Capacitors and capacitance — Paper 4 · A Level Physics 9702

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 Two capacitors P and Q, each of capacitance C, are connected in series with a battery of e.m.f. 9.0 V, as shown in Fig. 6.1. Q C switch S 9.0 V X Y P T R C C Fig. 6.1 A switch S is used to connect either a third capacitor T, also of capacitance C, or a resistor R, in parallel with capacitor P. (a) Switch S is in position X. Calculate (i) the combined capacitance, in terms of C, of the three capacitors, capacitance = … [2] (ii) the potential difference across capacitor Q. Explain your working. potential difference = … V [2] (b) Switch S is now moved to position Y. State what happens to the potential difference across capacitor P and across capacitor Q. capacitor P: … … … capacitor Q: … … … [4] [Total: 8]

8 marks

Mark scheme: 6(a)(i) 1 / T = 1 / (2C) + 1 / C C1 T = ⅔C or 0.67C A1 6(a)(ii) same charge on Q as on combination B1 so p.d. is 6.0 V B1 6(b) P: p.d. will decrease (from 3.0 V) B1 to zero B1 Q: p.d. will increase (from 6.0 V) B1 to 9.0 V B1

This question in 9702/41 Oct/Nov 2017

6 Two capacitors P and Q, each of capacitance C, are connected in series with a battery of e.m.f. 9.0 V, as shown in Fig. 6.1. Q C switch S 9.0 V X Y P T R C C Fig. 6.1 A switch S is used to connect either a third capacitor T, also of capacitance C, or a resistor R, in parallel with capacitor P. (a) Switch S is in position X. Calculate (i) the combined capacitance, in terms of C, of the three capacitors, capacitance = … [2] (ii) the potential difference across capacitor Q. Explain your working. potential difference = … V [2] (b) Switch S is now moved to position Y. State what happens to the potential difference across capacitor P and across capacitor Q. capacitor P: … … … capacitor Q: … … … [4] [Total: 8]

8 marks

Mark scheme: 6(a)(i) 1 / T = 1 / (2C) + 1 / C C1 T = ⅔C or 0.67C A1 6(a)(ii) same charge on Q as on combination B1 so p.d. is 6.0 V B1 6(b) P: p.d. will decrease (from 3.0 V) B1 to zero B1 Q: p.d. will increase (from 6.0 V) B1 to 9.0 V B1

This question in 9702/43 Oct/Nov 2017

7 (a) Explain what is meant by the capacitance of a parallel plate capacitor. … … … … [3] (b) A parallel plate capacitor C is connected into the circuit shown in Fig. 7.1. X Y S A 120 V C Fig. 7.1 When switch S is at position X, the battery of electromotive force 120 V and negligible internal resistance is connected to capacitor C. When switch S is at position Y, the capacitor C is discharged through the sensitive ammeter. The switch vibrates so that it is first in position X, then moves to position Y and then back to position X fifty times each second. The current recorded on the ammeter is 4.5 μA. Determine (i) the charge, in coulomb, passing through the ammeter in 1.0 s, charge = … C [1] (ii) the charge on one plate of the capacitor, each time that it is charged, charge = … C [1] (iii) the capacitance of capacitor C. capacitance = … F [2] (c) A second capacitor, having a capacitance equal to that of capacitor C, is now placed in series with C. Suggest and explain the effect on the current recorded on the ammeter. … … … [2] [Total: 9]

9 marks

Mark scheme: 7(a) (capacitance =) charge / potential M1 charge is (numerically equal to) charge on one plate A1 potential is potential difference between plates A1 7(b)(i) 4.5 × 10–6 C A1 7(b)(ii) 9.0 × 10–8 C A1 7(b)(iii) capacitance = (9.0 × 10–8) / 120 C1 = 7.5 × 10–10 F A1 7(c) total capacitance is halved B1 current is halved B1

