Energy stored in a capacitor — Paper 4 · A Level Physics 9702

5 A capacitor C is charged using a supply of e.m.f. 8.0 V. It is then discharged through a For resistor R. Examiner’s The circuit is shown in Fig. 5.1. Use 8.0V R C Fig. 5.1 The variation with time t of the potential difference V across the resistor R during the discharge of the capacitor is shown in Fig. 5.2. 8 V / V 6 4 2 0 0 0.5 1.0 1.5 2.0 t / s Fig. 5.2 (a) During the first 1.0 s of the discharge of the capacitor, 0.13 J of energy is transferred to the resistor R. Show that the capacitance of the capacitor C is 4500 µF. [3] (b) Some capacitors, each of capacitance 4500 µF with a maximum working voltage of 6 V, For are available. Examiner’s Use Draw an arrangement of these capacitors that could provide a total capacitance of 4500 µF for use in the circuit of Fig. 5.1. [2]

5 marks

Mark scheme: 5 (a) at t = 1.0 s, V = 2.5 V C1 energy = ½CV 2 C1 0.13 = ½ × C × (8.02 – 2.52) M1 C = 4500 µF A0 [3] (b) use of two capacitors in series in all branches of combination M1 connected into correct parallel arrangement A1 [2] GCE A/AS LEVEL – May/June 2008 9702 04

This question in 9702/41 May/June 2008

5 A solid metal sphere, of radius r, is insulated from its surroundings. The sphere has For charge +Q. Examiner’s This charge is on the surface of the sphere but it may be considered to be a point charge at Use its centre, as illustrated in Fig. 5.1. +Q r Fig. 5.1 (a) (i) Define capacitance. … … [1] (ii) Show that the capacitance C of the sphere is given by the expression C = 4πε0r. [1] (b) The sphere has radius 36 cm. Determine, for this sphere, (i) the capacitance, capacitance = … F [1] (ii) the charge required to raise the potential of the sphere from zero to 7.0 × 105 V. For Examiner’s Use charge = … C [1] (c) Suggest why your calculations in (b) for the metal sphere would not apply to a plastic sphere. … … … … [3] (d) A spark suddenly connects the metal sphere in (b) to the Earth, causing the potential of the sphere to be reduced from 7.0 × 105 V to 2.5 × 105 V. Calculate the energy dissipated in the spark. energy = … J [3]

10 marks

Mark scheme: 5 (a) (i) ratio of charge (on body) and its potential B1 [1] (do not allow reference to plates of a capacitor) (ii) (potential at surface of sphere =) V = Q / 4πε0r M1 C = Q / V = 4πε0r A0 [1] (b) (i) C = 4 × π × 8.85 × 10–12 × 0.36 = 4.0 × 10–11 F (allow 1 s.f.) A1 [1] (ii) Q = CV = 4.0 × 10–11 × 7.0 × 105 = 2.8 × 10–5 C A1 [1] (c) plastic is an insulator / not a conductor / has no free electrons B1 charges do not move (on an insulator) B1 either so no single value for the potential or charge cannot be considered to be at centre B1 [3] (d) either energy = ½CV2 or energy = ½QV and C = Q/V C1 energy = ½ × 4 × 10–11 × {(7.0 × 105)2 – (2.5 × 105)2)} C1 = 8.6 J A1 [3] GCE A/AS LEVEL – May/June 2009 9702 04

This question in 9702/41 May/June 2009

5 (a) State two functions of capacitors in electrical circuits. For Examiner’s 1. … Use 2. … [2] (b) Three capacitors, each marked ‘30 μF, 6 V max’, are arranged as shown in Fig. 5.1. A B Fig. 5.1 Determine, for the arrangement shown in Fig. 5.1, (i) the total capacitance, capacitance = … μF [2] (ii) the maximum potential difference that can safely be applied between points A and B. potential difference = … V [2] (c) A capacitor of capacitance 4700 μF is charged to a potential difference of 18 V. It is then For partially discharged through a resistor. The potential difference is reduced to 12 V. Examiner’s Calculate the energy dissipated in the resistor during the discharge. Use energy = … J [3]

9 marks

Mark scheme: 5 (a) e.g. ‘storage of charge’ / storage of energy blocking of direct current producing of electrical oscillations smoothing (any two, 1 mark each) B2 [2] (b) (i) capacitance of parallel combination = 60 µF C1 total capacitance = 20 µF A1 [2] (ii) p.d. across parallel combination = ½ × p.d. across single capacitor C1 maximum is 9V A1 [2] (c) either energy = ½CV 2 or energy = ½QV and Q = CV C1 energy = ½ × 4700 × 10–6 × (182 – 122) C1 = 0.42 J A1 [3] GCE AS/A LEVEL – May/June 2010 9702 42

