Electric fields and field lines — Paper 4 · A Level Physics 9702

5 An isolated conducting sphere of radius r is placed in air. It is given a charge +Q. This charge may be assumed to act as a point charge situated at the centre of the sphere. (a) (i) Define electric field strength. … … [1] (ii) State a formula for the electric field strength E at the surface of the sphere. Also, state the meaning of any other symbols used. … … … [2] (b) The maximum field strength at the surface of the sphere before electrical breakdown (sparking) occurs is 2.0 ×106 V m–1. The sphere has a radius r of 0.35 m. Calculate the maximum values of (i) the charge that can be stored on the sphere, charge = ………...……………… C [2] (ii) the potential at the surface of the sphere. potential = ………...……………… V [2] Use (c) Suggest the effect of the electric field on a single atom near the sphere’s surface as electrical breakdown of the air occurs. … … … [2]

9 marks

Mark scheme: 5 (a) (i) force per unit positive charge (ratio idea essential) B1 [1] (ii) E = Q / 4πε0r2 M1 ε0 being the permittivity of free space A1 [2] (b) (i) 2.0 × 106 = Q / (4π × 8.85 × 10-12 × 0.352) C1 Q = 2.7 × 10-5 C A1 [2] (ii) V = (2.7 × 10-5) / (4π × 8.85 × 10-12 × 0.35) C1 = 7.0 × 105 V A1 [2] (c) electrons are stripped off the atoms B1 electrons and positive ions move in opposite directions, (giving rise to a current) B1 [2]

This question in 9702/41 May/June 2006

8 A proton is moving with constant velocity v. It enters a uniform magnetic field that is normal to the initial direction of motion of the proton, as shown in Fig. 8.1. proton velocity v uniform magnetic field into plane of paper Fig. 8.1 A uniform electric field is applied in the same region as the magnetic field so that the proton passes undeviated through the fields. (a) On Fig. 8.1, draw an arrow labelled E to show the direction of the electric field. [1] (b) The proton is replaced by other particles. The electric and magnetic fields remain unchanged. State and explain the deviation, if any, of the following particles in the region of the fields. (i) an α-particle with initial velocity v … … … … [3] (ii) an electron with initial velocity 2v … … … … [3]

7 marks

Mark scheme: 8 (a) arrow labelled E pointing down the page B1 [1] (b) (i) Bqv = qE M1 forces are independent of mass and charge ‘cancels’ M1 so no deviation A1 [3] (ii) magnetic force > electric force M1 so deflects M1 ‘downwards’ A1 [3]

This question in 9702/41 May/June 2006

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

11 marks

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

This question in 9702/41 May/June 2007

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

11 marks

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

This question in 9702/41 Oct/Nov 2007

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

7 marks

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

This question in 9702/41 May/June 2008

7 Negatively-charged particles are moving through a vacuum in a parallel beam. The particles For have speed v. Examiner’s The particles enter a region of uniform magnetic field of flux density 930 μT. Initially, the Use particles are travelling at right-angles to the magnetic field. The path of a single particle is shown in Fig. 7.1. negatively-charged arc of radius 7.9 cm particles, speed v uniform magnetic field, flux density 930 μT Fig. 7.1 The negatively-charged particles follow a curved path of radius 7.9 cm in the magnetic field. A uniform electric field is then applied in the same region as the magnetic field. For an electric field strength of 12 kV m–1, the particles are undeviated as they pass through the region of the fields. (a) On Fig. 7.1, mark with an arrow the direction of the electric field. [1] (b) Calculate, for the negatively-charged particles, (i) the speed v, v = … m s–1 [3] charge (ii) the ratio . mass ratio = … C kg–1 [3]

7 marks

Mark scheme: 7 (a) arrow pointing up the page B1 [1] (b) (i) Eq = Bqv C1 v = (12 × 103) / (930 × 10–6) C1 = 1.3 × 107 m s–1 A1 [3] (ii) Bqv = mv 2 / r C1 q/m = (1.3 × 107) / (7.9 × 10–2 × 930 × 10–6) C1 = 1.8 × 1011 C kg–1 A1 [3]

