Electric potential — Paper 4 · A Level Physics 9702

1 The definitions of electric potential and of gravitational potential at a point have some similarity. (a) State one similarity between these two definitions. … … [1] (b) Explain why values of gravitational potential are always negative whereas values of electric potential may be positive or negative. … … … … … [4]

5 marks

Mark scheme: 1 (a) either ratio of work done to mass/charge or work done moving unit mass/charge from infinity or both have zero potential at infinity B1 [1] (b) gravitational forces are (always attractive) B1 electric forces can be attractive or repulsive B1 for gravitational, work got out as masses come together /mass moves from infinity B1 for electric, work done on charges if same sign, work got out if opposite sign as charges come together B1 [4]

This question in 9702/41 Oct/Nov 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) Define electric potential at a point. For Examiner’s … Use … … [2] (b) Two isolated point charges A and B are separated by a distance of 30.0 cm, as shown in Fig. 4.1. 30.0cm A B x Fig. 4.1 The charge at A is + 3.6 × 10–9 C. The variation with distance x from A along AB of the potential V is shown in Fig. 4.2. 600 V / V 400 200 0 0 5 10 15 20 25 30 x / cm –200 –400 –600 Fig. 4.2 (i) State the value of x at which the potential is zero. For Examiner’s x = … cm [1] Use (ii) Use your answer in (i) to determine the charge at B. charge = … C [3] (c) A small test charge is now moved along the line AB in (b) from x = 5.0 cm to x = 27 cm. State and explain the value of x at which the force on the test charge will be maximum. … … … … [3]

9 marks

Mark scheme: 4 (a) work done moving unit positive charge M1 from infinity to the point A1 [2] (b) (i) x = 18 cm A1 [1] (ii) VA + VB = 0 C1 (3.6 × 10–9) / (4πε0 × 18 × 10–2) + q / (4πε0 × 12 × 10–2) = 0 C1 q = –2.4 × 10–9 C A1 [3] (use of VA = VB giving 2.4 × 10 –9 C scores one mark) (c) field strength = (–) gradient of graph B1 force = charge × gradient / field strength or force ∝ gradient B1 force largest at x = 27 cm B1 [3]

This question in 9702/41 May/June 2008

5 (a) Define electric potential at a point. For Examiner’s … Use … … [2] (b) An α-particle is emitted from a radioactive source with kinetic energy of 4.8 MeV. The α-particle travels in a vacuum directly towards a gold (19779Au) nucleus, as illustrated in Fig. 5.1. path of gold α - particle nucleus Fig. 5.1 The α-particle and the gold nucleus may be considered to be point charges in an isolated system. (i) Explain why, as the α-particle approaches the gold nucleus, it comes to rest. … … … [2] (ii) For the closest approach of the α-particle to the gold nucleus determine 1. their separation, separation = … m [3] 2. the magnitude of the force on the α-particle. For Examiner’s Use force = … N [2]

9 marks

Mark scheme: 5 (a) work done per / on unit positive charge … M1 moving charge from infinity to the point … A1 [2] (b) (i) α-particle and gold nucleus repel each other … B1 all kinetic energy of α-particle converted into electric potential energy … B1 [2] (ii) 1 potential energy = (79 × 2 × {1.6 × 10-19}2) / (4π × 8.85 × 10-12 × d) … C1 kinetic energy = 4.8 × 1.6 × 10-13 = 7.68 × 10-13 J … C1 equating to give d = 4.7 × 10-14 m … A1 [3] (ii) 2 F = Qq / 4πε0d × 1 / d = 7.68 × 10-13 × 1 / (4.7 × 10-14) … C1 = 16 N … A1 [2] [Total: 9]

This question in 9702/41 Oct/Nov 2009

4 Two point charges A and B each have a charge of + 6.4 × 10–19 C. They are separated in a For vacuum by a distance of 12.0 μm, as shown in Fig. 4.1. Examiner’s Use 12.0 μm A P Q B +6.4 × 10–19 C +6.4 × 10–19 C 3.0 μm 3.0 μm Fig. 4.1 Points P and Q are situated on the line AB. Point P is 3.0 μm from charge A and point Q is 3.0 μm from charge B. (a) Calculate the force of repulsion between the charges A and B. force = … N [3] (b) Explain why, without any calculation, when a small test charge is moved from point P to point Q, the net work done is zero. … … … [2] (c) Calculate the work done by an electron in moving from the midpoint of line AB to point P. work done = … J [4]

9 marks

Mark scheme: 4 (a) force = q1q2 / 4πε0x2 C1 = (6.4 × 10–19)2 / (4π × 8.85 × 10–12 × {12 × 10–6}2) C1 = 2.56 × 10–17 N A1 [3] (b) potential at P is same as potential at Q B1 work done = q∆V M1 ∆V = 0 so zero work done A0 [2] (c) at midpoint, potential is 2 × (6.4 × 10–19) / (4πε0 × 6 × 10–6) C1 at P, potential is (6.4 × 10–19) / (4πε0 × 3 × 10–6) + (6.4 × 10–19) / (4πε0 × 9 × 10–6) C1 change in potential = (6.4 × 10–19) / (4πε0 × 9 × 10–6) energy = 1.6 × 10–19 × (6.4 × 10–19) / (4πε0 × 9 × 10–6) C1 = 1.0 × 10–22 J A1 [4]

This question in 9702/42 May/June 2010

4 Two point charges A and B each have a charge of + 6.4 × 10–19 C. They are separated in a For vacuum by a distance of 12.0 μm, as shown in Fig. 4.1. Examiner’s Use 12.0 μm A P Q B +6.4 × 10–19 C +6.4 × 10–19 C 3.0 μm 3.0 μm Fig. 4.1 Points P and Q are situated on the line AB. Point P is 3.0 μm from charge A and point Q is 3.0 μm from charge B. (a) Calculate the force of repulsion between the charges A and B. force = … N [3] (b) Explain why, without any calculation, when a small test charge is moved from point P to point Q, the net work done is zero. … … … [2] (c) Calculate the work done by an electron in moving from the midpoint of line AB to point P. work done = … J [4]

