Electric force between point charges — Paper 4 · A Level Physics 9702

3 Two charged points A and B are separated by a distance of 6.0 cm, as shown in Fig. 3.1. 6.0 cm A B d Fig. 3.1 The variation with distance d from A of the electric field strength E along the line AB is shown in Fig. 3.2. 20 E / kV m–1 15 10 5 0 0 2 4 6 d /cm position position of A of B Fig. 3.2 An electron is emitted with negligible speed from A and travels along AB. (a) State the relation between electric field strength E and potential V. … … [2] Examiner’s Use (b) The area below the line of the graph of Fig. 3.2 represents the potential difference between A and B. Use Fig. 3.2 to determine the potential difference between A and B. potential difference = …………………………. V [4] (c) Use your answer to (b) to calculate the speed of the electron as it reaches point B. speed = …………………………. m s–1 [2] (d) (i) Use Fig. 3.2 to determine the value of d at which the electron has maximum acceleration. d = …………………… cm [1] (ii) Without any further calculation, describe the variation with distance d of the acceleration of the electron. … … … [2]

11 marks

Mark scheme: 3 (a) field strength = potential gradient M1 correct sign OR directions discussed A1 [2] (b) area is 21.2 cm2 ± 0.4 cm2 C2 (if outside ± 0.4 cm2 but within ± 0.8 cm2, allow 1 mark) 1.0 cm2 represents (1.0 × 10–2 × 2.5 × 103 =) 25 V C1 potential difference = 530 V A1 [4] (c) ½mv2 = qV ½ × 9.1 × 10–31 × v2 = 1.6 × 10–19 × 530 C1 v = 1.37 × 107 ms–1 A1 [2] (d) (i) d = 0 B1 [1] (ii) acceleration decreases then increases B1 some quantitative analysis (e.g. minimum at 4.0 cm) B1 [2] (any suggestion that acceleration becomes zero or that there is a deceleration scores 0/2) GCE A/AS LEVEL – May/June 2007 9702 04 √ √

This question in 9702/41 May/June 2007

4 A small charged metal sphere is situated in an earthed metal box. Fig. 4.1 illustrates the electric field between the sphere and the metal box. A Fig. 4.1 (a) By reference to Fig. 4.1, state and explain (i) whether the sphere is positively or negatively charged, … … … [2] (ii) why it appears as if the charge on the sphere is concentrated at the centre of the sphere. … … [1] (b) On Fig. 4.1, draw an arrow to show the direction of the force on a stationary electron situated at point A. [2] Examiner’s Use (c) The radius r of the sphere is 2.4 cm. The magnitude of the charge q on the sphere is 0.76 nC. (i) Use the expression Q V = 4πε0r to calculate a value for the magnitude of the potential V at the surface of the sphere. V = … V [2] (ii) State the sign of the charge induced on the inside of the metal box. Hence explain whether the actual magnitude of the potential will be greater or smaller than the value calculated in (i). … … … … [3] (d) A lead sphere is placed in a lead box in free space, in a similar arrangement to that shown in Fig. 4.1. Explain why it is not possible for the gravitational field to have a similar shape to that of the electric field. … … … [1]

11 marks

Mark scheme: 4 (a) (i) either lines directed away from sphere or lines go from positive to negative or line shows direction of force on positive charge … M1 so positively charged … A1 [2] (ii) either all lines (appear to) radiate from centre or all lines are normal to surface of sphere … B1 [1] (b) tangent to curve … B1 in correct position and direction … B1 [2] (c) (i) V = (0.76 × 10-9) / (4π × 8.85 × 10-12 × 0.024) … C1 = 285 V … A1 [2] (ii) negative charge is induced on (inside of) box … M1 formula applies to isolated (point) charge OR less work done moving test charge from infinity … A1 so potential is lower … A1 [3] (d) either gravitational field is always attractive or field lines must be directed towards both box and sphere … B1 [1] GCE A/AS LEVEL – October/November 2007 9702 04

This question in 9702/41 Oct/Nov 2007

4 (a) Define electric potential at a point. For Examiner’s … Use … … [2] (b) Two isolated point charges A and B are separated by a distance of 30.0 cm, as shown in Fig. 4.1. 30.0cm A B x Fig. 4.1 The charge at A is + 3.6 × 10–9 C. The variation with distance x from A along AB of the potential V is shown in Fig. 4.2. 600 V / V 400 200 0 0 5 10 15 20 25 30 x / cm –200 –400 –600 Fig. 4.2 (i) State the value of x at which the potential is zero. For Examiner’s x = … cm [1] Use (ii) Use your answer in (i) to determine the charge at B. charge = … C [3] (c) A small test charge is now moved along the line AB in (b) from x = 5.0 cm to x = 27 cm. State and explain the value of x at which the force on the test charge will be maximum. … … … … [3]

9 marks

Mark scheme: 4 (a) work done moving unit positive charge M1 from infinity to the point A1 [2] (b) (i) x = 18 cm A1 [1] (ii) VA + VB = 0 C1 (3.6 × 10–9) / (4πε0 × 18 × 10–2) + q / (4πε0 × 12 × 10–2) = 0 C1 q = –2.4 × 10–9 C A1 [3] (use of VA = VB giving 2.4 × 10 –9 C scores one mark) (c) field strength = (–) gradient of graph B1 force = charge × gradient / field strength or force ∝ gradient B1 force largest at x = 27 cm B1 [3]

This question in 9702/41 May/June 2008

7 The Millikan oil-drop experiment enabled the charge on the electron to be determined. For Examiner’s (a) State a fundamental property of charge that was suggested by this experiment. Use … … [1] (b) Two parallel metal plates P and Q are situated in a vacuum. The plates are horizontal and separated by a distance of 5.4 mm, as illustrated in Fig. 7.1. plate Q 5.4mm plate P Fig. 7.1 The lower plate P is earthed. The potential difference between the plates can be varied. An oil droplet of mass 7.7 × 10–15 kg is observed to remain stationary between the plates when plate Q is at a potential of +850 V. (i) Suggest why plates P and Q must be parallel and horizontal. … … … [2] (ii) Calculate the charge, with its sign, on the oil droplet. charge = … C [3] (c) The procedure in (b) was repeated for three further oil droplets. The magnitude of For the charge on each of the droplets was found to be 3.2 × 10–19 C, 6.4 × 10–19 C and Examiner’s 3.2 × 10–19 C. Use Explain what value these data and your answer in (b)(ii) would suggest for the charge on the electron. … … … [1]

7 marks

Mark scheme: 7 (a) charge is quantised / discrete quantities B1 [1] (b) (i) parallel so that the electric field is uniform / constant B1 horizontal so that either oil drop will not drift sideways or field is vertical or electric force is equal to weight B1 [2] (ii) qE = mg C1 q × 850 / (5.4 × 10–3) = 7.7 × 10–15 × 9.8 C1 q = 4.8 × 10–19 C and is negative A1 [3] (c) charge changes by 1.6 × 10–19 C between droplets / integral multiples M1 so charge on electron is 1.6 × 10–19 C A0 [1]