This question in 9702/41 May/June 2010

4 (a) Define electric potential at a point. For Examiner’s … Use … … [2] (b) Two small spherical charged particles P and Q may be assumed to be point charges located at their centres. The particles are in a vacuum. Particle P is fixed in position. Particle Q is moved along the line joining the two charges, as illustrated in Fig. 4.1. particle P particle Q x Fig. 4.1 The variation with separation x of the electric potential energy EP of particle Q is shown in Fig. 4.2. 0 0 2 4 6 8 10 12 14 16 x / 10–10 m –1 –2 E / eV P –3 –4 Fig. 4.2 (i) State how the magnitude of the electric field strength is related to potential gradient. … … [1] (ii) Use your answer in (i) to show that the force on particle Q is proportional to the For gradient of the curve of Fig. 4.2. Examiner’s Use … … … [2] (c) The magnitude of the charge on each of the particles P and Q is 1.6 × 10–19 C. Calculate the separation of the particles at the point where particle Q has electric potential energy equal to –5.1 eV. separation = … m [4] (d) By reference to Fig. 4.2, state and explain (i) whether the two charges have the same, or opposite, sign, … … … [2] (ii) the effect, if any, on the shape of the graph of doubling the charge on particle P. … … … [2]

13 marks

Mark scheme: 4 (a) work done in bringing unit positive charge M1 from infinity (to that point) A1 [2] (b) (i) field strength is potential gradient B1 [1] (ii) field strength proportional to force (on particle Q) B1 potential gradient proportional to gradient of (potential energy) graph B1 so force is proportional to the gradient of the graph A0 [2] GCE AS/A LEVEL – May/June 2011 9702 41 (c) energy = 5.1 × 1.6 × 10–19 (J) C1 potential energy = Q1Q2 / 4πε0r C1 5.1 × 1.6 × 10–19 = (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × r C1 r = 2.8 × 10–10 m A1 [4] (d) (i) work is got out as x decreases M1 so opposite sign A1 [2] (ii) energy would be doubled B1 gradient would be increased B1 [2]

This question in 9702/41 May/June 2011

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

8 marks

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

This question in 9702/41 Oct/Nov 2011

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

8 marks

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

This question in 9702/42 Oct/Nov 2011

5 (a) Define electric field strength. For Examiner’s … Use … [1] (b) An isolated metal sphere is to be used to store charge at high potential. The charge stored may be assumed to be a point charge at the centre of the sphere. The sphere has a radius of 25 cm. Electrical breakdown (a spark) occurs in the air surrounding the sphere when the electric field strength at the surface of the sphere exceeds 1.8 × 104 V cm–1. (i) Show that the maximum charge that can be stored on the sphere is 12.5 μC. [2] (ii) Calculate the potential of the sphere for this maximum charge. potential = … V [2]

5 marks

Mark scheme: 5 (a) force per unit positive charge acting on a stationary charge B1 [1] (b) (i) E = Q / 4πε0r2 C1 Q = 1.8 × 104 × 102 × 4π × 8.85 × 10–12 × (25 × 10–2)2 M1 Q = 1.25 × 10–5 C = 12.5 µC A0 [2] (ii) V = Q / 4πε0r = (1.25 × 10–5) / (4π × 8.85 × 10–12 × 25 × 10–2) C1 = 4.5 × 105 V A1 [2] (Do not allow use of V = Er unless explained) GCE AS/A LEVEL – May/June 2012 9702 41

This question in 9702/41 May/June 2012

5 (a) Define electric field strength. For Examiner’s … Use … [1] (b) An isolated metal sphere is to be used to store charge at high potential. The charge stored may be assumed to be a point charge at the centre of the sphere. The sphere has a radius of 25 cm. Electrical breakdown (a spark) occurs in the air surrounding the sphere when the electric field strength at the surface of the sphere exceeds 1.8 × 104 V cm–1. (i) Show that the maximum charge that can be stored on the sphere is 12.5 μC. [2] (ii) Calculate the potential of the sphere for this maximum charge. potential = … V [2]

