.67 × 10–27 × 4.0 × 1012 = 6.7 (6.68) × 10–15N A1 5(b)(iii) 1. E = F / Q C1 = 6.68 × 10–15 / 1.6 × 10–19 = 4.2 (4.18) × 104 N C–1 A1 2. E = V / d C1 V = 4.18 × 104 × 0.045 = 1.9 × 103 V A1 Question Answer Marks 5(c) a = Eq / m or F = ma and F = Eq C1 ratio = − − − − × × × × × × × × 19 27 19 27 (2 1.6 10 ) (1.67 10 ) (1.6 10 ) (4 1.66 10 ) or × × 2 1 1 4 = 0.50 A1

This question in 9702/23 Oct/Nov 2017

5 (a) State what is meant by an electric field. … … [1] (b) A particle of mass m and charge q is in a uniform electric field of strength E. The particle has acceleration a due to the field. Show that Eq a = . m [2] (c) A stationary nucleus X decays by emitting an α-particle to form a nucleus of plutonium, 24094 Pu, as shown. α X 24094 Pu + (i) Determine the number of protons and the number of neutrons in nucleus X. number of protons = … number of neutrons = … [2] (ii) The total mass of the plutonium nucleus and the α-particle is less than that of nucleus X. Explain this difference in mass. … … … … [2] (iii) The plutonium nucleus and the α-particle are both accelerated by the same uniform electric field. Use the expression in (b) to determine the ratio acceleration of the α-particle . acceleration of the plutonium nucleus ratio = … [2] [Total: 9]

9 marks

Mark scheme: 5(a) region (of space) where a force acts on a (stationary) charge B1 5(b) E = F / Q B1 F = ma and (so) Eq a m = A1 5(c)(i) protons = 96 A1 neutrons = 148 A1 5(c)(ii) mass-energy is conserved/mass change is ‘seen’ as energy B1 energy released as gamma (radiation)/KE of α/KE of Pu B1 5(c)(iii) 9 4 2 4 0 × = 4 2 ratio or 1 9 2 7 2 7 1 9 1 0 6 0 . 1 9 4 1 0 6 6 . 1 2 4 0 1 0 6 6 . 1 4 − − − − × × × × × × × × × = 10 1.60 2 ratio C1 ratio = 1.3 A1

This question in 9702/21 Oct/Nov 2018

4 (a) Define electric field strength. … … [1] (b) Two very small metal spheres X and Y are connected by an insulating rod of length 72 mm. A side view of this arrangement is shown in Fig. 4.1. +3e uniform electric field, X field strength 5.0 × 104 V m–1 72 mm in vertically upwards direction θ horizontal Z θ SIDE rod VIEW Y –3e Fig. 4.1 (not to scale) Sphere X has a charge of +3e and sphere Y has a charge of –3e, where e is the elementary charge. The rod is held at its mid point Z at an angle θ to the horizontal. The rod and spheres have negligible mass and are in a uniform electric field. The electric field strength is 5.0 × 104 V m–1. The direction of this field is vertically upwards. (i) The electric field is produced by applying a potential difference of 4.0 kV between two charged parallel metal plates. 1. Calculate the separation between the plates. separation = … m [2] 2. Describe the arrangement of the two plates. Include in your answer a statement of the sign of the charge on each plate. You may draw on Fig. 4.1. … … … … [2] (ii) Determine the magnitude and direction of the force on sphere Y. magnitude = … N direction … [2] (iii) The electric forces acting on the two spheres form a couple. This couple acts on the rod with a torque of 6.2 × 10–16 N m. Calculate the angle θ of the rod to the horizontal. θ = … ° [2] [Total: 9]

9 marks

Mark scheme: 4(a) force per unit positive charge B1 4(b)(i) 1 E = V / d or E = ∆V / ∆d d = 4.0 × 103 / 5.0 × 104 C1 = 8.0 × 10–2 m A1 2 plates are (in) horizontal (plane) (above and below the rod) B1 top (plate) negative and bottom (plate) positive B1 4(b)(ii) magnitude = 5.0 × 104 × 3 × 1.6 × 10–19 = 2.4 × 10–14 N A1 direction is (vertically) downwards / down B1 Question Answer Marks 4(b)(iii) 6.2 × 10–16 = 2.4 × 10–14 × 72 × 10–3 × cosθ C1 θ = 69° A1

This question in 9702/22 Feb/March 2019

6 (a) State what is meant by a field line (line of force) in an electric field. … … [1] (b) An electric field has two different regions X and Y. The field strength in X is less than that in Y. Describe a difference between the pattern of field lines (lines of force) in X and in Y. … … [1] (c) A particle P has a mass of 0.15 u and a charge of −1e, where e is the elementary charge. (i) Particle P and an α-particle are in the same uniform electric field. Calculate the ratio magnitude of acceleration of particle P . magnitude of acceleration of α-particle ratio = … [3] (ii) Particle P is a hadron composed of only two quarks. One of them is a down (d) quark. By considering charge, determine a possible type (flavour) of the other quark. Explain your working. … … [3] [Total: 8]