This question in 9702/41 May/June 2018

6 (a) Explain what is meant by the capacitance of a parallel plate capacitor. … … … … [3] (b) Three parallel plate capacitors each have a capacitance of 6.0 μF. Draw circuit diagrams, one in each case, to show how the capacitors may be connected together to give a combined capacitance of (i) 9.0 μF, [1] (ii) 4.0 μF. [1] (c) Two capacitors of capacitances 3.0 μF and 2.0 μF are connected in series with a battery of electromotive force (e.m.f.) 8.0 V, as shown in Fig. 6.1. 3.0 μF 2.0 μF 8.0 V Fig. 6.1 (i) Calculate the combined capacitance of the capacitors. capacitance = … μF [1] (ii) Use your answer in (i) to determine, for the capacitor of capacitance 3.0 μF, 1. the charge on one plate of the capacitor, charge = … μC 2. the energy stored in the capacitor. energy = … J [4] [Total: 10]

10 marks

Mark scheme: 6(a) capacitance = charge / potential M1 charge is (numerically equal to) charge on one plate A1 potential is potential difference between plates A1 6(b)(i) two in series, in parallel with the other (correct symbols) A1 6(b)(ii) two in parallel connected to one in series (correct symbols) A1 6(c)(i) capacitance = 1.2 µF A1 6(c)(ii) 1. Q = CV C1 = 1.2 × 8.0 = 9.6 µC A1 2. E = ½QV and V = Q / C or E = ½CV2 and V = Q / C or E = ½Q2 / C C1 E = ½ (9.6 × 10–6)2 / (3.0 × 10–6) = 1.5 × 10–5 J A1

This question in 9702/42 May/June 2018

7 (a) Explain what is meant by the capacitance of a parallel plate capacitor. … … … … [3] (b) A parallel plate capacitor C is connected into the circuit shown in Fig. 7.1. X Y S A 120 V C Fig. 7.1 When switch S is at position X, the battery of electromotive force 120 V and negligible internal resistance is connected to capacitor C. When switch S is at position Y, the capacitor C is discharged through the sensitive ammeter. The switch vibrates so that it is first in position X, then moves to position Y and then back to position X fifty times each second. The current recorded on the ammeter is 4.5 μA. Determine (i) the charge, in coulomb, passing through the ammeter in 1.0 s, charge = … C [1] (ii) the charge on one plate of the capacitor, each time that it is charged, charge = … C [1] (iii) the capacitance of capacitor C. capacitance = … F [2] (c) A second capacitor, having a capacitance equal to that of capacitor C, is now placed in series with C. Suggest and explain the effect on the current recorded on the ammeter. … … … [2] [Total: 9]

9 marks

Mark scheme: 7(a) (capacitance =) charge / potential M1 charge is (numerically equal to) charge on one plate A1 potential is potential difference between plates A1 7(b)(i) 4.5 × 10–6 C A1 7(b)(ii) 9.0 × 10–8 C A1 7(b)(iii) capacitance = (9.0 × 10–8) / 120 C1 = 7.5 × 10–10 F A1 7(c) total capacitance is halved B1 current is halved B1

This question in 9702/43 May/June 2018

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) State two different functions of capacitors in electrical circuits. 1. … … 2. … … [2] (b) Three uncharged capacitors of capacitances C1, C2 and C3 are connected in series with a battery of electromotive force (e.m.f.) E and a switch, as shown in Fig. 6.1. E C1 C2 C3 plate P charge +q Fig. 6.1 When the switch is closed, there is a charge + q on plate P of the capacitor of capacitance C1. Show that the combined capacitance C of the three capacitors is given by the expression 1 1 1 1 = + + . C C1 C2 C3 [3] (c) A student has available four capacitors, each of capacitance 20 μF. Draw circuit diagrams, one in each case, to show how the student may connect some or all of the capacitors to produce a combined capacitance of: (i) 60 μF [1] (ii) 15 μF. [1] [Total: 7]

7 marks

Mark scheme: 6(a) Any valid two points e.g.: • to store (electrical) energy • smoothing/reduce ripple (on direct voltages/currents) • to block d.c. • timing/time delay (circuits) • in oscillator (circuits) • in tuning (circuits) • to prevent arcing/sparks B2 6(b) clear indication of equal charge on each capacitor B1 E = V1 + V2 + V3 and V = Q / C M1 completion of algebra leading to 1 / C = 1 / C1 + 1 / C2 + 1 / C3 A1 6(c)(i) three capacitors connected in parallel B1 6(c)(ii) parallel combination of three capacitors connected in series with one capacitor B1