5 marks

Mark scheme: 5 (a) force per unit positive charge acting on a stationary charge B1 [1] (b) (i) E = Q / 4πε0r2 C1 Q = 1.8 × 104 × 102 × 4π × 8.85 × 10–12 × (25 × 10–2)2 M1 Q = 1.25 × 10–5 C = 12.5 µC A0 [2] (ii) V = Q / 4πε0r = (1.25 × 10–5) / (4π × 8.85 × 10–12 × 25 × 10–2) C1 = 4.5 × 105 V A1 [2] (Do not allow use of V = Er unless explained) GCE AS/A LEVEL – May/June 2012 9702 43

This question in 9702/43 May/June 2012

3 (a) State what is meant by a line of force in For Examiner’s (i) a gravitational field, Use … … [1] (ii) an electric field. … … [2] (b) A charged metal sphere is isolated in space. State one similarity and one difference between the gravitational force field and the electric force field around the sphere. similarity: … … difference: … … … [3] (c) Two horizontal metal plates are separated by a distance of 1.8 cm in a vacuum. A potential difference of 270 V is maintained between the plates, as shown in Fig. 3.1. 0 V proton 1.8 cm +270 V Fig. 3.1 A proton is in the space between the plates. Explain quantitatively why, when predicting the motion of the proton between the plates, the gravitational field is not taken into consideration. [4]

10 marks

Mark scheme: 3 (a) (i) (tangent to line gives) direction of force on a (small test) mass B1 [1] (ii) (tangent to line gives) direction of force on a (small test) charge M1 charge is positive A1 [2] (b) similarity: e.g. radial fields lines normal to surface greater separation of lines with increased distance from sphere field strength ∝ 1 / (distance to centre of sphere)2 (allow any sensible answer) B1 difference: e.g. gravitational force (always) towards sphere B1 electric force direction depends on sign of charge on sphere / towards or away from sphere B1 e.g. gravitational field/force is attractive (B1) electric field/force is attractive or repulsive (B1) (allow any sensible comparison) [3] (c) gravitational force = 1.67 × 10–27 × 9.81 = 1.6 × 10–26 N A1 electric force = 1.6 × 10–19 × 270 / (1.8 × 10–2) C1 = 2.4 × 10–15 N A1 electric force very much greater than gravitational force B1 [4]

This question in 9702/43 Oct/Nov 2012

6 A particle has mass m and charge +q and is travelling with speed v through a vacuum. For The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as Examiner’s shown in Fig. 6.1. Use + V metal plate path of particle metal plate Fig. 6.1 The uniform electric field between the plates has magnitude 2.8 × 104 V m–1 and is zero outside the plates. The particle passes between the plates and emerges beyond them, as illustrated in Fig. 6.1. (a) Explain why the path of the particle in the electric field is not an arc of a circle. … … … [1] (b) A uniform magnetic field is now formed in the region between the metal plates. The magnetic field strength is adjusted so that the positively charged particle passes undeviated between the plates, as shown in Fig. 6.2. + V region of uniform electric and magnetic fields path of particle path of particle Fig. 6.2 (i) State and explain the direction of the magnetic field. For Examiner’s … Use … … [2] (ii) The particle has speed 4.7 × 105 m s–1. Calculate the magnitude of the magnetic flux density. Explain your working. magnetic fl ux density = … T [3] (c) The particle in (b) has mass m, charge +q and speed v. Without any further calculation, state the effect, if any, on the path of a particle that has (i) mass m, charge –q and speed v, … [1] (ii) mass m, charge +q and speed 2v, … [1] (iii) mass 2m, charge +q and speed v. … [1]

9 marks

Mark scheme: 6 (a) either constant speed parallel to plate or accelerated motion / force normal to plate / in direction field B1 so not circular A0 [1] (b) (i) direction of force due to magnetic field opposite to that due to electric field B1 magnetic field into plane of page B1 [2] (ii) force due to magnetic field = force due to electric field B1 Bqv = qE B = E / v C1 = (2.8 × 104) / (4.7 × 105) = 6.0 × 10–2 T A1 [3] (c) (i) no change / not deviated B1 [1] (ii) deviated upwards B1 [1] (iii) no change / not deviated B1 [1]

