Physics Term 2 STPM Chapter 13 Capacitors 45 13 6. Since a conductor is able to store charges, the concept of capacitance can also be applied to a conductor. The capacitance of a conductor is the ratio of the charge Q on the conductor to the electric potential V of the conductor. Capacitance of conductor = Charge on conductor ————————————––— Electric potential of conductor C = Q —V 7. The unit of capacitance is farad (F). The farad (1 F) is the capacitance of a capacitor that has a charge of one coulomb (1 C) on each plate when the potential difference between the plates is one volt (1 V). Capacitors with capacitance of a few PF are used in simple radio receiver circuits and large capacitors capacitance of a few MF are used in electrical appliances such as washing machines. Quick Check 1 1. A capacitor stores 24 mC of charge when the potential difference between the plates is 12 V. What is the capacitance of the capacitor? 2. A 50 PF capacitor is charged by connecting it to a 6.0 V battery. What is the charge in the capacitor when the potential difference across the capacitor is (a) 3.0 V? (b) 6.0 V? Sketch a graph to show how the charge Q in the capacitor varies with the potential difference V across the capacitor. 3. The capacitor of variable capacitance is connected to a 12 V supply. Its capacitance is 50 μF. After the capacitor is charged, the voltage supply is disconnected and the capacitance is changed to 100 μF. What is the new potential difference across the capacitor? 13.2 Parallel-Plate Capacitors Students should be able to: GHVFULEHWKHPHFKDQLVPRIFKDUJLQJDSDUDOOHOSODWHFDSDFLWRU XVHWKHIRUPXODC = Q V WRGHULYHC = H0 A d IRUWKHFDSDFLWDQFHRIDSDUDOOHOSODWHFDSDFLWRU Learning Outcomes 1. Figure 13.5 shows a parallel-plate capacitor which consists of two parallel metal plates each of area A and separated by a distance d in free space or vacuum. The capacitor is charged to a potential difference V. 2. The charge on each plate is Q. Using Gauss’s law, ε0 ) = 6Q ε0 (EA) = 6Q E = Q ε0 A ……… Figure 13.5 Q Q E V d 2016/P2/Q5
Physics Term 2 STPM Chapter 13 Capacitors 46 13 3. The electric fi eld E between the plates is also given by E = V d ……… Equating and , Q ε0 A = V d Capacitance, C = Q V = ε0 A d where ε0 = 8.85 u 10–12 F m–1 is the permittivity of free space between the plates Example 1 A capacitor is formed using two parallel metal plates, each measuring 15 cm u 20 cm separated by a distance of 0.50 cm. (a) Find the capacitance of the capacitor. (b) If the plate separation is increased to 1.0 cm, what is the new capacitance? (c) The new capacitor is connected to a battery of 12 V. What is the charge in the capacitor? Solution: (a) Capacitance, C = ε0 A—– d = (8.85 u 10–12)(0.15 u 0.20) (0.50 u 10–2) F = 5.31 u 10–11 F (b) Capacitance, C v 1 d , when d’ = 2d, C’ = 1 2 C = 2.66 u10–11 F (c) Charge, Q’ = C’V = (2.66 u10–11)(12) = 3.19 u10–10 C Quick Check 2 1. A parallel-plate capacitor has a capacitance of 6.0 nF. The plates are separated by a distance of 2.0 mm. What is the area of each plate? 2. The plates of a parallel-plate capacitor are separated by a distance of 5.00 mm. The area of each plate is 0.040 m2 . (a) Find the capacitance of the capacitor. (b) Insulation of the air between the capacitor breaks down if the electric field between the plate exceeds 3.0 u 106 V m–1. (i) What is the maximum potential difference that can be applied across the capacitor? (ii) What is the maximum charge that can be stored in the capacitor? 3. The capacitor used in the fl ash-light system of a camera has a capacitance of 24 mF. If the capacitor is charged by a constant current of 1.5 mA from a 6.0 V battery, what is the time taken to charge the capacitor to a potential difference of 6.0 V?
Physics Term 2 STPM Chapter 13 Capacitors 47 13 13.3 Dielectrics Students should be able to: GHÀQHUHODWLYHSHUPLWWLYLW\Hr GLHOHFWULFFRQVWDQW GHVFULEHWKHHIIHFWRIDGLHOHFWULFLQDSDUDOOHOSODWHFDSDFLWRU XVHWKHIRUPXODC = Hr H0 A d Learning Outcomes 1. The parallel-plate capacitor discussed in section 13.2 has free space or vacuum in between the plates. 2. The capacitance of a capacitor is greatly increased by having an insulator, known as dielectric, in between the plates. 3. The dielectric constant or relative permittivity εr of an insulator is defi ned as εr = capacitance of a parallel-plate capacitor with the insulator in between the plates capacitance of the parallel-plate capacitor with free space in between the plates = C C0 Hence, capacitance of a parallel-plate capacitor with an insulator of dielectric constant εr in between the plates is C = εr C0 = εr ε0 A d 4. Typical values of dielectric constant εr are Air: 1.0006, Paper: 3, Mica: 7, Paraffi n wax: 2.5 5. Since the dielectric constant for air εr = 1.0006 = 1.00 (to 3 signifi cant fi gures), parallel-plate capacitors with air between the plates can be taken as a good approximation to capacitors with free space or vacuum between the plates. 6. The action of an insulator is illustrated as in Figure 13.6. (a) The molecules of the insulator are polarised by the electric fi eld in between the plates. (b) This results in the surface of the insulator facing the positive plate being negatively charged, and the other side being positively charged. (c) A reverse electric field is set up. The resultant electric fi eld between the plates is reduced. (d) Since E = —V d , when E decreases, the potential difference V between the plates decreases. (e) If the capacitor is still connected to the battery, charges continue to flow into the capacitor until the potential difference across the capacitor equals the e.m.f. of the battery. (f) Since the capacitor is able to store more charge for the same potential difference, its capacitance increases. Figure 13.6 – + – + – + – + – + – + – + – + – + – + – + – + – + – + + + + + + + + + + + + + + + + – Dielectric Original field Reverse field Polarized molecule Battery – – – – – – – – – – – – – – – + – + – + – + – + – + – + 2012/P1/Q25, 2018/P2/Q3
Physics Term 2 STPM Chapter 13 Capacitors 48 13 7. On the other hand, if an air capacitor after being charged is disconnected from the battery, the insertion of an insulator between the plates (a) reduces the electric fi eld to —1 εr its initial value, (b) reduces the potential difference to —1 εr its initial value, (c) Since C = —Q V , when V decreases for the same charge Q on either plates, the capacitance C increases by a factor εr . Example 2 (a) A parallel-plate capacitor consists of two metal plates each of area 2.0 m2 separated by a distance of 5.0 mm in air. A potential difference of 1.0 u 104 V is applied across the capacitor. Calculate (i) the capacitance, (ii) the charge on each plate, (iii) the electric fi eld strength between the plates. (εr for air = 1.00) (b) The charged capacitor is then disconnected from the charging voltage and insulated so that the charge in the capacitor is constant. A piece of dielectric of thickness 5.0 mm and dielectric constant 5.0 is inserted in between the plates. Calculate (i) the electric fi eld strength between the plates, (ii) the potential difference across the capacitor, (iii) the capacitance. Solution: (a) (i) Capacitance, C = ε0 A—– d = 8.85 u 10–12 ———————u 2 5 u 10–3 = 3.54 u 10–9 F (ii) Charge on each plate Q = CV = (3.54 u 10–9) u (1 u 104 ) = 3.54 u 10–5 C (iii) Electric fi eld strength E = —V d = 1 u 104 —–—— 5 u 10–3 = 2.0 u 106 V m–1 (b) (i) Electric fi eld strength E’ = —E εr = 2.0 u 106 ————– 5 = 4.0 u 105 V m–1 Exam Tips 1. When a charged capacitor is disconnected from the charging voltage, the charge 8 remains constant. 2. If the capacitor remained connected to the charging voltage, the potential difference = remains unchanged.