8 marks

Mark scheme: 6(a) path/direction in which a (free) positive charge will move B1 6(b) (lines) closer together in Y/further apart in X B1 6(c)(i) a = Eq / m or F = Eq and F = ma C1 ratio = (1e / 0.15 u) × (4 u / 2e) or 1 / 0.15 × 4 / 2 C1 ratio = 13 A1 6(c)(ii) down quark charge is –(1 / 3)e C1 – (1 / 3)e + q = –1e so q = –(2 / 3)e A1 (–(2 / 3)e is) anti-up / u (quark) (allow charm or top antiquark) B1

This question in 9702/22 May/June 2019

4 Two vertical metal plates in a vacuum are separated by a distance of 0.12 m. Fig. 4.1 shows a side view of this arrangement. 0.080 m sand X particle 2.0 m 0 V + 900 V path of particle metal plate metal plate Y 0.12 m Fig. 4.1 (not to scale) Each plate has a length of 2.0 m. The potential difference between the plates is 900 V. The electric field between the plates is uniform. A negatively charged sand particle is released from rest at point X, which is a horizontal distance of 0.080 m from the top of the positively charged plate. The particle then travels in a straight line and collides with the positively charged plate at its lowest point Y, as illustrated in Fig. 4.1. (a) Describe the pattern of the field lines (lines of force) between the plates. … … … [2] (b) State the names of the two forces acting on the particle as it moves from X to Y. … [1] (c) By considering the vertical motion of the sand particle, show that the time taken for the particle to move from X to Y is 0.64 s. [2] (d) Calculate the horizontal component of the acceleration of the particle. horizontal component of acceleration = … m s−2 [2] (e) (i) Calculate the magnitude of the electric field strength. electric field strength = … N C−1 [2] (ii) The sand particle has mass m and charge q. Use your answers in (d) and (e)(i) to q determine the ratio m. ratio = … C kg−1 [2] q(f) Another particle has a smaller magnitude of the ratio than the sand particle. This particle is m also released from point X. For the movement of this particle, state the effect, if any, of the decreased magnitude of the ratio on: (i) the vertical component of the acceleration … [1] (ii) the horizontal component of the acceleration. … [1] [Total: 13]

13 marks

Mark scheme: 4(a) straight (horizontal) lines and from the +0.90 kV plate/to the 0 V plate B1 (lines are) equally spaced B1 4(b) weight/gravitational force and electric force B1 4(c) s = ½ at 2 or s = ut + ½at 2 and u = 0 C1 2.0 = ½ × 9.81 × t 2 so t = 0.64 s A1 4(d) 0.080 = ½ × a × 0.642 C1 a = 0.39 m s–2 A1 4(e)(i) E = (∆)V / (∆)d C1 E = 0.90 × 103 / 0.12 = 7.5 × 103 N C–1 A1 4(e)(ii) ma = Eq or F = ma and F = Eq C1 q / m = 0.39 / 7.5 × 103 = 5.2 × 10–5 C kg–1 A1 4(f)(i) no effect B1 4(f)(ii) decreases/smaller B1

This question in 9702/23 May/June 2019

7 (a) State a similarity and a difference between an up quark and an up antiquark. similarity: … difference: … [2] (b) Fig. 7.1 shows an electron in an electric field, in a vacuum, at an instant when the electron is stationary. electric field lines electron Fig. 7.1 (i) On Fig. 7.1, draw an arrow to show the direction of the electric force acting on the stationary electron. [1] (ii) The electric field causes the electron to move from its initial position. Describe and explain the acceleration of the electron due to the field, as the electron moves through the field. … … … … [2] (iii) A stationary α-particle is now placed in the same electric field at the same initial position that was occupied by the electron. Compare the initial electric force acting on the α-particle with the initial electric force that acted on the electron. … … … … [2] [Total: 7]

7 marks

Mark scheme: 7(a) similarity: same/equal mass or same/equal (magnitude of) charge or both fundamental (particles) B1 difference: opposite (sign of) charge or one is matter and the other is antimatter B1 7(b)(i) arrow points to the right B1 7(b)(ii) (electric) field strength increases or (electric) force increases B1 acceleration increases B1 7(b)(iii) force (on α-particle) has twice the magnitude (of force on electron) B1 force (on α-particle) is in opposite direction (to force on electron) B1