This question in 9702/43 Oct/Nov 2013

5 An isolated solid metal sphere of radius r is given a positive charge. The distance from the centre of the sphere is x. (a) The electric potential at the surface of the sphere is V0. On the axes of Fig. 5.1, sketch a graph to show the variation with distance x of the electric potential due to the charged sphere, for values of x from x = 0 to x = 4r. 1.00 V0 0.75 V0 potential 0.50 V0 0.25 V0 0 0 r 2r 3r 4r distance x Fig. 5.1 [3] (b) The electric field strength at the surface of the sphere is E0. On the axes of Fig. 5.2, sketch a graph to show the variation with distance x of the electric field strength due to the charged sphere, for values of x from x = 0 to x = 4r. 1.00 E0 0.75 E0 field strength 0.50 E0 0.25 E0 0 0 r 2r 3r 4r distance x Fig. 5.2 [3]

6 marks

Mark scheme: 5 (a) graph: straight line at constant potential = V0 from x = 0 to x = r B1 curve with decreasing gradient M1 passing through (2r, 0.50V0) and (4r, 0.25V0) A1 [3] (b) graph: straight line at E = 0 from x = 0 to x = r B1 curve with decreasing gradient from (r, E0) M1 passing through (2r, ¼E0) A1 [3] (for 3rd mark line must be drawn to x = 4r and must not touch x-axis)

This question in 9702/41 May/June 2014

6 (a) Explain the use of a uniform electric field and a uniform magnetic field for the selection of the velocity of a charged particle. You may draw a diagram if you wish. … … … … [3] (b) Ions, all of the same isotope, are travelling in a vacuum with a speed of 9.6 × 104 m s−1. The ions are incident normally on a uniform magnetic field of flux density 640 mT. The ions follow semicircular paths A and B before reaching a detector, as shown in Fig. 6.1. detector A B vacuum uniform magnetic field, flux density 640 mT Fig. 6.1 Data for the diameters of the paths are shown in Fig. 6.2. path diameter / cm A 6.2 B 12.4 Fig. 6.2 The ions in path B each have charge +1.6 × 10−19 C. (i) Determine the mass, in u, of the ions in path B. mass = … u [4] (ii) Suggest and explain quantitatively a reason for the difference in radii of the paths A and B of the ions. … … … … [3]

10 marks

Mark scheme: 6 (a) electric and magnetic fields normal to each other B1 either charged particle enters region normal to both fields or correct B direction w.r.t. E for zero deflection B1 for no deflection, v R E / B B1 [3] (no credit if magnetic field region clearly not overlapping with electric field region) GCE A LEVEL – May/June 2014 9702 42 (b) (i) m R Bqr / v C1 = = = R=(640 × 10–3 × 1.6 × 10–19 × 6.2 × 10–2) / (9.6 × 104) C1 = = = R=6.61 × 10–26 kg C1 = = = R=(6.61 × 10–26) / (1.66 × 10–27) u = = = R=40 u A1 [4] (ii) q / m ∝ 1 / r or m constant and q ∝ 1 / r B1 q / m for A is twice that for B B1 ions in path A have (same mass but) twice the charge (of ions in path B) B1 [3]

This question in 9702/42 May/June 2014

5 (a) Define electric potential at a point. … … … [2] (b) An isolated metal sphere is charged to a potential V . The charge on the sphere is q. The charge on the sphere may be considered to act as a point charge at the centre of the sphere. The variation with potential V of the charge q on the sphere is shown in Fig. 5.1. 6 q / 10–8 C 4 2 0 0 5 10 15 20 25 30 V / kV Fig. 5.1 Use Fig. 5.1 to determine (i) the radius of the sphere, radius = … m [2] (ii) the energy required to increase the potential of the sphere from zero to 24 kV. energy = … J [3] (c) The sphere in (b) discharges by causing sparks when the electric field strength at the surface of the sphere is greater than 2.0 × 106 V m−1. Use your answer in (b)(i) to calculate the maximum potential to which the sphere can be charged. potential = … V [3]

