Electric fields and field lines — 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

9 (a) State what is meant by a field of force. … … … [2] (b) Explain the use of a uniform magnetic field and a uniform electric field for the selection of the velocity of charged particles. You may draw a diagram if you wish. … … … … … … [4] (c) A beam of charged particles enters a region of uniform magnetic and electric fields, as illustrated in Fig. 9.1. region of uniform magnetic and electric fields path of particle mass m charge +q velocity v magnetic field into plane of paper Fig. 9.1 The direction of the magnetic field is into the plane of the paper. The velocity of the charged particles is normal to the magnetic field as the particles enter the field. A particle in the beam has mass m, charge +q and velocity v. The particle passes undeviated through the region of the two fields. On Fig. 9.1, sketch the path of a particle that has (i) mass m, charge +2q and velocity v (label this path Q), [1] (ii) mass m, charge +q and velocity slightly larger than v (label this path V). [2] [Total: 9]

9 marks

Mark scheme: 9(a) region (of space) B1 where an object/particle experiences a force B1 9(b) electric and magnetic fields normal to each other B1 velocity of particle normal to both fields B1 forces (on particle) due to fields are in opposite directions B1 forces are equal for particles with a particular speed/for a selected speed/for speed given by v = E(q) / B(q) B1 9(c)(i) path labelled Q shown undeviated B1 9(c)(ii) reasonable curve in field and no ‘kink’ on entering, labelled V B1 deviated ‘upwards’ 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

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

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) (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

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

4 (a) State what is represented by an electric field line. … … [2] (b) Two point charges P and Q are placed 0.120 m apart as shown in Fig. 4.1. 0.120 m P Q +4.0 nC –7.2 nC Fig. 4.1 (i) The charge of P is +4.0 nC and the charge of Q is –7.2 nC. Determine the distance from P of the point on the line joining the two charges where the electric potential is zero. distance = … m [2] (ii) State and explain, without calculation, whether the electric field strength is zero at the same point at which the electric potential is zero. … … … [1] (iii) An electron is positioned at point X, equidistant from both P and Q, as shown in Fig. 4.2. P Q X Fig. 4.2 On Fig. 4.2, draw an arrow to represent the direction of the resultant force acting on the electron. [1] [Total: 6]

6 marks

Mark scheme: 4(a) direction of force B1 force on a positive charge B1 4(b)(i) o Q V = 4 r πε 9 9 o o 4.0 10 7.2 10 + = 0 4 x 4 (0.120 x) − − × − × πε πε − ( ) 4 0.120 x = 7.2 x − C1 x = 0.043 m A1 4(b)(ii) fields are in the same direction so no B1 4(b)(iii) straight arrow drawn leftwards from X in direction between extended line joining Q and X and the horizontal B1

This question in 9702/42 Feb/March 2022

4 (a) State what is indicated by the direction of an electric field line. … … [2] (b) Fig. 4.1 shows a pair of parallel metal plates with a potential difference (p.d.) of 2400 V between them. + 2400 V metal plates 4.6 cm 0 V Fig. 4.1 The plates are separated by a distance of 4.6 cm. The plates are in a vacuum. (i) On Fig. 4.1, draw five lines to represent the electric field in the region between the plates. [3] (ii) Calculate the strength of the electric field between the plates. electric field strength = … N C–1 [2] (c) A moving proton enters the region between the plates from the left, as shown in Fig. 4.2. + 2400 V region of electric field proton 0 V Fig. 4.2 (i) The proton is deflected by the electric field. On Fig. 4.2, draw a line to show the path of the proton as it moves through and out of the region of the electric field. [2] (ii) A helium nucleus (42He) now enters the region of the electric field along the same initial path as the proton and travelling at the same initial speed. State and explain how the final speed of the helium nucleus compares with the final speed of the proton after leaving the region of the electric field. … … … … … [3] [Total: 12]