This question in 9702/41 May/June 2019

6 (a) State two different functions of capacitors in electrical circuits. 1. … … 2. … … [2] (b) Three uncharged capacitors of capacitances C1, C2 and C3 are connected in series with a battery of electromotive force (e.m.f.) E and a switch, as shown in Fig. 6.1. E C1 C2 C3 plate P charge +q Fig. 6.1 When the switch is closed, there is a charge + q on plate P of the capacitor of capacitance C1. Show that the combined capacitance C of the three capacitors is given by the expression 1 1 1 1 = + + . C C1 C2 C3 [3] (c) A student has available four capacitors, each of capacitance 20 μF. Draw circuit diagrams, one in each case, to show how the student may connect some or all of the capacitors to produce a combined capacitance of: (i) 60 μF [1] (ii) 15 μF. [1] [Total: 7]

7 marks

Mark scheme: 6(a) Any valid two points e.g.: • to store (electrical) energy • smoothing/reduce ripple (on direct voltages/currents) • to block d.c. • timing/time delay (circuits) • in oscillator (circuits) • in tuning (circuits) • to prevent arcing/sparks B2 6(b) clear indication of equal charge on each capacitor B1 E = V1 + V2 + V3 and V = Q / C M1 completion of algebra leading to 1 / C = 1 / C1 + 1 / C2 + 1 / C3 A1 6(c)(i) three capacitors connected in parallel B1 6(c)(ii) parallel combination of three capacitors connected in series with one capacitor B1

This question in 9702/43 May/June 2019

6 (a) (i) Define the capacitance of a parallel plate capacitor. … … … [2] (ii) State three functions of capacitors in electrical circuits. 1. … 2. … 3. … [3] (b) A student has available four capacitors, each of capacitance 24 μF. The capacitors are connected as shown in Fig. 6.1. 24 μF X 24 μF 24 μF 24 μF Y Fig. 6.1 Calculate the combined capacitance between the terminals X and Y. capacitance = … μF [2] [Total: 7]

7 marks

Mark scheme: 6(a)(i) charge per unit potential (difference) M1 charge on one plate and potential difference across the plates A1 6(a)(ii) any three points from: • smoothing • timing/(time) delay • tuning • oscillator • blocking d.c. • surge protection • temporary power supply B3 6(b) (capacitors in series have combined capacitance =) 8 μF C1 capacitance = 8 + 24 = 32 μF A1

This question in 9702/41 Oct/Nov 2020

6 (a) (i) Define the capacitance of a parallel plate capacitor. … … … [2] (ii) State three functions of capacitors in electrical circuits. 1. … 2. … 3. … [3] (b) A student has available three capacitors, each of capacitance 12 μF. Draw diagrams, one in each case, to show how the student connects the capacitors to give a combined capacitance between the terminals of: (i) 18 μF [1] (ii) 8 μF. [1] [Total: 7]

7 marks

Mark scheme: 6(a)(i) charge per unit potential (difference) M1 charge on one plate and potential difference between the plates A1 6(a)(ii) any three points from: • smoothing • timing/(time) delaying • tuning • oscillator • blocking d.c. • surge protection • temporary power supply B3 6(b)(i) parallel combination of two in series and a single capacitor B1 6(b)(ii) one capacitor in series with two in parallel B1

This question in 9702/42 Oct/Nov 2020

6 (a) (i) Define the capacitance of a parallel plate capacitor. … … … [2] (ii) State three functions of capacitors in electrical circuits. 1. … 2. … 3. … [3] (b) A student has available four capacitors, each of capacitance 24 μF. The capacitors are connected as shown in Fig. 6.1. 24 μF X 24 μF 24 μF 24 μF Y Fig. 6.1 Calculate the combined capacitance between the terminals X and Y. capacitance = … μF [2] [Total: 7]

7 marks

Mark scheme: 6(a)(i) charge per unit potential (difference) M1 charge on one plate and potential difference across the plates A1 6(a)(ii) any three points from: • smoothing • timing/(time) delay • tuning • oscillator • blocking d.c. • surge protection • temporary power supply B3 6(b) (capacitors in series have combined capacitance =) 8 μF C1 capacitance = 8 + 24 = 32 μF A1