10 marks

Mark scheme: 5 (a) work done / energy in moving unit positive charge M1 from infinity (to the point) A1 [2] (b) (i) V = q / 4πε0r at 16 kV, q = 3.0 × 10–8 C r = (3.0 × 10–8) / (4π × 8.85 × 10–12 × 16 × 103) C1 = 1.69 × 10–2 m (allow 2 s.f.) A1 [2] (allow any answer which rounds to 1.7 × 10–2) (ii) energy is / represented by area ‘below’ line C1 energy = ½qV = ½ × 24 × 103 × 4.5 × 10–8 C1 = 5.4 × 10–4 J A1 [3] (c) V = q / 4πε0r and E = q / 4πε0r 2 giving Er = V B1 2.0 × 106 × 1.7 × 10–2 = V C1 V = 3.4 × 104 V A1 [3]

This question in 9702/43 Oct/Nov 2014

7 (a) State what is meant by quantisation of charge. … … [1] (b) Charged parallel plates, as shown in Fig. 7.1, produce a uniform electric field between the plates. + beam of protons – Fig. 7.1 The electric field outside the region between the plates is zero. A uniform magnetic field is applied in the region between the plates so that a beam of protons passes undeviated between the plates. (i) State and explain the direction of the magnetic field between the plates. … … … [2] (ii) The magnetic flux density between the plates is now increased. On Fig. 7.1, sketch the path of the protons between the plates. [2]

5 marks

Mark scheme: 7 (a) either charge exists in discrete and equal quantities or multiples of elementary charge / e / 1.6 × 10–19 C B1 [1] (b) (i) force due to magnetic field must be upwards B1 B-field into the plane of the paper B1 [2] (ii) sketch showing: deflection consistent with force in (b)(i) B1 reasonable curve B1 [2]

This question in 9702/43 Oct/Nov 2014

6 (a) State the type of field, or fields, that may cause a force to be exerted on a particle that is (i) uncharged and moving, … [1] (ii) charged and stationary, … [1] (iii) charged and moving at right-angles to the field. … [2] (b) A particle X has mass 3.32 × 10−26 kg and charge +1.60 × 10−19 C. The particle is travelling in a vacuum with speed 7.60 × 104 m s−1. It enters a region of uniform magnetic field that is normal to the direction of travel of the particle. The particle travels in a semicircle of diameter 12.2 cm, as shown in Fig. 6.1. region of uniform magnetic field 12.2 cm path of particle X Fig. 6.1 For the uniform magnetic field, (i) state its direction, … … [1] (ii) calculate the magnetic flux density. magnetic flux density = … T [3] (c) A second particle Y has mass less than that of particle X in (b) and the same charge. It enters the region of uniform magnetic field in (b) with the same speed and along the same initial path as particle X. On Fig. 6.1, draw the path of particle Y in the region of the magnetic field. [1]

9 marks

Mark scheme: 6 (a) (i) gravitational B1 [1] (ii) gravitational and electric B1 [1] (iii) magnetic and one other field given B1 magnetic, graviational and electric B1 [2] (b) (i) out of (plane of) paper/page (not “upwards”) B1 [1] (ii) B = mv / qr C1 = (3.32 × 10–26 × 7.6 × 104) / (1.6 × 10–19 × 6.1 × 10–2) C1 = 0.26 T A1 [3] (c) sketch: semicircle with diameter < 12.2 cm B1 [1]

This question in 9702/41 May/June 2015

5 (a) Define electric potential at a point. … … … [2] (b) Two positively charged metal spheres A and B are situated in a vacuum, as shown in Fig. 5.1. sphere A sphere B P x Fig. 5.1 A point P lies on the line joining the centres of the two spheres and is a distance x from the surface of sphere A. The variation with x of the electric potential V due to the two charged spheres is shown in Fig. 5.2. 600 500 V / V 400 300 200 100 0 2 4 6 8 10 x / cm surface surface of A of B Fig. 5.2 (i) State how the magnitude of the electric field strength at any point P may be determined from the graph of Fig. 5.2. … … [1] (ii) Without any calculation, describe the force acting on a positively charged particle placed at point P for values of x from x = 0 to x = 10 cm. … … … … [3] (c) The positively charged particle in (b)(ii) has charge q and mass m given by the expression q = 4.8 × 107 C kg–1. m Initially, the particle is at rest on the surface of sphere A where x = 0. It then moves freely along the line joining the centres of the spheres until it reaches the surface of sphere B. (i) On Fig. 5.2, mark with the letter M the point where the charged particle has its maximum speed. [1] (ii) 1. Use Fig. 5.2 to determine the potential difference between the spheres. potential difference = … V [1] 2. Use your answer in (ii) part 1 to calculate the speed of the particle as it reaches the surface of sphere B. Explain your working. speed = … m s–1 [3]