9 marks

Mark scheme: 4 (a) force = q1q2 / 4πε0x2 C1 = (6.4 × 10–19)2 / (4π × 8.85 × 10–12 × {12 × 10–6}2) C1 = 2.56 × 10–17 N A1 [3] (b) potential at P is same as potential at Q B1 work done = q∆V M1 ∆V = 0 so zero work done A0 [2] (c) at midpoint, potential is 2 × (6.4 × 10–19) / (4πε0 × 6 × 10–6) C1 at P, potential is (6.4 × 10–19) / (4πε0 × 3 × 10–6) + (6.4 × 10–19) / (4πε0 × 9 × 10–6) C1 change in potential = (6.4 × 10–19) / (4πε0 × 9 × 10–6) energy = 1.6 × 10–19 × (6.4 × 10–19) / (4πε0 × 9 × 10–6) C1 = 1.0 × 10–22 J A1 [4]

This question in 9702/43 May/June 2010

4 (a) Define capacitance. For Examiner’s … Use … [1] (b) An isolated metal sphere has a radius r. When charged to a potential V, the charge on the sphere is q. The charge may be considered to act as a point charge at the centre of the sphere. (i) State an expression, in terms of r and q, for the potential V of the sphere. … [1] (ii) This isolated sphere has capacitance. Use your answers in (a) and (b)(i) to show that the capacitance of the sphere is proportional to its radius. [1] (c) The sphere in (b) has a capacitance of 6.8 pF and is charged to a potential of 220 V. Calculate (i) the radius of the sphere, radius = … m [3] (ii) the charge, in coulomb, on the sphere. For Examiner’s Use charge = … C [1] (d) A second uncharged metal sphere is brought up to the sphere in (c) so that they touch. The combined capacitance of the two spheres is 18 pF. Calculate (i) the potential of the two spheres, potential = … V [1] (ii) the change in the total energy stored on the spheres when they touch. change = … J [3]

11 marks

Mark scheme: 4 (a) charge / potential (difference) (ratio must be clear) B1 [1] (b) (i) V = Q / 4πε0r B1 [1] (ii) C = Q / V = 4πε0r and 4πε0 is constant M1 so C ∝ r A0 [1] (c) (i) r = C / 4πε0r C1 r = (6.8 × 10–12) / (4π × 8.85 × 10–12) C1 = 6.1 × 10–2 m A1 [3] (ii) Q = CV = 6.8 × 10–12 × 220 = 1.5 × 10–9 C A1 [1] (d) (i) V = Q/C = (1.5 × 10–9) / (18 × 10–12) = 83 V A1 [1] (ii) either energy = ½CV 2 C1 ∆E = ½ × 6.8 × 10–12 × 2202 – ½ × 18 × 10–12 × 832 C1 = 1.65 × 10–7 – 6.2 × 10–8 = 1.03 × 10–7 J A1 [3] or energy = ½QV (C1) ∆E = ½ × 1.5 × 10–9 × 220 – ½ × 1.5 × 10–9 × 83 (C1) = 1.03 × 10–7 J (A1) GCE AS/A LEVEL – October/November 2010 9702 41

This question in 9702/41 Oct/Nov 2010

4 (a) Define capacitance. For Examiner’s … Use … [1] (b) An isolated metal sphere has a radius r. When charged to a potential V, the charge on the sphere is q. The charge may be considered to act as a point charge at the centre of the sphere. (i) State an expression, in terms of r and q, for the potential V of the sphere. … [1] (ii) This isolated sphere has capacitance. Use your answers in (a) and (b)(i) to show that the capacitance of the sphere is proportional to its radius. [1] (c) The sphere in (b) has a capacitance of 6.8 pF and is charged to a potential of 220 V. Calculate (i) the radius of the sphere, radius = … m [3] (ii) the charge, in coulomb, on the sphere. For Examiner’s Use charge = … C [1] (d) A second uncharged metal sphere is brought up to the sphere in (c) so that they touch. The combined capacitance of the two spheres is 18 pF. Calculate (i) the potential of the two spheres, potential = … V [1] (ii) the change in the total energy stored on the spheres when they touch. change = … J [3]

11 marks

Mark scheme: 4 (a) charge / potential (difference) (ratio must be clear) B1 [1] (b) (i) V = Q / 4πε0r B1 [1] (ii) C = Q / V = 4πε0r and 4πε0 is constant M1 so C ∝ r A0 [1] (c) (i) r = C / 4πε0r C1 r = (6.8 × 10–12) / (4π × 8.85 × 10–12) C1 = 6.1 × 10–2 m A1 [3] (ii) Q = CV = 6.8 × 10–12 × 220 = 1.5 × 10–9 C A1 [1] (d) (i) V = Q/C = (1.5 × 10–9) / (18 × 10–12) = 83 V A1 [1] (ii) either energy = ½CV 2 C1 ∆E = ½ × 6.8 × 10–12 × 2202 – ½ × 18 × 10–12 × 832 C1 = 1.65 × 10–7 – 6.2 × 10–8 = 1.03 × 10–7 J A1 [3] or energy = ½QV (C1) ∆E = ½ × 1.5 × 10–9 × 220 – ½ × 1.5 × 10–9 × 83 (C1) = 1.03 × 10–7 J (A1) GCE AS/A LEVEL – October/November 2010 9702 42

This question in 9702/42 Oct/Nov 2010

4 (a) (i) State what is meant by electric potential at a point. For Examiner’s … Use … … [2] (ii) Define capacitance. … … [1] (b) The variation of the potential V of an isolated metal sphere with charge Q on its surface is shown in Fig. 4.1. 200 150 V / kV 100 50 0 0 0.5 1.0 1.5 2.0 2.5 3.0 Q / µC Fig. 4.1 An isolated metal sphere has capacitance. For Examiner’s Use Fig. 4.1 to determine Use (i) the capacitance of the sphere, capacitance = … F [2] (ii) the electric potential energy stored on the sphere when charged to a potential of 150 kV. energy = … J [2] (c) A spark reduces the potential of the sphere from 150 kV to 75 kV. Calculate the energy lost from the sphere. energy = … J [2]

9 marks

Mark scheme: 4 (a) (i) work done moving unit positive charge M1 from infinity to the point A1 [2] (ii) charge / potential (difference) (ratio must be clear) B1 [1] (b) (i) capacitance = (2.7 × 10–6) / (150 × 103) C1 (allow any appropriate values) capacitance = 1.8 × 10–11 (allow 1.8 ±0.05) A1 [2] (ii) either energy = ½CV 2 or energy = ½QV and Q = CV C1 energy = ½ × 1.8 × 10–11 × (150 × 103)2 or ½ × 2.7 × 10–6 × 150 × 103 = 0.20 J A1 [2] (c) either since energy ∝ V 2, capacitor has (½)2 of its energy left or full formula treatment C1 energy lost = 0.15 J A1 [2] GCE A LEVEL – October/November 2010 9702 43

