Electric field of a point charge — Paper 4 · A Level Physics 9702

6 (a) State one similarity and one difference between the electric field lines and the gravitational field lines around an isolated positively charged metal sphere. similarity … … difference … … [2] (b) A positive point charge +Q is positioned at a fixed point X and an identical positive point charge is positioned at a fixed point Y, as shown in Fig. 6.1. X A B Y +Q +Q 2.5 cm 2.5 cm 10.0 cm Fig. 6.1 The charges are separated in a vacuum by a distance of 10.0 cm. Points A and B are on the line XY. Point A is a distance of 2.5 cm from X and point B is a distance of 2.5 cm from Y. The electric field strength at point A is 4.1 × 10–5 V m–1. (i) Calculate charge +Q. +Q = … C [3] (ii) On Fig. 6.2, sketch the variation of the electric field strength E with distance d from A to B, along the line AB. 5 E / 10–5 V m–1 4 3 2 1 0 0 1 2 3 4 5 d / cm –1 –2 –3 –4 –5 Fig. 6.2 [2] (iii) A small positive charge is placed at A. The electric field causes this charge to move from rest along the line AB. Describe the acceleration of the charge as it moves from A to B. … … … … [2] [Total: 9]

9 marks

Mark scheme: 6(a) similarity: lines are radial / greater separation of lines with increased distance from the sphere B1 difference: gravitational lines directed towards sphere and electric lines directed away from sphere B1 6(b)(i) E = Q / 4πε0r 2 or E = kQ / r 2 with k defined / substituted in C1 4.1 × 10–5 = [Q / (4π × 8.85 ×10–12 × 0.0252)] – [Q / (4π × 8.85 × 10–12 × 0.0752)] C1 Q = 3.2 × 10–18 C A1 6(b)(ii) smooth curve with gradient decreasing starting at (0, 4.1 × 10–5) to d-axis at (2.5, 0) B1 smooth curve with gradient increasing from (2.5, 0) ending at (5, – 4.1 × 10–5) B1 6(b)(iii) acceleration decreases (to zero at mid-point) B1 then acceleration increases in the opposite direction / increasing negative acceleration B1

This question in 9702/42 Feb/March 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 9702/41 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

6 (a) For any point outside a spherical conductor, the charge on the sphere may be considered to act as a point charge at its centre. By reference to electric field lines, explain this. … … … … [2] (b) An isolated spherical conductor has charge q, as shown in Fig. 6.1. x sphere, charge q P Fig. 6.1 Point P is a movable point that, at any one time, is a distance x from the centre of the sphere. The variation with distance x of the electric potential V at point P due to the charge on the sphere is shown in Fig. 6.2. 14 12 V / 103 V 10 8 6 4 2 0 0 2 4 6 8 10 12 x / cm Fig. 6.2 Use Fig. 6.2 to determine (i) the electric field strength E at point P where x = 6.0 cm, E = … N C–1 [3] (ii) the radius R of the sphere. Explain your answer. R = … cm [2] [Total: 7]

7 marks

Mark scheme: 6(a) electric field lines are radial/normal to surface (of sphere) B1 electric field lines appear to originate from centre (of sphere) B1 6(b)(i) tangent drawn at x = 6.0 cm and gradient calculation attempted C1 E = 9.0 × 104 N C–1 (1 mark if in range ±1.2; 2 marks if in range ±0.6) A2 or correct pair of values of V and x read from curved part of graph and substituted into V = q / 4πε0x (C1) to give q = 3.6 × 10–8 C (C1) (then E = q / 4πε0x2 and x = 6 cm gives) E = 9.0 × 104 N C–1 (A1) or (E = q / 4πε0x2 and V = q / 4πε0x and so) E = V / x (C1) giving E = 5.4 × 103 / 0.060 (C1) = 9.0 × 104 N C–1 (A1) 6(b)(ii) (R =) 2.5 cm B1 potential inside a conductor is constant or field strength inside a conductor zero (so gradient is zero) B1

This question in 9702/42 Oct/Nov 2017

6 (a) State what is meant by electric field strength. … … [1] (b) An isolated metal sphere A of radius 26 cm is positively charged. Sphere A is shown in Fig. 6.1. charged sphere A 26 cm Fig. 6.1 Electrical breakdown (a spark) occurs when the electric field strength at the surface of the sphere exceeds 2.0 × 104 V m–1. Calculate the maximum charge Q that can be stored on the sphere. Q = … C [2] (c) A second isolated metal sphere B, also with charge +Q, has a radius of 52 cm. Calculate the additional charge, in terms of Q, that may be stored on this sphere before electrical breakdown occurs. additional charge = … [2] [Total: 5]