11 marks

Mark scheme: 5 (a) work done bringing/moving per unit positive charge M1 from infinity (to the point) A1 [2] (b) (i) slope/gradient (of the line/graph/tangent) B1 [1] (allow dV / dx, but not ∆V / ∆x or V / x) (allow potential gradient) (negative sign not required) (ii) maximum at surface of sphere A or at x = 0 (cm) B1 zero at x = 6 (cm) B1 then increases but in opposite direction B1 [3] (any mention of attraction max. 2/3) (c) (i) M shown between x = 5.5 cm and x = 6.5 cm B1 [1] (ii) 1. ∆V = (570 – 230) = 340 V (allow 330 V to 340 V) A1 [1] 2. q(∆)V = ½mv 2 or change/loss in PE = change/gain in KE or ∆EK = ∆EP B1 4.8 × 107 × 340 = ½v 2 C1 v2 = 3.26 × 1010 v = 1.8 × 105 m s–1 (not 1 s.f.) A1 [3]

This question in 9702/42 May/June 2015

11 One channel of communication is by the use of a coaxial cable. Such a cable is illustrated in Fig. 11.1. protective covering inner copper wire plastic insulation A Fig. 11.1 (a) (i) Suggest the material from which the component labelled A on Fig. 11.1 is made. … [1] (ii) Suggest two functions of the component labelled A. 1. … … 2. … … [2] (b) When a signal travels along the coaxial cable, it is attenuated. (i) State the meaning of attenuation. … … [1] (ii) State and explain why attenuation is frequently measured in decibels (dB). … … … [2] (c) A television aerial is connected to a receiver using a coaxial cable of length 11 m. The attenuation per unit length of the cable is 190 dB km–1. Calculate the ratio output signal from coaxial cable . input signal to coaxial cable ratio = … [3] Please turn over for Question 12.

9 marks

Mark scheme: 11 (a) (i) metal (allow specific example of a metal) B1 [1] (ii) e.g. provides ‘return’ for the signal shields inner core from interference/reduces cross-talk/reduces noise increased security (any two sensible suggestions, 1 each) B2 [2] (b) (i) (gradual) loss of power/intensity/amplitude B1 [1] (ii) dB is a log scale B1 either large (range of) numbers are easier to handle (on a log scale) or compounding attenuations/amplifications is easier B1 [2] (c) attenuation = 190 × 11 × 10–3 = 2.09 dB C1 –2.09 = 10 lg(POUT / PIN) C1 ratio = 0.62 A1 [3]

This question in 9702/42 May/June 2015

5 A charged metal sphere is isolated in space. Measurements of the electric potential V are made for different distances x from the centre of the sphere. The variation with distance x of the potential V is shown in Fig. 5.1. 4.0 3.0 V / 103 V 2.0 1.0 0 0 2.0 4.0 6.0 8.0 10.0 x / cm Fig. 5.1 (a) Use Fig. 5.1 to determine the electric field strength, in N C−1, at a point where x = 4.0 cm. Explain your working. electric field strength = … N C−1 [3] (b) The charge on the sphere is 8.0 × 10−9 C. (i) Use Fig. 5.1 to state the electric potential at the surface of the sphere. potential = … V [1] (ii) The sphere acts as a capacitor. Determine the capacitance of the sphere. capacitance = … F [2]

6 marks

Mark scheme: 5 (a) (magnitude of electric field strength is the potential gradient B1 use of gradient at x = 4.0 cm M1 gradient = 4.5 × 104 N C–1 (allow ± 0.3 × 10 4) A1 or Q Q V V = and E = 2 leading to E = (B1) 4πε 0 x 4 πε 0 x x E = 1.8 × 103 / 0.04 (M1) = 4.5 × 104 N C–1 (A1) [3] (b) (i) 3.6 × 103 V A1 [1] (ii) capacitance = Q / V C1 = (8.0 × 10–9) / (3.6 × 103) = 2.2 × 10–12 F A1 [2]