This question in 9702/43 Oct/Nov 2020

6 (a) State a similarity between the gravitational field lines around a point mass and the electric field lines around a point charge. … … [1] (b) The variation with radius r of the electric field strength E due to an isolated charged sphere in a vacuum is shown in Fig. 6.1. 1.3 1.2 1.1 E / 105 V m–1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 6 r / cm Fig. 6.1 Use data from Fig. 6.1 to: (i) state the radius of the sphere radius = … cm [1] (ii) calculate the charge on the sphere. charge = … C [2] (c) Using the formula for the electric potential due to an isolated point charge, determine the capacitance of the sphere in (b). capacitance = … F [3] [Total: 7]

7 marks

Mark scheme: 6(a) (both have) radial field lines B1 6(b)(i) 2.1 cm B1 6(b)(ii) 2 4 o Q E r πε = e.g. r = 2.1 cm, E = 1.30 × 105 V m–1 2 4 o Q r E πε = 12 2 5 4 8.85 10 0.021 1.30 10 π − = × × × × × × C1 9 6.4 10 C − = × A1 Question Answer Marks 6(c) Q C V = either 4 o Q V r πε = leading to 4 o C r πε = C1 12 4 8.85 10 0.021 C π − = × × × × C1 ( ) 12 2.3 10 C − = × F A1 or 4 o Q V r πε = 9 12 6.4 10 4 8.85 10 0.021 π − − × = × × × × 2740 V = 9 6.4 10 2740 C − × = (C1) 12 2.3 10 F − = × (A1)

This question in 9702/42 Feb/March 2021

7 (a) State what is meant by the capacitance of a parallel plate capacitor. … … … [2] (b) A capacitor of capacitance C is connected into the circuit shown in Fig. 7.1. A B sensitive + ammeter V A – C Fig. 7.1 When the two-way switch is in position A, the capacitor is charged so that the potential difference across it is V. The switch moves to position B and the capacitor fully discharges through the sensitive ammeter. The switch moves repeatedly between A and B so that the capacitor charges and then discharges with frequency f. (i) Show that the average current I in the ammeter is given by the expression I = fCV. [2] (ii) For a potential difference V of 150 V and a frequency f of 60 Hz, the average current in the ammeter is 4.8 μA. Calculate the capacitance, in pF, of the capacitor. capacitance = … pF [2] (c) A second capacitor, having the same capacitance as the capacitor in (b), is connected into the circuit of Fig. 7.1. The two capacitors are connected in series. State and explain the new reading on the ammeter. new reading = … μA … … … [3] [Total: 9]

9 marks

Mark scheme: 7(a) charge / potential M1 charge is on one plate, potential is p.d. between the plates A1 7(b)(i) I = Q / t M1 charge = CV and time = 1 / f leading to I = fCV A1 7(b)(ii) 4.8 × 10–6 = 150 × 60 × C C1 C = 530 pF A1 7(c) (total) capacitance is halved B1 charge (for each cycle/discharge) is halved or since f and V are constant, current is proportional to capacitance B1 current = 2.4 μA B1

This question in 9702/41 May/June 2021

6 (a) Two flat metal plates are held a small distance apart by means of insulating pads, as shown in Fig. 6.1. metal plate insulating pad metal plate Fig. 6.1 Explain how the plates could act as a capacitor. … … … [2] (b) The arrangement in Fig. 6.1 has capacitance C. The arrangement is connected into the circuit of Fig. 6.2. A B sensitive ammeter V A C Fig. 6.2 When the two-way switch is moved to position A, the capacitor is charged so that the potential difference across it is V. When the switch moves to position B, the capacitor fully discharges through the sensitive ammeter. The switch moves repeatedly between A and B so that the capacitor charges and then discharges with frequency f. (i) Show that the average current I in the ammeter is given by I = CVf. [2] (ii) For a potential difference V of 180 V and a frequency f of switching of 50 Hz, the average current I in the ammeter is 2.5 μA. Calculate the capacitance, in pF, of the parallel plates. capacitance = … pF [2] (c) A second capacitor is connected into the circuit of Fig. 6.2. The two capacitors are connected in parallel. State and explain the change, if any, in the average current in the ammeter. … … … [2] [Total: 8]

8 marks

Mark scheme: 6(a) potential difference applied between the plates M1 causes charge separation (between the plates) or causes energy to be stored (between the plates) A1 6(b)(i) I = Q / t M1 clear substitution of Q = CV and f = 1 / t, leading to I = fCV A1 6(b)(ii) 2.5 × 10–6 = 50 × C × 180 C1 C = 280 pF A1 6(c) (total) capacitance increases B1 greater charge (for each cycle/discharge) so greater (average) current or V and f are constant so (average) current increases or I is (directly) proportional to C so (average) current increases B1