This question in 9702/41 May/June 2008

5 Two deuterium (21H) nuclei are travelling directly towards one another. When their separation For is large compared with their diameters, they each have speed v as illustrated in Fig. 5.1. Examiner’s Use v v deuterium deuterium nucleus nucleus Fig. 5.1 The diameter of a deuterium nucleus is 1.1 × 10–14 m. (a) Use energy considerations to show that the initial speed v of the deuterium nuclei must be approximately 2.5 × 106 m s–1 in order that they may come into contact. Explain your working. [3] (b) For a fusion reaction to occur, the deuterium nuclei must come into contact. Assuming that deuterium behaves as an ideal gas, deduce a value for the temperature of the deuterium such that the nuclei have an r.m.s. speed equal to the speed calculated in (a). temperature = … K [4] (c) Comment on your answer to (b). … … [1]

8 marks

Mark scheme: 5 (a) change/loss in kinetic energy = change/gain in electric potential energy B1 2 × ½mv2 = q2 / 4πε0r C1 2 × ½ × 2 × 1.67 × 10–27 × v2 = (1.6 × 10–19)2 / (4π × 8.85 × 10–12 × 1.1 × 10–14) M1 v = 2.5 × 106 m s–1 A0 [3] (b) pV = ½Nm<c2> and pV = NkT C1 ½ m<c2> = 3 2 kT (award 1 mark of first two if <c2> not used) C1 ½ × 2 × 1.67 × 10–27 × (2.5 × 106)2 = 3 2 × 1.38 × 10–23 × T C1 T = 5 × 108 K A1 [4] (c) e.g. this is very high temperature temperature found in stars (any sensible comment, 1 mark) (if T < 106 K, should comment that too low for fusion to occur) B1 [1]

This question in 9702/41 Oct/Nov 2008

4 Two point charges A and B each have a charge of + 6.4 × 10–19 C. They are separated in a For vacuum by a distance of 12.0 μm, as shown in Fig. 4.1. Examiner’s Use 12.0 μm A P Q B +6.4 × 10–19 C +6.4 × 10–19 C 3.0 μm 3.0 μm Fig. 4.1 Points P and Q are situated on the line AB. Point P is 3.0 μm from charge A and point Q is 3.0 μm from charge B. (a) Calculate the force of repulsion between the charges A and B. force = … N [3] (b) Explain why, without any calculation, when a small test charge is moved from point P to point Q, the net work done is zero. … … … [2] (c) Calculate the work done by an electron in moving from the midpoint of line AB to point P. work done = … J [4]

9 marks

Mark scheme: 4 (a) force = q1q2 / 4πε0x2 C1 = (6.4 × 10–19)2 / (4π × 8.85 × 10–12 × {12 × 10–6}2) C1 = 2.56 × 10–17 N A1 [3] (b) potential at P is same as potential at Q B1 work done = q∆V M1 ∆V = 0 so zero work done A0 [2] (c) at midpoint, potential is 2 × (6.4 × 10–19) / (4πε0 × 6 × 10–6) C1 at P, potential is (6.4 × 10–19) / (4πε0 × 3 × 10–6) + (6.4 × 10–19) / (4πε0 × 9 × 10–6) C1 change in potential = (6.4 × 10–19) / (4πε0 × 9 × 10–6) energy = 1.6 × 10–19 × (6.4 × 10–19) / (4πε0 × 9 × 10–6) C1 = 1.0 × 10–22 J A1 [4]

This question in 9702/42 May/June 2010

4 Two point charges A and B each have a charge of + 6.4 × 10–19 C. They are separated in a For vacuum by a distance of 12.0 μm, as shown in Fig. 4.1. Examiner’s Use 12.0 μm A P Q B +6.4 × 10–19 C +6.4 × 10–19 C 3.0 μm 3.0 μm Fig. 4.1 Points P and Q are situated on the line AB. Point P is 3.0 μm from charge A and point Q is 3.0 μm from charge B. (a) Calculate the force of repulsion between the charges A and B. force = … N [3] (b) Explain why, without any calculation, when a small test charge is moved from point P to point Q, the net work done is zero. … … … [2] (c) Calculate the work done by an electron in moving from the midpoint of line AB to point P. work done = … J [4]

9 marks

Mark scheme: 4 (a) force = q1q2 / 4πε0x2 C1 = (6.4 × 10–19)2 / (4π × 8.85 × 10–12 × {12 × 10–6}2) C1 = 2.56 × 10–17 N A1 [3] (b) potential at P is same as potential at Q B1 work done = q∆V M1 ∆V = 0 so zero work done A0 [2] (c) at midpoint, potential is 2 × (6.4 × 10–19) / (4πε0 × 6 × 10–6) C1 at P, potential is (6.4 × 10–19) / (4πε0 × 3 × 10–6) + (6.4 × 10–19) / (4πε0 × 9 × 10–6) C1 change in potential = (6.4 × 10–19) / (4πε0 × 9 × 10–6) energy = 1.6 × 10–19 × (6.4 × 10–19) / (4πε0 × 9 × 10–6) C1 = 1.0 × 10–22 J A1 [4]

This question in 9702/43 May/June 2010

4 Two small charged metal spheres A and B are situated in a vacuum. The distance between For the centres of the spheres is 12.0 cm, as shown in Fig. 4.1. Examiner’s Use 12.0 cm sphere A P sphere B x Fig. 4.1 (not to scale) The charge on each sphere may be assumed to be a point charge at the centre of the sphere. Point P is a movable point that lies on the line joining the centres of the spheres and is distance x from the centre of sphere A. The variation with distance x of the electric field strength E at point P is shown in Fig. 4.2. 150 E / 106 N C–1 100 50 0 0 2 4 6 8 10 12 x / cm –50 –100 –150 –200 Fig. 4.2 (a) State the evidence provided by Fig. 4.2 for the statements that For Examiner’s (i) the spheres are conductors, Use … … [1] (ii) the charges on the spheres are either both positive or both negative. … … … [2] (b) (i) State the relation between electric field strength E and potential gradient at a point. … … [1] (ii) Use Fig. 4.2 to state and explain the distance x at which the rate of change of potential with distance is 1. maximum, … … … [2] 2. minimum. … … … [2]

8 marks

Mark scheme: 4 (a) (i) zero field (strength) inside spheres B1 [1] (ii) either field strength is zero or the fields are in opposite directions M1 at a point between the spheres A1 [2] (b) (i) field strength is (–) potential gradient (not V/x) B1 [1] (ii) 1. field strength has maximum value B1 at x = 11.4 cm B1 [2] 2. field strength is zero B1 either at x = 7.9 cm (allow ±0.3 cm) or at 0 to 1.4 cm or 11.4 cm to 12 cm B1 [2]

This question in 9702/41 Oct/Nov 2011

5 Positively charged particles are travelling in a vacuum through three narrow slits S1, S2 and For Examiner’s S3, as shown in Fig. 5.1. Use S1 S2 S3 beam of charged particles direction of electric field Fig. 5.1 Each particle has speed v and charge q. There is a uniform magnetic field of flux density B and a uniform electric field of field strength E in the region between the slits S2 and S3. (a) State the expression for the force F acting on a charged particle due to (i) the magnetic field, … [1] (ii) the electric field. … [1] (b) The electric field acts downwards in the plane of the paper, as shown in Fig. 5.1. State and explain the direction of the magnetic field so that the positively charged particles may pass undeviated through the region between slits S2 and S3. … … … [2]