This question in 9702/43 May/June 2015

6 (a) State the type of field, or fields, that may cause a force to be exerted on a particle that is (i) uncharged and moving, … [1] (ii) charged and stationary, … [1] (iii) charged and moving at right-angles to the field. … [2] (b) A particle X has mass 3.32 × 10−26 kg and charge +1.60 × 10−19 C. The particle is travelling in a vacuum with speed 7.60 × 104 m s−1. It enters a region of uniform magnetic field that is normal to the direction of travel of the particle. The particle travels in a semicircle of diameter 12.2 cm, as shown in Fig. 6.1. region of uniform magnetic field 12.2 cm path of particle X Fig. 6.1 For the uniform magnetic field, (i) state its direction, … … [1] (ii) calculate the magnetic flux density. magnetic flux density = … T [3] (c) A second particle Y has mass less than that of particle X in (b) and the same charge. It enters the region of uniform magnetic field in (b) with the same speed and along the same initial path as particle X. On Fig. 6.1, draw the path of particle Y in the region of the magnetic field. [1]

9 marks

Mark scheme: 6 (a) (i) gravitational B1 [1] (ii) gravitational and electric B1 [1] (iii) magnetic and one other field given B1 magnetic, graviational and electric B1 [2] (b) (i) out of (plane of) paper/page (not “upwards”) B1 [1] (ii) B = mv / qr C1 = (3.32 × 10–26 × 7.6 × 104) / (1.6 × 10–19 × 6.1 × 10–2) C1 = 0.26 T A1 [3] (c) sketch: semicircle with diameter < 12.2 cm B1 [1]

This question in 9702/43 May/June 2015

5 A positively charged solid metal sphere is isolated in space. The electric field strength E is measured for different distances x from the centre of the sphere. The variation with x of the field strength E is shown in Fig. 5.1. 100 80 E / N C–1 60 40 20 0 0 5 10 15 20 25 x / cm Fig. 5.1 (a) Suggest why, for values of x less than 4.0 cm, the electric field strength is zero. … … … [2] (b) A point charge of +8.5 × 10−9 C moves from a point where x = 7.0 cm to a point where x = 5.0 cm. Use Fig. 5.1 to estimate the change in electric potential energy of this point charge. energy = … J [3]

5 marks

Mark scheme: 5 (a) charges in metal do not move B1 no (resultant) force on charges so no (electric) field B1 [2] (allow 1/2 for “no field inside sphere”) (b) either average field strength = ½ (28 + 54) N C–1 C1 average force = 8.5 × 10–9 × ½ (28 + 54) C1 = 3.49 × 10–7 N change in potential energy = 3.49 × 10–7 × 2.0 × 10–2 = 7.0 × 10–9 J (allow 1 s.f.) A1 (allow range 54 ± 1) or (for a point charge) V = Ex (C1) ∆V = (54 × 5.0 × 10–2) – (28 × 7.0 × 10–2) (C1) change in potential energy = 8.5 × 10–9 × (2.70 – 1.96) = 6.3 × 10–9 J (allow 1 s.f.) (A1) (allow range 54 ± 1) or ∆V is area under curve (C1) ∆V = 0.74 V (C1) change in potential energy = 8.5 × 10–9 × 0.74 = 6.3 × 10–9 J (allow 1 s.f.) (A1) [3] (allow range 0.70 to 0.84)