This question in 9702/43 Oct/Nov 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 (a) Define electric potential at a point. For Examiner’s … Use … … [2] (b) A charged particle is accelerated from rest in a vacuum through a potential difference V. Show that the final speed v of the particle is given by the expression ⎛ ⎞ 2Vq v = ⎜ ⎟ ⎝ m ⎠ q where is the ratio of the charge to the mass (the specific charge) of the particle. m [2] (c) A particle with specific charge +9.58 × 107 C kg–1 is moving in a vacuum towards a fixed metal sphere, as illustrated in Fig. 4.1. metal sphere 2.5 × 105 m s–1 potential +470 V particle specific charge +9.58 × 107 C kg–1 Fig. 4.1 The initial speed of the particle is 2.5 × 105 m s–1 when it is a long distance from the sphere. The sphere is positively charged and has a potential of +470 V. Use the expression in (b) to determine whether the particle will reach the surface of the sphere. [3]

7 marks

Mark scheme: 4 (a) work done moving unit positive charge M1 from infinity (to the point) A1 [2] (b) (gain in) kinetic energy = change in potential energy B1 ½mv2 = qV leading to v = (2Vq/m)½ B1 [2] (c) either (2.5 × 105)2 = 2 × V × 9.58 × 107 C1 V = 330 V M1 this is less than 470 V and so ‘no’ A1 [3] or v = (2 × 470 × 9.58 × 107) (C1) v = 3.0 × 105 m s–1 (M1) this is greater than 2.5 × 105 m s–1 and so ‘no’ (A1) or (2.5 × 105)2 = 2 × 470 × (q/m) (C1) (q/m) = 6.6 × 107 C kg–1 (M1) this is less than 9.58 × 107 C kg–1 and so ‘no’ (A1) GCE AS/A LEVEL – May/June 2013 9702 41

This question in 9702/41 May/June 2013

4 (a) An insulated metal sphere of radius R is situated in a vacuum. The charge q on the For sphere may be considered to be a point charge at the centre of the sphere. Examiner’s Use (i) State a formula, in terms of R and q, for the potential V on the surface of the sphere. … [1] (ii) Define capacitance and hence show that the capacitance C of the sphere is given by the expression C = 4πε0R. [1] (b) An isolated metal sphere has radius 45 cm. (i) Use the expression in (a)(ii) to calculate the capacitance, in picofarad, of the sphere. capacitance = … pF [2] (ii) The sphere is charged to a potential of 9.0 × 105 V. A spark occurs, partially discharging the sphere so that its potential is reduced to 3.6 × 105 V. Determine the energy of the spark. energy = … J [3]

7 marks

Mark scheme: 4 (a) (i) V = q / 4πε0R B1 [1] (ii) (capacitance is) ratio of charge and potential or q/V M1 C = q/V = 4πε0R A0 [1] (b) (i) C = 4π × 8.85 × 10–12 × 0.45 C1 = 50 pF A1 [2] (ii) either energy = ½ CV2 or energy = ½ QV and Q = CV C1 energy of spark = ½ × 50 × 10–12 {(9.0 × 105)2 – (3.6 × 105)2} C1 = 17 J A1 [3]

This question in 9702/42 May/June 2013

4 (a) Define electric potential at a point. For Examiner’s … Use … … [2] (b) A charged particle is accelerated from rest in a vacuum through a potential difference V. Show that the final speed v of the particle is given by the expression ⎛ ⎞ 2Vq v = ⎜ ⎟ ⎝ m ⎠ q where is the ratio of the charge to the mass (the specific charge) of the particle. m [2] (c) A particle with specific charge +9.58 × 107 C kg–1 is moving in a vacuum towards a fixed metal sphere, as illustrated in Fig. 4.1. metal sphere 2.5 × 105 m s–1 potential +470 V particle specific charge +9.58 × 107 C kg–1 Fig. 4.1 The initial speed of the particle is 2.5 × 105 m s–1 when it is a long distance from the sphere. The sphere is positively charged and has a potential of +470 V. Use the expression in (b) to determine whether the particle will reach the surface of the sphere. [3]

7 marks

Mark scheme: 4 (a) work done moving unit positive charge M1 from infinity (to the point) A1 [2] (b) (gain in) kinetic energy = change in potential energy B1 ½mv2 = qV leading to v = (2Vq/m)½ B1 [2] (c) either (2.5 × 105)2 = 2 × V × 9.58 × 107 C1 V = 330 V M1 this is less than 470 V and so ‘no’ A1 [3] or v = (2 × 470 × 9.58 × 107) (C1) v = 3.0 × 105 m s–1 (M1) this is greater than 2.5 × 105 m s–1 and so ‘no’ (A1) or (2.5 × 105)2 = 2 × 470 × (q/m) (C1) (q/m) = 6.6 × 107 C kg–1 (M1) this is less than 9.58 × 107 C kg–1 and so ‘no’ (A1) GCE AS/A LEVEL – May/June 2013 9702 43

This question in 9702/43 May/June 2013

3 (a) Define electric potential at a point. For Examiner’s … Use … … [2] (b) Two point charges A and B are separated by a distance of 20 nm in a vacuum, as illustrated in Fig. 3.1. 20 nm P B A x Fig. 3.1 A point P is a distance x from A along the line AB. The variation with distance x of the electric potential VA due to charge A alone is shown in Fig. 3.2. 0.8 potential V / V VA VB 0.6 0.4 0.2 0 2 4 6 8 10 12 14 16 18 x / nm Fig. 3.2 The variation with distance x of the electric potential VB due to charge B alone is also shown in Fig. 3.2. (i) State and explain whether the charges A and B are of the same, or opposite, sign. For Examiner’s … Use … … [2] (ii) By reference to Fig. 3.2, state how the combined electric potential due to both charges may be determined. … … [1] (iii) Without any calculation, use Fig. 3.2 to estimate the distance x at which the combined electric potential of the two charges is a minimum. x = … nm [1] (iv) The point P is a distance x = 10 nm from A. An α-particle has kinetic energy EK when at infinity. Use Fig. 3.2 to determine the minimum value of EK such that the α-particle may travel from infinity to point P. EK = … J [3]