4 marks

Mark scheme: 5 (a) (i) Bqv(sinθ) or Bqv(cosθ) B1 [1] (ii) qE B1 [1] (b) FB must be opposite in direction to FE B1 so magnetic field into plane of paper B1 [2] GCE AS/A LEVEL – October/November 2011 9702 41

This question in 9702/41 Oct/Nov 2011

4 Two small charged metal spheres A and B are situated in a vacuum. The distance between For the centres of the spheres is 12.0 cm, as shown in Fig. 4.1. Examiner’s Use 12.0 cm sphere A P sphere B x Fig. 4.1 (not to scale) The charge on each sphere may be assumed to be a point charge at the centre of the sphere. Point P is a movable point that lies on the line joining the centres of the spheres and is distance x from the centre of sphere A. The variation with distance x of the electric field strength E at point P is shown in Fig. 4.2. 150 E / 106 N C–1 100 50 0 0 2 4 6 8 10 12 x / cm –50 –100 –150 –200 Fig. 4.2 (a) State the evidence provided by Fig. 4.2 for the statements that For Examiner’s (i) the spheres are conductors, Use … … [1] (ii) the charges on the spheres are either both positive or both negative. … … … [2] (b) (i) State the relation between electric field strength E and potential gradient at a point. … … [1] (ii) Use Fig. 4.2 to state and explain the distance x at which the rate of change of potential with distance is 1. maximum, … … … [2] 2. minimum. … … … [2]

8 marks

Mark scheme: 4 (a) (i) zero field (strength) inside spheres B1 [1] (ii) either field strength is zero or the fields are in opposite directions M1 at a point between the spheres A1 [2] (b) (i) field strength is (–) potential gradient (not V/x) B1 [1] (ii) 1. field strength has maximum value B1 at x = 11.4 cm B1 [2] 2. field strength is zero B1 either at x = 7.9 cm (allow ±0.3 cm) or at 0 to 1.4 cm or 11.4 cm to 12 cm B1 [2]

This question in 9702/42 Oct/Nov 2011

5 Positively charged particles are travelling in a vacuum through three narrow slits S1, S2 and For Examiner’s S3, as shown in Fig. 5.1. Use S1 S2 S3 beam of charged particles direction of electric field Fig. 5.1 Each particle has speed v and charge q. There is a uniform magnetic field of flux density B and a uniform electric field of field strength E in the region between the slits S2 and S3. (a) State the expression for the force F acting on a charged particle due to (i) the magnetic field, … [1] (ii) the electric field. … [1] (b) The electric field acts downwards in the plane of the paper, as shown in Fig. 5.1. State and explain the direction of the magnetic field so that the positively charged particles may pass undeviated through the region between slits S2 and S3. … … … [2]

4 marks

Mark scheme: 5 (a) (i) Bqv(sinθ) or Bqv(cosθ) B1 [1] (ii) qE B1 [1] (b) FB must be opposite in direction to FE B1 so magnetic field into plane of paper B1 [2] GCE AS/A LEVEL – October/November 2011 9702 42

This question in 9702/42 Oct/Nov 2011

4 A charged point mass is situated in a vacuum. A proton travels directly towards the mass, as For illustrated in Fig. 4.1. Examiner’s Use charged proton point mass r Fig. 4.1 When the separation of the mass and the proton is r, the electric potential energy of the system is UP . The variation with r of the potential energy UP is shown in Fig. 4.2. r / cm 0 2 4 6 8 10 0 –10 UP / 10–26 J –20 –30 –40 –50 Fig. 4.2 (a) (i) Use Fig. 4.2 to state and explain whether the mass is charged positively or For negatively. Examiner’s Use … … … [2] (ii) The gradient at a point on the graph of Fig. 4.2 is G. Show that the electric field strength E at this point due to the charged point mass is given by the expression Eq = G where q is the charge at this point. … … … [2] (b) Use the expression in (a)(ii) and Fig. 4.2 to determine the electric field strength at a distance of 4.0 cm from the charged point mass. field strength = … V m–1 [4]

8 marks

Mark scheme: 4 (a) (i) as r decreases, energy decreases/work got out (due to) M1 attraction so point mass is negatively charged A1 [2] (ii) electric potential energy = charge × electric potential B1 electric field strength is potential gradient B1 field strength = gradient of potential energy graph/charge A0 [2] (b) tangent drawn at (4.0, 14.5) B1 gradient = 3.6 × 10–24 A2 (for < ±0.3 allow 2 marks, for < ±0.6 allow 1 mark) field strength = (3.6 × 10–24) / (1.6 × 10–19) = 2.3 × 10–5 V m–1 (allow ecf from gradient value) A1 [4] (one point solution for gradient leading to 2.3 × 10–5 Vm–1 scores 1 mark only) GCE AS/A LEVEL – May/June 2012 9702 42

This question in 9702/42 May/June 2012

4 An α-particle and a proton are at rest a distance 20 μm apart in a vacuum, as illustrated in For Fig. 4.1. Examiner’s Use 20 +m _-particle proton P x Fig. 4.1 (a) (i) State Coulomb’s law. … … … [2] (ii) The α-particle and the proton may be considered to be point charges. Calculate the electric force between the α-particle and the proton. force = … N [2] (b) (i) Define electric field strength. … … … [2] (ii) A point P is distance x from the α-particle along the line joining the α-particle to the For Examiner’s proton (see Fig. 4.1). The variation with distance x of the electric field strength Eα due to the α-particle alone is shown in Fig. 4.2. Use 300 E_ 200 electric field strength / V m–1 100 0 0 2 4 6 8 10 12 14 16 x / +m EP –100 –200 –300 Fig. 4.2 The variation with distance x of the electric field strength EP due to the proton alone is also shown in Fig. 4.2. 1. Explain why the two separate electric fields have opposite signs. … … … [2] 2. On Fig. 4.2, sketch the variation with x of the combined electric field due to the α-particle and the proton for values of x from 4 μm to 16 μm. [3]

11 marks

Mark scheme: 4 (a) (i) force proportional to product of (two) charges and inversely proportional to square of separation M1 reference to point charges A1 [2] (ii) F = 2 × (1.6 × 10–19)2 / {4π × 8.85 × 10–12 × (20 × 10–6)2} C1 F = 1.15 × 10–18 N A1 [2] (b) (i) force per unit charge M1 on either a stationary charge or a positive charge A1 [2] (ii) 1. electric field is a vector quantity electric fields are in opposite directions charges repel Any two of the above, 1 each B2 [2] 2. graph: line always between given lines M1 crosses x-axis between 11.0 µm and 12.3 µm A1 reasonable shape for curve A1 [3]