12 marks

Mark scheme: 4(a) (field line indicates) direction of force B1 force on a positive charge B1 4(b)(i) one straight line perpendicular to plates, starting on one plate and finishing on the other B1 five straight lines perpendicular to plates between the plates, uniformly spaced B1 downwards arrows on lines B1 4(b)(ii) E = V / d C1 = 2400 / 0.046 A1 = 5.2  104 N C–1 4(c)(i) smooth curve in region of field and straight line outside field B1 direction of deflection shown as downwards in region of field B1 4(c)(ii) helium nucleus has double the charge but four times the mass B1 velocity parallel to plates same and acceleration perpendicular to plates smaller (for helium) B1 final speed is lower (for helium) B1

This question in 9702/41 Oct/Nov 2022

4 (a) State what is indicated by the direction of an electric field line. … … [2] (b) Fig. 4.1 shows a pair of parallel metal plates with a potential difference (p.d.) of 2400 V between them. + 2400 V metal plates 4.6 cm 0 V Fig. 4.1 The plates are separated by a distance of 4.6 cm. The plates are in a vacuum. (i) On Fig. 4.1, draw five lines to represent the electric field in the region between the plates. [3] (ii) Calculate the strength of the electric field between the plates. electric field strength = … N C–1 [2] (c) A moving proton enters the region between the plates from the left, as shown in Fig. 4.2. + 2400 V region of electric field proton 0 V Fig. 4.2 (i) The proton is deflected by the electric field. On Fig. 4.2, draw a line to show the path of the proton as it moves through and out of the region of the electric field. [2] (ii) A helium nucleus (42He) now enters the region of the electric field along the same initial path as the proton and travelling at the same initial speed. State and explain how the final speed of the helium nucleus compares with the final speed of the proton after leaving the region of the electric field. … … … … … [3] [Total: 12]

12 marks

Mark scheme: 4(a) (field line indicates) direction of force B1 force on a positive charge B1 4(b)(i) one straight line perpendicular to plates, starting on one plate and finishing on the other B1 five straight lines perpendicular to plates between the plates, uniformly spaced B1 downwards arrows on lines B1 4(b)(ii) E = V / d C1 = 2400 / 0.046 A1 = 5.2  104 N C–1 4(c)(i) smooth curve in region of field and straight line outside field B1 direction of deflection shown as downwards in region of field B1 4(c)(ii) helium nucleus has double the charge but four times the mass B1 velocity parallel to plates same and acceleration perpendicular to plates smaller (for helium) B1 final speed is lower (for helium) B1

This question in 9702/43 Oct/Nov 2022

1 (a) (i) Define gravitational field. … … [1] (ii) Define electric field. … … [1] (iii) State one similarity and one difference between the gravitational potential due to a point mass and the electric potential due to a point charge. similarity: … … difference: … … [2] (b) An isolated uniform conducting sphere has mass M and charge Q. The gravitational field strength at the surface of the sphere is g. The electric field strength at the surface of the sphere is E. (i) Show that M g = α Q E where α is a constant. [3] (ii) Show that the numerical value of α is 1.35 × 1020 kg2 C–2. [1] (c) Assume that the Earth is a uniform conducting sphere of mass 5.98 × 1024 kg. The surface of the Earth carries a charge of – 4.80 × 105 C that is evenly distributed. (i) Use the information in (b) to determine the electric field strength at the surface of the Earth. Give a unit with your answer. electric field strength = … unit … [2] (ii) State how the direction of the electric field at the surface of the Earth compares with the direction of the gravitational field. … [1] [Total: 11]