This question in 9702/43 Oct/Nov 2015

9 A particle of charge +q and mass m is travelling with a constant speed of 1.6 × 105 m s–1 in a vacuum. The particle enters a uniform magnetic field of flux density 9.7 × 10–2 T, as shown in Fig. 9.1. uniform magnetic field out of the plane of particle the paper charge +q flux density 9.7 × 10–2 T mass m speed 1.6 × 105 m s–1 path of particle Fig. 9.1 The magnetic field direction is perpendicular to the initial velocity of the particle and perpendicular to, and out of, the plane of the paper. A uniform electric field is applied in the same region as the magnetic field so that the particle passes undeviated through the fields. (a) State and explain the direction of the electric field. … … … [2] (b) Calculate the magnitude of the electric field strength. Explain your working. electric field strength = … V m–1 [3] (c) The electric field is now removed so that the positively-charged particle follows a curved path in the magnetic field. This path is an arc of a circle of radius 4.0 cm. q Calculate, for the particle, the ratio m. ratio = … C kg–1 [3] (d) The particle has a charge of 3e where e is the elementary charge. (i) Use your answer in (c) to determine the mass, in u, of the particle. mass = … u [2] (ii) The particle is the nucleus of an atom. State the number of protons and the number of neutrons in this nucleus. number of protons = … number of neutrons = … [1] [Total: 11]

11 marks

Mark scheme: 9 (a) direction of force due to electric field opposite to force due to magnetic field B1 electric field is up the page B1 [2] (b) force due to electric field = force due to magnetic field or Eq = Bqv B1 E = Bv C1 = 9.7 × 10–2 × 1.6 × 105 = 1.6 (1.55) × 104 V m–1 A1 [3] (c) q / m = v / Br C1 = 1.6 × 105 / (9.7 ×10–2 × 4.0 × 10–2) C1 = 4.1 (4.12) × 107 C kg–1 A1 [3] (d) (i) m = (3 × 1.60 × 10–19) / (4.12 × 107) C1 m = 1.16 × 10–26 / 1.66 × 10–27 = 7(.0) u (allow 7.1 u) A1 [2] (ii) 3 protons, 4 neutrons A1 [1]

This question in 9702/42 Feb/March 2016

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

6 marks

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

This question in 9702/42 May/June 2016

5 Two small solid metal spheres A and B have equal radii and are in a vacuum. Their centres are 15 cm apart. Sphere A has charge +3.0 pC and sphere B has charge +12 pC. The arrangement is illustrated in Fig. 5.1. sphere A sphere B P charge + 3.0 pC charge + 12 pC 5.0 cm 15 cm Fig. 5.1 Point P lies on the line joining the centres of the spheres and is a distance of 5.0 cm from the centre of sphere A. (a) Suggest why the electric field strength in both spheres is zero. … … … [2] (b) Show that the electric field strength is zero at point P. Explain your working. [3] (c) Calculate the electric potential at point P. electric potential = … V [2] (d) A silver-107 nucleus (10747 Ag) has speed v when it is a long distance from point P. Use your answer in (c) to calculate the minimum value of speed v such that the nucleus can reach point P. speed = … m s−1 [3] [Total: 10]

10 marks

Mark scheme: 5 (a) in an electric field, charges (in a conductor) would move B1 no movement of charge so zero field strength or charge moves until F = 0 / E = 0 B1 [2] or charges in metal do not move (B1) no (resultant) force on charges so no (electric) field (B1) (b) at P, EA = (3.0 × 10–12) / [4πε0(5.0 × 10–2)2] (= 10.79 N C–1) M1 at P, EB = (12 × 10–12) / [4πε0(10 × 10–2)2] (= 10.79 N C–1) M1 or (3.0 × 10–12) / [4πε0(5.0 × 10–2)2] – (12 × 10–12) / [4πε0(10 × 10–2)2] = 0 or (3.0 × 10–12) / [4πε0(5.0 × 10–2)2] = (12 × 10–12) / [4πε0(10 × 10–2)2] (M2) fields due to charged spheres are (equal and) opposite in direction, so E = 0 A1 [3] (c) potential = 8.99 × 109 {(3.0 × 10–12) / (5.0 × 10–2) + (12 × 10–12) / (10 × 10–2)} C1 = 1.62 V A1 [2] (d) ½mv2 = qV EK = ½ × 107 × 1.66 × 10–27 × v 2 C1 qV = 47 × 1.60 × 10–19 × 1.62 C1 v 2 = 1.37 × 108 v = 1.2 × 104 m s–1 A1 [3]

This question in 9702/43 Oct/Nov 2016

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