9 marks

Mark scheme: 3 (a) work done bringing unit positive charge M1 from infinity (to the point) A1 [2] (b) (i) either both potentials are positive / same sign M1 so same sign A1 [2] or gradients are positive & negative (so fields in opposite directions) (M1) so same sign (A1) (ii) the individual potentials are summed B1 [1] (iii) allow value of x between 10 nm and 13 nm A1 [1] (iv) V = 0.43 V (allow 0.42 V → 0.44 V) M1 energy = 2 × 1.6 × 10–19 × 0.43 A1 = 1.4 × 10–19 J A1 [3] GCE A LEVEL – October/November 2013 9702 43

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

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 9702/43 May/June 2014

5 (a) Define electric potential at a point. … … … [2] (b) An isolated solid metal sphere is positively charged. The variation of the potential V with distance x from the centre of the sphere is shown in Fig. 5.1. 200 160 V / V 120 80 40 0 0 2 4 6 8 10 x / cm Fig. 5.1 Use Fig. 5.1 to suggest (i) why the radius of the sphere cannot be greater than 1.0 cm, … … [1] (ii) that the charge on the sphere behaves as if it were a point charge. [3] (c) Assuming that the charge on the sphere does behave as a point charge, use data from Fig. 5.1 to determine the charge on the sphere. charge = … C [2]

8 marks

Mark scheme: 5 (a) work done in moving unit positive charge M1 from infinity (to the point) A1 [2] (b) (i) inside the sphere, the potential would be constant B1 [1] (ii) for point charge, Vx is constant B1 co-ordinates clear and determines two values of Vx at least 4 cm apart M1 conclusion made clear A1 [3] (c) q = 4πε0Vx q = 4π × 8.85 × 10–12 × 180 × 1.0 × 10–2 M1 = 2.0 × 10–10 C A1 [2]

This question in 9702/41 Oct/Nov 2014

5 (a) Define electric potential at a point. … … … [2] (b) An isolated solid metal sphere is positively charged. The variation of the potential V with distance x from the centre of the sphere is shown in Fig. 5.1. 200 160 V / V 120 80 40 0 0 2 4 6 8 10 x / cm Fig. 5.1 Use Fig. 5.1 to suggest (i) why the radius of the sphere cannot be greater than 1.0 cm, … … [1] (ii) that the charge on the sphere behaves as if it were a point charge. [3] (c) Assuming that the charge on the sphere does behave as a point charge, use data from Fig. 5.1 to determine the charge on the sphere. charge = … C [2]

8 marks

Mark scheme: 5 (a) work done in moving unit positive charge M1 from infinity (to the point) A1 [2] (b) (i) inside the sphere, the potential would be constant B1 [1] (ii) for point charge, Vx is constant B1 co-ordinates clear and determines two values of Vx at least 4 cm apart M1 conclusion made clear A1 [3] (c) q = 4πε0Vx q = 4π × 8.85 × 10–12 × 180 × 1.0 × 10–2 M1 = 2.0 × 10–10 C A1 [2]

This question in 9702/42 Oct/Nov 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

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

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

6 A solid metal sphere of radius R is isolated in space. The sphere is positively charged so that the electric potential at its surface is VS. The electric field strength at the surface is ES. (a) On the axes of Fig. 6.1, show the variation of the electric potential with distance x from the centre of the sphere for values of x from x = 0 to x = 3R. 1.0 Vs 0.8 Vs potential 0.6 Vs 0.4 Vs 0.2 Vs 0 0 R 2R 3R distance x Fig. 6.1 [3] (b) On the axes of Fig. 6.2, show the variation of the electric field strength with distance x from the centre of the sphere for values of x from x = 0 to x = 3R. 1.0 Es 0.8 Es field strength 0.6 Es 0.4 Es 0.2 Es 0 0 R 2R 3R distance x Fig. 6.2 [3] [Total: 6]

6 marks

Mark scheme: 6 (a) sketch: from x = 0 to x = R, potential is constant at VS B1 smooth curve through (R, VS) and (2R, 0.5VS) B1 smooth curve continues to (3R, 0.33VS) B1 [3] (b) sketch: from x = 0 to x = R, field strength is zero B1 smooth curve through (R, E) and (2R, 0.25E) B1 smooth curve continues to (3R, 0.11E) B1 [3]

This question in 9702/41 May/June 2016

6 A solid metal sphere of radius R is isolated in space. The sphere is positively charged so that the electric potential at its surface is VS. The electric field strength at the surface is ES. (a) On the axes of Fig. 6.1, show the variation of the electric potential with distance x from the centre of the sphere for values of x from x = 0 to x = 3R. 1.0 Vs 0.8 Vs potential 0.6 Vs 0.4 Vs 0.2 Vs 0 0 R 2R 3R distance x Fig. 6.1 [3] (b) On the axes of Fig. 6.2, show the variation of the electric field strength with distance x from the centre of the sphere for values of x from x = 0 to x = 3R. 1.0 Es 0.8 Es field strength 0.6 Es 0.4 Es 0.2 Es 0 0 R 2R 3R distance x Fig. 6.2 [3] [Total: 6]

6 marks

Mark scheme: 6 (a) sketch: from x = 0 to x = R, potential is constant at VS B1 smooth curve through (R, VS) and (2R, 0.5VS) B1 smooth curve continues to (3R, 0.33VS) B1 [3] (b) sketch: from x = 0 to x = R, field strength is zero B1 smooth curve through (R, E) and (2R, 0.25E) B1 smooth curve continues to (3R, 0.11E) B1 [3]

This question in 9702/43 May/June 2016

6 Two solid metal spheres A and B, each of radius 1.5 cm, are situated in a vacuum. Their centres are separated by a distance of 20.0 cm, as shown in Fig. 6.1. 1.5 cm 1.5 cm 20.0 cm P sphere A sphere B x Fig. 6.1 (not to scale) Both spheres are positively charged. Point P lies on the line joining the centres of the two spheres, at a 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. 6.2. 50 40 30 E / N C–1 20 10 0 0 2 4 6 8 10 12 14 16 18 20 –10 x / cm –20 –30 –40 –50 Fig. 6.2 (a) Use Fig. 6.2 to determine the ratio magnitude of charge on sphere A . magnitude of charge on sphere B Explain your working. ratio = … [3] (b) The variation with distance x of the electric potential V at point P is shown in Fig. 6.3. 0.8 0.7 0.6 V / V 0.5 0.4 0.3 0.2 0.1 0 0 2 4 6 8 10 12 14 16 18 20 x / cm Fig. 6.3 An α-particle is initially at rest on the surface of sphere A. The α-particle moves along the line joining the centres of the two spheres. Determine, for the α-particle as it moves between the two spheres, (i) its maximum speed, maximum speed = … m s–1 [3] (ii) its speed on reaching the surface of sphere B. speed = … m s–1 [2] [Total: 8]

