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

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