This question in 9702/22 Oct/Nov 2020

2 A charged oil drop is in a vacuum between two horizontal metal plates. A uniform electric field is produced between the plates by applying a potential difference of 1340 V across them, as shown in Fig. 2.1. top metal plate + 1340 V oil drop, 1.4 × 10–2 m weight 4.6 × 10–14 N uniform electric field bottom metal plate 0 V Fig. 2.1 The separation of the plates is 1.4 × 10–2 m. The oil drop of weight 4.6 × 10–14 N remains stationary at a point mid-way between the plates. (a) (i) Calculate the magnitude of the electric field strength. electric field strength = … N C–1 [2] (ii) Determine the magnitude and the sign of the charge on the oil drop. magnitude of charge = … C sign of charge … [3] (b) The electric potentials of the plates are instantaneously reversed so that the top plate is at a potential of 0 V and the bottom plate is at a potential of +1340 V. This change causes the oil drop to start moving downwards. (i) Compare the new pattern of the electric field lines between the plates with the original pattern. … … [2] (ii) Determine the magnitude of the resultant force acting on the oil drop. resultant force = … N [1] (iii) Show that the magnitude of the acceleration of the oil drop is 20 m s–2. [2] (iv) Assume that the radius of the oil drop is negligible. Use the information in (b)(iii) to calculate the time taken for the oil drop to move to the bottom metal plate from its initial position mid-way between the plates. time = … s [2] (c) The oil drop in (b) starts to move at time t = 0. The distance of the oil drop from the bottom plate is x. On Fig. 2.2, sketch the variation with time t of distance x for the movement of the drop from its initial position until it hits the surface of the bottom plate. Numerical values of t are not required. 0.7 x / 10–2 m 0 0 t Fig. 2.2 [2] [Total: 14]

14 marks

Mark scheme: 2(a)(i) E = (Δ)V / (Δ)d C1 = 1340 / 1.4 × 10–2 = 9.6 × 104 N C–1 A1 2(a)(ii) F = Eq or q(Δ)V / (Δ)d C1 q = 4.6 × 10–14 / 9.6 × 104 or 4.6 × 10–14 × 1.4 × 10–2 / 1340 = 4.8 × 10–19 C A1 sign of charge: negative B1 2(b)(i) (adjacent field) lines have same separation (for both patterns) B1 (direction of lines changes from) downwards to upwards B1 Question Answer Marks 2(b)(ii) resultant force = 4.6 × 10–14 + (9.6 × 104 × 4.8 × 10–19) = 4.6 × 10–14 + 4.6 × 10–14 = 9.2 × 10–14 N A1 2(b)(iii) (a =) F / m or 2W / m or 2g B1 a = 9.2 × 10–14 / (4.6 × 10–14 / 9.81) = 20 (m s–2) or a = 2 × 9.81 = 20 (m s–2) A1 2(b)(iv) s = ut + ½at2 (1.4 × 10–2 / 2) = ½ × 20 × t2 C1 t = 2.6 × 10–2 s A1 2(c) line from (0, 0.7 × 10–2) to a non-zero point on the t-axis M1 magnitude of gradient of line increases A1

This question in 9702/22 Oct/Nov 2021

4 An α-particle moves in a straight line through a vacuum with a constant speed of 4.1 × 106 m s–1. The α-particle enters a uniform electric field at point A, as shown in Fig. 4.1. uniform electric field α-particle, A B speed 4.1 × 106 m s–1 Fig. 4.1 The α-particle continues to move in the same straight line until it is brought to rest at point B by the electric field. The deceleration of the α-particle by the electric field is 2.7 × 1014 m s–2. (a) State the direction of the electric field. … [1] (b) Calculate the distance AB. distance = … m [2] (c) Calculate the electric field strength. electric field strength = … V m–1 [3] (d) The α-particle is at point A at time t = 0. On Fig. 4.2, sketch the variation with time t of the momentum of the α-particle as it travels from point A to point B. Numerical values are not required. momentum 00 t Fig. 4.2 [1] (e) State the name of the quantity that is represented by the gradient of the graph in (d). … [1] (f) A β– particle now enters the electric field along the same initial path as the α-particle and with the same initial speed of 4.1 × 106 m s–1. (i) Calculate the kinetic energy, in J, of the β– particle at point A. kinetic energy = … J [3] (ii) State and explain the differences between the electric force on the β– particle in the electric field and the electric force on the α-particle in the electric field. … … … … … [3] (iii) The β– particle is produced by the decay of a nucleus. State the name of another lepton that is produced at the same time as the β– particle. … [1] [Total: 15]

15 marks

Mark scheme: 4(a) to the left/from the right/from B to A/opposite (direction) to (α-particle) velocity B1 4(b) v2 = u2 + 2as s = (4.1 × 106)2 / (2 × 2.7 × 1014) C1 = 0.031 m A1 4(c) E = F / Q or E = ma / Q C1 = (4 × 1.66 × 10–27 × 2.7 × 1014) / (2 × 1.60 × 10–19) C1 = 5.6 × 106 V m–1 A1 4(d) straight line with negative gradient that intercepts both the momentum and t axes B1 4(e) force (on α-particle) B1 4(f)(i) E = ½mv2 C1 = ½ × 9.11 × 10–31 × (4.1 × 106)2 C1 = 7.7 × 10–18 J A1 4(f)(ii) particles have opposite charges B1 (so) forces (on charges) are opposite (directions) B1 β– has less/half the charge so less/half the force B1 4(f)(iii) (electron) antineutrino B1

This question in 9702/23 Oct/Nov 2021