This question in 9702/42 May/June 2010

5 (a) State two functions of capacitors in electrical circuits. For Examiner’s 1. … Use 2. … [2] (b) Three capacitors, each marked ‘30 μF, 6 V max’, are arranged as shown in Fig. 5.1. A B Fig. 5.1 Determine, for the arrangement shown in Fig. 5.1, (i) the total capacitance, capacitance = … μF [2] (ii) the maximum potential difference that can safely be applied between points A and B. potential difference = … V [2] (c) A capacitor of capacitance 4700 μF is charged to a potential difference of 18 V. It is then For partially discharged through a resistor. The potential difference is reduced to 12 V. Examiner’s Calculate the energy dissipated in the resistor during the discharge. Use energy = … J [3]

9 marks

Mark scheme: 5 (a) e.g. ‘storage of charge’ / storage of energy blocking of direct current producing of electrical oscillations smoothing (any two, 1 mark each) B2 [2] (b) (i) capacitance of parallel combination = 60 µF C1 total capacitance = 20 µF A1 [2] (ii) p.d. across parallel combination = ½ × p.d. across single capacitor C1 maximum is 9V A1 [2] (c) either energy = ½CV 2 or energy = ½QV and Q = CV C1 energy = ½ × 4700 × 10–6 × (182 – 122) C1 = 0.42 J A1 [3] GCE AS/A LEVEL – May/June 2010 9702 43

This question in 9702/43 May/June 2010

3 A capacitor consists of two metal plates separated by an insulator, as shown in Fig. 3.1. For Examiner’s Use metal plate insulator metal plate Fig. 3.1 The potential difference between the plates is V. The variation with V of the magnitude of the charge Q on one plate is shown in Fig. 3.2. 20 15 Q / mC 10 5 00 5 10 15 V / V Fig. 3.2 (a) Explain why the capacitor stores energy but not charge. … … … … [3] (b) Use Fig. 3.2 to determine For Examiner’s (i) the capacitance of the capacitor, Use capacitance = … μF [2] (ii) the loss in energy stored in the capacitor when the potential difference V is reduced from 10.0 V to 7.5 V. energy = … mJ [2] (c) Three capacitors X, Y and Z, each of capacitance 10 μF, are connected as shown in For Fig. 3.3. Examiner’s Use Y X A B Z Fig. 3.3 Initially, the capacitors are uncharged. A potential difference of 12 V is applied between points A and B. Determine the magnitude of the charge on one plate of capacitor X. charge = … μC [3]

10 marks

Mark scheme: 3 (a) charges on plates are equal and opposite M1 so no resultant charge A1 energy stored because there is charge separation B1 [3] (b) (i) capacitance = Q / V C1 = (18 × 10–3) / 10 = 1800 µF A1 [2] (ii) use of area under graph or energy = ½CV2 C1 energy = 2.5 × 15.7 × 10–3 or energy = ½ × 1800 × 10–6 × (102 – 7.52) = 39 mJ A1 [2] (c) combined capacitance of Y & Z = 20 µF or total capacitance = 6.67 µF C1 p.d. across capacitor X = 8 V or p.d. across combination = 12 V C1 charge = 10 × 10–6 × 8 or 6.67 × 10–6 × 12 = 80 µC A1 [3] GCE AS/A LEVEL – May/June 2011 9702 42

This question in 9702/42 May/June 2011

3 A capacitor consists of two metal plates separated by an insulator, as shown in Fig. 3.1. For Examiner’s Use metal plate insulator metal plate Fig. 3.1 The potential difference between the plates is V. The variation with V of the magnitude of the charge Q on one plate is shown in Fig. 3.2. 20 15 Q / mC 10 5 00 5 10 15 V / V Fig. 3.2 (a) Explain why the capacitor stores energy but not charge. … … … … [3] (b) Use Fig. 3.2 to determine For Examiner’s (i) the capacitance of the capacitor, Use capacitance = … μF [2] (ii) the loss in energy stored in the capacitor when the potential difference V is reduced from 10.0 V to 7.5 V. energy = … mJ [2] (c) Three capacitors X, Y and Z, each of capacitance 10 μF, are connected as shown in For Fig. 3.3. Examiner’s Use Y X A B Z Fig. 3.3 Initially, the capacitors are uncharged. A potential difference of 12 V is applied between points A and B. Determine the magnitude of the charge on one plate of capacitor X. charge = … μC [3]

10 marks

Mark scheme: 3 (a) charges on plates are equal and opposite M1 so no resultant charge A1 energy stored because there is charge separation B1 [3] (b) (i) capacitance = Q / V C1 = (18 × 10–3) / 10 = 1800 µF A1 [2] (ii) use of area under graph or energy = ½CV2 C1 energy = 2.5 × 15.7 × 10–3 or energy = ½ × 1800 × 10–6 × (102 – 7.52) = 39 mJ A1 [2] (c) combined capacitance of Y & Z = 20 µF or total capacitance = 6.67 µF C1 p.d. across capacitor X = 8 V or p.d. across combination = 12 V C1 charge = 10 × 10–6 × 8 or 6.67 × 10–6 × 12 = 80 µC A1 [3] GCE AS/A LEVEL – May/June 2011 9702 43