5 marks

Mark scheme: 6(a) force per unit charge B1 6(b) E = Q / (4πε0r2) C1 2.0 × 104 = Q / (4π × 8.85 × 10–12 × 0.262) charge = 1.5 × 10–7 C A1 6(c) charge (= Q [52 / 26]2) = 4Q C1 additional charge = 3Q A1

This question in 9702/41 May/June 2018

6 (a) State what is meant by electric field strength. … … [1] (b) An isolated metal sphere A of radius 26 cm is positively charged. Sphere A is shown in Fig. 6.1. charged sphere A 26 cm Fig. 6.1 Electrical breakdown (a spark) occurs when the electric field strength at the surface of the sphere exceeds 2.0 × 104 V m–1. Calculate the maximum charge Q that can be stored on the sphere. Q = … C [2] (c) A second isolated metal sphere B, also with charge +Q, has a radius of 52 cm. Calculate the additional charge, in terms of Q, that may be stored on this sphere before electrical breakdown occurs. additional charge = … [2] [Total: 5]

5 marks

Mark scheme: 6(a) force per unit charge B1 6(b) E = Q / (4πε0r2) C1 2.0 × 104 = Q / (4π × 8.85 × 10–12 × 0.262) charge = 1.5 × 10–7 C A1 6(c) charge (= Q [52 / 26]2) = 4Q C1 additional charge = 3Q A1

This question in 9702/43 May/June 2018

5 (a) State what is meant by an electric field. … … [1] (b) An isolated solid metal sphere has radius R. The charge on the sphere is +Q and the electric field strength at its surface is E. On Fig. 5.1, draw a line to show the variation of the electric field strength with distance x from the centre of the solid sphere for values of x from x = 0 to x = 3R. 1.00E 0.75E electric field strength 0.50E 0.25E 0 0 R 2R 3R distance x Fig. 5.1 [4] (c) The sphere in (b) has radius R = 0.26 m. 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 maximum charge that can be stored on the sphere before electrical breakdown occurs. charge = … C [3]

8 marks

Mark scheme: 5(a) region where charge experiences an (electric) force B1 5(b) graph: field strength zero from x = 0 to x = R B1 curve with negative gradient, decreasing from x = R to x = 3R B1 line passes through field strength E at x = R, B1 line passes through field strength 0.25E at x = 2R and field strength 0.11E at x = 3R B1 Question Answer Marks 5(c) field strength = q / 4πϵ0x2 C1 2.0 × 106 = q / (4 × π × 8.85 × 10–12 × 0.262) C1 q = 1.5 × 10–5 C A1

This question in 9702/42 Feb/March 2019

5 (a) State what is meant by electric field strength. … … … [2] (b) Two point charges A and B are situated a distance 15 cm apart in a vacuum, as illustrated in Fig. 5.1. A P B x 15 cm Fig. 5.1 Point P lies on the line joining the charges and is a distance x from charge A. The variation with distance x of the electric field strength E at point P is shown in Fig. 5.2. 10 8 E / 103 N C–1 6 4 2 0 0 2 4 6 8 10 12 14 x / cm –2 –4 –6 Fig. 5.2 (i) By reference to the direction of the electric field, state and explain whether the charges A and B have the same, or opposite, signs. … … … [2] (ii) State why, although charge A is a point charge, the electric field strength between x = 3 cm and x = 7 cm does not obey an inverse-square law. … … [1] (iii) Use Fig. 5.2 to determine the ratio magnitude of charge A . magnitude of charge B ratio = … [3] [Total: 8]

8 marks

Mark scheme: 5(a) force per unit charge B1 (force on) positive charge B1 5(b)(i) field changes direction (between A and B)/field is zero at a point (between A and B) M1 so charges have same sign A1 5(b)(ii) Any one from: • field is (also) influenced by charge B • charge A is not isolated/is not the only charge present • field is due to two/both charges • field is the resultant of two fields B1 5(b)(iii) E = Q / (4πε0x2) C1 at x = 10 cm, EA = EB C1 QA / 102 = QB / 52 QA / QB = 4.0 A1

This question in 9702/41 May/June 2019

6 (a) State what is meant by electric potential at a point. … … … [2] (b) Two parallel metal plates A and B are held a distance d apart in a vacuum, as illustrated in Fig. 6.1. plate B +V0 x P d 0 V plate A Fig. 6.1 Plate A is earthed and plate B is at a potential of +V0. Point P is situated in the centre region between the plates at a distance x from plate B. The potential at point P is V. On Fig. 6.2, show the variation with x of the potential V for values of x from x = 0 to x = d. +V0 potential V 00 d distance x Fig. 6.2 [3] (c) Two isolated solid metal spheres M and N, each of radius R, are situated in a vacuum. Their centres are a distance D apart, as illustrated in Fig. 6.3. D sphere M sphere N charge +Q charge +Q P R R y Fig. 6.3 Each sphere has charge +Q. Point P lies on the line joining the centres of the two spheres, and is a distance y from the centre of sphere M. On Fig. 6.4, show the variation with distance y of the electric potential at point P, for values of y from y = 0 to y = D. + potential 0 0 R (D – R) D y – Fig. 6.4 [4] [Total: 9]