8 marks

Mark scheme: 6 (a) E = 0 or EA = (–)EB (at x = 11 cm) B1 QA / x2 = QB / (20 – x)2 = 112 / 92 C1 QA / QB or ratio = 1.5 A1 [3] or E ∝ Q because r same or E = Q / 4πε0r2 and r same (B1) QA / QB = 48 / 32 (C1) QA / QB or ratio = 1.5 (A1) (b) (i) for max. speed, ∆V = (0.76 – 0.18) V or ∆V = 0.58 V C1 q∆V = ½mv2 2 × (1.60 × 10–19) × 0.58 = ½ × 4 × 1.66 × 10–27 × v2 C1 v2 = 5.59 × 107 v = 7.5 × 103 m s–1 A1 [3] (ii) ∆V = 0.22 V C1 2 × (1.60 × 10–19) × 0.22 = ½ × 4 × 1.66 × 10–27 × v2 v2 = 2.12 × 107 v = 4.6 × 103 m s–1 A1 [2]

This question in 9702/42 Oct/Nov 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 Coulomb’s law. … … … [2] (b) Two charged metal spheres A and B are situated in a vacuum, as illustrated in Fig. 6.1. 6.0 cm sphere A sphere B P x Fig. 6.1 The shortest distance between the surfaces of the spheres is 6.0 cm. A movable point P lies along the line joining the centres of the two spheres, a distance x from the surface of sphere A. The variation with distance x of the electric field strength E at point P is shown in Fig. 6.2. 10 E / 103 V m–1 5 0 0 1 2 3 4 5 6 x / cm –5 –10 –15 Fig. 6.2 (i) Use Fig. 6.2 to explain whether the two spheres have charges of the same, or opposite, sign. … … … … [2] (ii) A proton is at point P where x = 5.0 cm. Use data from Fig. 6.2 to determine the acceleration of the proton. acceleration = … m s–2 [3] (c) Use data from Fig. 6.2 to state the value of x at which the rate of change of electric potential is maximum. Give the reason for the value you have chosen. … … … [2] [Total: 9]

9 marks

Mark scheme: 6(a) force proportional to product of charges and inversely proportional to the square of the separation M1 reference to point charges A1 6(b)(i) (near to each sphere,) fields are in opposite directions or point (between spheres) where fields are equal and opposite or point (between spheres) where field strength is zero M1 so same (sign of charge) A1 6(b)(ii) (at x = 5.0 cm,) E = 3.0 × 103 V m–1 and a = qE / m C1 E = (1.60 × 10–19 × 3.0 × 103) / (1.67 × 10–27) C1 = 2.9 × 1011 m s–2 A1 6(c) field strength or E is potential gradient or field strength is rate of change of (electric) potential M1 (field strength) maximum at x = 6 cm A1

This question in 9702/42 May/June 2017

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

7 marks

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

This question in 9702/42 Oct/Nov 2017

7 (a) State what is meant by electric potential at a point. … … … [2] (b) The centres of two charged metal spheres A and B are separated by a distance of 44.0 cm, as shown in Fig. 7.1. 44.0 cm sphere A sphere B P x Fig. 7.1 (not to scale) A moveable point P lies on the line joining the centres of the two spheres. Point P is a distance x from the centre of sphere A. The variation with distance x of the electric potential V at point P is shown in Fig. 7.2. 2.2 V / 104 V 2.0 1.8 1.6 1.4 1.2 0 10 20 30 40 50 x / cm Fig. 7.2 (i) Use Fig. 7.2 to state and explain whether the two spheres have charges of the same, or opposite, sign. … … … [1] (ii) A positively-charged particle is at rest on the surface of sphere A. The particle moves freely from the surface of sphere A to the surface of sphere B. 1. Describe qualitatively the variation, if any, with distance x of the speed of the particle as it moves from x = 12 cm to x = 25 cm … … passes through x = 26 cm … … moves from x = 27 cm to x = 31 cm … … reaches x = 32 cm … … [4] 2. The particle has charge 3.2 × 10–19 C and mass 6.6 × 10–27 kg. Calculate the maximum speed of the particle. speed = … m s–1 [2] [Total: 9]

9 marks

Mark scheme: 7(a) work done per unit charge B1 (work done) moving positive charge from infinity (to the point) B1 7(b)(i) potential always same sign / potential is always positive so same sign of charge B1 Question Answer Marks 7(b)(ii) 1 from x = 12 cm to x = 25 cm: speed increases and from x = 27 cm to x = 31 cm: speed decreases B1 (from x = 12 cm to x = 25 cm: speed increases) at decreasing rate or (from x = 27 cm to x = 31 cm: speed decreases) at increasing rate B1 at x = 26 cm: speed maximum B1 at 32 cm: speed still decreasing B1 2 q ∆V = ½mv2 3.2 × 10–19 × (2.14 – 1.43) × 104 = ½ × 6.6 × 10–27 × v2 v2 = 6.88 × 1011 C1 v = 8.3 × 105 m s–1 (8.30) A1

This question in 9702/42 Feb/March 2018

6 (a) (i) Define electric potential at a point. … … … [2] (ii) State the relationship between electric potential and electric field strength at a point. … … … [2] (b) Two parallel metal plates A and B are situated a distance 1.2 cm apart in a vacuum, as shown in Fig. 6.1. –75 V plate B helium nucleus 1.2 cm x 0 V plate A Fig. 6.1 Plate A is earthed and plate B is at a potential of –75 V. A helium nucleus is situated between the plates, a distance x from plate A. Initially, the helium nucleus is at rest on plate A where x = 0. (i) The helium nucleus is free to move between the plates. By considering energy changes of the helium nucleus, explain why the speed at which it reaches plate B is independent of the separation of the plates. … … … … [2] (ii) As the helium nucleus (42He) moves from plate A towards plate B, its distance x from plate A increases. Calculate the speed of the nucleus after it has moved a distance x = 0.40 cm from plate A. speed = … m s–1 [3] [Total: 9]