This question in 9702/41 Oct/Nov 2013

4 An α-particle and a proton are at rest a distance 20 μm apart in a vacuum, as illustrated in For Fig. 4.1. Examiner’s Use 20 +m _-particle proton P x Fig. 4.1 (a) (i) State Coulomb’s law. … … … [2] (ii) The α-particle and the proton may be considered to be point charges. Calculate the electric force between the α-particle and the proton. force = … N [2] (b) (i) Define electric field strength. … … … [2] (ii) A point P is distance x from the α-particle along the line joining the α-particle to the For Examiner’s proton (see Fig. 4.1). The variation with distance x of the electric field strength Eα due to the α-particle alone is shown in Fig. 4.2. Use 300 E_ 200 electric field strength / V m–1 100 0 0 2 4 6 8 10 12 14 16 x / +m EP –100 –200 –300 Fig. 4.2 The variation with distance x of the electric field strength EP due to the proton alone is also shown in Fig. 4.2. 1. Explain why the two separate electric fields have opposite signs. … … … [2] 2. On Fig. 4.2, sketch the variation with x of the combined electric field due to the α-particle and the proton for values of x from 4 μm to 16 μm. [3]

11 marks

Mark scheme: 4 (a) (i) force proportional to product of (two) charges and inversely proportional to square of separation M1 reference to point charges A1 [2] (ii) F = 2 × (1.6 × 10–19)2 / {4π × 8.85 × 10–12 × (20 × 10–6)2} C1 F = 1.15 × 10–18 N A1 [2] (b) (i) force per unit charge M1 on either a stationary charge or a positive charge A1 [2] (ii) 1. electric field is a vector quantity electric fields are in opposite directions charges repel Any two of the above, 1 each B2 [2] 2. graph: line always between given lines M1 crosses x-axis between 11.0 µm and 12.3 µm A1 reasonable shape for curve A1 [3]

This question in 9702/42 Oct/Nov 2013

8 (a) State what is meant by quantisation of charge. … … [1] (b) A student carries out an experiment to determine the elementary charge. A charged oil drop is positioned between two horizontal metal plates, as shown in Fig. 8.1. + 680 V oil drop 7.0 mm Fig. 8.1 The plates are separated by a distance of 7.0 mm. The lower plate is earthed. The potential of the upper plate is gradually increased until the drop is held stationary. The potential for the drop to be stationary is 680 V. The weight of the oil drop, allowing for the upthrust of the air, is 4.8 × 10−14 N. Calculate the value for the charge on the oil drop. charge = … C [2] (c) The student repeats the experiment and determines the following values for the charge on oil drops. 3.3 × 10−19 C 4.9 × 10−19 C 9.7 × 10−19 C 3.4 × 10−19 C Use these values to suggest a value for the elementary charge. Explain your working. elementary charge = … C [2]

5 marks

Mark scheme: 8 (a) discrete and equal amounts (of charge) B1 [1] allow: discrete amounts of 1.6 × 10–19C/elementary charge/e integral multiples of 1.6 × 10–19C/elementary charge/e (b) weight = qV / d 4.8 × 10–14 = (q × 680)/(7.0 × 10–3) C1 q = 4.9 × 10–19 C A1 [2] (c) elementary charge = 1.6 × 10–19 C (allow 1.6 × 10–19 C to 1.7 × 10–19 C ) M0 either the values are (approximately) multiples of this or it is a common factor C1 it is the highest common factor A1 [2]

This question in 9702/41 May/June 2014

4 A helium nucleus contains two protons. In a model of the helium nucleus, each proton is considered to be a charged point mass. The separation of these point masses is assumed to be 2.0 × 10−15 m. (a) For the two protons in this model, calculate (i) the electrostatic force, electrostatic force = … N [2] (ii) the gravitational force. gravitational force = … N [2] (b) Using your answers in (a), suggest why (i) there must be some other force between the protons in the nucleus, … … … … [3] (ii) this additional force must have a short range. … … … [2]

9 marks

Mark scheme: 4 (a) (i) FE R Q1Q2 / 4πε0r2 C1 = = = = R 8.99 × 109 × (1.6 × 10–19)2 / (2.0 × 10–15)2 R 58 N A1 [2] (ii) FG R Gm1m2 / r2 C1 = = = R 6.67 × 10–11 × (1.67 × 10–27)2 / (2.0 × 10–15)2 = = = R 4.7 × 10–35 N A1 [2] (b) (i) force of repulsion (much) greater than force of attraction B1 must be some other force of attraction M1 to hold nucleus together A1 [3] (Do not allow if FG > FE in (a) or one of the forces not calculated in (a)) (ii) outside nucleus there is repulsion between protons B1 either attractive force must act only in nucleus or if not short range, all nuclei would stick together B1 [2]

This question in 9702/42 May/June 2014

8 (a) State what is meant by quantisation of charge. … … [1] (b) A student carries out an experiment to determine the elementary charge. A charged oil drop is positioned between two horizontal metal plates, as shown in Fig. 8.1. + 680 V oil drop 7.0 mm Fig. 8.1 The plates are separated by a distance of 7.0 mm. The lower plate is earthed. The potential of the upper plate is gradually increased until the drop is held stationary. The potential for the drop to be stationary is 680 V. The weight of the oil drop, allowing for the upthrust of the air, is 4.8 × 10−14 N. Calculate the value for the charge on the oil drop. charge = … C [2] (c) The student repeats the experiment and determines the following values for the charge on oil drops. 3.3 × 10−19 C 4.9 × 10−19 C 9.7 × 10−19 C 3.4 × 10−19 C Use these values to suggest a value for the elementary charge. Explain your working. elementary charge = … C [2]

5 marks

Mark scheme: 8 (a) discrete and equal amounts (of charge) B1 [1] allow: discrete amounts of 1.6 × 10–19C/elementary charge/e integral multiples of 1.6 × 10–19C/elementary charge/e (b) weight = qV / d 4.8 × 10–14 = (q × 680)/(7.0 × 10–3) C1 q = 4.9 × 10–19 C A1 [2] (c) elementary charge = 1.6 × 10–19 C (allow 1.6 × 10–19 C to 1.7 × 10–19 C ) M0 either the values are (approximately) multiples of this or it is a common factor C1 it is the highest common factor A1 [2]

This question in 9702/43 May/June 2014

5 A charged particle P is situated in a vacuum at a distance x from the centre of a charged conducting sphere of radius r, as illustrated in Fig. 5.1. P r x Fig. 5.1 For the particle P outside the conducting sphere, the charge on the sphere may be assumed to be a point charge at its centre. (a) (i) State Coulomb’s law. … … … [2] (ii) The sphere and the particle P are both charged positively. 1. State the direction of the force acting on particle P. … [1] 2. State the position of particle P for the force to be maximum. … [1] 3. Determine the ratio force on particle P at x = r . force on particle P at x = 4r ratio = … [2] (b) When the charge on the sphere is 6.0 × 10–7 C, the electric field strength at the surface of the sphere is 1.5 × 106 V m–1. Electrical breakdown (a spark) occurs when the electric field strength at the surface of the sphere exceeds 2.0 × 106 V m–1. Determine the additional charge that may be added to the sphere before breakdown occurs. charge = … C [3]

9 marks

Mark scheme: 5 (a) (i) force proportional to the product of the two/point charges B1 and inversely proportional to the square of their separation B1 [2] (ii) 1. force radially away from sphere/to right/to east B1 [1] 2. (maximum) at/on surface of sphere or x = r B1 [1] 3. F ∝ 1 / x2 or F = q1q2 / (4πε 0x2) C1 ratio = 16 A1 [2] (b) E = q / (4πε0x2) or E ∝ q C1 maximum charge = (2.0 / 1.5) × 6.0 × 10–7 C1 = 8.0 × 10–7 C additional charge = 2.0 × 10–7 C A1 [3]