This question in 9702/42 May/June 2021

7 (a) State what is meant by the capacitance of a parallel plate capacitor. … … … [2] (b) A capacitor of capacitance C is connected into the circuit shown in Fig. 7.1. A B sensitive + ammeter V A – C Fig. 7.1 When the two-way switch is in position A, the capacitor is charged so that the potential difference across it is V. The switch moves to position B and the capacitor fully discharges through the sensitive ammeter. The switch moves repeatedly between A and B so that the capacitor charges and then discharges with frequency f. (i) Show that the average current I in the ammeter is given by the expression I = fCV. [2] (ii) For a potential difference V of 150 V and a frequency f of 60 Hz, the average current in the ammeter is 4.8 μA. Calculate the capacitance, in pF, of the capacitor. capacitance = … pF [2] (c) A second capacitor, having the same capacitance as the capacitor in (b), is connected into the circuit of Fig. 7.1. The two capacitors are connected in series. State and explain the new reading on the ammeter. new reading = … μA … … … [3] [Total: 9]

9 marks

Mark scheme: 7(a) charge / potential M1 charge is on one plate, potential is p.d. between the plates A1 7(b)(i) I = Q / t M1 charge = CV and time = 1 / f leading to I = fCV A1 7(b)(ii) 4.8 × 10–6 = 150 × 60 × C C1 C = 530 pF A1 7(c) (total) capacitance is halved B1 charge (for each cycle/discharge) is halved or since f and V are constant, current is proportional to capacitance B1 current = 2.4 μA B1

This question in 9702/43 May/June 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 9702/41 Oct/Nov 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

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 (a) Define the capacitance of a parallel plate capacitor. … … … [2] (b) Two capacitors, of capacitances C1 and C2, are connected in parallel to a power supply of electromotive force (e.m.f.) E, as shown in Fig. 5.1. E C1 C2 Fig. 5.1 Show that the combined capacitance CT of the two capacitors is given by CT = C1 + C2. Explain your reasoning. You may draw on Fig. 5.1 if you wish. [3] (c) Two capacitors of capacitances 22 μF and 47 μF, and a resistor of resistance 2.7 MΩ, are connected into the circuit of Fig. 5.2. 12 V X S 2.7 MΩ Y 22 μF 47 μF Fig. 5.2 The battery has an e.m.f. of 12 V. (i) Show that the combined capacitance of the two capacitors is 15 μF. [1] (ii) The two-way switch S is initially at position X, so that the capacitors are fully charged. Use the information in (c)(i) to calculate the total energy stored in the two capacitors. total energy = … J [2] (iii) The two-way switch is now moved to position Y. Determine the time taken for the potential difference (p.d.) across the 22 μF capacitor to become 6.0 V. time = … s [3] [Total: 11]

11 marks

Mark scheme: 5(a) charge / potential (difference) M1 charge is charge on one plate, and potential is p.d. across the plates A1 5(b) p.d. across both capacitors = E B1 QT = Q1 + Q2 B1 CTE = C1E + C2E hence CT = C1 + C2 B1 5(c)(i) [(1 / 22) + (1 / 47)]–1 = 15 F A1 5(c)(ii) energy = ½CV 2 C1 = ½  15  10–6  122 = 1.1  10–3 J A1 5(c)(iii) initial p.d. (across 22 F) = 12  (15 / 22) = 8.2 V or final p.d. across both capacitors = 6.0  (22 / 15) = 8.8 V C1 V = V0 exp [– t / (2.7  106  15  10–6)] C1 6.0 = 8.2 exp [– t / (2.7  106  15  10–6)] or 8.8 = 12 exp [– t / (2.7  106  15  10–6)] t = 13 s A1