9 marks

Mark scheme: 6(a) work done per unit charge B1 (work done) moving positive charge from infinity B1 6(b) straight line with non-zero gradient from x = 0 to x = d B1 line with gradient of constant sign and end-points between which ∆V = V0 and ∆x = d B1 line passes through (d, 0) and (0, +V0) with negative gradient throughout B1 6(c) V constant (and non-zero) from 0 → R and from (D – R) → D B1 equal (non-zero) values of (magnitude of) V at R and (D – R). B1 curve (with a minimum) from R to (D – R) with V always positive B1 minimum at mid-point of curve B1

This question in 9702/42 May/June 2019

5 (a) State what is meant by electric field strength. … … … [2] (b) Two point charges A and B are situated a distance 15 cm apart in a vacuum, as illustrated in Fig. 5.1. A P B x 15 cm Fig. 5.1 Point P lies on the line joining the charges and is a distance x from charge A. The variation with distance x of the electric field strength E at point P is shown in Fig. 5.2. 10 8 E / 103 N C–1 6 4 2 0 0 2 4 6 8 10 12 14 x / cm –2 –4 –6 Fig. 5.2 (i) By reference to the direction of the electric field, state and explain whether the charges A and B have the same, or opposite, signs. … … … [2] (ii) State why, although charge A is a point charge, the electric field strength between x = 3 cm and x = 7 cm does not obey an inverse-square law. … … [1] (iii) Use Fig. 5.2 to determine the ratio magnitude of charge A . magnitude of charge B ratio = … [3] [Total: 8]

8 marks

Mark scheme: 5(a) force per unit charge B1 (force on) positive charge B1 5(b)(i) field changes direction (between A and B)/field is zero at a point (between A and B) M1 so charges have same sign A1 5(b)(ii) Any one from: • field is (also) influenced by charge B • charge A is not isolated/is not the only charge present • field is due to two/both charges • field is the resultant of two fields B1 5(b)(iii) E = Q / (4πε0x2) C1 at x = 10 cm, EA = EB C1 QA / 102 = QB / 52 QA / QB = 4.0 A1

This question in 9702/43 May/June 2019

6 (a) State an expression for the electric field strength E at a distance r from a point charge Q in a vacuum. State the name of any other symbol used. … … … [2] (b) Two point charges A and B are situated a distance 10.0 cm apart in a vacuum, as illustrated in Fig. 6.1. charge A charge B P x 10.0 cm Fig. 6.1 A point P lies on the line joining the charges A and B. Point P is a distance x from A. The variation with distance x of the electric field strength E at point P is shown in Fig. 6.2. 2.5 E / 10–2 N C–1 2.0 1.5 1.0 0 2 4 6 8 10 x / cm Fig. 6.2 State and explain whether the charges A and B: (i) have the same, or opposite, signs … … … [2] (ii) have the same, or different, magnitudes. … … … [2] (c) An electron is situated at point P. Without calculation, state and explain the variation in the magnitude of the acceleration of the electron as it moves from the position where x = 3 cm to the position where x = 7 cm. … … … … … … [4] [Total: 10]

10 marks

Mark scheme: 6(a) M1 where ε0 is permittivity (of free space) A1 6(b)(i) field does not change direction/field does not become zero M1 so (charges have) opposite (sign) A1 6(b)(ii) minimum is at the midpoint (between the charges) M1 so (magnitudes are the) same A1 6(c) force = field strength × charge and force = mass × acceleration or acceleration is proportional to field strength B1 (from x = 3.0 cm) to x = 5.0 cm: acceleration decreases B1 at x = 5.0 cm: acceleration is a minimum B1 from x = 5.0 cm (to x = 7.0 cm): acceleration increases B1