This question in 9702/41 Oct/Nov 2015

5 A charged particle P is situated in a vacuum at a distance x from the centre of a charged conducting sphere of radius r, as illustrated in Fig. 5.1. P r x Fig. 5.1 For the particle P outside the conducting sphere, the charge on the sphere may be assumed to be a point charge at its centre. (a) (i) State Coulomb’s law. … … … [2] (ii) The sphere and the particle P are both charged positively. 1. State the direction of the force acting on particle P. … [1] 2. State the position of particle P for the force to be maximum. … [1] 3. Determine the ratio force on particle P at x = r . force on particle P at x = 4r ratio = … [2] (b) When the charge on the sphere is 6.0 × 10–7 C, the electric field strength at the surface of the sphere is 1.5 × 106 V m–1. Electrical breakdown (a spark) occurs when the electric field strength at the surface of the sphere exceeds 2.0 × 106 V m–1. Determine the additional charge that may be added to the sphere before breakdown occurs. charge = … C [3]

9 marks

Mark scheme: 5 (a) (i) force proportional to the product of the two/point charges B1 and inversely proportional to the square of their separation B1 [2] (ii) 1. force radially away from sphere/to right/to east B1 [1] 2. (maximum) at/on surface of sphere or x = r B1 [1] 3. F ∝ 1 / x2 or F = q1q2 / (4πε 0x2) C1 ratio = 16 A1 [2] (b) E = q / (4πε0x2) or E ∝ q C1 maximum charge = (2.0 / 1.5) × 6.0 × 10–7 C1 = 8.0 × 10–7 C additional charge = 2.0 × 10–7 C A1 [3]

This question in 9702/42 Oct/Nov 2015

6 (a) By reference to electric field lines, explain why, for points outside an isolated spherical conductor, the charge on the sphere may be considered to act as a point charge at its centre. … … … … [2] (b) Two isolated protons are separated in a vacuum by a distance x. (i) Calculate the ratio electric force between the two protons . gravitational force between the two protons ratio = … [3] (ii) By reference to your answer in (i), suggest why gravitational forces are not considered when calculating the force between charged particles. … … [1] [Total: 6]

6 marks

Mark scheme: 6 (a) lines perpendicular to surface or lines are radial M1 lines appear to come from centre A1 [2] (b) (i) FE = (1.6 × 10–19)2 / 4πε0x2 C1 FG = G × (1.67 × 10–27)2 / x2 C1 FE / FG = (1.6 × 10–19)2 × (8.99 × 109) / [(1.67×10–27)2 × (6.67×10–11)] = 1.2 (1.24) × 1036 A1 [3] (ii) FE ≫ FG B1 [1]

This question in 9702/42 May/June 2016

5 An α-particle is travelling in a vacuum towards the centre of a gold nucleus, as illustrated in Fig. 5.1. gold nucleus α-particle × 10–13 J charge 79e energy 7.7 Fig. 5.1 The gold nucleus has charge 79e. The gold nucleus and the α-particle may be assumed to behave as point charges. At a large distance from the gold nucleus, the α-particle has energy 7.7 × 10–13 J. (a) The α-particle does not collide with the gold nucleus. Show that the radius of the gold nucleus must be less than 4.7 × 10–14 m. [3] (b) Determine the acceleration of the α-particle for a separation of 4.7 × 10–14 m between the centres of the gold nucleus and of the α-particle. acceleration = … m s–2 [3] (c) In an α-particle scattering experiment, the beam of α-particles is incident on a very thin gold foil. Suggest why the gold foil must be very thin. … … [1] [Total: 7]

7 marks

Mark scheme: 5(a) or 7.7 × 10–13 = Qq / 4πε0r C1 7.7 × 10–13 = 8.99 × 109 × 79 × 2 × (1.60 × 10–19)2/ r M1 r = 4.7 × 10–14 m r is closest distance of approach so radius less than this A1 5(b) force = Qq / 4πε0r 2 = 4u × a C1 8.99 × 109 × 79 × 2 × (1.60 × 10–19)2/ (4.7 × 10–14)2 = 4 × 1.66 × 10–27 × a C1 a = 2.5 × 1027 m s–2 A1 5(c) so that single interactions between nucleus and α-particle can be studied or so that multiple deflections with nucleus do not occur B1

This question in 9702/41 May/June 2017

6 (a) State Coulomb’s law. … … … [2] (b) Two charged metal spheres A and B are situated in a vacuum, as illustrated in Fig. 6.1. 6.0 cm sphere A sphere B P x Fig. 6.1 The shortest distance between the surfaces of the spheres is 6.0 cm. A movable point P lies along the line joining the centres of the two spheres, a distance x from the surface of sphere A. The variation with distance x of the electric field strength E at point P is shown in Fig. 6.2. 10 E / 103 V m–1 5 0 0 1 2 3 4 5 6 x / cm –5 –10 –15 Fig. 6.2 (i) Use Fig. 6.2 to explain whether the two spheres have charges of the same, or opposite, sign. … … … … [2] (ii) A proton is at point P where x = 5.0 cm. Use data from Fig. 6.2 to determine the acceleration of the proton. acceleration = … m s–2 [3] (c) Use data from Fig. 6.2 to state the value of x at which the rate of change of electric potential is maximum. Give the reason for the value you have chosen. … … … [2] [Total: 9]

9 marks

Mark scheme: 6(a) force proportional to product of charges and inversely proportional to the square of the separation M1 reference to point charges A1 6(b)(i) (near to each sphere,) fields are in opposite directions or point (between spheres) where fields are equal and opposite or point (between spheres) where field strength is zero M1 so same (sign of charge) A1 6(b)(ii) (at x = 5.0 cm,) E = 3.0 × 103 V m–1 and a = qE / m C1 E = (1.60 × 10–19 × 3.0 × 103) / (1.67 × 10–27) C1 = 2.9 × 1011 m s–2 A1 6(c) field strength or E is potential gradient or field strength is rate of change of (electric) potential M1 (field strength) maximum at x = 6 cm A1

This question in 9702/42 May/June 2017

5 An α-particle is travelling in a vacuum towards the centre of a gold nucleus, as illustrated in Fig. 5.1. gold nucleus α-particle × 10–13 J charge 79e energy 7.7 Fig. 5.1 The gold nucleus has charge 79e. The gold nucleus and the α-particle may be assumed to behave as point charges. At a large distance from the gold nucleus, the α-particle has energy 7.7 × 10–13 J. (a) The α-particle does not collide with the gold nucleus. Show that the radius of the gold nucleus must be less than 4.7 × 10–14 m. [3] (b) Determine the acceleration of the α-particle for a separation of 4.7 × 10–14 m between the centres of the gold nucleus and of the α-particle. acceleration = … m s–2 [3] (c) In an α-particle scattering experiment, the beam of α-particles is incident on a very thin gold foil. Suggest why the gold foil must be very thin. … … [1] [Total: 7]

