Electric fields

4(a) direction of force B1 force on a positive charge B1 4(b)(i) Q C1 V = 4πε r o 4.0×10 −9 −7.2×10 −9 + = 0 4πε x 4πε (0.120 − x) o o ( ) 4 0.120 − x = 7.2 x 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 © UCLES 2022 Page 9 of 17

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 B1 or force does not change the speed of the particles or force has constant magnitude 2(b)(ii) F = e2 / 4x2 C1 0 = (1.60  10–19)2 / [4  8.85  10–12  (2  1.59  10–10)2] A1 = 2.28  10–9 N 2(b)(iii) F = mr2 and  = 2 / T C1 or F = mv2 / r and v = 2r / T F = 42mr / T2 C1 T = √ [42  9.11  10–31  1.59  10–10 / (2.28  10–9)] = 1.58  10–15 s A1 2(c)(i)  electron and positron interact B2  positron is anti-particle of electron  (pair) annihilation occurs Any two points, 1 mark each mass of the electron and positron converted into photon energy B1 2(c)(ii) PET scanning B1 © UCLES 2022 Page 8 of 16

1(a) force per unit mass B1 1(b)(i) lines drawn are radial from the surface B1 arrows show pointing towards planet B1 1(b)(ii) field lines show force (on satellite) is towards centre of planet B1 or velocity of satellite is perpendicular to field lines (gravitational) force perpendicular to velocity causes centripetal acceleration B1 1(c)(i) T = 24 hours C1 a = r 2 and  = 2 / T C1 or a = v2 / r and v = 2r /T or a = 42r / T 2 a = (42  6.4  106) / (24  60  60)2 A1 = 0.034 m s–2 1(c)(ii) identification of the two forces acting on the object as gravitational force and (normal) contact force M1 gravitational force and normal contact force are in opposite directions, and their resultant causes the (centripetal) A1 acceleration © UCLES 2022 Page 6 of 16

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x C1 0 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) E = Q2 / 4x C1 P 0 = (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 © UCLES 2022 Page 10 of 16

1(a) work done per unit mass B1 work (done on mass) moving mass from infinity (to the point) B1 1(b)(i) –3.55  107 J kg–1 B1 1(b)(ii) GM B1  =− r −3.55 1 07  4 800 000 M = – 6.6710−11 = 2.55  1024 kg 1(b)(iii) GM  C1 g = or g =− r2 r 6.67 1 0−11  2.55 1 024 3.55 1 07 A1 = or = 48000002 4800000 = 7.4 N kg–1 1(b)(iv) r in range 2.60  107 to 2.65  107m C1 mv2 GMm 2r GMm 2 C1 = and v = or mr 2 = and  = r r2 T r2 T 42r 3 42  ( 2.65 1 07)3 C1 T2 = = = 4.20  109 GM 6.67 1 0−11 2.55 1 024 T = 64 800s A1 = 18 hours © UCLES 2023 Page 6 of 18 1(c) similarity – any one point from B1 • inversely proportional to distance (from point) • points of equal potential lie on concentric spheres • zero at infinite distance difference – any one point from B1 • gravitational potential is (always) negative • electric potential can be positive or negative Question Answer Marks

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 F = E 48.8510−120.0802 ( = 8.63  10–3 N) either angle to vertical = sin–1 0.080 / 1.2 C1 ( = 3.82°) weight = F / tan 3.82 = 8.63  10–3 / tan 3.82 C1 E ( = 0.129 N) mass = 0.129 / 9.81 A1 = 0.013 kg or T sin  = mg and T cos  = F or tan  = mg / F (C1) E E tan  = 1.2 / 0.080 (C1) m = (1.2  8.63  10–3) / (0.080  9.81) (A1) = 0.013 kg © UCLES 2023 Page 10 of 18 4(b)(iii) QQ 9610−96410−9 A1 E = 1 2 = p 4r 48.8510−120.080 o = 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 © UCLES 2023 Page 11 of 18

1(a)(i) force per unit mass B1 1(a)(ii) force per unit positive charge B1 1(a)(iii) similarity: B1  inversely proportional to distance (from point)  points of equal potential lie on concentric spheres  zero at infinite distance Any point, 1 mark difference: B1  gravitational potential is (always) negative  electric potential can be positive or negative Any point, 1 mark 1(b)(i) g = GM / r2 M1 E = Q / 4r2 M1 0 algebra showing the elimination of r leading to M / Q = (1 / 4G) (g / E) A1 0 1(b)(ii)  = 1 / (4  6.67  10–11  8.85  10–12) = 1.35  1020 (kg2 C–2) A1 or  = (8.99  109) / (6.67  10–11) = 1.35  1020 (kg2 C–2) 1(c)(i) E = gQ / M C1 = (1.35  1020  9.81  4.80  105) / (5.98  1024) = 106 N C–1 or 106 V m–1 A1 1(c)(ii) same (direction) B1 © UCLES 2023 Page 6 of 16