Physics Term 2 STPM Chapter 13 Capacitors 49 13 1. A parallel-plate capacitor is charged in air. It is then electrically isolated and lowered into a liquid dielectric. Which of the following sets of changes is correct? A Both the capacitance and the charges on the plates increase. B Both the capacitance and the charges on the plates decrease. C The capacitance increases and the potential difference across the plates decreases. D The capacitance decreases and the potential difference across the plates increases. 2. The electric fi eld between the plates of an isolated air-spaced parallel-plate capacitor is E. Without disconnecting the charging voltage, an insulator of relative permittivity 2 is inserted in between the plates. What is the fi nal electric fi eld between the plates? A 1 –– 2 E C 2E B E D 2E 3. A rolled paper capacitor is made from strips of metal foil of dimensions 2 cm u 40 cm separated by paper of relative permittivity 2 and thickness 0.002 cm. Estimate its capacitance. 13.4 Capacitors in Series and in Parallel Students should be able to: GHULYHDQGXVHWKHIRUPXODHIRUHIIHFWLYHFDSDFLWDQFHRIFDSDFLWRUVLQVHULHVDQGLQSDUDOOHO Learning Outcome 1. Figure 13.7 shows three capacitors of capacitance C1 , C2 and C3 connected in parallel to a battery. 2. Charges fl ow into the capacitors until the potential difference across each capacitor is equal to the charging voltage V. Hence the potential differences across the capacitors are the same, V. 3. The charge in each capacitor is given by Q1 = C1 V, Q2 = C2 V, Q3 = C3 V The total charge, Q = Q1 + Q2 + Q3 = C1 V + C2 V + C3 V = (C1 + C2 + C3 )V Figure 13.7 Q2 Q3 V Q1 C2 C3 C1 Battery (ii) Potential difference V’ = —V εr = 1 u 104 —–——5 = 2.0 u 103 V (iii) Capacitance C’ = εr C = 5 u3.54 u 10–9 = 1.77 u 10–8 F Quick Check 3 2012/P1/Q24, 2013/P2/Q3, 2014/P2/Q3, 2017/P2/Q4
Physics Term 2 STPM Chapter 13 Capacitors 50 13 The three capacitors can be replaced by a single capacitor of capacitance C, if the charge in the equivalent capacitor is Q = CV Hence CV = (C1 + C2 + C3 )V Equivalent capacitance, C = C1 + C2 + C3 This formula for the equivalent capacitance can be extended for any number of capacitors connected in parallel. Capacitors in Series 1. Figure 13.8 shows the three capacitors of capacitance C1 , C2 , and C3 connected in series to a charging voltage V. 2. Electrons flow from the negative terminal of the charging voltage and charge plate Y of the capacitor C3 negative. 3. The charge –Q on plate Y induced a charge +Q on the plate X which in turn causes the plate N to be charged –Q and so on. Hence the charges on all the capacitors are the same, i.e. Q. 4. The potential difference across each capacitor is given by V1 = Q —C1 , V2 = Q —C2 , V3 = Q —C3 The total potential difference, V = V1 + V2 + V3 = Q —C1 + Q —C2 + Q —C3 = Q(—1 C1 + —1 C2 + —1 C3 ) 5. Since only a charge of Q is transferred from the charging voltage to the capacitors, the equivalent capacitor would have a charge of Q when connected to the same charging voltage V, and V = Q —C C = equivalent capacitance Hence, Q —C = Q(—1 C1 + —1 C2 + —1 C3 ) 1 —C = — 1 C1 + — 1 C2 + — 1 C3 Exam Tips For capacitors in parallel ÷VDPHSRWHQWLDOGLIIHUHQFHDFURVVDOOFDSDFLWRUV ÷FKDUJHLQHDFKFDSDFLWRUv capacitance. ÷HTXLYDOHQWFDSDFLWDQFH* = sum of capacitance Figure 13.8 +Q –Q C1 C2 C3 V1 V2 V V3 +Q –Q +Q –Q AB MN XY Exam Tips When two capacitors *1 and *2 are connected in series, ÷ WKHFKDUJHVLQERWKFDSDFLWRUVDQGLQWKHHTXLYDOHQWFDSDFLWRU DUHHTXDO ÷ SRWHQWLDOGLIIHUHQFHDFURVV*1 :=1 = ( C2 C1 + C2 )= potential difference across *2 : =2 = ( C1 C1 + C2 )=
Physics Term 2 STPM Chapter 13 Capacitors 51 13 Example 3 A capacitor made from two thin, flat metal sheets separated by an insulating material has a capacitance C. Each metal sheet is then cut into four identical sheets, which are used to make the capacitor as shown. The separation between the interleaved sheets is fi lled with the same insulating material. Find in terms of C, the capacitance of the reconstructed capacitor. Solution: If A = area of each large metal sheet d = separation between sheets then C = εr ε0 A ——– d The reconstructed capacitor consists of 7 capacitors connected in parallel. Four plates are to one terminal, and another four to the other terminal. Each capacitor is of area —A 4 , hence of capacitance = —C 4 Hence capacitance of 7 capacitors each of capacitance —C 4 in parallel C ’ = 7 u(—C 4 )= — 7 4 C Example 4 A parallel-plate capacitor consisting of two metal plates separated by a distance x has capacitance C. Each metal plate is cut into three identical small plates and connected to form the system of capacitors shown in the diagram. The separation between plates remains as x. Deduce in terms of C, the effective capacitance of the arrangement shown in the diagram. Solution: –Q –Q –Q –Q –4Q +Q +Q +Q +Q +4Q – – – – – – – – – – – – – – – – – + 12345 + + + + + + + + + + + + + + + + + + + + + + + + + – – – – – – – – – Figure (a) When the arrangement is connected to a voltage supply, the charges on the plates are as shown in Figure (a). Since the number of spaces between plates is 5, there are 5 capacitors. Exam Tips 7R GHWHUPLQH WKH QXPEHU RI FDSDFLWRUV LQ WKH TXHVWLRQffl &RXQW WKH QXPEHU RI VSDFHV EHWZHHQ two plates x Metal plate x x – – – – – – – – – + + + + – – – – + + + + + + + + –4Q –Q –Q –Q –Q +Q +Q +Q +Q +4Q – – – – + + + + + – – – – + + + + – – – – + + + + Figure (b)
Physics Term 2 STPM Chapter 13 Capacitors 52 13 Four of the capacitors have one plate each connected to one common terminal, and the other plate to another common terminal. Hence these four capacitors are in parallel. The charge on the fi fth capacitor 4Q = sum of charges on the four capacitors. Hence the fi fth capacitor is in series with the four capacitors. The area of each small plate = —A 3 . Hence capacitance of each small capacitor = —C 3 . The equivalent capacitance of 4 such capacitors in parallel = 4 u (—C 3 )= —4 3 C The effective capacitance of the arrangement C’ is given by —1 C’ = ——1 —4 3 C + ——1 —1 3 C = ——— 3 + 12 4C C’ = 4 –—15 C Example 5 Two capacitors of capacitance 2 PF and 4 PF, initially uncharged are connected in series with a 12 V battery. Calculate (a) the equivalent capacitance, (b) the charge on each capacitor, (c) the potential difference across each capacitor. Solution: (a) The equivalent capacitance C is given by —1 C = —1 2 + —1 4 = — 3 4 C = — 4 3 PF (b) Charges on both capacitors are equal = charge on equivalent capacitor = (—4 3 PF) u12 V = 16 PC (c) Potential difference across 2 PF capacitor V1 = Q —C1 = 16 PC ——— 2 PF = 8 V Potential difference across 4 PF capacitor V 2 = V – V1 = (12 – 8) V = 4 V 12 V 2 μF 4 μF V1 V2