7 marks

Mark scheme: 5(a) or 7.7 × 10–13 = Qq / 4πε0r C1 7.7 × 10–13 = 8.99 × 109 × 79 × 2 × (1.60 × 10–19)2/ r M1 r = 4.7 × 10–14 m r is closest distance of approach so radius less than this A1 5(b) force = Qq / 4πε0r 2 = 4u × a C1 8.99 × 109 × 79 × 2 × (1.60 × 10–19)2/ (4.7 × 10–14)2 = 4 × 1.66 × 10–27 × a C1 a = 2.5 × 1027 m s–2 A1 5(c) so that single interactions between nucleus and α-particle can be studied or so that multiple deflections with nucleus do not occur B1

This question in 9702/43 May/June 2017

5 (a) (i) State Coulomb’s law for the force between two point charges. … … [1] (ii) Two point charges are situated in a vacuum and separated by a distance R. The force between the charges is FC. On Fig. 5.1, sketch a graph to show the variation of the force F between the charges with separation x for values of x from x = R to x = 4R. 1.0 Fc 0.8 Fc F 0.6 Fc 0.4 Fc 0.2 Fc 0 R 2R 3R 4R x [3] Fig. 5.1 (b) Two coils C and D are placed close to one another, as shown in Fig. 5.2. coil C coil D I e.m.f. E V Fig. 5.2 The variation with time t of the current I in coil C is shown in Fig. 5.3. On Fig. 5.4, show the variation with time t of the e.m.f. E induced in coil D for time t = 0 to time t = t5. I 0 0 t1 t2 t3 t4 t5 t Fig. 5.3 E 0 0 t1 t2 t3 t4 t5 t Fig. 5.4 [4] [Total: 8]

8 marks

Mark scheme: 5(a)(i) force proportional to product of charges and inversely proportional to square of separation A1 5(a)(ii) curve starting at (R, FC) B1 passing through (2R, 0.25FC) B1 passing through (4R, 0.06FC) B1 5(b) graph: E = 0 when current constant (0 to t1, t2 to t3, t4 to t5) B1 stepped from t1 to t2 and t3 to t4 B1 (steps) in opposite directions B1 later one larger in magnitude B1

This question in 9702/41 Oct/Nov 2017

5 (a) (i) State Coulomb’s law for the force between two point charges. … … [1] (ii) Two point charges are situated in a vacuum and separated by a distance R. The force between the charges is FC. On Fig. 5.1, sketch a graph to show the variation of the force F between the charges with separation x for values of x from x = R to x = 4R. 1.0 Fc 0.8 Fc F 0.6 Fc 0.4 Fc 0.2 Fc 0 R 2R 3R 4R x [3] Fig. 5.1 (b) Two coils C and D are placed close to one another, as shown in Fig. 5.2. coil C coil D I e.m.f. E V Fig. 5.2 The variation with time t of the current I in coil C is shown in Fig. 5.3. On Fig. 5.4, show the variation with time t of the e.m.f. E induced in coil D for time t = 0 to time t = t5. I 0 0 t1 t2 t3 t4 t5 t Fig. 5.3 E 0 0 t1 t2 t3 t4 t5 t Fig. 5.4 [4] [Total: 8]

8 marks

Mark scheme: 5(a)(i) force proportional to product of charges and inversely proportional to square of separation A1 5(a)(ii) curve starting at (R, FC) B1 passing through (2R, 0.25FC) B1 passing through (4R, 0.06FC) B1 5(b) graph: E = 0 when current constant (0 to t1, t2 to t3, t4 to t5) B1 stepped from t1 to t2 and t3 to t4 B1 (steps) in opposite directions B1 later one larger in magnitude B1

This question in 9702/43 Oct/Nov 2017

9 (a) Define what is meant by electric potential at a point. … … … [2] (b) In an α-particle scattering experiment, α-particles are directed towards a thin film of gold, as illustrated in Fig. 9.1. gold film beam of α-particles Fig. 9.1 The apparatus is in a vacuum. The gold-197 (19779 Au) nuclei in the film may be considered to be fixed point charges. The α-particles emitted from the source each have an energy of 4.8 MeV. Calculate: (i) the initial kinetic energy EK, in J, of an α-particle emitted from the source EK = … J [1] (ii) the distance d of closest approach of an α-particle to a gold nucleus. d = … m [4] (c) Use your answer in (b)(ii) to comment on the possible diameter of a gold nucleus. … … [1] [Total: 8]

8 marks

Mark scheme: 9(a) work done per unit charge B1 (work done) moving positive charge from infinity B1 9(b)(i) energy = 4.8 × 1.60 × 10–13 = 7.7 × 10–13 J A1 9(b)(ii) EP = Qq / 4πε0d C1 Q = 79e and q = 2e C1 7.68 × 10–13 = (79 × 2 × {1.60 × 10–19}2 / (4π × 8.85 × 10–12 × d) C1 d = 4.7 × 10–14 m A1 9(c) (diameter must be) less than/equal to 10–13 or 10–14 m B1

This question in 9702/42 Oct/Nov 2019

2 (a) State Coulomb’s law. … … … [2] (b) Positronium is a system in which an electron and a positron orbit, with the same period, around their common centre of mass, as shown in Fig. 2.1. centre of mass r electron positron Fig. 2.1 (not to scale) The radius r of the orbit of both particles is 1.59 × 10–10 m. (i) Explain how the electric force between the electron and the positron causes the path of the moving particles to be circular. … … … [2] (ii) Show that the magnitude of the electric force between the electron and the positron is 2.28 × 10–9 N. [2] (iii) Use the information in (b)(ii) to determine the period of the circular orbit of the two particles. period = … s [3] (c) Positronium is highly unstable, and after a very short period of time it becomes gamma radiation. (i) Describe how gamma radiation is formed from the two particles in positronium. … … … … [3] (ii) State one medical application of the process described in (c)(i). … [1] [Total: 13]

13 marks

Mark scheme: 2(a) (electric) force is (directly) proportional to product of charges B1 force (between point charges) is inversely proportional to the square of their separation B1 2(b)(i) (electric) force is perpendicular to velocity (of particles) B1 force (perpendicular to velocity) causes centripetal acceleration or force does not change the speed of the particles or force has constant magnitude B1 2(b)(ii) F = e2 / 40x2 C1 = (1.60  10–19)2 / [4  8.85  10–12  (2  1.59  10–10)2] = 2.28  10–9 N A1 2(b)(iii) F = mr2 and  = 2 / T or F = mv2 / r and v = 2r / T C1 F = 42mr / T2 T = √ [42  9.11  10–31  1.59  10–10 / (2.28  10–9)] C1 = 1.58  10–15 s A1 2(c)(i)  electron and positron interact  positron is anti-particle of electron  (pair) annihilation occurs Any two points, 1 mark each B2 mass of the electron and positron converted into photon energy B1 2(c)(ii) PET scanning B1