11 marks

Mark scheme: 1(a)(i) force per unit mass B1 1(a)(ii) force per unit positive charge B1 1(a)(iii) similarity:  inversely proportional to distance (from point)  points of equal potential lie on concentric spheres  zero at infinite distance Any point, 1 mark B1 difference:  gravitational potential is (always) negative  electric potential can be positive or negative Any point, 1 mark B1 1(b)(i) g = GM / r2 M1 E = Q / 40r2 M1 algebra showing the elimination of r leading to M / Q = (1 / 4G0) (g / E) A1 1(b)(ii)  = 1 / (4  6.67  10–11  8.85  10–12) = 1.35  1020 (kg2 C–2) or  = (8.99  109) / (6.67  10–11) = 1.35  1020 (kg2 C–2) A1 1(c)(i) E = gQ / M = (1.35  1020  9.81  4.80  105) / (5.98  1024) C1 = 106 N C–1 or 106 V m–1 A1 1(c)(ii) same (direction) B1

This question in 9702/41 May/June 2023

1 (a) (i) Define gravitational field. … … [1] (ii) Define electric field. … … [1] (iii) State one similarity and one difference between the gravitational potential due to a point mass and the electric potential due to a point charge. similarity: … … difference: … … [2] (b) An isolated uniform conducting sphere has mass M and charge Q. The gravitational field strength at the surface of the sphere is g. The electric field strength at the surface of the sphere is E. (i) Show that M g = α Q E where α is a constant. [3] (ii) Show that the numerical value of α is 1.35 × 1020 kg2 C–2. [1] (c) Assume that the Earth is a uniform conducting sphere of mass 5.98 × 1024 kg. The surface of the Earth carries a charge of – 4.80 × 105 C that is evenly distributed. (i) Use the information in (b) to determine the electric field strength at the surface of the Earth. Give a unit with your answer. electric field strength = … unit … [2] (ii) State how the direction of the electric field at the surface of the Earth compares with the direction of the gravitational field. … [1] [Total: 11]

11 marks

Mark scheme: 1(a)(i) force per unit mass B1 1(a)(ii) force per unit positive charge B1 1(a)(iii) similarity:  inversely proportional to distance (from point)  points of equal potential lie on concentric spheres  zero at infinite distance Any point, 1 mark B1 difference:  gravitational potential is (always) negative  electric potential can be positive or negative Any point, 1 mark B1 1(b)(i) g = GM / r2 M1 E = Q / 40r2 M1 algebra showing the elimination of r leading to M / Q = (1 / 4G0) (g / E) A1 1(b)(ii)  = 1 / (4  6.67  10–11  8.85  10–12) = 1.35  1020 (kg2 C–2) or  = (8.99  109) / (6.67  10–11) = 1.35  1020 (kg2 C–2) A1 1(c)(i) E = gQ / M = (1.35  1020  9.81  4.80  105) / (5.98  1024) C1 = 106 N C–1 or 106 V m–1 A1 1(c)(ii) same (direction) B1

This question in 9702/43 May/June 2023

5 (a) Define electric field. … … … [2] (b) Fig. 5.1 shows two parallel conducting plates that are in a vacuum. The plates are separated by a distance of 6.7 cm and have a potential difference (p.d.) of 430 V between them. +430 V conducting plate electron, speed 6.7 cm 2.6 × 107 m s–1 conducting plate 0 V Fig. 5.1 (i) On Fig. 5.1, draw four field lines to represent the electric field between the plates. [2] (ii) Determine the strength E of the electric field between the plates. E = … N C–1 [2] (iii) An electron travels at a speed of 2.6 × 107 m s–1 towards the region between the plates, as shown in Fig. 5.1. On Fig. 5.1, draw the path of the electron as it moves between and beyond the plates. [2] (c) A uniform magnetic field is now applied in the region of the electric field in Fig. 5.1, so that the electron in (b)(iii) travels undeviated through the region. (i) Determine the direction of the uniform magnetic field. … [1] (ii) Explain, with reference to the forces exerted by the two fields on the electron, why the path of the electron is undeviated. … … … [2] (iii) Determine the flux density B of the uniform magnetic field. Give a unit with your answer. B = … unit … [2] [Total: 13]