Physics Term 2 STPM Chapter 13 Capacitors 53 13 Example 6 (a) Explain why the concept of capacitance is not applicable for an insulator which is charged. (b) Discuss how the capacitance of a metal plate is affected by (i) the presence of a earthed metal plate that is parallel to it, (ii) a piece of insulator placed between the two metal plates. (c) A 10 PF capacitor is charged to a potential difference of 103 V, and then disconnected from the voltage supply. An uncharged 0.2 PF capacitor is connected in parallel to the 10 PF capacitor, then disconnected and discharged. The process of connecting the 0.2 PF capacitor to the 10 PF capacitor, disconnecting and discharging is repeated ten times. Calculate (i) the fraction of the charge in the large capacitor that remains in the large capacitor after the fi rst connection, (ii) the value of the fraction after the tenth connection, (iii) the fi nal potential difference across the large capacitor. Solution: (a) The concept of capacitance is not applicable to a charged insulator because charge cannot fl ow into or out of an insulator. Hence charge cannot be channelled into an insulator to be stored. (b) (i) When an earthed metal plate is arranged parallel to the charged plate, the electric potential V of the charged plate decreases. Using C = Q —V when V decreases, the capacitance C increases. (ii) When a piece of insulator is inserted in between the two parallel plates, the capacitance becomes C’ = εr C where εr = dielectric constant of the insulator Since εr 1, the capacitance C’ C (c) (i) Initial charge in the 10 PF capacitor Q = CV = (10 u 10–6) u (103 ) = 1.0 u 10–2 C When the 0.2 PF capacitor is connected in parallel (fi gure), the equivalent capacitance C = (C1 + C2 ) = (10 + 0.2) PF = 10.2 PF The new potential difference V1 = Q ——–— (C1 + C2 ) Charge remaining in 10 PF capacitor, Q1 = C1 V1 = C1 Q ——–— (C1 + C2 ) so, — Q1 Q = C ——–—1 (C1 + C2 ) = 10 ——10.2 = 0.98 10 +F 0.2 +F Q1 Q1v V1
Physics Term 2 STPM Chapter 13 Capacitors 54 13 (ii) Similarly, after the second connection, — Q2 Q1 = C ——–– 1 C1 + C2 [Q1 = ( C ——–– 1 C1 + C2 )Q] Q2 = ( C ——–– 1 C1 + C2 ) 2 Q After the 10th connection, Q10 = ( C ——–– 1 C1 + C2 ) 10 Q —– Q10 Q = (0.98)10 = 0.817 (iii) Q10 = 0.817 Q Q = C1 V = 0.817 (10 u 10–6 u 103 ) = 8.17 u 10–3 C From Q10 = CV10 V10 = 8.17 u10–3 ————–– 10 u10–6 = 8.17 u 102 V Quick Check 4 1. Both the parallel-plate capacitors shown in diagram (a) and diagram (b) have the same dimensions. Mica Air V1 V2 Figure (a) Q1 Q2 Mica Air Figure (b) Which of the following pairs of comparison between V1 and V2 , Q1 and Q2 is correct? A V1 < V2 , Q1 > Q2 C V1 > V2 , Q1 = Q2 B V1 < V2 , Q1 < Q2 D V1 > V2 , Q > Q2 2. A capacitor X of capacitance C is connected to a voltage V. The charge on the capacitor is Q. Without removing the voltage, a second capacitor Y of the same capacitance is connected in parallel capacitor X. What is the potential difference and charge on the capacitor Y? Potential difference Charge A 1 2V 1 2Q B 1 2V Q C V 1 2Q D V Q 3. A 10 PF capacitor and a 20 PF capacitor are connected to a 3.0 kV voltage as shown below. 3 kV 10 +F 20 +F
Physics Term 2 STPM Chapter 13 Capacitors 55 13 What is the charge on the 10 PF capacitor? A 4.5 mC C 15 mC B 10 mC D 20 mC 4. A capacitor has capacitance of 4 PF. When another capacitor is connected to it, the effective capacitance is 2 PF. The capacitance of the second capacitor is A 4 PF and it is connected in series B 4 PF and it is connected in parallel C 2 PF and it is connected in parallel D 2 PF and it is connected in series 5. A 2 PF capacitor charged initially by a 30 V source is disconnected from the source and then connected to an uncharged 1 PF capacitor. What is the final potential difference across the capacitors? A 10 V C 20 V B 15 V D 25 V 6. Three capacitors are arranged as shown in the diagram. X Y 2 μF 3 μF 6 μF What is the effective capacitance between X and Y? A 1 PF C 6 PF B 2 PF D 11 PF 7. A pair of capacitors of the same capacitance are connected in series to a voltage V. The charge stored in the circuit is Q1 . Another pair of capacitors of the same capacitance are connected in parallel to the same voltage. The charge stored in the circuit is Q2 . Which of the followings is correct? A Q1 = 1 4Q2 C Q1 = 2Q2 B Q1 = 1 2Q2 D Q1 = 4Q2 8. Four identical capacitors of capacitance 2 PF each are connected as shown below. (a) (b) (c) (d) Calculate (i) the effective capacitance, (ii) the charge in each capacitor when the arrangement is connected to a 12 V supply. 9. You are provided with a number of identical capacitors each marked 2 PF, 10 V. How would you arrange these capacitors to obtain (a) an affective capacitance of 1 PF connected to a 20 V supply, (b) an effective capacitance of 2 PF connected to a 20 V supply. 10. (a) Two capacitors of capacitance 0.10 PF and 0.20 PF are connected in series to a 100 V supply. Calculate the charge in each capacitor. (b) Without discharging the capacitors, they are disconnected from the supply and then connected in parallel so that plates with the same sign are connected to each other. What is the final potential difference across the capacitors? 11. (a) What is meant by the capacitance of a capacitor? (b) Two capacitors of capacitance C1 and C2 are separately charged to potentials V1 and V2 respectively. The capacitors are then connected by connecting the positively charged plates together and the negatively charged plates together. Derive an expression for the final potential difference V across the combination.
Physics Term 2 STPM Chapter 13 Capacitors 56 13 13.5 Energy Stored in a Charged Capacitor Students should be able to: XVHWKHIRUPXODHU = 1 2 QV, U = 1 2 Q2 C and U = 1 2 CV2 GHULYDWLRQVDUHQRWUHTXLUHG Learning Outcomes 1. When a capacitor is charged using a voltage supply V, charge is transferred from the voltage supply to the capacitor until the potential difference across the capacitor is equal to the charging voltage V as shown in Figure 13.9. 2. Work done by the supply to charge the capacitor is QV. 3. Energy stored in the charged capacitor is U = 1 2 QV (Q = CV) = 1 2 CV2 (V = Q C ) = 1 2 Q2 C Energy is stored in the electric fi eld between the plates of the capacitor. 4. The difference in the work done by the applied voltage and the energy stored in the charged capacitor is (QV – 1 2 QV) = 1 2 QV. The energy difference is dissipated as heat from the connecting wires due to the charging current. Info Physics &DSDFLWRUVDUHXVHGWRVWRUHHOHFWULFDOHQHUJ\0DQ\HOHFWULFGHYLFHVVXFKDVFDPHUDÁDVKXQLWDQGGHÀEULOODWRUKDYHFDSDFLWRUV LQWKHP$GHÀEULOODWRULVXVHGWRUHYLYHDSDWLHQWZKRVXIIHUVDKHDUWDWWDFN Example 7 A 160 PF capacitor is charged to a potential difference of 200 V. It is then connected across a discharge tube which conducts until the potential difference across it has fallen to 100 V. What is the energy dissipated in the tube? Solution: Energy dissipated = loss of energy from capacitor = — 1 2 CV0 2 – —1 2 CV1 2 = —1 2 C(V0 2 – V1 2 ) = —1 2 (160 u 10–6)(2002 – 1002 ) = 2.4 J Figure 13.9 Q Q V V 2007/P1/Q28, 2007/P2/Q5, 2010/P1/Q26, 2010/P1/Q27, 2014/P2/Q4
Physics Term 2 STPM Chapter 13 Capacitors 57 13 Example 8 In the diagram, C1 is a fi xed capacitor of capacitance C0 , and C2 is a variable capacitor. The e.m.f. of the battery is V0 . (a) Initially, the switch is closed and C2 is adjusted so that its capacitance is also C0 . Find in terms of C0 and V0 , the total energy stored in the two capacitors. (b) The switch is then opened, and C2 is adjusted so that its capacitance becomes —1 4 C0 . (i) What is the resulting potential difference across the capacitors? (ii) How much work is done against electrical forces in reducing the capacitance of C2 ? Solution: The capacitors C1 and C2 are in parallel, hence the potential difference across both capacitors is V0 . (a) Total energy stored = — 1 2 C0 V0 2 + — 1 2 C0 V0 2 = C0 V0 2 (b) (i) Total initial charge on both capacitors = C0 V0 + C0 V0 = 2C0 V0 When C2 = —1 4 C0 , the effective capacitance C = C0 + —1 4 C0 = —5 4 C0 If V = fi nal potential difference, then total fi nal charge = total initial charge (— 5 4 C0)V = 2C0 V0 V = — 8 5 V0 = 1.6V0 (ii) Work done = increase in energy stored by capacitors = —1 2 (—5 4 C0)(1.6 V0 )2 – C0 V0 2 = 0.6C0 V0 2 Example 9 (a) The diagram shows a parallel-plate capacitor. Copy the fi gure and show the distribution of charges and the electric fi eld pattern between the plates when the plate X is charged negative. Indicate the direction of conventional current fl ow in the wire connecting plate Y to the Earth while X is being charged. Show also the electrostatic force on plate X. (b) X is disconnected from the charging system. Explain why the force on X is independent of its distance from Y, neglecting edge effect. V0 C2 C1 X Y Earth