1(a)(i) force per unit mass B1 1(a)(ii) force per unit positive charge B1 1(a)(iii) similarity: B1  inversely proportional to distance (from point)  points of equal potential lie on concentric spheres  zero at infinite distance Any point, 1 mark difference: B1  gravitational potential is (always) negative  electric potential can be positive or negative Any point, 1 mark 1(b)(i) g = GM / r2 M1 E = Q / 4r2 M1 0 algebra showing the elimination of r leading to M / Q = (1 / 4G) (g / E) A1 0 1(b)(ii)  = 1 / (4  6.67  10–11  8.85  10–12) = 1.35  1020 (kg2 C–2) A1 or  = (8.99  109) / (6.67  10–11) = 1.35  1020 (kg2 C–2) 1(c)(i) E = gQ / M C1 = (1.35  1020  9.81  4.80  105) / (5.98  1024) = 106 N C–1 or 106 V m–1 A1 1(c)(ii) same (direction) B1 © UCLES 2023 Page 6 of 16

1(a)(i) direction of the force acting on a (test) mass placed at the point B1 1(a)(ii) change in height negligible compared with radius (of Earth) B1 (so) field lines are (effectively) parallel B1 1(b)(i) Y = GM / R2 M1 G is the gravitational constant A1 1(b)(ii) gravitational force is (always) attractive B1 or gravitational force (always) acts towards the centre of the sphere force is in opposite direction to displacement B1 or at a point to the right of the centre, force acts to the left or at a point to the left of the centre, force acts to the right 1(b)(iii) sketch: smooth curve with decreasing positive gradient, starting at (R, –Y) and reaching 3R with g still negative B1 or smooth curve with increasing positive gradient, ending at (–R, Y) and reaching –3R with g still positive both of the above curves, in correct quadrants B1 curve passing through (2R, 0.25Y) and (3R, 0.11Y) B1 © UCLES 2023 Page 7 of 16

5(a) work done per unit charge B1 work (done on charge) moving positive charge from infinity (to the point) B1 5(b) Any three points from: B3 Up to 2 points from: • radius of sphere X is 2.0 m • radius of sphere Y is 4.0 m • radius of Y is double the radius of X Up to 2 points from: • charge on X is negative • charge on Y is positive • spheres carry opposite charges Up to 1 point from: • magnitudes of charges on the spheres are equal 5(c) particle is attracted to X or repelled from Y B1 or resultant force on particle is towards X / away from Y / to the left particle accelerates towards X / away from Y / to the left B1 (magnitude of) acceleration of particle increases B1 © UCLES 2023 Page 11 of 16

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x2 C1 0 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 © UCLES 2023 Page 10 of 15

1(a)(i) direction of the force acting on a (test) mass placed at the point B1 1(a)(ii) change in height negligible compared with radius (of Earth) B1 (so) field lines are (effectively) parallel B1 1(b)(i) Y = GM / R2 M1 G is the gravitational constant A1 1(b)(ii) gravitational force is (always) attractive B1 or gravitational force (always) acts towards the centre of the sphere force is in opposite direction to displacement B1 or at a point to the right of the centre, force acts to the left or at a point to the left of the centre, force acts to the right 1(b)(iii) sketch: smooth curve with decreasing positive gradient, starting at (R, –Y) and reaching 3R with g still negative B1 or smooth curve with increasing positive gradient, ending at (–R, Y) and reaching –3R with g still positive both of the above curves, in correct quadrants B1 curve passing through (2R, 0.25Y) and (3R, 0.11Y) B1 © UCLES 2023 Page 7 of 16

5(a) work done per unit charge B1 work (done on charge) moving positive charge from infinity (to the point) B1 5(b) Any three points from: B3 Up to 2 points from: • radius of sphere X is 2.0 m • radius of sphere Y is 4.0 m • radius of Y is double the radius of X Up to 2 points from: • charge on X is negative • charge on Y is positive • spheres carry opposite charges Up to 1 point from: • magnitudes of charges on the spheres are equal 5(c) particle is attracted to X or repelled from Y B1 or resultant force on particle is towards X / away from Y / to the left particle accelerates towards X / away from Y / to the left B1 (magnitude of) acceleration of particle increases B1 © UCLES 2023 Page 11 of 16

5(a) work done per unit charge B1 work (done) moving positive charge from infinity (to the point) B1 5(b) Any three points from: B3 Up to 2 points from:  radius of sphere X is 0.30 m  radius of sphere Y is 0.10 m  radius of X is treble the radius of Y Up to 2 points from:  charge on X is positive  charge on Y is positive  spheres X and Y carry charges of the same sign Up to 1 point from:  (magnitudes of) charges on the spheres are equal  charges on the spheres have the same magnitude 5(c) proton remains at rest (in the position of release) M1 potential energy of proton is (already) at its minimum A1 or (electric) forces (from spheres) on proton are equal and opposite or no resultant (electric) force on proton or resultant electric field strength (at proton) is zero © Cambridge University Press & Assessment 2024 Page 10 of 16