This question in 9702/41 Oct/Nov 2019

6 (a) State an expression for the electric field strength E at a distance r from a point charge Q in a vacuum. State the name of any other symbol used. … … … [2] (b) Two point charges A and B are situated a distance 10.0 cm apart in a vacuum, as illustrated in Fig. 6.1. charge A charge B P x 10.0 cm Fig. 6.1 A point P lies on the line joining the charges A and B. Point P is a distance x from A. The variation with distance x of the electric field strength E at point P is shown in Fig. 6.2. 2.5 E / 10–2 N C–1 2.0 1.5 1.0 0 2 4 6 8 10 x / cm Fig. 6.2 State and explain whether the charges A and B: (i) have the same, or opposite, signs … … … [2] (ii) have the same, or different, magnitudes. … … … [2] (c) An electron is situated at point P. Without calculation, state and explain the variation in the magnitude of the acceleration of the electron as it moves from the position where x = 3 cm to the position where x = 7 cm. … … … … … … [4] [Total: 10]

10 marks

Mark scheme: 6(a) M1 where ε0 is permittivity (of free space) A1 6(b)(i) field does not change direction/field does not become zero M1 so (charges have) opposite (sign) A1 6(b)(ii) minimum is at the midpoint (between the charges) M1 so (magnitudes are the) same A1 6(c) force = field strength × charge and force = mass × acceleration or acceleration is proportional to field strength B1 (from x = 3.0 cm) to x = 5.0 cm: acceleration decreases B1 at x = 5.0 cm: acceleration is a minimum B1 from x = 5.0 cm (to x = 7.0 cm): acceleration increases B1

This question in 9702/43 Oct/Nov 2019

6 Two positively charged identical metal spheres A and B have their centres separated by a distance of 24 cm, as shown in Fig. 6.1. 24 cm x sphere A sphere B Fig. 6.1 (not to scale) The variation with distance x from the centre of A of the electric field strength E due to the two spheres, along the line joining their centres, is represented in Fig. 6.2. 9 8 E / 104 N C–1 7 6 5 4 3 2 1 0 0 2 4 6 8 10 12 14 16 18 20 22 24 – 1 x / cm – 2 Fig. 6.2 (a) State the radius of the two spheres. radius = … cm [1] (b) The charge on sphere A is 3.6 × 10−9 C. Determine the charge QB on sphere B. Assume that spheres A and B can be treated as point charges at their centres. Explain your working. QB = … C [3] (c) (i) Sphere B is removed. Use information from (b) to determine the electric potential on the surface of sphere A. electric potential = … V [2] (ii) Calculate the capacitance of sphere A. capacitance = … F [2] [Total: 8]

8 marks

Mark scheme: 6(a) 2.0 cm B1 6(b) At 16 (cm) from A the electric fields are equal or EA = EB B1 E = Q / 4πεor2 QA / (4πεorA2) = QB / (4πεorB2) 3.6 × 10-9 / 0.162 = QB / 0.082 C1 QB = 9.0 × 10–10 C A1 6(c)(i) V = Q / 4πεorA V = 3.6 × 10–9 / (4 × π × 8.85 × 10–12 × 0.020) C1 V = 1600 V A1 6(c)(ii) C = Q / V = 3.6 × 10–9 / 1600 C1 = 2.3 × 10–12 F A1

This question in 9702/42 Feb/March 2020

5 (a) State one similarity and one difference between the fields of force produced by an isolated point charge and by an isolated point mass. similarity: … … difference: … … [2] (b) An isolated solid metal sphere A of radius R has charge +Q, as illustrated in Fig. 5.1. R P 2R sphere A charge +Q Fig. 5.1 A point P is distance 2R from the surface of the sphere. Determine an expression that includes the terms R and Q for the electric field strength E at point P. E = … [2] (c) A second identical solid metal sphere B is now placed near sphere A. The centres of the spheres are separated by a distance 6R, as shown in Fig. 5.2. R R P sphere A sphere B charge +Q charge –Q 6R Fig. 5.2 Point P lies midway between spheres A and B. Sphere B has charge –Q. Explain why: (i) the magnitude of the electric field strength at P is given by the sum of the magnitudes of the field strengths due to each sphere … … [1] (ii) the electric field strength at point P due to the charged metal spheres is not, in practice, equal to 2E, where E is the electric field strength determined in (b). … … … [2] [Total: 7]

7 marks

Mark scheme: 5(a) similarity: both are radial or both have inverse square (variations) B1 difference: direction is always/only towards the mass or direction can be towards or away from charge B1 5(b) field strength = Q / 4πε0x2 C1 E = Q / 36πε0R 2 A1 5(c)(i) fields (due to each sphere) are in same direction B1 5(c)(ii) charges on spheres attract/affect each other or charge distribution on each sphere distorted by the other sphere or charges on the surface of the spheres move B1 spheres are not point charges (at their centres) B1