Physics Term 2 STPM Chapter 13 Capacitors 58 13 (c) Write down expressions for (i) the capacitance of a parallel-plate capacitor of plate area A, and separation d. (ii) the energy stored by a parallel plate capacitor in terms of the plate area A, separation d and charge q. (d) (i) By considering energy changes when the plate X is moved, with charge q remaining constant, deduce an of expression for the force on the plates. (ii) Calculate this force for plates of area 100 cm2 carrying charge of 1 PC, in vacuum. Solution: (a) X Y – – – – – + + + + + X Y Current – – – – – + + + + + X F Y F = electrostatic force (b) The electric fi eld strength E between the plates is uniform and is given by E = —σ ε0 where σ = charge density = q —A Electrostatic force on plate X, F = qE = qσ —– ε0 which is independent of the distance d between the plates X and Y. (c) (i) Capacitance of parallel-plate capacitor, C = ε0 A—– d (ii) Energy stored in a charged capacitor = — 1 2 q2 —C = — 1 2 q2 —— ( ε0 A—– d ) = q2 d —–– 2ε0 A (d) (i) Suppose that the plate X is moved so that the plate separation changes from d to d1 (d1 < d). Work done by force F = Change in energy stored F(d – d1 ) = —1 2 q2 d —–– ε0 A – —1 2 q2 d —––1 ε0 A F = q2 —–– 2ε0 A (ii) When q = 1 μC = 1 u 10–6 C A = 100 cm2 = 100 u 10–4 m2 Force, F = (1 u 10–6)2 ———————————— 2(8.85 u 10–12) (100 u 10–4) = 5.6 N
Physics Term 2 STPM Chapter 13 Capacitors 59 13 Example 10 (a) A uniform copper plate of thickness x is inserted between the plates of a parallel-plate capacitor as shown in the diagram. The separation between the plates of the capacitor is d and the area of each plate is A. Derive an expression, in terms of A, d, x and ε0 , the permittivity of free space, for the capacitance of the arrangement shown. d x (b) The area of each plate of a parallel-plate capacitor is 0.200 m2 and the plate separation is 8.00 mm. The capacitor is charged by a d.c. supply of 500 V. (i) Find the capacitance of the capacitor and the charge on each plate of the capacitor. (ii) The capacitor is then disconnected from the d.c. supply. A metal plate of thickness 2.00 mm is inserted in between the plates of the capacitor to produce the arrangement as shown in the fi gure. What is the capacitance of the arrangement, and fi nd the work done in the process. Solution: (a) Figure (a) Figure (b) x C d1 + d2 +++++ –––––– ++++++ –––––– C1 d1 d2 C2 ++++++ –––––– ++++++ –––––– The arrangement shown in Figure (a) is equivalent to two capacitors C1 and C2 in series as shown in Figure (b). C1 = ε0 A —–– d1 and C2 = ε0 A —–– d2 The effective capacitance C is given by —1 C = —1 C1 + — 1 C2 = d1 ––– ε0 A + d2 ––– ε0 A = d1 + d2 –––—– ε0 A (d1 + d2 = d – x) = d – x –––– ε0 A C = ε0 A –––– d – x
Physics Term 2 STPM Chapter 13 Capacitors 60 13 (b) (i) Capacitance, C = ε0 A —–– d = (8.85 u 10–12)(0.200) —————————– (8.00 u 10–3) = 2.21 u 10–10 F Charge, Q = CV = (2.21 u 10–10)(500) = 1.11 u 10–7 C (ii) Using C = ε0 A –––– d – x Capacitance, C = (8.85 u 10–12)(0.200) ————————— (8.00 – 2.00) u 10–3 = 2.95 u 10–10 F Work done = Change in energy stored by capacitor = —1 2 Q2 —C – —1 2 Q2 —C1 = —1 2 (1.11 u 10–7)2 (————— 1 2.21 u 10–10 – ————— 1 2.95 u 10–10 ) = 6.99 u 10–6 J Quick Check 5 1. Three capacitors of capacitance C1 , C2 and C3 are connected in series to a battery. If C1 C2 C3 , which statement is correct? A The effective capacitance is C1 + C2 + C3 . B The charge on capacitor C3 is the greatest. C The potential difference across capacitor C1 is the greatest. D The three capacitors stored the same amount of energy. 2. The energy stored in a parallel-plate capacitor is U when it is charged to a potential difference V. An identical parallel-plate capacitor, but with an insulator of dielectric constant εr is also charged to a potential difference V. The energy stored in this capacitor is A εr U B εr 2 U C —– U εr D —– U εr 2 3. A charged capacitor X is connected in parallel to an identical but uncharged capacitor Y. Which of the following statements about the capacitors is true? A The potential difference across capacitor X remains unchanged, but the total energy decreases. B The total charge and energy are conserved. C The charge in X decreases but its potential difference remains unchanged. D The total charge is conserved but the potential difference across X decreases. 4. Two capacitors of capacitance C1 and C2 are charged to potential difference of V1 and V2 respectively. If the two positive plates are connected and the two negative plates of the capacitors are connected, the system of capacitors A gains energy but loses charge B loses both energy and charge C loses energy but charge remains constant D loses charge but energy remains constant
Physics Term 2 STPM Chapter 13 Capacitors 61 13 5. The diagram shows two ways of connecting two identical capacitors of capacitance C to a voltage supply V. C C V C V C Circuit 1 Circuit 2 What is the value of total energy stored in the capacitors in circuit 1 ————————————————————— total energy stored in the capacitors in circuit 2? A —1 4 C 2 B — 1 2 D 4 6. The charge in a capacitor is increased from Q1 to Q2 . In which of the following graphs of potential difference V against charge Q, does the shaded area represent the increase in energy stored in the capacitor? A C V 0 Q1 Q2 Q V V2 V1 0 Q1 Q2 Q B D V 0 Q1 Q2 Q V V2 V1 0 Q1 Q2 Q 7. A photographic flash unit consists of a xenon-filled tube energised by the discharge of a capacitor initially charged to a potential difference of 1 000 V. The average power delivered to the flash tube is 2 000 W and the flash lasts 0.020 s. The capacitance of the capacitor is A 40 PF C 80 mF B 80 PF D 160 mF 8. A 3 PF capacitor and a 6 PF capacitor are connected in series to a 10 V battery. What is the value of energy stored in 3 the PF capacitor ———––—————–—————— energy stored in 6 PF capacitor ? A — 1 4 C 2 B — 1 2 D 4 9. A 0.1 PF capacitor is charged by connecting it to a battery of 50 V. Calculate (a) the charge transferred from the battery to the capacitor, (b) the energy stored in the charged capacitor, (c) the energy supplied by the battery, (d) the heat dissipated in the connecting wires. 10. A 10 PF capacitor is charged to a potential difference of 500 V. It is then disconnected from the voltage supply and connected in parallel to a 40 PF capacitor, initially uncharged. Calculate the total energy stored in the capacitors (a) before (b) after connection. Explain the conservation of energy during the process. 11. Two capacitors of capacitance C1 and C2 are charged to potential difference of V1 and V2 respectively. The charged capacitors are then connected to obtain a larger capacitance C. (a) Derive an expression for the potential difference V across the larger capacitor in terms of C1 , C2 , V1 and V2 . (b) If C1 = C2 = 2 PF and V1 = 200 V, V2 = 0, calculate the loss of energy during the process.