This question in 9702/42 May/June 2022

5 A capacitor of capacitance 470 μF is connected to a battery of electromotive force (e.m.f.) 24 V in the circuit of Fig. 5.1. X Y S 24 V V 470 μF P Q 5.6 kΩ 5.6 kΩ Fig. 5.1 The two-way switch S is initially at position X. P and Q are identical long straight wires, each with a resistance of 5.6 kΩ. These wires are placed near to, and parallel to, each other. Wire Q is connected to a voltmeter. At time t = 0, switch S is moved to position Y so that the capacitor discharges through wire P. (a) (i) Calculate the charge Q0 on the capacitor at time t = 0. Q0 = … C [2] (ii) Calculate the current I0 in wire P at time t = 0. I0 = … A [1] (iii) Calculate the time constant τ of the discharge circuit. τ = … s [2] (iv) On Fig. 5.2, sketch a line to show the variation with t of the current I in wire P as the capacitor discharges. I0 I 0 0 t Fig. 5.2 [2] (b) (i) Explain why there is an induced e.m.f. across wire Q during the discharge of the capacitor. … … … … [3] (ii) On Fig. 5.3, sketch a line to suggest the variation with t of the voltmeter reading V. V 0 0 t Fig. 5.3 [1] [Total: 11]

11 marks

Mark scheme: 5(a)(i) Q = CV C1 Q0 = 24  470  10–6 A1 = 0.011 C 5(a)(ii) I0 = 24 / 5600 A1 = 4.3  10–3 A 5(a)(iii) = RC C1 = 5600  470  10–6 A1 = 2.6 s 5(a)(iv) line with negative gradient throughout passing through (0, I0) B1 exponential decay curve asymptotic to t-axis B1 5(b)(i) current in wire P gives rise to a magnetic field B1 as current (in P) changes, wire Q cuts (magnetic) flux (of wire P) B1 cutting magnetic flux causes induced e.m.f. (across Q) B1 5(b)(ii) sketch shows line with a negative gradient throughout B1

This question in 9702/41 Oct/Nov 2022

6 A capacitor of capacitance C and a resistor of resistance R are connected as shown in Fig. 6.1. C R Fig. 6.1 Initially, the capacitor is charged and the switch is open. The switch is closed at time t = 0. Fig. 6.2 and Fig. 6.3 show, respectively, the variations with t of the charge Q on the capacitor and the potential difference (p.d.) V across the resistor. 1.0 10 Q / mC V / V 0.5 5 0 0 0 5 10 15 0 5 10 15 t / s t / s Fig. 6.2 Fig. 6.3 (a) Explain the shape of the line in Fig. 6.3 representing the variation of V with t. … … … … … [3] (b) Use Fig. 6.2 to show that the time constant of the circuit in Fig. 6.1 is 5.5 s. [3] (c) Use Fig. 6.2, Fig. 6.3 and the information in (b) to determine: (i) capacitance C, in μF C = … μF [2] (ii) resistance R, in kΩ. R = … kΩ [2] [Total: 10]

10 marks

Mark scheme: 6(a) • p.d. across resistor = p.d. across capacitor B2 • current (in resistor) proportional to p.d. across it • current causes capacitor to lose charge • charge (on capacitor) proportional to p.d. so p.d. decreases Any two points, 1 mark each rate of change of p.d. decreases as p.d. decreases B1 6(b) Q0 = 0.90 mC and at t = one time constant, Q = Q0 exp (–1) B1 at t = one time constant, Q = 0.90 exp (–1) = 0.33 mC M1 evidence of graph reading: when Q = 0.33 mC, t = 5.5 s A1 or evidence of two correct sets of readings for Q and t from the graph (B1) correct substitution of Q and t values into Q2 = Q1 exp [(t1 – t2) / ] (M1) calculation to give = 5.5 s (A1) or read-off of half-life as 3.75 s (B1) use of Q = Q0 exp (–t / ) to show that = half-life / ln 2 (M1) = 3.75 / ln 2 = 5.4 s (A1) 6(b) or tangent drawn on Q–t graph and value of Q at exact same time as tangent read from graph (M1) gradient of tangent correctly calculated (A1) = Q / gradient used to correctly calculate a value for as 5.5 s (A1) 6(c)(i) C = Q / V C1 = [(0.90  10–3) / 7.5] = 1.2  10–4 C A1 = 120 F 6(c)(ii) R = τ / C C1 = 5.5 / (1.2  10–4) (= 45 800 ) A1 = 46 k