9 marks

Mark scheme: 6(a)(i) work done per unit charge B1 work done moving positive charge from infinity (to the point) B1 6(a)(ii) field strength = potential gradient M1 ‘–’ sign included or directions discussed A1 6(b)(i) gain in kinetic energy (= loss in potential energy) = charge × p.d. or qV = ½mv2 M1 so v is independent of separation (because separation not in expressions) A1 Question Answer Marks 6(b)(ii) (at x = 0.40 cm), potential = (–) 75 × 0.40 / 1.2 (= (–) 25 V) C1 ½mv2 = qV ½ × 4 × 1.66 × 10–27 × v2 = 2 × 1.60 × 10–19 × 25 C1 or a = Vq / dm and v2 = 2as (C1) v2 = (2 × 75 × 2 × 1.60 × 10–19 × 0.40 × 10–2) / (1.2 × 10–2 × 4 × 1.66 × 10–27) (C1) v = 4.9 × 104 m s–1 A1

This question in 9702/41 Oct/Nov 2018

6 (a) State (i) what is meant by the electric potential at a point, … … … [2] (ii) the relationship between electric potential at a point and electric field strength at the point. … … … [2] (b) Two similar solid metal spheres A and B, each of radius R, are situated in a vacuum such that the separation of their centres is D, as shown in Fig. 6.1. D x P sphere A sphere B R R charge +Q charge +q Fig. 6.1 The charge +Q on sphere A is larger than the charge +q on sphere B. A movable point P is located on the line joining the centres of the two spheres. The point P is a distance x from the centre of sphere A. On Fig. 6.2, sketch a graph to show the variation with x of the electric potential V between the centres of the two spheres. V 0 0 D x surface of surface of sphere A sphere B Fig. 6.2 [4] [Total: 8]

8 marks

Mark scheme: 6(a)(i) work done per unit charge B1 work done moving positive charge from infinity (to the point) B1 6(a)(ii) field strength = potential gradient M1 negative sign included or directions discussed A1 6(b) horizontal straight lines, at non-zero potential, within the spheres B1 magnitude of potential greater at surface of sphere A than at surface of sphere B B1 concave curve between A and B, with a minimum nearer to B B1 lines show V positive all the way from 0 to D B1

This question in 9702/42 Oct/Nov 2018

6 (a) (i) Define electric potential at a point. … … … [2] (ii) State the relationship between electric potential and electric field strength at a point. … … … [2] (b) Two parallel metal plates A and B are situated a distance 1.2 cm apart in a vacuum, as shown in Fig. 6.1. –75 V plate B helium nucleus 1.2 cm x 0 V plate A Fig. 6.1 Plate A is earthed and plate B is at a potential of –75 V. A helium nucleus is situated between the plates, a distance x from plate A. Initially, the helium nucleus is at rest on plate A where x = 0. (i) The helium nucleus is free to move between the plates. By considering energy changes of the helium nucleus, explain why the speed at which it reaches plate B is independent of the separation of the plates. … … … … [2] (ii) As the helium nucleus (42He) moves from plate A towards plate B, its distance x from plate A increases. Calculate the speed of the nucleus after it has moved a distance x = 0.40 cm from plate A. speed = … m s–1 [3] [Total: 9]

9 marks

Mark scheme: 6(a)(i) work done per unit charge B1 work done moving positive charge from infinity (to the point) B1 6(a)(ii) field strength = potential gradient M1 ‘–’ sign included or directions discussed A1 6(b)(i) gain in kinetic energy (= loss in potential energy) = charge × p.d. or qV = ½mv2 M1 so v is independent of separation (because separation not in expressions) A1 Question Answer Marks 6(b)(ii) (at x = 0.40 cm), potential = (–) 75 × 0.40 / 1.2 (= (–) 25 V) C1 ½mv2 = qV ½ × 4 × 1.66 × 10–27 × v2 = 2 × 1.60 × 10–19 × 25 C1 or a = Vq / dm and v2 = 2as (C1) v2 = (2 × 75 × 2 × 1.60 × 10–19 × 0.40 × 10–2) / (1.2 × 10–2 × 4 × 1.66 × 10–27) (C1) v = 4.9 × 104 m s–1 A1

This question in 9702/43 Oct/Nov 2018

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

9 marks

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

This question in 9702/42 May/June 2019

9 (a) Define what is meant by electric potential at a point. … … … [2] (b) In an α-particle scattering experiment, α-particles are directed towards a thin film of gold, as illustrated in Fig. 9.1. gold film beam of α-particles Fig. 9.1 The apparatus is in a vacuum. The gold-197 (19779 Au) nuclei in the film may be considered to be fixed point charges. The α-particles emitted from the source each have an energy of 4.8 MeV. Calculate: (i) the initial kinetic energy EK, in J, of an α-particle emitted from the source EK = … J [1] (ii) the distance d of closest approach of an α-particle to a gold nucleus. d = … m [4] (c) Use your answer in (b)(ii) to comment on the possible diameter of a gold nucleus. … … [1] [Total: 8]

8 marks

Mark scheme: 9(a) work done per unit charge B1 (work done) moving positive charge from infinity B1 9(b)(i) energy = 4.8 × 1.60 × 10–13 = 7.7 × 10–13 J A1 9(b)(ii) EP = Qq / 4πε0d C1 Q = 79e and q = 2e C1 7.68 × 10–13 = (79 × 2 × {1.60 × 10–19}2 / (4π × 8.85 × 10–12 × d) C1 d = 4.7 × 10–14 m A1 9(c) (diameter must be) less than/equal to 10–13 or 10–14 m B1

This question in 9702/42 Oct/Nov 2019

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

8 marks

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

This question in 9702/42 Feb/March 2020

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

9 marks

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

This question in 9702/41 Oct/Nov 2020

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

9 marks

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

This question in 9702/43 Oct/Nov 2020

5 (a) An isolated metal sphere of radius r is charged so that the electric potential at its surface is V0. On Fig. 5.1, sketch the variation with distance x from the centre of the sphere of the electric potential. Your graph should extend from x = 0 to x = 3r. 1.0 V0 electric potential 0.5 V0 0 0 r 2r 3r x Fig. 5.1 [3] (b) Photons having wavelength λ are incident on a metal surface. The maximum wavelength for which there is emission of electrons is λ 0. λ 0 For photons of wavelength , the maximum kinetic energy of the emitted electrons is EMAX. 2 On Fig. 5.2, sketch the variation with wavelength λ of the maximum kinetic energy for values λ 0 of wavelength between λ = and λ = λ 0. 3 3 EMAX energy 2 EMAX EMAX 0 0 λ λ λ 0 0 0 3 2 λ Fig. 5.2 [3] (c) A pure sample of a radioactive isotope contains N0 nuclei. The half-life of the isotope is T12. The product of the radioactive decay is stable. The variation with time t of the number N of nuclei of the radioactive isotope is shown in Fig. 5.3. N0 number N0 2 N 0 0 T time t Fig. 5.3 On Fig. 5.3: ● label, on the time axis, the time t = 1.0T12 and the time t = 2.0T12 ● sketch the variation with time t of the number of nuclei of the decay product for time t = 0 to time t = T. [3] [Total: 9]