Physics Term 2 STPM Chapter 13 Capacitors 62 13 12. The capacitance of a variable capacitor can be varied between 1 u 10–10 F and 5 u 10–10 F by turning a knob which is connected to the movable plates of the capacitor. The capacitance is first fixed at 5 u 10–10 F and the capacitor is charged using a battery of e.m.f. 200 V. (a) What is the charge in the capacitor? (b) The battery is disconnected and the capacitance changed to 1 u 10–10 F. (i) What is the final potential difference across the capacitor? (ii) Find the mechanical work done against the electrostatic forces to change the capacitance. 13. (a) Derive the formula for the capacitance of a parallel-plate capacitor. Identify the symbols you use. (b) Explain the effect of a dielectric in between the plates of a parallel-plate capacitor on the capacitance of the capacitor. 14. (a) A parallel-plate capacitor with an insulator of dielectric constant 6.0 in between the plates has a capacitance of 120 pF. It is charged to a potential difference of 100 V. Calculate (i) the charge on each plate of the capacitor, (ii) the energy stored in the capacitor. (b) The charged capacitor is isolated and the insulator between the plates is pulled so that only half the space between the plates is filled with dielectric as shown in the diagram. Air Insulator If the air-filled and insulator-filled sections are considered as two separate capacitors, calculate (i) the new value of the capacitance, (ii) the energy stored, Explain how energy is conserved. 15. Two capacitors of capacitance 8.0 mF and 6 mF are charged to a potential difference of 6.0 V and 12 V respectively. The charged capacitors are disconnected from the charging voltage, and connected as shown in diagram (a). + + – – + + – – S 8 mF 6 mF (a) + + – – – – + + S 8 mF 6 mF (b) (a) Calculate (i) the potential difference across each capacitor, (ii) the energy loss after the switch S is closed. (b) The capacitors are then recharged, and then connected as shown in Figure (b). Calculate (i) the potential difference across each capacitor, (ii) the energy loss after the switch S is closed.
Physics Term 2 STPM Chapter 13 Capacitors 63 13 13.6 Charging and Discharging of a Capacitor Students should be able to: GHVFULEHWKHFKDUJLQJDQGGLVFKDUJLQJSURFHVVRIDFDSDFLWRUWKURXJKDUHVLVWRU GHÀQHWKHWLPHFRQVWDQWDQGXVHWKHIRUPXODτ = RC GHULYHDQGXVHWKHIRUPXODHQ = Q0 11 – e– t W 2, V = V011 – e– t W 2 and I = I 0 e – t W IRUFKDUJLQJDFDSDFLWRUWKURXJKDUHVLVWRU GHULYHDQGXVHWKHIRUPXODHQ = Q0 e – t W , V = V0 e – t W and I = I 0 e – t W IRUGLVFKDUJLQJDFDSDFLWRUWKURXJKDUHVLVWRU VROYHSUREOHPVLQYROYLQJFKDUJLQJDQGGLVFKDUJLQJRIDFDSDFLWRUWKURXJKDUHVLVWRU Learning Outcomes Charging a Capacitor through a High Resistance 1. Figure 13.10 shows an uncharged capacitor of capacitance C connected in series with a resistor of high resistance R to a battery of e.m.f. E. 2. At time t = 0, when the switch S is closed the potential difference across the capacitor, V = 0 since there is no charge yet in the capacitor. Hence, the potential difference across the resistor is equal to the e.m.f. E of the battery. The current I 0 in the circuit is given by E = I0 R I0 = —E R 3. At a time = t after the switch S is closed, the potential difference across the capacitor = V and the charge in the capacitor, Q = CV. The potential difference across the resistor = IR where I = current in the circuit. Also, I = dQ—– dt (Current is positive because charge Q in the capacitor is increasing) = C dV —– dt (Q = CV) Hence e.m.f. of battery, E = V + IR = V + CR dV—– dt E – V = CR dV—– dt ∫ V 0 ——– dV E – V = —– 1 CR ∫ t 0 dt [–ln (E – V)]V 0 = [—–t CR ] t 0 –ln (E – V) + ln E = —–t CR ln ( ——– E E – V ) = —–t CR ( ——– E – V E ) = e t – —– CR V = E(1 – e t – —– CR) The potential difference V across the capacitor increases exponentially to a value equal to E, the e.m.f. of the battery. When V = E, the current I = 0. I R IR V S E C Figure 13.10 Exam Tips When a capacitor is charged through a high resistance, its potential difference =, and charge 8 PUJYLHZL EXW WKH FXUUHQW LQ WKH circuit KLJYLHZLZ. 2013/P2/Q18, 2015/P2/Q4, Q16, 2016/P2/Q19, 2017/P2/Q18
Physics Term 2 STPM Chapter 13 Capacitors 64 13 4. The charge in the capacitor, Q = CV = CE(1 – e t – —– CR ) Q = Q0 (1 – e t – —– CR) when Q0 = CE, the final charge in the capacitor. The charge Q increases exponentially with time, in the same manner as the potential difference V. 5. The current I in the circuit is given by I = —– dQ dt Q = Q0 (1 – e t – —– CR ) = —– d dt (Q0 (1 – e t – —– CR )) = Q0(—– 1 CR )e t – —– CR (Q0 = CE) = —– CE CR e t – —– CR (—E R = I0) I = I0 e t – —– CR The current in the circuit decreases exponentially with time as the capacitor gets charged up. 6. Figure 13.12 shows how the three quantities, current, I, charge Q and potential difference V varies with time t during the charging of a capacitor through a high resistance. 7. The time constant, τ = CR of a charging circuit is the time taken for the potential difference V across the capacitor (or the charge Q in the capacitor) to increase to (1 – —1 e ) of its final value. 8. Alternatively, the time constant of a charging circuit is the time taken for the current I in the circuit to decrease to —1 e its initial value. Discharging a Capacitor Through a High Resistance 1. Suppose that a capacitor of capacitance C is charged to a potential difference V0 , then the charge in the capacitor, Q0 = CV0 2. To discharge the capacitor, it is connected to a high resistance R (Figure 13.13). At the instant the switch is closed (t = 0), potential difference = potential difference across the resistor across the capacitor Then, I0 R = V0 The initial current, I0 = —– V0 R τ I0 e E R I 0 I0 = – – t Figure 13.11 Switch + – C I V R Figure 13.13 Figure 13.12 τ Q0 Q0 = CE e Q 0 1 (1– –) t τ E E e V 0 1 (1– –) t
Physics Term 2 STPM Chapter 13 Capacitors 65 13 3. As the capacitor discharges, the potential difference V, the charge Q in the capacitor and the current I in the circuit decrease. 4. Suppose at time = t, charge in the capacitor, Q = CV Then current, I = – —– dQ dt The negative sign denotes that the current I in the circuit is due to the decrease of the charge in the capacitor. Hence, I = – —– d dt (CV) (Q = CV) = –C—– dV dt Also, V = IR = –CR—– dV dt – —– 1 CR ∫ t 0 dt = ∫ V V0 —– dV V – —–t CR = ln — V – V0 ? — V – V0 = e t – —– CR Potential difference V = V0 e t – —– CR Charge, Q = CV Q = CV0 e t – —– CR (Q0 = CV0 ) Q = Q0 e t – —– CR Current, I = – —– dQ dt = —– Q0 CR e t – —– CR = CV —–0 CR e t – —– CR (Q0 = CV0 ) = V —–0 R e t – —– CR I = I0 e t – —– CR (I0 = V —–0 R , the current when t = 0) Exam Tips :KHQDFDSDFLWRUGLVFKDUJHV V = V0 e t – —– CR Q = Q0 e t – —– CR I = I 0 e t – —– CR I 0 = — V0 R Info Physics 0DQ\ HOHFWULFDO DSSOLDQFHV VXFK DV ZDVKLQJ PDFKLQH DQG WHOHYLVLRQ XVH ODUJH FDSDFLWRUV 'R QRWWRXFKWKH FLUFXLWU\ RIWKHVH DSSOLDQFHV HYHQ DIWHU VZLWFKLQJ RIIWKH SRZHU VXSSO\7KH FDSDFLWRUV LQ WKH DSSOLDQFHV DUH FKDUJHG DQG FDQ SURGXFH D QDVW\ HOHFWULF VKRFN$OORZ WKH FDSDFLWRUV WR GLVFKDUJH EHIRUH VWDUWLQJ DQ\ UHSDLULQJMRE VIDEO ;`WLZVM*HWHJP[VY