This question in 9702/42 May/June 2022

5 (a) Define electric potential at a point. … … … [2] (b) An isolated conducting sphere is charged. Fig. 5.1 shows the variation of the potential V due to the sphere with displacement x from its centre. 0 – 0.3 – 0.2 – 0.1 0 0.1 0.2 0.3 x / m – 250 V / V – 500 – 750 – 1000 Fig. 5.1 Use Fig. 5.1 to determine: (i) the radius of the sphere radius = … m [1] (ii) the charge on the sphere. charge = … C [2] (c) Two spheres are identical to the sphere in (b). Each sphere has the same charge as the sphere in (b). The spheres are held in a vacuum so that their centres are separated by a distance of 0.46 m. Assume that the charge on each sphere is a point charge at the centre of the sphere. (i) Calculate the electric potential energy EP of the two spheres. EP = … J [2] (ii) The two spheres are now released simultaneously so that they are free to move. Describe and explain the subsequent motion of the spheres. … … … … [3] [Total: 10]

10 marks

Mark scheme: 5(a) work done per unit charge B1 work done (on charge) in moving positive charge from infinity (to the point) B1 5(b)(i) radius = 0.060 m A1 5(b)(ii) V = Q / 40x C1 Q = (–) 850  4  8.85  10–12  0.060 or Q = (–) 850  0.060 / 8.99  109 (any correct pair of V and x values from curve) Q = – 5.7  10–9 C A1 5(c)(i) EP = Q2 / 40x C1 = (5.67  10–9)2 / (4  8.85  10–12  0.46) = 6.3  10–7 J A1 5(c)(ii) • force is repulsive so spheres move apart B3 • force in direction of motion so speed increases • potential energy converted to kinetic energy so speed increases • force decreases with distance so acceleration decreases • momentum is conserved (at zero) (and masses are equal) so velocities are always equal and opposite Any three points, 1 mark each

This question in 9702/42 Oct/Nov 2022

4 (a) State Coulomb’s law. … … … … [2] (b) A charged sphere X is supported on an insulating stand. A second charged sphere Y is suspended by an insulating thread so that sphere Y is in equilibrium at the position shown in Fig. 4.1. vertical line 1.2 m thread sphere X sphere Y charge +96 nC charge +64 nC 0.080 m stand Fig. 4.1 The charge on sphere X is +96 nC and the charge on sphere Y is +64 nC. Assume that the spheres behave as point charges. The length of the thread is 1.2 m and the centres of sphere X and sphere Y are separated horizontally by a distance of 0.080 m. (i) On Fig. 4.2, draw and label all the forces acting on sphere Y. Fig. 4.2 [1] (ii) Determine the mass of sphere Y. mass = … kg [4] (iii) Calculate the total electric potential energy stored between X and Y. energy = … J [1] (c) An electron enters the region between two parallel plates P and Q, that are separated by a distance of 18 mm, as shown in Fig. 4.3. plate P +250 V path of electron 18 mm plate Q Fig. 4.3 The space between the plates is a vacuum. The potential difference between the plates is 250 V. The electric field may be assumed to be uniform in the region between the plates and zero outside this region. (i) State the direction of the electric force on the electron when between the plates. … [1] (ii) Determine the magnitude of the force acting on the electron due to the electric field. force = … N [2] (iii) Explain why the electron does not follow a circular path. … … [1] [Total: 12]

12 marks

Mark scheme: 4(a) (electric) force is (directly) proportional to product of charges B1 force (between point charges) is inversely proportional to the square of their separation B1 4(b)(i) arrows showing tension upwards in direction of string, electric force horizontally to the right and weight vertically B1 downwards and all three labelled 4(b)(ii) 96  10 −9  64  10 −9 C1 FE = 4   8.85  10 −12  0.080 2 ( = 8.63  10–3 N) either angle to vertical = sin–1 0.080 / 1.2 C1 ( = 3.82°) weight = FE / tan 3.82 = 8.63  10–3 / tan 3.82 C1 ( = 0.129 N) mass = 0.129 / 9.81 A1 = 0.013 kg or T sin = mg and T cos = FE or tan = mg / FE (C1) tan = 1.2 / 0.080 (C1) m = (1.2  8.63  10–3) / (0.080  9.81) (A1) = 0.013 kg 4(b)(iii) QQ1 2 96  10 −9  64  10 −9 A1 E p = = −12 4o r 4   8.85  10  0.080 = 6.9  10–4 J 4(c)(i) towards the top of the page / towards plate P B1 4(c)(ii) F = QE and E = V / d C1 F = 1.6  10–19  250 / 0.018 A1 = 2.2  10–15 N 4(c)(iii) either the force is not (always) perpendicular to the velocity B1 or the force is always in the same direction

This question in 9702/42 Feb/March 2023

5 (a) State Coulomb’s law. … … … [2] (b) Two identical oil droplets are in a vacuum. The centres of the droplets are a distance of 3.8 × 10–6 m apart. The droplets have equal charge and exert an electric force on each other of magnitude 6.3 × 10–17 N. Determine the magnitude of the charge on each droplet. charge = … C [2] (c) One of the oil droplets in (b) is now placed between two horizontal metal plates, as shown in Fig. 5.1. + 1200 V oil droplet metal plates 5.2 cm 0 V Fig. 5.1 (not to scale) A potential difference (p.d.) of 1200 V is applied between the plates, with the top plate at the higher potential. The oil droplet is stationary and in equilibrium. (i) State the sign of the charge on the oil droplet. … [1] (ii) On Fig. 5.1, draw four lines to represent the electric field between the plates. [3] (iii) The distance between the plates is 5.2 cm. Determine the mass of the oil droplet. mass = … kg [3] [Total: 11]

11 marks

Mark scheme: 5(a) (electric) force is (directly) proportional to product of charges B1 (electric) force (between point charges) is inversely proportional to the square of their separation B1 5(b) F = Q2 / 40x2 C1 6.3  10–17 = Q2 / [4  8.85  10–12  (3.8  10–6)2] charge = 3.2  10–19 C A1 5(c)(i) negative B1 5(c)(ii) four straight lines perpendicular to the plates, starting on one plate and finishing on the other B1 lines equally spaced B1 arrows indicating direction downwards B1 5(c)(iii) E = V / d C1 mg = EQ C1 mass = (1200  3.2  10–19) / (9.81  0.052) A1 = 7.5  10–16 kg

This question in 9702/42 Oct/Nov 2023

6 (a) State Coulomb’s law. … … … [2] (b) Fig. 6.1 shows an isolated hollow conducting sphere that is positively charged. + + + + + + + + Fig. 6.1 On Fig. 6.1, draw field lines to represent the electric field outside the sphere. [3] (c) Fig. 6.2 shows the variation of the electric field strength E with distance x from the centre of the sphere in (b). 3 E / 105 N C–1 2 1 0 0 2 4 6 8 x / cm Fig. 6.2 (i) Determine the radius, in cm, of the sphere. radius = … cm [1] (ii) Calculate the charge on the sphere. charge = … C [3] (iii) Suggest an explanation for the fact that the electric field inside the sphere is zero. … … … [1] [Total: 10]