9 marks

Mark scheme: 5(a) from x = 0 to x = r: horizontal line at V = 1.0V0 B1 from x = r to x = 3r: curve with negative gradient of decreasing magnitude starting at (r, 1.0V0) B1 line passing through (2r, ½V0) and (3r, ⅓V0) B1 5(b) line with negative gradient from λ = ⅓λ0 to λ = λ0 B1 line passing through (λ0, 0) B1 curve with negative gradient of decreasing magnitude passing through (½λ0, EMAX) and (⅓λ0, 2EMAX) B1 5(c) 1.0T½ shown at ½N0 and 2.0T½ shown at ¼N0 B1 line starting at (0, 0) and reaching (T, N0–N) B1 line starting at (0, 0) and reaching original curve at (1.0T½, ½N0) B1

This question in 9702/42 May/June 2021

6 (a) Define electric potential. … … … [2] (b) An isolated conducting sphere in a vacuum has radius r and is initially uncharged. It is then charged by friction so that it carries a final charge Q. This charge can be considered to be acting at the centre of the sphere. By considering the electric potential at its surface, show that the capacitance C of the sphere is given by C = 4πε0r where ε0 is the permittivity of free space. [2] (c) The dome of an electrostatic generator is a spherical conductor of radius 13 cm. It is initially charged so that the electric potential at the surface is 4.5 kV. A smaller isolated sphere of radius 5.2 cm, initially uncharged, is brought near to the dome. Sparking causes a current between the two spheres until they reach the same potential. Assume that any charge on a sphere may be considered to act as a point charge at its centre. Calculate the charge that is transferred between the two spheres. charge = … C [3] [Total: 7]

7 marks

Mark scheme: 6(a) work done per unit charge B1 (work done in) moving positive charge from infinity B1 6(b) C = Q / V C1 V = Q / (4πε0r) and so C = Q / [Q / (4πε0r)] = 4πε0r A1 6(c) Q = 4πε0rV = 4π × 8.85 × 10–12 × 0.13 × 4500 ( = 6.5 × 10–8 C) C1 (Q – q) / 13 = q / 5.2 C1 5.2Q – 5.2q = 13q, so q = (5.2 / 18.2)Q q = (5.2 / 18.2) × 6.5 × 10–8 = 1.9 × 10–8 C A1 or VT = QT / CT = 6.5 × 10–8 / [4π × 8.85 × 10–12 × (0.13 + 0.052)] ( = 3210 V) (C1) q = 4π × 8.85 × 10–12 × 0.052 × 3210 = 1.9 × 10–8 C (A1)

This question in 9702/42 Oct/Nov 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

5 (a) Define electric potential at a point. … … … [2] (b) An isolated conducting sphere is charged. Fig. 5.1 shows the variation of the potential V due to the sphere with displacement x from its centre. 0 – 0.3 – 0.2 – 0.1 0 0.1 0.2 0.3 x / m – 250 V / V – 500 – 750 – 1000 Fig. 5.1 Use Fig. 5.1 to determine: (i) the radius of the sphere radius = … m [1] (ii) the charge on the sphere. charge = … C [2] (c) Two spheres are identical to the sphere in (b). Each sphere has the same charge as the sphere in (b). The spheres are held in a vacuum so that their centres are separated by a distance of 0.46 m. Assume that the charge on each sphere is a point charge at the centre of the sphere. (i) Calculate the electric potential energy EP of the two spheres. EP = … J [2] (ii) The two spheres are now released simultaneously so that they are free to move. Describe and explain the subsequent motion of the spheres. … … … … [3] [Total: 10]

10 marks

Mark scheme: 5(a) work done per unit charge B1 work done (on charge) in moving positive charge from infinity (to the point) B1 5(b)(i) radius = 0.060 m A1 5(b)(ii) V = Q / 40x C1 Q = (–) 850  4  8.85  10–12  0.060 or Q = (–) 850  0.060 / 8.99  109 (any correct pair of V and x values from curve) Q = – 5.7  10–9 C A1 5(c)(i) EP = Q2 / 40x C1 = (5.67  10–9)2 / (4  8.85  10–12  0.46) = 6.3  10–7 J A1 5(c)(ii) • force is repulsive so spheres move apart B3 • force in direction of motion so speed increases • potential energy converted to kinetic energy so speed increases • force decreases with distance so acceleration decreases • momentum is conserved (at zero) (and masses are equal) so velocities are always equal and opposite Any three points, 1 mark each

This question in 9702/42 Oct/Nov 2022

5 (a) Define electric potential at a point. … … … [2] (b) Two isolated charged metal spheres X and Y are situated near to each other in a vacuum with their centres a distance of 24 m apart. Point P is at a variable distance x from the centre of sphere X on the line joining the centres of the spheres. Fig. 5.1 shows the variation with x of the electric potential V due to the spheres at point P. V 0 0 4 8 12 16 20 24 x / m Fig. 5.1 State three conclusions that can be drawn about the spheres from Fig. 5.1. The conclusions may be qualitative or quantitative. 1 … … 2 … … 3 … … [3] (c) A positively charged particle is placed at point P in (b), such that x = 12 m. The particle is released. Describe and explain the subsequent motion of the particle. … … … … … [3] [Total: 8]

8 marks

Mark scheme: 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

This question in 9702/41 Oct/Nov 2023

5 (a) Define electric potential at a point. … … … [2] (b) Two isolated charged metal spheres X and Y are situated near to each other in a vacuum with their centres a distance of 24 m apart. Point P is at a variable distance x from the centre of sphere X on the line joining the centres of the spheres. Fig. 5.1 shows the variation with x of the electric potential V due to the spheres at point P. V 0 0 4 8 12 16 20 24 x / m Fig. 5.1 State three conclusions that can be drawn about the spheres from Fig. 5.1. The conclusions may be qualitative or quantitative. 1 … … 2 … … 3 … … [3] (c) A positively charged particle is placed at point P in (b), such that x = 12 m. The particle is released. Describe and explain the subsequent motion of the particle. … … … … … [3] [Total: 8]