This question in 9702/43 May/June 2011

4 (a) State two functions of capacitors in electrical circuits. For Examiner’s 1. … Use … 2. … … [2] (b) Three uncharged capacitors of capacitance C1, C2 and C3 are connected in series, as shown in Fig. 4.1. plate A C1 C2 C3 Fig. 4.1 A charge of +Q is put on plate A of the capacitor of capacitance C1. (i) State and explain the charges that will be observed on the other plates of the capacitors. You may draw on Fig. 4.1 if you wish. … … … [2] (ii) Use your answer in (i) to derive an expression for the combined capacitance of the capacitors. [2] (c) A capacitor of capacitance 12 μF is charged using a battery of e.m.f. 9.0 V, as shown in For Fig. 4.2. Examiner’s Use S1 S2 12 μF 20 μF 9.0 V Fig. 4.2 Switch S1 is closed and switch S2 is open. (i) The capacitor is now disconnected from the battery by opening S1. Calculate the energy stored in the capacitor. energy = … J [2] (ii) The 12 μF capacitor is now connected to an uncharged capacitor of capacitance 20 μF by closing S2. Switch S1 remains open. The total energy now stored in the two capacitors is 1.82 × 10–4 J. Suggest why this value is different from your answer in (i). … … [1]

9 marks

Mark scheme: 4 (a) e.g. storing energy separating charge blocking d.c. producing electrical oscillations tuning circuits smoothing preventing sparks timing circuits (any two sensible suggestions, 1 each, max 2) B2 [2] (b) (i) –Q (induced) on opposite plate of C1 B1 by charge conservation, charges are –Q, +Q, –Q, +Q, –Q B1 [2] (ii) total p.d. V = V1 + V2 + V3 B1 Q/C = Q/C1 + Q/C2 + Q/C3 B1 1/C = 1/C1 + 1/C2 + 1/C3 A0 [2] (c) (i) energy = ½CV 2 or energy = ½ QV and C = Q/V C1 = ½ × 12 × 10–6 × 9.02 = 4.9 × 10–4 J A1 [2] (ii) energy dissipated in (resistance of) wire/as a spark B1 [1] GCE AS/A LEVEL – October/November 2011 9702 43

This question in 9702/43 Oct/Nov 2011

5 (a) (i) Define capacitance. For Examiner’s … Use … [1] (ii) A capacitor is made of two metal plates, insulated from one another, as shown in Fig. 5.1. insulation metal plate Fig. 5.1 Explain why the capacitor is said to store energy but not charge. … … … … … [4] (b) Three uncharged capacitors X, Y and Z, each of capacitance 12 μF, are connected as shown in Fig. 5.2. Y X 12 ȝF A B Z 12 ȝF 12 ȝF Fig. 5.2 A potential difference of 9.0 V is applied between points A and B. (i) Calculate the combined capacitance of the capacitors X, Y and Z. For Examiner’s Use capacitance = … μF [2] (ii) Explain why, when the potential difference of 9.0 V is applied, the charge on one plate of capacitor X is 72 μC. … … … [2] (iii) Determine 1. the potential difference across capacitor X, potential difference = … V [1] 2. the charge on one plate of capacitor Y. charge = … μC [2]

12 marks

Mark scheme: 5 (a) (i) ratio of charge and potential (difference) / voltage (ratio must be clear) B1 [1] (ii) capacitor has equal magnitudes of (+)ve and (-)ve charge B1 total charge on capacitor is zero (so does not store charge) B1 (+)ve and (-)ve charges to be separated M1 work done to achieve this so stores energy A1 [4] (b) (i) capacitance of Y and Z together is 24 µF C1 1 / C = 1 / 24 + 1 / 12 C = 8.0 µF (allow 1 s.f.) A1 [2] (ii) some discussion as to why all charge of one sign on one plate of X B1 Q = (CV =) 8.0 × 10–6 × 9.0 M1 = 72 µC A0 [2] (iii) 1. V = (72 × 10–6) / (12 × 10–6) = 6.0 V (allow 1 s.f.) (allow 72/12) A1 [1] 2. either Q = 12 × 10–6 × 3.0 or charge is shared between Y and Z C1 charge = 36 µC A1 [2] Must have correct voltage in (iii)1 if just quote of 36 µC in (iii)2.