Physics Term 2 STPM Chapter 13 Capacitors 66 13 5. Hence (a) the potential difference V, (b) the charge Q and, (c) the current I decrease exponentially with time t as shown by the graphs in Figure 13.14. Figure 13.14 V V0 0 t Q Q0 0 t I I0 0 t (a) V = V0 e t – —– CR (b) Q = Q0 e t – —– CR (c) I = I0 e t – —– CR 6. The product CR is known as the time constant of the discharge circuit. Time constant, τ = CR 7. The time constant of a discharge circuit for a capacitor is the time taken for the charge (or potential difference, or current) in the capacitor to decrease to 1 –– e of the initial charge. The unit for time constant is second (s). 8. The definition is deduced from the equation Q = Q0 e t – —– CR When t = CR, Q = Q0 e CR – —– CR = 1 –– eQ0 < 0.37Q0 9. The value of the time constant denotes how fast a capacitor discharges (Figure 13.15). Obviously a charged capacitor would take a longer time to discharge and the time constant would be larger, if s its capacitance C is larger. A large capacitor contains more charge. s the value of the resistance R in the circuit is bigger. A larger resistance results in a smaller discharge current. 10. A natural phenomenon which is due to discharge of a body is lightning. (a) When the charge accumulated by a cloud increases, the electric field between two charged clouds of different signs, or between the charged cloud and the earth increases until the insulation of the air breaks down. (b) Motion of the ions discharges the clouds and produces lightning. Figure 13.15 Figure 13.16 Q0 Q0 –e Q0 e2 e3 CR 0 2CR 3CR – Q0 – Q t τ1 τ2 > τ1 0 Q0 Q
Physics Term 2 STPM Chapter 13 Capacitors 67 13 Experimental Determination of Time Constant of a Discharge Circuit 1. The circuit is set up as shown in Figure 13.17. 2. The two-way switch S is closed at A to charge the capacitor C. 3. The switch S is then closed at B. The capacitor discharges through the high resistance R2 . 4. The current in the circuit I is measured by the microammeter whose reading is noted at suitable intervals, say every 20 seconds until the reading is almost zero. 5. A graph of current I against time t is plotted (Figure 13.18). τ 2τ I / μA I0 I0 e – I0 e – 2 0 Time / s Figure 13.18 6. From the graph, the time taken for the current to decay (a) from I0 to I —0 e , which is the time constant τ, (b) from I0 to I —0 e2 , which is 2τ are noted and the average value of the time constant calculated. Example 11 A charged 100 PF capacitor is discharged through a resistor of resistance 10 k:. What is the ratio charge in capacitor after 1 second ——————————————— initial charge in capacitor ? Solution: Time constant, τ = CR = (100 u 10–6)(10 u 103 ) = 1.0 s Using Q = Q0 e t – —– CR , t = 1.0 s Q–––Q0 = e 1.0 – —– 1.0 = 1 –– e = 0.368 High resistance R2 R1 C A S B μA Figure 13.17
Physics Term 2 STPM Chapter 13 Capacitors 68 13 Example 12 The diagram shows the exponential decay of the potential difference across a charged capacitor as the capacitor discharges through a resistor of resistance 2 M:. What is the capacitance of the capacitor? V0 V0 e –2 0 Potential difference 5 10 20 Time/ s 15 Solution: Time taken for the potential difference V across the capacitor to decay from V0 to 1 –– e V0 is the time constant, τ = CR. Time taken for V to decay from 1 –– e V0 to 1 –– e(1 –– e V0) or V —0 e2 is also τ. Hence, the total time taken for V to decay from V0 to V —0 e2 is 2τ. 2τ = 10 s (from graph) 2C u (2 u 106 ) = 10 C = 2.5 u 10–6 F = 2.5 PF Example 13 In a television set, a capacitor is charged to a potential difference of 20 kV. As a safety precaution, a resistor is connected across the capacitor when the television is switched off. The capacitance of the capacitor is 10 PF and the resistance of the resistor is 1 M:. How long does it take after the television is switched off for the potential difference across the capacitor to drop to 4.5 u 10–5 time its initial value, so that the capacitor is safe to touch with the hand? Solution: Time constant, τ = CR = (10 u 10–6)(1 u 106 ) = 10 s Using V = V0 e t – —– CR , (4.5 u 10–5)V0 = V0 e t – —– 10 —t 10 = ln ————– 1 4.5 u 10–5 t = 100 s
Physics Term 2 STPM Chapter 13 Capacitors 69 13 Quick Check 6 1. An uncharged capacitor C is connected to the circuit shown in the diagram. V S R C Which of the graphs represents the variation of the current I in the circuit with time t after the switch S is closed? A C 0 t I 0 t I B D 0 t I 0 t I 2. An uncharged capacitor of capacitance 2 PF, a 25 k: resistor, a switch and a 100 V battery are connected in series. What is the initial current when the switch is closed? A 0 C 40 mA B 4 mA D 4 A 3. When a capacitor discharges through a resistor, the time constant of the circuit is the time after A 1 –– e of the initial charge fl ows out of the capacitor. B 1 –– 2 of the initial charge fl ows out of the capacitor. C 1 –– e of the initial charge remains in the capacitor. D 1 –– 2 of the initial charge remains in the capacitor. 4. A 1 PF capacitor that contains 1 u 10–5 C of charge is connected to a 10 : resistor using a switch S. S 1 μF 10 Ω What is the value of the current in the circuit at the instant the switch S is closed? A 0 C 1.0 A B 10–5 A D 10 A 5. In Circuit 1, a charged capacitor discharges when the switch is closed. In Circuit 2, an uncharged capacitor is charged when the switch is closed. V3 A V1 V2 Circuit 1 Circuit 2 When the switches are closed, one of the four meters gives a reading θ which varies with time as shown below. Time θ 0 Which meter records in this way? A A C V2 B V1 D V3
Physics Term 2 STPM Chapter 13 Capacitors 70 13 6. A 20 PF capacitor, initially uncharged is charged by a constant current of 10 mA. What is the time taken for the potential difference across the capacitor to reach 300 V? A 6.0 u 10–4 s B 0.60 s C 15 s D 1.5 u 105 s 7. A capacitor discharges through a resistor made of a length of wire. The time constant of the circuit is τ1 . When the resistor is substituted with another wire made of the same metal but of twice the diameter and twice the length of the first wire, the time constant of the circuit is τ2 . What is the ratio of τ2 : τ1 ? A 4 : 1 B 2 : 1 C 1 : 2 D 1 : 4 8. A 1 M: resistor, a 10 μF capacitor initially uncharged, a switch and a 100 V d.c. supply are connected in series. Which of the following gives the value of the current I and the potential difference V across the capacitor immediately after the switch is closed? I/PA V/V A 0 0 B 0 100 C 100 0 D 10 10 9. The circuit shown is used to discharge a 10 PF capacitor through a variable resistor R. The current is kept constant at 20 PA by continuously adjusting R. R 10 +F +A By how much does the potential difference across the capacitor fall in 40 s? A 1.3 u 10–2 V C 20 V B 14 V D 80 V 10. A capacitor is charged using the circuit shown below. 4 MΩ 3 MΩ 7 +F What is the time constant of the circuit? A 12.0 s C 28.0 s B 21.0 s D 30.0 s 11. The diagram shows the circuit used to determine the time of contact between a ball and a bat when the ball at the end of a thread is released. Both the ball and bat are coated with a conducting paint. 2.0 kΩ 6.0 V 1.2 +F To y-plates of C.R.O. S Bat 0 With the ball displaced from the bat, the capacitor is charged by closing the switch S. The switch S is then opened and the ball released. The trace on the screen of the cathode-ray oscilloscope is as shown in the diagram. (a) Explain the form of the trace shown. (b) The sensitivity of the y-plates of the C.R.O. is 1.0 V per division, but the timebase is not given. (i) Calculate the value of the ratio final potential difference across the capacitor ———————————————–————— initial potential difference across the capacitor (ii) Calculate the time that the ball is in contact with the bat.