13 marks

Mark scheme: 5(a) force per unit charge B1 force on positive charge B1 5(b)(i) four straight vertical parallel lines, approximately evenly spaced B1 arrows downwards B1 5(b)(ii) E = V / d C1 E = 430 / 0.067 = 6.4  103 N C–1 A1 5(b)(iii) smooth curve within plates and straight lines outside plates B1 direction of deflection shown as upwards B1 5(c)(i) into the page B1 5(c)(ii) forces are in opposite directions B1 (undeviated) when (magnitudes of) forces are equal B1 5(c)(iii) Eq = Bqv C1 B = E / v = (6.4  103) / (2.6  107) = 2.5  10–4 T A1

This question in 9702/41 May/June 2024

5 (a) Define electric field. … … … [2] (b) Fig. 5.1 shows two parallel conducting plates that are in a vacuum. The plates are separated by a distance of 6.7 cm and have a potential difference (p.d.) of 430 V between them. +430 V conducting plate electron, speed 6.7 cm 2.6 × 107 m s–1 conducting plate 0 V Fig. 5.1 (i) On Fig. 5.1, draw four field lines to represent the electric field between the plates. [2] (ii) Determine the strength E of the electric field between the plates. E = … N C–1 [2] (iii) An electron travels at a speed of 2.6 × 107 m s–1 towards the region between the plates, as shown in Fig. 5.1. On Fig. 5.1, draw the path of the electron as it moves between and beyond the plates. [2] (c) A uniform magnetic field is now applied in the region of the electric field in Fig. 5.1, so that the electron in (b)(iii) travels undeviated through the region. (i) Determine the direction of the uniform magnetic field. … [1] (ii) Explain, with reference to the forces exerted by the two fields on the electron, why the path of the electron is undeviated. … … … [2] (iii) Determine the flux density B of the uniform magnetic field. Give a unit with your answer. B = … unit … [2] [Total: 13]

13 marks

Mark scheme: 5(a) force per unit charge B1 force on positive charge B1 5(b)(i) four straight vertical parallel lines, approximately evenly spaced B1 arrows downwards B1 5(b)(ii) E = V / d C1 E = 430 / 0.067 = 6.4  103 N C–1 A1 5(b)(iii) smooth curve within plates and straight lines outside plates B1 direction of deflection shown as upwards B1 5(c)(i) into the page B1 5(c)(ii) forces are in opposite directions B1 (undeviated) when (magnitudes of) forces are equal B1 5(c)(iii) Eq = Bqv C1 B = E / v = (6.4  103) / (2.6  107) = 2.5  10–4 T A1

This question in 9702/43 May/June 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

6 (a) Define electric field at a point. … … [1] (b) An isolated conducting sphere in a vacuum has a capacitance of 69 pF. The charge on the sphere is +83 pC. (i) On Fig. 6.1, draw field lines to represent the electric field outside the sphere due to the charge on the sphere. Fig. 6.1 [2] (ii) Calculate the electric potential at the surface of the sphere. electric potential = … V [2] (iii) Determine the radius of the sphere. radius = … m [2] (iv) Calculate the electric field strength E at the surface of the sphere. Give a unit with your answer. E = … unit … [2] (c) The sphere in (b) is discharged by connecting it to earth (0 V) through a resistor of resistance 120 MΩ. Calculate the time taken for the charge to fall to 26 pC. time = … s [2] [Total: 11]

11 marks

Mark scheme: 6(a) force per unit positive charge B1 6(b)(i) radial lines B1 arrows pointing away from the sphere B1 6(b)(ii) C = Q / V C1 V = 83 / 69 A1 = (+)1.2 V 6(b)(iii) V = Q / 4ε0r C1 r = (83  10–12) / (4  8.85  10–12  1.2) = 0.62 m A1 6(b)(iv) E = Q / 4ε0r2 C1 = (83  10–12) / (4  8.85  10–12  0.622) A1 = 1.9 N C–1 6(c) 26 = 83 exp [– t / (120  106  69  10–12)] C1 t = 9.6  10–3 s A1

This question in 9702/42 Oct/Nov 2025