This question in 9702/41 Oct/Nov 2012

5 (a) (i) Define capacitance. For Examiner’s … Use … [1] (ii) A capacitor is made of two metal plates, insulated from one another, as shown in Fig. 5.1. insulation metal plate Fig. 5.1 Explain why the capacitor is said to store energy but not charge. … … … … … [4] (b) Three uncharged capacitors X, Y and Z, each of capacitance 12 μF, are connected as shown in Fig. 5.2. Y X 12 ȝF A B Z 12 ȝF 12 ȝF Fig. 5.2 A potential difference of 9.0 V is applied between points A and B. (i) Calculate the combined capacitance of the capacitors X, Y and Z. For Examiner’s Use capacitance = … μF [2] (ii) Explain why, when the potential difference of 9.0 V is applied, the charge on one plate of capacitor X is 72 μC. … … … [2] (iii) Determine 1. the potential difference across capacitor X, potential difference = … V [1] 2. the charge on one plate of capacitor Y. charge = … μC [2]

12 marks

Mark scheme: 5 (a) (i) ratio of charge and potential (difference) / voltage (ratio must be clear) B1 [1] (ii) capacitor has equal magnitudes of (+)ve and (-)ve charge B1 total charge on capacitor is zero (so does not store charge) B1 (+)ve and (-)ve charges to be separated M1 work done to achieve this so stores energy A1 [4] (b) (i) capacitance of Y and Z together is 24 µF C1 1 / C = 1 / 24 + 1 / 12 C = 8.0 µF (allow 1 s.f.) A1 [2] (ii) some discussion as to why all charge of one sign on one plate of X B1 Q = (CV =) 8.0 × 10–6 × 9.0 M1 = 72 µC A0 [2] (iii) 1. V = (72 × 10–6) / (12 × 10–6) = 6.0 V (allow 1 s.f.) (allow 72/12) A1 [1] 2. either Q = 12 × 10–6 × 3.0 or charge is shared between Y and Z C1 charge = 36 µC A1 [2] Must have correct voltage in (iii)1 if just quote of 36 µC in (iii)2.

This question in 9702/42 Oct/Nov 2012

7 A student sets up the circuit shown in Fig. 7.1 to measure the charge on a capacitor C for different values of potential difference across the capacitor. C meter to V measure charge Fig. 7.1 The variation with potential difference V of the charge Q stored on the capacitor is shown in Fig. 7.2. 15 Q / mC 10 5 0 0 1 2 3 4 5 6 V / V Fig. 7.2 (a) State and explain how Fig. 7.2 indicates that there is a systematic error in the readings of one of the meters. … … … [2] (b) Use Fig. 7.2 to determine the capacitance, in μF, of capacitor C. capacitance = … μF [3] (c) Use your answer in (b) to determine the additional energy stored in the capacitor C when the potential difference across it is increased from 6.0 V to 9.0 V. energy = … J [3] [Total: 8]

8 marks

Mark scheme: 7 (a) line has non-zero intercept/line does not pass through origin B1 charge is/should be proportional to potential (difference) or charge is/should be zero when p.d. is zero (therefore there is a systematic error) B1 [2] (b) reasonable attempt at line of best fit B1 use of gradient of line of best fit clear M1 C = 2800 µF (allow ± 200 µF) A1 [3] (c) energy = ½ CV 2 or energy = ½ QV and C = Q / V C1 ∆ energy = ½ × 2800 × 10–6 × (9.02 – 6.02) C1 = 6.3 × 10–2 J A1 [3]

This question in 9702/41 May/June 2016

7 A student sets up the circuit shown in Fig. 7.1 to measure the charge on a capacitor C for different values of potential difference across the capacitor. C meter to V measure charge Fig. 7.1 The variation with potential difference V of the charge Q stored on the capacitor is shown in Fig. 7.2. 15 Q / mC 10 5 0 0 1 2 3 4 5 6 V / V Fig. 7.2 (a) State and explain how Fig. 7.2 indicates that there is a systematic error in the readings of one of the meters. … … … [2] (b) Use Fig. 7.2 to determine the capacitance, in μF, of capacitor C. capacitance = … μF [3] (c) Use your answer in (b) to determine the additional energy stored in the capacitor C when the potential difference across it is increased from 6.0 V to 9.0 V. energy = … J [3] [Total: 8]

8 marks

Mark scheme: 7 (a) line has non-zero intercept/line does not pass through origin B1 charge is/should be proportional to potential (difference) or charge is/should be zero when p.d. is zero (therefore there is a systematic error) B1 [2] (b) reasonable attempt at line of best fit B1 use of gradient of line of best fit clear M1 C = 2800 µF (allow ± 200 µF) A1 [3] (c) energy = ½ CV 2 or energy = ½ QV and C = Q / V C1 ∆ energy = ½ × 2800 × 10–6 × (9.02 – 6.02) C1 = 6.3 × 10–2 J A1 [3]

This question in 9702/43 May/June 2016

7 A capacitor consists of two parallel metal plates, separated by an insulator, as shown in Fig. 7.1. insulator metal plates Fig. 7.1 (a) Suggest why, when the capacitor is connected across the terminals of a battery, the capacitor stores energy, not charge. … … … [2] (b) Define the capacitance of the capacitor. … … … [2] (c) The capacitor is charged so that the potential difference between its plates is V0. The capacitor is then connected across a resistor for a short time. It is then disconnected. 1 The energy stored in the capacitor is reduced to of its initial value. 16 Determine, in terms of V0, the potential difference across the capacitor. potential difference = … [2] [Total: 6]