This question in 9702/41 May/June 2020

7 A metal sphere of radius R is isolated in space. Point P is a distance x from the centre of the sphere, as illustrated in Fig. 7.1. R P x Fig. 7.1 The variation with distance x of the electric field strength E due to the charge on the sphere is shown in Fig. 7.2. 20 15 E / 105 V m–1 10 5 0 0 2 4 6 8 10 12 x / cm Fig. 7.2 (a) State what is meant by electric field strength. … … … [2] (b) (i) Use Fig. 7.2 to determine the radius R of the sphere. Explain your working. R = … cm [2] (ii) Use Fig. 7.2 to determine the charge Q on the sphere. Q = … C [3] (c) An α‑particle is situated a distance 8.0 cm from the centre of the sphere. Calculate the acceleration of the α‑particle. acceleration = … m s–2 [3] [Total: 10]

10 marks

Mark scheme: 7(a) force per unit charge M1 (force on) positive charge A1 7(b)(i) no electric field inside a conductor B1 R = 4.5 cm A1 7(b)(ii) E = Q / (4πε0x2) C1 clear correct read-off of a pair of values of E and x C1 e.g. Q = 18 × 105 × 4π × 8.85 × 10–12 × (4.5 × 10–2)2 = 4.0 × 10–7 C or 4.1 × 10–7 C A1 7(c) At 8.0 cm, E = 5.75 × 105 V m–1 C1 F = Eq and a = F / m C1 F = (5.75 × 105 × 2 × 1.6 × 10–19) / (4 × 1.66 × 10–27) = 2.8 × 1013 m s–2 A1

This question in 9702/42 May/June 2020

5 (a) State one similarity and one difference between the fields of force produced by an isolated point charge and by an isolated point mass. similarity: … … difference: … … [2] (b) An isolated solid metal sphere A of radius R has charge +Q, as illustrated in Fig. 5.1. R P 2R sphere A charge +Q Fig. 5.1 A point P is distance 2R from the surface of the sphere. Determine an expression that includes the terms R and Q for the electric field strength E at point P. E = … [2] (c) A second identical solid metal sphere B is now placed near sphere A. The centres of the spheres are separated by a distance 6R, as shown in Fig. 5.2. R R P sphere A sphere B charge +Q charge –Q 6R Fig. 5.2 Point P lies midway between spheres A and B. Sphere B has charge –Q. Explain why: (i) the magnitude of the electric field strength at P is given by the sum of the magnitudes of the field strengths due to each sphere … … [1] (ii) the electric field strength at point P due to the charged metal spheres is not, in practice, equal to 2E, where E is the electric field strength determined in (b). … … … [2] [Total: 7]

7 marks

Mark scheme: 5(a) similarity: both are radial or both have inverse square (variations) B1 difference: direction is always/only towards the mass or direction can be towards or away from charge B1 5(b) field strength = Q / 4πε0x2 C1 E = Q / 36πε0R 2 A1 5(c)(i) fields (due to each sphere) are in same direction B1 5(c)(ii) charges on spheres attract/affect each other or charge distribution on each sphere distorted by the other sphere or charges on the surface of the spheres move B1 spheres are not point charges (at their centres) B1

This question in 9702/43 May/June 2020

5 (a) Define electric potential at a point. … … … [2] (b) Two point charges A and B are separated by a distance of 12.0 cm in a vacuum, as illustrated in Fig. 5.1. x charge A P charge B 12.0 cm Fig. 5.1 The charge of A is +2.0 × 10–9 C. A point P lies on the line joining charges A and B. Its distance from charge A is x. The variation with distance x of the electric potential V at point P is shown in Fig. 5.2. 20 V / 102 V 10 0 0 2 4 6 8 10 12 x / cm –10 –20 –30 –40 Fig. 5.2 Use Fig. 5.2 to determine: (i) the charge of B charge = … C [3] (ii) the change in electric potential when point P moves from the position where x = 9.0 cm to the position where x = 3.0 cm. change = … V [1] (c) An α-particle moves along the line joining point charges A and B in Fig. 5.1. The α-particle moves from the position where x = 9.0 cm and just reaches the position where x = 3.0 cm. Use your answer in (b)(ii) to calculate the speed v of the α-particle at the position where x = 9.0 cm. v = … m s–1 [3] [Total: 9]

9 marks

Mark scheme: 5(a) work done per unit charge B1 (work done on charge) moving positive charge from infinity B1 5(b)(i) (2.0 × 10–9) / 4πε0(4.0 × 10–2) + Q / 4πε0(8.0 × 10–2) = 0 C1 Q = 4.0 × 10–9 C A1 Q given with negative sign B1 5(b)(ii) change = 1200 V A1 5(c) ½mv2 = qV C1 ½ × 4 × 1.66 × 10–27 × v2 = 2 × 1.60 × 10–19 × 1200 C1 v = 3.4 × 105 m s–1 A1