This question in 9702/42 Oct/Nov 2022

5 A capacitor of capacitance 470 μF is connected to a battery of electromotive force (e.m.f.) 24 V in the circuit of Fig. 5.1. X Y S 24 V V 470 μF P Q 5.6 kΩ 5.6 kΩ Fig. 5.1 The two-way switch S is initially at position X. P and Q are identical long straight wires, each with a resistance of 5.6 kΩ. These wires are placed near to, and parallel to, each other. Wire Q is connected to a voltmeter. At time t = 0, switch S is moved to position Y so that the capacitor discharges through wire P. (a) (i) Calculate the charge Q0 on the capacitor at time t = 0. Q0 = … C [2] (ii) Calculate the current I0 in wire P at time t = 0. I0 = … A [1] (iii) Calculate the time constant τ of the discharge circuit. τ = … s [2] (iv) On Fig. 5.2, sketch a line to show the variation with t of the current I in wire P as the capacitor discharges. I0 I 0 0 t Fig. 5.2 [2] (b) (i) Explain why there is an induced e.m.f. across wire Q during the discharge of the capacitor. … … … … [3] (ii) On Fig. 5.3, sketch a line to suggest the variation with t of the voltmeter reading V. V 0 0 t Fig. 5.3 [1] [Total: 11]

11 marks

Mark scheme: 5(a)(i) Q = CV C1 Q0 = 24  470  10–6 A1 = 0.011 C 5(a)(ii) I0 = 24 / 5600 A1 = 4.3  10–3 A 5(a)(iii) = RC C1 = 5600  470  10–6 A1 = 2.6 s 5(a)(iv) line with negative gradient throughout passing through (0, I0) B1 exponential decay curve asymptotic to t-axis B1 5(b)(i) current in wire P gives rise to a magnetic field B1 as current (in P) changes, wire Q cuts (magnetic) flux (of wire P) B1 cutting magnetic flux causes induced e.m.f. (across Q) B1 5(b)(ii) sketch shows line with a negative gradient throughout B1

This question in 9702/43 Oct/Nov 2022

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 Fig. 6.1 shows a capacitor of capacitance C connected in series with a resistor of resistance R. C R Fig. 6.1 Initially the switch is open and there is a p.d. of 12 V across the capacitor. At time t = 0, the switch is closed so that there is a current I in the resistor. Fig. 6.2 shows the variation of I with t. 0.2 I / mA 0.1 0 0 2 4 6 8 t / s Fig. 6.2 (a) Explain the shape of the line in Fig. 6.2. … … … … … [3] (b) Use Fig. 6.2 to determine: (i) resistance R R = … Ω [2] (ii) the time constant τ of the circuit in Fig. 6.1. τ = … s [3] (c) Use your answers in (b) to determine capacitance C. C = … F [2] [Total: 10]

10 marks

Mark scheme: 6(a) p.d. across capacitor proportional to charge on capacitor  p.d. across capacitor = p.d. across resistor  current in resistor proportional to p.d. across resistor  current in resistor = rate of decrease of charge on capacitor Any two points, 1 mark each B2 charge proportional to current so rate of decrease of current decreases as current decreases (therefore exponential shape) B1 6(b)(i) R = V / I = 12 / (0.13  10–3) C1 = 9.2  104  A1 6(b)(ii) correct read-off of at least one pair of values for I and t C1 attempted read-off of t when I = 0.048 mA or substitution of a correct pair of values of I and t into I = 0.13 exp (– t / ) C1  = 4.3 s A1 6(c)  = RC C1 C =  / R = 4.3 / (9.2  104) = 4.7  10–5 F A1

This question in 9702/41 May/June 2024

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

6 Fig. 6.1 shows a capacitor of capacitance C connected in series with a resistor of resistance R. C R Fig. 6.1 Initially the switch is open and there is a p.d. of 12 V across the capacitor. At time t = 0, the switch is closed so that there is a current I in the resistor. Fig. 6.2 shows the variation of I with t. 0.2 I / mA 0.1 0 0 2 4 6 8 t / s Fig. 6.2 (a) Explain the shape of the line in Fig. 6.2. … … … … … [3] (b) Use Fig. 6.2 to determine: (i) resistance R R = … Ω [2] (ii) the time constant τ of the circuit in Fig. 6.1. τ = … s [3] (c) Use your answers in (b) to determine capacitance C. C = … F [2] [Total: 10]