6 marks

Mark scheme: 7(a) equal and opposite charges on the plates so no resultant charge B1 +ve and –ve charges separated so energy stored B1 7(b) charge / potential difference M1 reference to charge on one plate and p.d. between plates A1 7(c) energy = ½ CV2 or energy = ½ QV and C = Q / V C1 (1 / 16) × ½ CV0 2 = ½ CV2 V = ¼ V0 A1

This question in 9702/42 May/June 2017

6 (a) Define the capacitance of a parallel-plate capacitor. … … … [2] (b) A student has three capacitors. Two of the capacitors have a capacitance of 4.0 μF and one has a capacitance of 8.0 μF. Draw labelled circuit diagrams, one in each case, to show how the three capacitors may be connected to give a total capacitance of: (i) 1.6 μF [1] (ii) 10 μF. [1] (c) A capacitor C of capacitance 47 μF is connected across the output terminals of a bridge rectifier, as shown in Fig. 6.1. C bridge R rectifier 47 μF Fig. 6.1 The variation with time t of the potential difference V across the resistor R is shown in Fig. 6.2. 10 8 V / V 6 4 2 0 0 t1 t2 time t Fig. 6.2 Use data from Fig. 6.2 to determine the energy transfer from the capacitor C to the resistor R between time t1 and time t2. energy = … J [3] [Total: 7]

7 marks

Mark scheme: 6(a) charge / potential (difference) M1 charge on one plate, p.d. between the plates A1 6(b)(i) all three capacitors connected in series B1 6(b)(ii) 8 ( µF) in parallel with the two 4 (µF) capacitors connected in series B1 6(c) discharge from 7.0 V to 4.0 V C1 Either energy = ½CV2 or energy = ½ QV and C = Q / V C1 energy = ½ × 47 × 10–6 × (72 – 42) = 7.8 × 10–4 J A1

This question in 9702/42 Feb/March 2019

6 (a) A capacitor consists of two parallel metal plates, separated by air, at a variable distance x apart, as shown in Fig. 6.1. The capacitance C is inversely proportional to x. x metal plates Fig. 6.1 The capacitor is charged by a supply so that there is a potential difference (p.d.) V between the plates. State expressions, in terms of C and V, for the charge Q on one of the plates and for the energy E stored in the capacitor. Q = … E = … [1] (b) The charged capacitor in (a) is now disconnected from the supply. The plates of the capacitor are initially separated by distance L. They are then moved closer together by a distance D, as shown in Fig. 6.2. D new position original position L Fig. 6.2 State expressions, in terms of C, V, L and D, for: (i) the new capacitance CN CN = … [1] (ii) the new charge QN on one of the plates QN = … [1] (iii) the new p.d. VN between the plates. VN = … [1] (c) Explain whether reducing the separation of the plates in (b) results in an increase or decrease in the energy stored in the capacitor. … … … [1] [Total: 5]

5 marks

Mark scheme: 6(a) Q = CV and E = ½CV2 B1 6(b)(i) CN = CL / (L – D) B1 6(b)(ii) (charge is unchanged by moving the plates so) QN = CV B1 6(b)(iii) VN = QN / CN = (CV) / [CL / (L – D)] = V(L – D) / L B1 6(c) oppositely charged plates attract, so energy stored decreases B1

This question in 9702/41 Oct/Nov 2021

6 (a) A capacitor consists of two parallel metal plates, separated by air, at a variable distance x apart, as shown in Fig. 6.1. The capacitance C is inversely proportional to x. x metal plates Fig. 6.1 The capacitor is charged by a supply so that there is a potential difference (p.d.) V between the plates. State expressions, in terms of C and V, for the charge Q on one of the plates and for the energy E stored in the capacitor. Q = … E = … [1] (b) The charged capacitor in (a) is now disconnected from the supply. The plates of the capacitor are initially separated by distance L. They are then moved closer together by a distance D, as shown in Fig. 6.2. D new position original position L Fig. 6.2 State expressions, in terms of C, V, L and D, for: (i) the new capacitance CN CN = … [1] (ii) the new charge QN on one of the plates QN = … [1] (iii) the new p.d. VN between the plates. VN = … [1] (c) Explain whether reducing the separation of the plates in (b) results in an increase or decrease in the energy stored in the capacitor. … … … [1] [Total: 5]

5 marks

Mark scheme: 6(a) Q = CV and E = ½CV2 B1 6(b)(i) CN = CL / (L – D) B1 6(b)(ii) (charge is unchanged by moving the plates so) QN = CV B1 6(b)(iii) VN = QN / CN = (CV) / [CL / (L – D)] = V(L – D) / L B1 6(c) oppositely charged plates attract, so energy stored decreases B1