This question in 9702/41 Oct/Nov 2020

5 (a) (i) State what is meant by a field of force. … … … [2] (ii) State one similarity and one difference between the electric field due to a point charge and the gravitational field due to a point mass. similarity: … … … difference: … … … [2] (b) An isolated solid metal sphere of radius 0.15 m is situated in a vacuum, as illustrated in Fig. 5.1. 0.15 m P x Fig. 5.1 The electric field strength at the surface of the sphere is 84 V m–1. Determine: (i) the charge Q on the sphere Q = … C [2] (ii) the electric field strength at point P, a distance x = 0.45 m from the centre of the sphere. electric field strength = … V m–1 [2] (c) Use information from (b) to show, on the axes of Fig. 5.2, the variation of the electric field strength E with distance x from the centre of the sphere for values of x from x = 0 to x = 0.45 m. 100 80 E / V m–1 60 40 20 0 0 0.1 0.2 0.3 0.4 0.5 x / m Fig. 5.2 [3] [Total: 11]

11 marks

Mark scheme: 5(a)(i) region (of space) B1 where a particle experiences a force B1 5(a)(ii) similarity – any one point from: • both have an inverse square variation • both decrease with distance • both are radial B1 difference – any one point from: • gravitational field always towards (the mass) • electric field can be towards or away from (the charge) B1 5(b)(i) E = Q / 4πε0x2 C1 Q = 4π × 8.85 × 10–12 × 84 × 0.152 = 2.1 × 10–10 C A1 5(b)(ii) E = 84 × (0.15 / 0.45)2 or E = (2.1 × 10–10) / (4π × 8.85 × 10–12 × 0.452) C1 E = 9.3 V m–1 A1 5(c) line at E = 0 from x = 0 to x = 0.15 m B1 smooth curve with decreasing negative gradient throughout, from x = 0.15 m to x = 0.45 m, passing through (0.15, 84) B1 line passing through (0.45, 9.3) B1

This question in 9702/42 Oct/Nov 2020

5 (a) Define electric potential at a point. … … … [2] (b) Two point charges A and B are separated by a distance of 12.0 cm in a vacuum, as illustrated in Fig. 5.1. x charge A P charge B 12.0 cm Fig. 5.1 The charge of A is +2.0 × 10–9 C. A point P lies on the line joining charges A and B. Its distance from charge A is x. The variation with distance x of the electric potential V at point P is shown in Fig. 5.2. 20 V / 102 V 10 0 0 2 4 6 8 10 12 x / cm –10 –20 –30 –40 Fig. 5.2 Use Fig. 5.2 to determine: (i) the charge of B charge = … C [3] (ii) the change in electric potential when point P moves from the position where x = 9.0 cm to the position where x = 3.0 cm. change = … V [1] (c) An α-particle moves along the line joining point charges A and B in Fig. 5.1. The α-particle moves from the position where x = 9.0 cm and just reaches the position where x = 3.0 cm. Use your answer in (b)(ii) to calculate the speed v of the α-particle at the position where x = 9.0 cm. v = … m s–1 [3] [Total: 9]

9 marks

Mark scheme: 5(a) work done per unit charge B1 (work done on charge) moving positive charge from infinity B1 5(b)(i) (2.0 × 10–9) / 4πε0(4.0 × 10–2) + Q / 4πε0(8.0 × 10–2) = 0 C1 Q = 4.0 × 10–9 C A1 Q given with negative sign B1 5(b)(ii) change = 1200 V A1 5(c) ½mv2 = qV C1 ½ × 4 × 1.66 × 10–27 × v2 = 2 × 1.60 × 10–19 × 1200 C1 v = 3.4 × 105 m s–1 A1

This question in 9702/43 Oct/Nov 2020

6 (a) State a similarity between the gravitational field lines around a point mass and the electric field lines around a point charge. … … [1] (b) The variation with radius r of the electric field strength E due to an isolated charged sphere in a vacuum is shown in Fig. 6.1. 1.3 1.2 1.1 E / 105 V m–1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 6 r / cm Fig. 6.1 Use data from Fig. 6.1 to: (i) state the radius of the sphere radius = … cm [1] (ii) calculate the charge on the sphere. charge = … C [2] (c) Using the formula for the electric potential due to an isolated point charge, determine the capacitance of the sphere in (b). capacitance = … F [3] [Total: 7]