10 marks

Mark scheme: 6(a) (electric) force is (directly) proportional to product of charges B1 force (between point charges) is inversely proportional to the square of their separation B1 6(b) at least four straight, radial lines to/from surface of sphere B1 at least four straight radial lines drawn, approximately equally spaced B1 arrows pointing away from the surface of the sphere B1 6(c)(i) radius = 3.2 cm A1 6(c)(ii) E = Q / (40x2) C1 Q = e.g. 2.2  105  4  8.85  10–12  0.0322 C1 = 2.5  10–8 C A1 6(c)(iii) • the (positive) charge is all the way around the surface B1 • a charge placed inside the sphere is pulled equally in all directions • if the field was not zero, the charges would move (until field is zero) • electric field lines go from positive charge to negative charge, and there are no negative charges inside the sphere Any point, 1 mark

This question in 9702/42 Oct/Nov 2024

2 A helium atom may be modelled as a nucleus surrounded by two electrons in diametrically opposite circular orbits, each of radius 170 pm, as shown in Fig. 2.1. orbit of electrons 170 pm electron electron nucleus Fig. 2.1 (a) State Coulomb’s law. … … … [2] (b) (i) State the charge on the nucleus, in terms of the elementary charge e. charge = … e [1] (ii) Show that the electric force between the nucleus and one of the electrons is 1.6 × 10–8 N. [1] (c) Assume that the force in (b)(ii) is the only force on the electrons. (i) Calculate the speed of the orbiting electrons. speed = … m s–1 [2] (ii) Calculate the period of the orbit of the electrons. period = … s [2] (d) In practice, the orbit of each electron is affected by the presence of the other electron. (i) For the position of one of the electrons, determine the ratio electric field strength due to the other electron . electric field strength due to the nucleus ratio = … [2] (ii) Use your answer in (d)(i) to suggest and explain how the orbit of the electron is affected by the presence of the other electron. … … [1] [Total: 11]

11 marks

Mark scheme: 2(a) (electric) force is (directly) proportional to product of charges B1 force (between point charges) is inversely proportional to the square of their separation B1 2(b)(i) charge = (+)2e A1 2(b)(ii) F = 2 × (1.60 × 10–19)2 / [4 × 8.85 × 10–12 × (170 × 10–12)2] = 1.6 × 10–8 N A1 2(c)(i) F = mv2 / r C1 v = [ (1.6 × 10–8 × 170 × 10–12) / (9.11 × 10–31) ]½ A1 = 1.7 × 106 m s–1 2(c)(ii) F = mr2 and = 2 / T C1 F = 42mr / T2 T = [ (42 × 9.11 × 10–31 × 170 × 10–12) / (1.6 × 10–8) ]½ A1 = 6.2 × 10–16 s or v = 2r / T (C1) T = (2 × 170 × 10–12) / (1.73 × 106) (A1) = 6.2 × 10–16 s 2(d)(i) E  Q / r2 C1 ratio = [1.60 × 10–19 × (170 × 10–12)2] / [3.2 × 10–19 × (340 × 10–12)2] A1 = 0.13 2(d)(ii) resultant force slightly less (than 1.6 × 10–8 N) so speed lower B1 or resultant force slightly less (than 1.6 × 10–8 N) so period greater

This question in 9702/41 May/June 2025

2 A helium atom may be modelled as a nucleus surrounded by two electrons in diametrically opposite circular orbits, each of radius 170 pm, as shown in Fig. 2.1. orbit of electrons 170 pm electron electron nucleus Fig. 2.1 (a) State Coulomb’s law. … … … [2] (b) (i) State the charge on the nucleus, in terms of the elementary charge e. charge = … e [1] (ii) Show that the electric force between the nucleus and one of the electrons is 1.6 × 10–8 N. [1] (c) Assume that the force in (b)(ii) is the only force on the electrons. (i) Calculate the speed of the orbiting electrons. speed = … m s–1 [2] (ii) Calculate the period of the orbit of the electrons. period = … s [2] (d) In practice, the orbit of each electron is affected by the presence of the other electron. (i) For the position of one of the electrons, determine the ratio electric field strength due to the other electron . electric field strength due to the nucleus ratio = … [2] (ii) Use your answer in (d)(i) to suggest and explain how the orbit of the electron is affected by the presence of the other electron. … … [1] [Total: 11]

11 marks

Mark scheme: 2(a) (electric) force is (directly) proportional to product of charges B1 force (between point charges) is inversely proportional to the square of their separation B1 2(b)(i) charge = (+)2e A1 2(b)(ii) F = 2 × (1.60 × 10–19)2 / [4 × 8.85 × 10–12 × (170 × 10–12)2] = 1.6 × 10–8 N A1 2(c)(i) F = mv2 / r C1 v = [ (1.6 × 10–8 × 170 × 10–12) / (9.11 × 10–31) ]½ A1 = 1.7 × 106 m s–1 2(c)(ii) F = mr2 and = 2 / T C1 F = 42mr / T2 T = [ (42 × 9.11 × 10–31 × 170 × 10–12) / (1.6 × 10–8) ]½ A1 = 6.2 × 10–16 s or v = 2r / T (C1) T = (2 × 170 × 10–12) / (1.73 × 106) (A1) = 6.2 × 10–16 s 2(d)(i) E  Q / r2 C1 ratio = [1.60 × 10–19 × (170 × 10–12)2] / [3.2 × 10–19 × (340 × 10–12)2] A1 = 0.13 2(d)(ii) resultant force slightly less (than 1.6 × 10–8 N) so speed lower B1 or resultant force slightly less (than 1.6 × 10–8 N) so period greater

This question in 9702/43 May/June 2025

5 (a) Explain why the electric potential near an isolated proton is positive. … … … … … [3] (b) An isolated metal sphere is positively charged and has radius R, as shown in Fig. 5.1. sphere + + + + R + + P X Y + + Q + + + + x Fig. 5.1 Line XY passes through the centre of the sphere. Point P lies on line XY at a variable displacement x from the centre of the sphere. Point Q is at a fixed position that is not on line XY. The electric field strength at the surface of the sphere is E0. (i) On Fig. 5.1, draw an arrow at point Q to show the direction of the electric field at that point. [1] (ii) On Fig. 5.2, sketch the variation of the electric field E at point P with x for values of x between x = –3R and x = 3R. Do not include the region inside the sphere between x = –R and x = R. E0 E ½E0 0 –3R –2R –R 0 R 2R 3R x –½E0 –E0 Fig. 5.2 [3] (c) The proton and the electron in a hydrogen atom are separated by a distance of 5.3 × 10–11 m. Calculate the electric potential energy of the proton and the electron. electric potential energy = … J [2] [Total: 9]

9 marks

Mark scheme: 5(a) potential is (defined as) zero at infinity B1 proton has a positive charge and so repels another positive charge B1 work is done on two (positive) charges to move them towards each other B1 or work is done by two (positive) charges as they move apart from each other 5(b)(i) arrow drawn through Q in a WSW direction directly away from the centre of the sphere B1 5(b)(ii) curve in at least one quadrant passing through (R, E0) and (2R, ¼E0) B1 curve between –3R and –R of increasing magnitude of gradient B1 and curve between R and 3R of decreasing magnitude of gradient two lines drawn, one in the top right quadrant, the other in the bottom left quadrant B1 5(c) EP = – (1.60  10–19)2 / [4  8.85  10–12  (5.3  10–11)] C1 = –4.3  10–18 J A1

This question in 9702/44 Oct/Nov 2025