8 marks

Mark scheme: 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

This question in 9702/43 Oct/Nov 2023

5 (a) Define electric potential at a point. … … … [2] (b) Two isolated charged metal spheres X and Y are near to each other in a vacuum. The centres of the spheres are 1.2 m apart, as shown in Fig. 5.1. X Y P x 1.2 m Fig. 5.1 (not to scale) Point P is on the line joining the centres of spheres X and Y and is at a variable distance x from the centre of X. Fig. 5.2 shows the variation with x of the total electric potential V due to the two spheres. V 0 0 0.2 0.4 0.6 0.8 1.0 1.2 x / m Fig. 5.2 State three conclusions that may be drawn about the spheres from Fig. 5.2. The conclusions may be qualitative or quantitative. 1 … … 2 … … 3 … … [3] (c) A proton is held at rest on the line joining the centres of the spheres in (b) at the position where x = 0.60 m. The proton is released. Describe and explain, without calculation, the subsequent motion of the proton. … … … [2] [Total: 7]

7 marks

Mark scheme: 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: 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 B3 5(c) proton remains at rest (in the position of release) M1 potential energy of proton is (already) at its minimum 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 A1

This question in 9702/42 May/June 2024

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

10 marks

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

This question in 9702/41 Oct/Nov 2024

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

10 marks

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

This question in 9702/43 Oct/Nov 2024

5 (a) Define electric potential at a point. … … … [2] (b) An isolated solid metal sphere of radius r is given a positive charge. The potential at the surface of the sphere is 9.0 × 104 V. At a distance of 3r from the centre of the sphere, the electric field strength is 2.0 × 105 N C–1. (i) Determine the electric field strength at the surface of the sphere. electric field strength = … N C–1 [2] (ii) Show that the radius of the sphere is 5.0 cm. [2] (iii) Calculate the charge on the sphere. charge = … C [2] (iv) Use your answer in (b)(iii) to determine the capacitance of the sphere. capacitance = … F [2] [Total: 10]

10 marks

Mark scheme: 5(a) work done per unit charge B1 work done (on charge) in moving positive charge from infinity (to the point) B1 5(b)(i) electric field strength inversely proportional to distance2 C1 E = 32  2.0  105 A1 = 1.8  106 N C–1 5(b)(ii) V = Q / 40r and E = Q / 40r2 (so E = V / r) B1 r = (9.0  104) / (1.8  106) = 0.050 m = 5.0 cm A1 5(b)(iii) Q = 40Vr C1 = 4  8.85  10–12  9.0  104  0.050 = 5.0  10–7 C A1 5(b)(iv) C = Q / V C1 = (5.0  10–7) / (9.0  104) A1 = 5.6  10–12 F

This question in 9702/44 May/June 2025

5 (a) Define electric potential at a point. … … … [2] (b) A hydrogen atom may be considered to consist of a proton and an electron separated by a distance of 120 pm, as shown in Fig. 5.1. proton P electron x 120 pm Fig. 5.1 The two particles may be considered as point charges. Point P lies on the line joining the electron and the proton and is at a variable distance x from the proton. (i) Show that the electric potential V at point P when x = 10 pm is equal to 130 V. [2] (ii) Calculate, to two significant figures, V when x = 30 pm. V = … V [2] (iii) On Fig. 5.1, draw a cross (×) at one position, other than infinity, where the electric potential is zero. [1] (iv) On Fig. 5.2, sketch the variation of V with x between x = 10 pm and x = 110 pm. 160 V / V 80 0 10 30 50 70 90 110 x / pm –80 –160 Fig. 5.2 [3] [Total: 10]

10 marks

Mark scheme: 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

This question in 9702/41 Oct/Nov 2025

5 (a) Define electric potential at a point. … … … [2] (b) A hydrogen atom may be considered to consist of a proton and an electron separated by a distance of 120 pm, as shown in Fig. 5.1. proton P electron x 120 pm Fig. 5.1 The two particles may be considered as point charges. Point P lies on the line joining the electron and the proton and is at a variable distance x from the proton. (i) Show that the electric potential V at point P when x = 10 pm is equal to 130 V. [2] (ii) Calculate, to two significant figures, V when x = 30 pm. V = … V [2] (iii) On Fig. 5.1, draw a cross (×) at one position, other than infinity, where the electric potential is zero. [1] (iv) On Fig. 5.2, sketch the variation of V with x between x = 10 pm and x = 110 pm. 160 V / V 80 0 10 30 50 70 90 110 x / pm –80 –160 Fig. 5.2 [3] [Total: 10]

10 marks

Mark scheme: 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

This question in 9702/43 Oct/Nov 2025

5 (a) Explain why the electric potential near an isolated proton is positive. … … … … … [3] (b) An isolated metal sphere is positively charged and has radius R, as shown in Fig. 5.1. sphere + + + + R + + P X Y + + Q + + + + x Fig. 5.1 Line XY passes through the centre of the sphere. Point P lies on line XY at a variable displacement x from the centre of the sphere. Point Q is at a fixed position that is not on line XY. The electric field strength at the surface of the sphere is E0. (i) On Fig. 5.1, draw an arrow at point Q to show the direction of the electric field at that point. [1] (ii) On Fig. 5.2, sketch the variation of the electric field E at point P with x for values of x between x = –3R and x = 3R. Do not include the region inside the sphere between x = –R and x = R. E0 E ½E0 0 –3R –2R –R 0 R 2R 3R x –½E0 –E0 Fig. 5.2 [3] (c) The proton and the electron in a hydrogen atom are separated by a distance of 5.3 × 10–11 m. Calculate the electric potential energy of the proton and the electron. electric potential energy = … J [2] [Total: 9]

9 marks

Mark scheme: 5(a) potential is (defined as) zero at infinity B1 proton has a positive charge and so repels another positive charge B1 work is done on two (positive) charges to move them towards each other B1 or work is done by two (positive) charges as they move apart from each other 5(b)(i) arrow drawn through Q in a WSW direction directly away from the centre of the sphere B1 5(b)(ii) curve in at least one quadrant passing through (R, E0) and (2R, ¼E0) B1 curve between –3R and –R of increasing magnitude of gradient B1 and curve between R and 3R of decreasing magnitude of gradient two lines drawn, one in the top right quadrant, the other in the bottom left quadrant B1 5(c) EP = – (1.60  10–19)2 / [4  8.85  10–12  (5.3  10–11)] C1 = –4.3  10–18 J A1

This question in 9702/44 Oct/Nov 2025