7 marks

Mark scheme: 6(a) (both have) radial field lines B1 6(b)(i) 2.1 cm B1 6(b)(ii) 2 4 o Q E r πε = e.g. r = 2.1 cm, E = 1.30 × 105 V m–1 2 4 o Q r E πε = 12 2 5 4 8.85 10 0.021 1.30 10 π − = × × × × × × C1 9 6.4 10 C − = × A1 Question Answer Marks 6(c) Q C V = either 4 o Q V r πε = leading to 4 o C r πε = C1 12 4 8.85 10 0.021 C π − = × × × × C1 ( ) 12 2.3 10 C − = × F A1 or 4 o Q V r πε = 9 12 6.4 10 4 8.85 10 0.021 π − − × = × × × × 2740 V = 9 6.4 10 2740 C − × = (C1) 12 2.3 10 F − = × (A1)

This question in 9702/42 Feb/March 2021

6 (a) An isolated metal sphere of radius r is charged so that the electric field strength at its surface is E0. On Fig. 6.1, sketch the variation of the electric field strength E with distance x from the centre of the sphere. Your sketch should extend from x = 0 to x = 3r. E0 field strength E 0 0 r 2r 3r distance x Fig. 6.1 [3] (b) The de Broglie wavelength of a particle is λ 0 when its momentum is p0. On Fig. 6.2, sketch the variation with momentum p of the de Broglie wavelength λ of the p0 particle for values of momentum from to p0. 2 2λ0 wavelength λ λ 0 0 0 p0 p0 2 momentum p Fig. 6.2 [2] (c) A radioactive isotope decays with a half-life of 15 s to form a stable product. A fresh sample of the radioactive isotope at time t = 0 contains N0 nuclei and no nuclei of the stable product. On Fig. 6.3, sketch the variation with t of the number n of nuclei of the stable product for time t = 0 to time t = 45 s. N0 number n 0.5 N0 0 0 15 30 45 time t / s Fig. 6.3 [3] [Total: 8]

8 marks

Mark scheme: 6(a) from x = 0 to x = r: E = 0 B1 from x = r to x = 3r: curve with negative gradient of decreasing magnitude passing through (r, E0) B1 line passing through (2r, E0 / 4) and (3r, E0 / 9) B1 6(b) from p = p0 / 2 to p = p0: curve with negative gradient of decreasing magnitude passing through (p0, λ0) B1 line passing through (½p0, 2λ0) B1 6(c) from t = 0 to t = 45 s: curve with positive gradient of decreasing magnitude starting at (0, 0) B1 line passing through (15, ½N0) B1 line passing through (30, 0.75N0) and (45, 0.88N0) B1

This question in 9702/41 May/June 2021

6 (a) An isolated metal sphere of radius r is charged so that the electric field strength at its surface is E0. On Fig. 6.1, sketch the variation of the electric field strength E with distance x from the centre of the sphere. Your sketch should extend from x = 0 to x = 3r. E0 field strength E 0 0 r 2r 3r distance x Fig. 6.1 [3] (b) The de Broglie wavelength of a particle is λ 0 when its momentum is p0. On Fig. 6.2, sketch the variation with momentum p of the de Broglie wavelength λ of the p0 particle for values of momentum from to p0. 2 2λ0 wavelength λ λ 0 0 0 p0 p0 2 momentum p Fig. 6.2 [2] (c) A radioactive isotope decays with a half-life of 15 s to form a stable product. A fresh sample of the radioactive isotope at time t = 0 contains N0 nuclei and no nuclei of the stable product. On Fig. 6.3, sketch the variation with t of the number n of nuclei of the stable product for time t = 0 to time t = 45 s. N0 number n 0.5 N0 0 0 15 30 45 time t / s Fig. 6.3 [3] [Total: 8]

8 marks

Mark scheme: 6(a) from x = 0 to x = r: E = 0 B1 from x = r to x = 3r: curve with negative gradient of decreasing magnitude passing through (r, E0) B1 line passing through (2r, E0 / 4) and (3r, E0 / 9) B1 6(b) from p = p0 / 2 to p = p0: curve with negative gradient of decreasing magnitude passing through (p0, λ0) B1 line passing through (½p0, 2λ0) B1 6(c) from t = 0 to t = 45 s: curve with positive gradient of decreasing magnitude starting at (0, 0) B1 line passing through (15, ½N0) B1 line passing through (30, 0.75N0) and (45, 0.88N0) B1

This question in 9702/43 May/June 2021

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

5 (a) State the relationship between electric field and electric potential. … … … [2] (b) Two charged isolated insulating spheres X and Y are near to each other, as shown in Fig. 5.1. X Y P Fig. 5.1 P is a point on the line joining the centres of the spheres. Explain why it is not possible for the total electric potential and the resultant electric field to simultaneously be zero at point P. … … … … … [3] (c) The magnitudes of the charges on spheres X and Y in Fig. 5.1 are Q and 2Q respectively. The spheres may be considered as point charges at their centres. Point P is a distance x from the centre of sphere X. The electric potential at point P is zero. (i) Show that the distance y of point P from the centre of sphere Y is equal to 2x. [2] (ii) State an expression, in terms of Q, x and the permittivity of free space ε0, for the electric field strength EX at P due to sphere X. EX = … [1] (iii) Determine an expression, in terms of Q, x and ε0, for the resultant electric field strength E at point P due to the two spheres. E = … [2] [Total: 10]