This question in 9702/43 Oct/Nov 2021

5 The variation with potential difference V of the charge Q on one of the plates of a capacitor is shown in Fig. 5.1. 1.8 1.6 Q / 10–4 C 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 V / V Fig. 5.1 The capacitor is connected to an 8.0 V power supply and two resistors R and S as shown in Fig. 5.2. 8.0 V R 25 kΩ X Y S 220 kΩ Fig. 5.2 The resistance of R is 25 kΩ and the resistance of S is 220 kΩ. The switch can be in either position X or position Y. (a) The switch is in position X so that the capacitor is fully charged. Calculate the energy E stored in the capacitor. E = … J [2] (b) The switch is now moved to position Y. (i) Show that the time constant of the discharge circuit is 3.3 s. [2] (ii) The fully charged capacitor in (a) stores energy E. Determine the time t taken for the stored energy to decrease from E to E / 9. t = … s [4] (c) A second identical capacitor is connected in parallel with the first capacitor. State and explain the change, if any, to the time constant of the discharge circuit. … … … [2] [Total: 10]

10 marks

Mark scheme: 5(a) (energy stored =) area under line or ½ QV = ½ × 8.0 × 1.2 × 10-4 = 4.8 × 10–4 J A1 5(b)(i) (τ=) RC C1 (τ=) 220 × 103 × (1.2 × 10-4/8.0) = 3.3 s A1 5(b)(ii) E ∝ V2 C1 (so time to) Vo / 3 tRC o V = V e − C1 t o 3.3 o V = V e 3 − t3.3 1 = e 3 − C1 t = 3.6 s A1 5(c) (total) capacitance is doubled M1 time constant is doubled A1

This question in 9702/42 Feb/March 2022

5 A capacitor, a battery of electromotive force (e.m.f.) 12 V, a resistor R and a two-way switch are connected in the circuit shown in Fig. 5.1. R T S 12 V Fig. 5.1 The switch is initially in position S. When the capacitor is fully charged, the switch is moved to position T so that the capacitor discharges. At time t after the switch is moved the charge on the capacitor is Q. The variation with t of ln (Q / μC) is shown in Fig. 5.2. 3 ln (Q /μC) 2 1 0 0 1 2 3 4 5 t / s Fig. 5.2 (a) Show that the capacitance of the capacitor is 1.5 μF. [3] (b) Determine the resistance of R. resistance = … Ω [3] (c) Calculate the energy stored in the capacitor at time t = 0. energy = … J [2] (d) A second identical resistor is now connected in parallel with R. The switch is initially in position S. When the capacitor is fully charged, the switch is moved to position T so that the capacitor discharges. At time t after the switch is moved the charge on the capacitor is Q. On Fig. 5.2, sketch a line to show the variation of ln (Q / μC) with t between time t = 0 and time t = 5.0 s. [2] [Total: 10]

10 marks

Mark scheme: 5(a) from graph ln Q = 2.9 B1 (so Q = 18.2 C) C = Q / V C1 = 18.2 / 12 = 1.5 F A1 5(b) gradient = –0.25 C1 gradient = –1 / RC C1 R = 1 / (0.25  1.5  10–6) A1 = 2.7  106  Q − tCR − t (C1) or = e or ln Q – ln Q0 = Q0 CR −5.2 −6 (C1) 4.95 (1.5  10 R ) e.g. = e or 1.6 – 2.9 = 5.2 / (1.5 × 10–6R) 18.2 R = 2.7  106  (A1) 5(c) W = ½ QV C1 = ½  18.2  10–6  12 A1 = 1.1  10–4 J or W = ½ CV2 (C1) = ½  1.5  10–6  122 (A1) = 1.1  10–4 J or W = ½ Q2 / C (C1) = ½  (18.2  10–6)2 / 1.5  10–6 (A1) = 1.1  10–4 J 5(d) straight line with different negative gradient starting from (0, 2.9) M1 straight line between t = 0 and at least t = 5.0 s with twice the gradient of the original line A1

This question in 9702/42 Feb/March 2023

5 Two capacitors A and B are connected into the circuit shown in Fig. 5.1. X A S Y B Fig. 5.1 Capacitor A has capacitance C and capacitor B has capacitance 3C. The electromotive force (e.m.f.) of the cell is V. The two-way switch S is initially at position X, and capacitor B is initially uncharged. (a) State, in terms of V and C, expressions for: (i) the initial charge QA on the plates of capacitor A QA = … [1] (ii) the initial energy EA stored in capacitor A. EA = … [1] (b) The two-way switch S is now moved to position Y. (i) State and explain what happens to the charge that was initially on the plates of capacitor A. … … … [2] (ii) Show that the final potential difference (p.d.) VB across capacitor B is given by V VB = . 4 Explain your reasoning. [3] (iii) Determine an expression, in terms of V and C, for the decrease ΔE in the total energy that is stored in the capacitors as a result of the change of the position of the switch. ΔE = … [2] [Total: 9]

9 marks

Mark scheme: 5(a)(i) QA = CV A1 5(a)(ii) EA = ½CV2 A1 5(b)(i) some of the charge transfers to (the plates of) capacitor B B1 transfer is because the p.d.s across the capacitors are not equal or transfer stops when the p.d.s across the capacitors become equal B1 5(b)(ii) VA = VB M1 charge on A + charge on B = CV M1 CVB + 3CVB = CV leading to VB = V / 4 A1 or CT = 4C (M1) QT = CV (M1) VB = CV / 4C = V / 4 (A1) 5(b)(iii) E = ½CV2 – nCV2, where n is a multiple that is less than ½ or total final energy = ½  4C  (V / 4)2 = ⅛CV2 C1 E = ½CV2 – ⅛CV2 = ⅜CV2 A1