Physics Term 2 STPM Chapter 13 Capacitors 71 13 12. A student uses the circuit shown below to determine the speed of a bullet which has a mass of 0.5 g and a speed of approximately 40 m s–1. The bullet fi rst breaks the aluminium foil at X, and after travelling through a measured distance breaks the aluminium foil at Y. The time of motion of the bullet between X and Y is measured from the readings of the voltmeter. 6.0 V Bullet Aluminium foil X Y C R V (a) What is the potential difference across the capacitor just before the aluminium foil at X is broken? (b) Describe and explain how the reading of the voltmeter varies when the bullet travels between X and Y. (c) If the distance between X and Y is 10 cm, which of the following combinations of capacitance C and resistance R is most suitable? Explain your choice. Capacitor 0.01 PF 0.1 PF 1 PF 10 PF Resistor 220 : 1 k: 5 k: 10 k: (d) State the measurements required and explain how the measurements are used to calculate the time of motion of the bullet between X and Y. (e) Explain why the voltmeter resistance should be high. Important Formulae 1. Defi nition of capacitance, C = Q —V 2. Capacitance of a parallel-plate capacitor, C = εr ε0 A —––– d 3. Capacitors in series: —1 C = —1 C1 + —1 C2 + —1 C3 Capacitors in parallel: C = C1 + C2 + C3 4. Energy stored in a capacitor, U = 1 ––2 CV2 = 1 ––2 QV = 1 ––2 Q2 —V 5. When a capacitor is charged: V = E(1 – e t – —– CR ) Q = Q0 (1 – e t – —– CR ) I = I 0 e t – —– CR 6. When a capacitor discharges: V = V0 e t – —– CR Q = Q0 e t – —– CR I = I 0 e t – —– CR I 0 = —V R STPM PRACTICE 13 1. A parallel-plate air capacitor is charged when the separation between the plates is d. The charge on the capacitor is Q0 . The charging voltage is removed and the separation x between the plates is slowly increased. Which graph correctly shows how the charge Q in the capacitor changes when the separation x is increased? A Q 0 d x Q0 C Q 0 d x Q0 B Q 0 d x Q0 D Q 0 d x Q0 2. A capacitor is charged by connecting it in series with a resistor and a battery of e.m.f. E. Switch C E V
Physics Term 2 STPM Chapter 13 Capacitors 72 13 Which graph shows the variation of the potential difference V across the resistor with time t after the switch is closed? A E 0 V t C 0 E V t B 0 E V t D 0 E V t 3. A charged capacitor of capacitance C is connected to two resistors and a switchs as shown in the diagram. The switch S is closed at time t = 0. The reading V of the voltmeter is recorded at various times t. The graph of ln V against t is shown below. 3.0 k 6.0 k V C S In V t (s) 0 0.8 40 What is the capacitance of the capacitor? A 5.6 mF C 25.0 mF B 16.7.0 mF D 33.3 mF 4. A parallel-plate capacitor is charged by connecting it to a battery. The energy stored in the capacitor is U0 . With the battery still connected, the space between the plates is filled with an insulator of dielectric constant 4.0. What is the energy stored in the capacitor? A 1 16U0 C 4U0 B 1 4 U0 D 16U0 5. A charged capacitor of capacitance C is connected to a resistors and a switch. A voltmeter is used to measure the potential difference V across the capacitor. The switch is closed at time t = 0. The reading V of the voltmeter is recorded at various times t. The graph of ln V against t is shown below. In V t (s) 0 0.4 20 What is the time constant of the circuit? A 20 ms C 20 s B 50 ms D 50 s 6. All the capacitors in the arrangements below are identical. The effective capacitance between X and Y is the least in A X Y B X Y C X Y D X Y
Physics Term 2 STPM Chapter 13 Capacitors 73 13 7. A resistance and a capacitor are connected to a voltage V as shown in the circuit. I C V R When the current in the circuit is I, the charge on the capacitor is Q. The voltage V is A IR C IR + Q C B Q C D IR – Q C 8. The capacitance of a parallel-plate capacitor can be increased A by increasing the potential difference between the plates. B by increasing the charge on the plates. C by increasing the separation between the plates. D by using a dielectric of higher permittivity. 9. A 4.0 PF capacitor is charged to a voltage of 12.0 V and a 6.0 PF capacitor is charged to a voltage of 6.0 V. The charged capacitors are connected as shown in the diagram. 12.0 V S + + – – 6.0 V 6.0 F 4.0 F What are the potential differences across the capacitors after the switch S is closed? Potential Potential difference difference across 4.0 PF across 6.0 PF capacitor capacitor A 6.0 V 12.0 V B 8.4 V 8.4 V C 9.0 V 9.0 V D 12.0 V 6.0 V 10. Two parallel-plate capacitors are connected in series to a battery. A slab of dielectric is then inserted into one of the capacitors. Which of the changes described below is correct? A The total charge on both capacitor is reduced. B The total energy stored is reduced. C The effective capacitance is increased. D The potential difference across both capacitors is increased. 11. A parallel-plate capacitor is charged using a battery. The battery is then removed and a piece of dielectric is inserted in between the plates of the capacitor. The changes in the capacitance and energy stored in the capacitor are correctly described in Capacitance Energy stored A Increases Increases B Increases Decreases C Decreases Increases D Decreases Decreases 12. The diagram below shows a charged capacitor C1 connected to another capacitor C2 which is not charged. C1 S C2 The switch S is then closed. What happens to the total charge in capacitors C1 and C2, the potential difference across capacitor C1 , and the total energy in the capacitors C1 and C2 ? Total charge in capacitors C1 and C2 Potential difference across capacitor C1 Total energy in capacitors C1 and C2 A No change Decrease No change B No change Decrease Decrease C Decrease No change Decrease D Decrease No change No change
Physics Term 2 STPM Chapter 13 Capacitors 74 13 13. The circuit shows a 3 PF capacitor charged by a battery of e.m.f. 6 V with the switch at P. The switch is then connected to Q. The charge is then shared with the 6 PF capacitor with the switch at Q. Q 6 +F 6 V P 3 +F What is the charge that remains in the 3 PF capacitor? A 6 PC C 12 PC B 9 PC D 18 PC 14. A 4 PF capacitor is charged to a charge of 200 PC, and a 3 μF capacitor is charged to a charge of 300 PC. Both the capacitors are connected so that the two positive plates are connected together and the two negative plates are connected together. What is the total loss of energy in the process? A 2.0 u 10–3 J C 2.0 u 10–2 J B 1.8 u 10–2 J D 5.3 u 10–2 J 15. A capacitor is charged to a potential difference of 4 V. It then discharges through a high resistance R. The time taken for the potential difference across the capacitor to decrease from 4 V to 2 V is 10 s. The capacitor is then charged to a potential difference of 24 V and it then discharges through the same high resistance R. What is the time taken for the potential difference across the capacitor to decrease from 12 V to 3 V? A 10 s C 30 s B 20 s D 40 s 16. A piece of dielectric is between the plates of a charged capacitor as shown in the diagram. The separation between the plates is d. Dielectric + + + + + – – – – – 0 d x The variation of electric field strength E with distance x from the positive plate is correctly shown in A C E d 0 x E d 0 x B D E d 0 x E d 0 x 17. Four capacitors of equal capacitance C are connected in difference arrangements. The maximum effective capacitance is Cmax and the minimum effective capacitance is Cmin. What is the value of Cmax/Cmin? A 2 C 8 B 4 D 16 18. Two charged capacitors each of capacitance C are discharged using the circuit shown which contains two resistors of the same resistance R. C C R R What is the time constant of the circuit? A 1 ––4 RC C RC B 1 ––2 RC D 2RC 19. The circuit shown is used to charge a capacitor C which was initially uncharged. S V0 R C
Physics Term 2 STPM Chapter 13 Capacitors 75 13 Which of the graphs shows correctly the variation of the potential difference V across resistor R with time t after the switch S is closed? A C V0 V 0 t V0 V 0 t B D V0 V 0 t V0 V 0 t 20. The diagram shows the circuit used to charge and then discharge a capacitor C. The switch S is first connected to X at time t = 0, and then to Y at time t = t 1 . V0 Y X S C R Which of the graphs correctly shows the variation of the potential difference V across the resistor R with time t? A C V V0 t1 –V0 0 t B D V V0 t1 –V0 0 t V V0 t1 –V0 0 t V V0 t1 –V0 0 t 21. A capacitor of capacitance C is charged through a high resistance R until the charge in the capacitor is Q0 . Which of the following gives the current in the circuit after a time t from the start of the charging? A Q0 e t – —– CR C Q0 –—CR e t – —– CR B —— CQ0 R e t – —– CR D Q0 t –—CR e t – —– CR 22. Two capacitors are connected in series as shown in the diagram. C1 12 V 1.0 +F C2 2.0 +F What is the charge carried by each capacitor? Charge on C1 /PC Charge on C2 /PC A 4.0 8.0 B 8.0 4.0 C 8.0 8.0 D 12.0 12.0 23. A 3.0 PF capacitor and a 4.0 PF capacitor are connected in series with a 12.0 V battery. (a) What is the equivalent capacitance? (b) What is the charge on each capacitor? (c) With the capacitors still connected to the battery, a dielectric is inserted between the plates of one of the capacitors. Discuss any change to the charge stored in each capacitor. 24. Four capacitors of capacitance 50 PF, 100 PF, 150 PF and 200 PF are connected to a battery as shown in the diagram. The charge on the 100 PF is 700 PC. X Y 150 PF 200 PF 50 PF 100 PF (a) Find the charge on the (i) 50 PF capacitor, and (ii) 200 PF capacitor (b) What is the potential difference across XY?