10 marks

Mark scheme: 5(a) (electric) field equals (electric) potential gradient M1 reference to minus sign A1 5(b) • for potential to be zero, one potential must be positive and the other potential must be negative B3 • for potential to be zero, the charges must have opposite sign • for field to be zero, the fields (due to X and Y) must be in opposite directions • for field to be zero, the charges must have the same sign • the signs of the charges cannot (simultaneously) be both the same and opposite (so not possible) Any three points, 1 mark each 5(c)(i) VX = (–) Q / 4ε0x and VY = (–) 2Q / 4ε0y C1 (VX + VY = 0 so) Q / 4ε0x = 2Q / 4ε0y leading to y = 2x A1 5(c)(ii) EX = Q / 4ε0x2 A1 5(c)(iii) EY = 2Q / 4ε0(2x)2 C1 ( = Q / 8ε0x2) (opposite charges so fields in same direction so magnitudes add): A1 E = (Q / 4ε0x2) + (Q / 8ε0x2) = 3Q / 8ε0x2

This question in 9702/41 Oct/Nov 2024

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

5 (a) State the relationship between electric field and electric potential. … … … [2] (b) Two charged isolated insulating spheres X and Y are near to each other, as shown in Fig. 5.1. X Y P Fig. 5.1 P is a point on the line joining the centres of the spheres. Explain why it is not possible for the total electric potential and the resultant electric field to simultaneously be zero at point P. … … … … … [3] (c) The magnitudes of the charges on spheres X and Y in Fig. 5.1 are Q and 2Q respectively. The spheres may be considered as point charges at their centres. Point P is a distance x from the centre of sphere X. The electric potential at point P is zero. (i) Show that the distance y of point P from the centre of sphere Y is equal to 2x. [2] (ii) State an expression, in terms of Q, x and the permittivity of free space ε0, for the electric field strength EX at P due to sphere X. EX = … [1] (iii) Determine an expression, in terms of Q, x and ε0, for the resultant electric field strength E at point P due to the two spheres. E = … [2] [Total: 10]

10 marks

Mark scheme: 5(a) (electric) field equals (electric) potential gradient M1 reference to minus sign A1 5(b) • for potential to be zero, one potential must be positive and the other potential must be negative B3 • for potential to be zero, the charges must have opposite sign • for field to be zero, the fields (due to X and Y) must be in opposite directions • for field to be zero, the charges must have the same sign • the signs of the charges cannot (simultaneously) be both the same and opposite (so not possible) Any three points, 1 mark each 5(c)(i) VX = (–) Q / 4ε0x and VY = (–) 2Q / 4ε0y C1 (VX + VY = 0 so) Q / 4ε0x = 2Q / 4ε0y leading to y = 2x A1 5(c)(ii) EX = Q / 4ε0x2 A1 5(c)(iii) EY = 2Q / 4ε0(2x)2 C1 ( = Q / 8ε0x2) (opposite charges so fields in same direction so magnitudes add): A1 E = (Q / 4ε0x2) + (Q / 8ε0x2) = 3Q / 8ε0x2

This question in 9702/43 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) Define electric potential at a point. … … … [2] (b) An isolated solid metal sphere of radius r is given a positive charge. The potential at the surface of the sphere is 9.0 × 104 V. At a distance of 3r from the centre of the sphere, the electric field strength is 2.0 × 105 N C–1. (i) Determine the electric field strength at the surface of the sphere. electric field strength = … N C–1 [2] (ii) Show that the radius of the sphere is 5.0 cm. [2] (iii) Calculate the charge on the sphere. charge = … C [2] (iv) Use your answer in (b)(iii) to determine the capacitance of the sphere. capacitance = … F [2] [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) electric field strength inversely proportional to distance2 C1 E = 32  2.0  105 A1 = 1.8  106 N C–1 5(b)(ii) V = Q / 40r and E = Q / 40r2 (so E = V / r) B1 r = (9.0  104) / (1.8  106) = 0.050 m = 5.0 cm A1 5(b)(iii) Q = 40Vr C1 = 4  8.85  10–12  9.0  104  0.050 = 5.0  10–7 C A1 5(b)(iv) C = Q / V C1 = (5.0  10–7) / (9.0  104) A1 = 5.6  10–12 F

This question in 9702/44 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