This question in 9702/42 May/June 2023

6 A capacitor C is charged so that the potential difference (p.d.) V across its terminals is 8.0 V. The capacitor is connected into the circuit of Fig. 6.1. C 8.0 V R Fig. 6.1 The switch is initially open. The switch is closed at time t = 0. (a) Fig. 6.2 shows the variation of V with the charge Q on the plates of capacitor C as the capacitor discharges. 8 V / V 4 0 0 200 400 600 Q / μC Fig. 6.2 (i) Show that the energy stored in capacitor C at time t = 0 is 1.8 mJ. [2] (ii) Determine the capacitance of capacitor C. Give a unit with your answer. capacitance = … unit … [2] V(b) Fig. 6.3 shows the variation with t of –ln  8.0 V. 2.0 V –ln 1 8.0 V2 1.0 0 0 2 4 6 8 t / s Fig. 6.3 V (i) Show that, when t is equal to one time constant, the value of –ln is equal to 1.0.  8.0 V [2] (ii) Determine the time constant τ of the circuit in Fig. 6.1. τ = … s [1] (iii) Calculate the resistance of resistor R. resistance = … Ω [2] [Total: 9]

9 marks

Mark scheme: 6(a)(i) energy stored = area under graph C1 = ½  450  10–6  8.0 = 1.8  10–3 J or 1.8 mJ A1 6(a)(ii) C = Q / V or E = ½CV2 C1 C = (450  10–6) / 8.0 or (2  1.8  10–3) / 8.02 A1 = 5.6  10–5 F 6(b)(i) V = V0 exp (– t / RC) and = RC C1 V = V0 exp (– t / ) A1 V0 = 8.0 V, and at one time constant, t =  V / 8.0 = exp (– / ), so ln (V / 8.0) = –1.0 or –ln (V / 8.0) = 1.0 6(b)(ii) [t read from graph at –ln (V / 8.0) = 1.0]: = 3.2 s A1 6(b)(iii)  = RC C1 R = 3.2 / (5.6  10–5) A1 = 5.7  104 

This question in 9702/42 Oct/Nov 2023

6 (a) Two capacitors X and Y are connected in series to a power supply of voltage V, as shown in Fig. 6.1. V X Y Fig. 6.1 The capacitance of X is CX and the capacitance of Y is CY. Derive an expression, in terms of CX and CY, for the combined capacitance CT of the capacitors in this circuit. Explain your reasoning. [3] (b) Two capacitors P and Q are connected in parallel to a power supply of voltage V. The capacitance of P is 200 μF. The capacitance CQ of Q can be varied between 0 and 400 μF. When CQ = 0, the total energy stored in the capacitors is 2.5 mJ. (i) Show that the supply voltage V is 5.0 V. [2] (ii) Calculate the total energy, in mJ, stored in the capacitors when CQ has its maximum value. total energy = … mJ [3] (iii) On Fig. 6.2, sketch the variation of the total energy E stored in the capacitors with CQ, as CQ varies from 0 to 400 μF. 10.0 E / mJ 7.5 5.0 2.5 0 0 100 200 300 400 CQ / μF Fig. 6.2 [2] [Total: 10]

10 marks

Mark scheme: 6(a) equal charge on both capacitors B1 VX + VY = V M1 (Q / CX) + (Q / CY) = (Q / CT) leading to (1 / CX) + (1 / CY) = (1 / CT) or (VX/ Q) + (VY/ Q) = (V / Q) leading to (1 / CX) + (1 / CY) = (1 / CT) A1 6(b)(i) E = ½CV 2 C1 V = √[(2  2.5  10–3) / (200  10–6)] = 5.0 V A1 6(b)(ii) total capacitance = 600 F C1 E = ½  600  10–6  5.02 ( = 7.5  10–3 J) C1 = 7.5 mJ A1 6(b)(iii) line with positive gradient starting at (0, 2.5) B1 straight line passing through (400, 7.5) B1

This question in 9702/42 May/June 2024

7 (a) Define the capacitance of a parallel‑plate capacitor. … … … [2] (b) An initially uncharged capacitor X, of capacitance C, is gradually charged so that the final potential difference (p.d.) between its plates is V and the final charge is Q. (i) On Fig. 7.1, sketch the variation of charge with p.d. for capacitor X as the p.d. increases from 0 to V. Q charge 0 0 V p.d. Fig. 7.1 [2] (ii) Determine an expression, in terms of Q and V, for the work W done on capacitor X during the charging process. Explain your reasoning. W = … [2] (c) Another capacitor Y is initially uncharged. The fully charged capacitor X in (b) is now connected to capacitor Y, as shown in Fig. 7.2. X Y Fig. 7.2 The capacitance of capacitor Y is 3C. (i) Complete Table 7.1 to show expressions, in terms of Q and V, for the final p.d.s across, and the final charges on, the two capacitors. Use the space below for any working that you need. Table 7.1 X Y final p.d. final charge [3] (ii) State whether the total energy stored in the two capacitors is less than, the same as, or greater than the energy initially stored in capacitor X. … [1] [Total: 10]