10 marks

Mark scheme: 6(a) p.d. across capacitor proportional to charge on capacitor  p.d. across capacitor = p.d. across resistor  current in resistor proportional to p.d. across resistor  current in resistor = rate of decrease of charge on capacitor Any two points, 1 mark each B2 charge proportional to current so rate of decrease of current decreases as current decreases (therefore exponential shape) B1 6(b)(i) R = V / I = 12 / (0.13  10–3) C1 = 9.2  104  A1 6(b)(ii) correct read-off of at least one pair of values for I and t C1 attempted read-off of t when I = 0.048 mA or substitution of a correct pair of values of I and t into I = 0.13 exp (– t / ) C1  = 4.3 s A1 6(c)  = RC C1 C =  / R = 4.3 / (9.2  104) = 4.7  10–5 F A1

This question in 9702/43 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

5 (a) A capacitor of capacitance C1 is connected in series with a second capacitor of capacitance C2. Show that the combined capacitance C of the two capacitors is given by 1 1 1 = + . C C1 C2 [2] (b) Three identical capacitors, each of capacitance C, are connected in a network as shown in Fig. 5.1. C X Y C C Fig. 5.1 The variation of the charge Q with the potential difference (p.d.) V between the terminals X and Y is shown in Fig. 5.2. 400 Q / μC 200 0 0 2 4 6 V / V Fig. 5.2 Show that C is equal to 44 µF. [3] (c) The capacitor network in Fig. 5.1 is charged and then connected to a resistor of resistance 54 kΩ. The capacitor network discharges through the resistor. (i) Determine the time constant τ of the circuit. Give a unit with your answer. τ = … unit … [2] (ii) Determine the time taken for the discharge current to reduce to 15% of the initial discharge current. time = … s [2] [Total: 9]

9 marks

Mark scheme: 5(a) Q = Q1 = Q2 and V = V1 + V2 M1 V = Q / C so: A1 Q / C = Q / C1 + Q / C2 leading to 1 / C = 1 / C1 + 1 / C2 5(b) total capacitance = C + ½C = (3 / 2)C C1 total capacitance = gradient C1 = 400  10–6 / 6.0 either: C = (2  400  10–6) / (3  6.0) = 4.4  10–5 F = 44 F A1 or: C = (2  400) / (3  6.0) = 44 F 5(c)(i) τ = RC C1 = 54  103  (3/2)  44  10–6 A1 = 3.6 s 5(c)(ii) 0.15 = exp(–t / 3.6) C1 t = 6.8 s A1

This question in 9702/42 Feb/March 2025

7 Fig. 7.1 shows a circuit containing a capacitor of capacitance C and a resistor of resistance R. C R Fig. 7.1 Initially, the switch is open and the potential difference (p.d.) across the capacitor is 12 V. The switch is closed at time t = 0 and the capacitor discharges through the resistor. Fig. 7.2 shows the variation of the charge Q on the capacitor with the p.d. VC across the capacitor as the capacitor discharges. Fig. 7.3 shows the variation of the current I in the resistor with the p.d. VR across the resistor as the capacitor discharges. 8 2 Q / mC I / mA 4 1 0 0 0 4 8 12 0 4 8 12 VC / V VR / V Fig. 7.2 Fig. 7.3 (a) State the relationship between VC and VR. … [1] (b) Determine: (i) the capacitance C, in μF C = … μF [2] (ii) the resistance R, in kΩ R = … kΩ [2] (iii) the time constant τ of the circuit. τ = … s [2] (c) Use Fig. 7.2, Fig. 7.3 and your answer in (a) to explain why the variation of Q with t is exponential in nature. … … … … … [3] [Total: 10]

10 marks

Mark scheme: 7(a) VC = VR B1 7(b)(i) C = Q / V C1 = (7.2 × 10–3) / 12 (= 6.0 × 10–4 F) A1 = 600 F 7(b)(ii) R = V / I C1 = 12 / (1.5 × 10–3) (= 8000 ) A1 = 8.0 k 7(b)(iii)  = RC C1 = 8000 × 6.0 × 10–4 A1 = 4.8 s 7(c) Any two points from: B2 • charge and current are both (directly) proportional to voltage • charge is (directly) proportional to current • current is the rate of change of charge Q is proportional to the rate of change of Q (so exponential variation) B1

This question in 9702/42 May/June 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 9702/41 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