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) V = (–) Q / 4ε x and V = (–) 2Q / 4ε y C1 X 0 Y 0 (V + V = 0 so) Q / 4ε x = 2Q / 4ε y leading to y = 2x A1 X Y 0 0 5(c)(ii) E = Q / 4ε x2 A1 X 0 5(c)(iii) E = 2Q / 4ε (2x)2 C1 Y 0 ( = Q / 8ε x2) 0 (opposite charges so fields in same direction so magnitudes add): A1 E = (Q / 4ε x2) + (Q / 8ε x2) 0 0 = 3Q / 8ε x2 0 © Cambridge University Press & Assessment 2024 Page 10 of 15

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x2) C1 0 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 © Cambridge University Press & Assessment 2024 Page 10 of 14

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) V = (–) Q / 4ε x and V = (–) 2Q / 4ε y C1 X 0 Y 0 (V + V = 0 so) Q / 4ε x = 2Q / 4ε y leading to y = 2x A1 X Y 0 0 5(c)(ii) E = Q / 4ε x2 A1 X 0 5(c)(iii) E = 2Q / 4ε (2x)2 C1 Y 0 ( = Q / 8ε x2) 0 (opposite charges so fields in same direction so magnitudes add): A1 E = (Q / 4ε x2) + (Q / 8ε x2) 0 0 = 3Q / 8ε x2 0 © Cambridge University Press & Assessment 2024 Page 10 of 15

2(a) sketch: B1 line from x = R to x = 4R entirely in the negative  region curve with continuously decreasing magnitude and with gradient of continuously decreasing magnitude, starting at (R, ) B1 line passing through (2R, ½) and (4R, ¼) B1 2(b) horizontal straight line from t = 0 to t = 24 hours B1 line starting at (0, –) B1 © Cambridge University Press & Assessment 2025 Page 7 of 14 2(c) straight line with non-zero gradient from 0 to d B1 line with negative gradient from (0, V) to (d, 0) B1 Question Answer Marks

6(a) plate X marked as negative and plate Y marked as positive B1 6(b)(i) E = V / x C1 = (58 × 103) / 0.041 A1 = 1.4 × 106 N C–1 6(b)(ii) ma = eE C1 a = (1.60 × 10–19 × 1.41 × 106) / (9.11 × 10–31) A1 = 2.5 × 1017 m s–2 6(c)(i) eV = hc /  C1 or eV = hf and f = c /  (1.60 × 10–19 × 58 × 103) = (6.63 × 10–34 × 3.00 × 108) /  M1 clear conversion from m to pm leading to  = 21 pm A1 6(c)(ii) X-rays B1 6(c)(iii) Any two points from: B2 • waves are passed into structure and transmitted waves detected • different parts of the structure absorb different fractions of energy • difference in detected / transmitted intensities used (to form image) © Cambridge University Press & Assessment 2025 Page 14 of 18

5(a) work done per unit charge B1 work (done) moving positive charge from infinity (to the point) B1 5(b)(i) potential (due to proton) = (1.60  10–19) / (4  8.85  10–12  10  10–12) C1 or potential (due to electron) = (–1.60  10–19) / (4  8.85  10–12  110  10–12) V = [(1.60  10–19) / (4  8.85  10–12)]  [(10–1 – 110–1)  1012] = 130 V A1 5(b)(ii) V = [(1.60  10–19) / (4  8.85  10–12)]  [(30–1 – 90–1)  1012] C1 = (+) 32 V A1 5(b)(iii) cross drawn midway between the electron and the proton B1 5(b)(iv) line from (10, +130) to (110, –130) B1 curve getting shallower until x = 60 pm, crossing V = 0 at (60, 0) and then getting steeper after x = 60 pm B1 curve passing through (30, ±32) and (90, ±32) B1 © Cambridge University Press & Assessment 2025 Page 12 of 17

5(a) work done per unit charge B1 work (done) moving positive charge from infinity (to the point) B1 5(b)(i) potential (due to proton) = (1.60  10–19) / (4  8.85  10–12  10  10–12) C1 or potential (due to electron) = (–1.60  10–19) / (4  8.85  10–12  110  10–12) V = [(1.60  10–19) / (4  8.85  10–12)]  [(10–1 – 110–1)  1012] = 130 V A1 5(b)(ii) V = [(1.60  10–19) / (4  8.85  10–12)]  [(30–1 – 90–1)  1012] C1 = (+) 32 V A1 5(b)(iii) cross drawn midway between the electron and the proton B1 5(b)(iv) line from (10, +130) to (110, –130) B1 curve getting shallower until x = 60 pm, crossing V = 0 at (60, 0) and then getting steeper after x = 60 pm B1 curve passing through (30, ±32) and (90, ±32) B1 © Cambridge University Press & Assessment 2025 Page 12 of 17