10 marks

Mark scheme: 7(a) charge / potential (difference) M1 charge is charge on one plate, and potential is p.d. between the plates A1 7(b)(i) straight line starting at the origin B1 line with positive gradient ending at (V, Q) B1 7(b)(ii) work done is the area under the graph B1 W = ½QV A1 7(c)(i) final p.d. shown as V / 4 for both capacitors B1 final charges add together to give Q B1 charge on Y = 3  charge on X (and both charges shown as a multiple of Q) B1 Fully correct answer: X Y final p.d. V / 4 V / 4 final charge Q / 4 3Q / 4 7(c)(ii) less than B1

This question in 9702/42 Oct/Nov 2024

6 (a) Two parallel plate capacitors C1 and C2 are connected to a supply that has a potential difference (p.d.) VS. The capacitors may be connected in series or in parallel. The supply provides charge QS and the plates of the two capacitors acquire charges Q1 and Q2 respectively. The p.d.s across the plates of the capacitors are V1 and V2 respectively. Complete Table 6.1 to indicate how QS, Q1 and Q2 relate to each other, and how VS, V1 and V2 relate to each other, for series and parallel connections of the capacitors to the supply. Table 6.1 relationship between charges relationship between p.d.s series parallel [4] (b) An isolated capacitor of capacitance 470 μF stores 19 mJ of energy. (i) Calculate the p.d. across the capacitor. p.d. = … V [2] (ii) Calculate the charge on the capacitor. charge = … C [2] (iii) The capacitor is now connected in parallel with a capacitor of capacitance 180 μF that is initially uncharged. Determine the total energy, in mJ, now stored in the two capacitors. energy = … mJ [3] [Total: 11]

11 marks

Mark scheme: 6(a) series charges: QS = Q1 = Q2 B1 series p.d.s: VS = V1 + V2 B1 parallel charges: QS = Q1 + Q2 B1 parallel p.d.s: VS = V1 = V2 B1 6(b)(i) E = ½ CV2 C1 p.d. = [(2  19  10–3) / (470  10–6)]½ A1 = 9.0 V 6(b)(ii) E = Q2 / 2C or C = Q / V C1 Q = (19 × 10–3 × 2 × 470 × 10–6)½ A1 or Q = 470 × 10–6 × 9.0 Q = 4.2  10–3 C 6(b)(iii) total charge unchanged C1 total capacitance = (470 + 180)  10–6 (F) C1 E = Q2 / 2C = (4.23  10–3)2 / (2  650  10–6) (= 0.014 J) A1 E = 14 mJ

This question in 9702/41 Oct/Nov 2025

6 (a) Two parallel plate capacitors C1 and C2 are connected to a supply that has a potential difference (p.d.) VS. The capacitors may be connected in series or in parallel. The supply provides charge QS and the plates of the two capacitors acquire charges Q1 and Q2 respectively. The p.d.s across the plates of the capacitors are V1 and V2 respectively. Complete Table 6.1 to indicate how QS, Q1 and Q2 relate to each other, and how VS, V1 and V2 relate to each other, for series and parallel connections of the capacitors to the supply. Table 6.1 relationship between charges relationship between p.d.s series parallel [4] (b) An isolated capacitor of capacitance 470 μF stores 19 mJ of energy. (i) Calculate the p.d. across the capacitor. p.d. = … V [2] (ii) Calculate the charge on the capacitor. charge = … C [2] (iii) The capacitor is now connected in parallel with a capacitor of capacitance 180 μF that is initially uncharged. Determine the total energy, in mJ, now stored in the two capacitors. energy = … mJ [3] [Total: 11]

11 marks

Mark scheme: 6(a) series charges: QS = Q1 = Q2 B1 series p.d.s: VS = V1 + V2 B1 parallel charges: QS = Q1 + Q2 B1 parallel p.d.s: VS = V1 = V2 B1 6(b)(i) E = ½ CV2 C1 p.d. = [(2  19  10–3) / (470  10–6)]½ A1 = 9.0 V 6(b)(ii) E = Q2 / 2C or C = Q / V C1 Q = (19 × 10–3 × 2 × 470 × 10–6)½ A1 or Q = 470 × 10–6 × 9.0 Q = 4.2  10–3 C 6(b)(iii) total charge unchanged C1 total capacitance = (470 + 180)  10–6 (F) C1 E = Q2 / 2C = (4.23  10–3)2 / (2  650  10–6) (= 0.014 J) A1 E = 14 mJ

This question in 9702/43 Oct/Nov 2025