Physics Term 2 STPM Chapter 13 Capacitors 76 13 25. (a) A parallel-plate air capacitor is charged to a voltage V. The charging voltage is then removed and the space between the plates is filled with a dielectric. Describe the changes to (i) the capacitance of the capacitor (ii) the charge and (iii) the potential difference across the capacitor. (b) A parallel-plate air capacitor of capacitance 20 μF is connected in series with a 12.0 battery and another capacitor B of the same dimensions but the space between the plates is filled with an insulator of dielectric constant 7.0. Determine (i) the effective capacitance, (ii) the energy stored in each capacitor. (c) The capacitor of a camera flash unit has a capacitance of 50 μF and is charged to 6.0 V. A flash is produced when the capacitor discharges through a resistor of 60 k:. The flash stops after the potential difference across the capacitor falls below 4.5 V. Calculate (i) the time constant of the discharge circuit, (ii) the duration of the flash. If a resistor of smaller resistance is used in the discharge circuit, discuss the effect on the brightness of the flash. 26. (a) (i) Define capacitance of a capacitor. (ii) What factors determine the capacitance of a parallel plate air capacitor? How is the amount of charge stored in the capacitor affected if the space between the plates is filled with mica? (b) A 20 PF capacitor and 30 μF capacitor are connected in series with a 6.0 battery. (i) How much charge flows from the battery? (ii) Find the potential difference across the 20 μF capacitor. (c) A 300 PF capacitor is charged to 6.0 V and discharged through a 500 : resistor. (i) What is the time constant of the discharge circuit? (ii) How much energy is dissipated from the resistor after 0.20 s? 27. A parallel-plate capacitor is charged to a potential difference of 12 V and then disconnected from the power supply. A simple pendulum which consists of a metal bob hangs from a nylon string at the mid-point between the plates. The bob then swings towards the negative plate. Each time the bob touches the negative plate 10% of the charge that remains on the plate is transferred to the pendulum bob. The charge on the bob is neutralized when the bob touches the positive plate. Calculate the potential difference between the plates of the capacitor after 6 oscillations of the bob. 28. (a) Two capacitors of capacitance C1 and C2 are connected in series. Derive the expression for the effective capacitance. (b) Capacitors of capacitance C1 = 12 μF and C2 = 24 μF are connected in series with a 12.0 V battery. Determine (i) the charge on each capacitor, (ii) the total energy stored in the capacitors, (iii) the energy supplied by the battery. Comment on the difference in the answers to (ii) and (iii) above. (c) The battery is then disconnected and the free ends of the connecting wires are connected to a 5.0 k: resistor. (i) Calculate the initial current in the circuit. (ii) Find the energy dissipated from the resistor, 29. A capacitor is marked 100 PF, 20 V. (a) Explain what is meant by a capacitance of 100 PF. (b) How much charge is stored by the capacitor when the potential difference across it is 20 V.
Physics Term 2 STPM Chapter 13 Capacitors 77 13 (c) Calculate the energy stored by the capacitor when it is charged to a potential difference of 20 V. (d) Suggest why the maximum voltage is marked on the capacitor. 30. (a) You are provided with four 100 V batteries and 12 capacitors each marked ’2.0 F, 100 V maximum’. With the aid of circuit diagrams, show how would you connect some of the components provided to obtain (i) a capacitance of 1 F connected to a 100 V supply, (ii) a capacitance of 1 F connected to a 400 V supply, (iii) a capacitance of 2 F connected to a 300 V supply. (b) Calculate the energy stored in each of the system of capacitor in (b). 31. (a) Explain what is meant by the time constant of a circuit used to discharge a capacitor. Time/s 5 15 25 35 45 55 65 75 Current I/PA 260 220 180 150 125 104 86 72 (b) A charged capacitor C is connected in series with a microammeter of negligible resistance, a 100 k: resistor and a switch. The microammeter readings are recorded after the switch is closed as shown below. (i) By drawing a suitable graph, determine the time constant of the circuit. (ii) Calculate the capacitance of the capacitor. (c) The diagram shows a 2 μF capacitor connected to a 10 V d.c. supply, a 2 M: resistor and switches K1 and K2 . K1 K2 2 MΩ 2 +F 10 V – + (i) Calculate the charge in the capacitor when switch K1 is closed and switch K2 is open. (ii) Switch K1 is then opened and switch K2 closed. Calculate the charge in the capacitor 8.0 s after switch K2 is closed. 32. (a) (i) Defi ne farad. (ii) Write the formula for the capacitance of a parallel-plate capacitor with an insulator as dielectric. Identify all the symbols in your formula. (b) A parallel-plate capacitor has plates each of area A and separated by a dielectric of thickness d and dielectric constant εr and resistivity ρ. (i) Write an expression for the resistance of the dielectric. (ii) Hence, show that the time constant for the discharge of the capacitor is τ = ε0 εrρ. (c) The dielectric constant of paper is 3.7 and its value of ρ is 1.0 u1010 : m. Find the fraction of the initial charge that remains in the capacitor (i) 0.33 s (ii) 1.0 s after it is disconnected from the voltage supply.
Physics Term 2 STPM Chapter 13 Capacitors 78 13 1 1. C = Q V = 24 mC 12 V = 2 mF 2. (a) Q = CV = (50 PF)(3.0 V) = 150 PC (b) Q = (50 PF)(6.0 V) = 300 PC Q V 0 3. Q = (100)V1 = (50)(12) V1 = 6 V 2 1. C = ε0 A d A = Cd ε0 = (6.0n) (2.0 u 10–3) 8.85 u 10–12 = 1.36 m2 2. (a) C = ε0 A d = (8.85 u 10–12)(0.04) 5 u 10–3 = 7.08 u10–11 F (b) (i) E = V d , Vmax = Emax d d = (3.0 u 106 )(5 u 10–3) = 1.50 u 104 V (ii) Qmax = CVmax = 1.06 u 10–6 C 3. Final charge, Q = CV = (24 m)(6.0 V) = 144 mC Time taken, t = Q I = 96 s 3 1. C 2. B 3. 1.42 u 10–8 F C = 2 u ( εr ε0 A —––– d ) 4 1. A 2. D: Capacitor X: Q = CV Capacitors X and Y in parallel. Same V, same Q. 3. D 4. A 5. C 6. D 7. A: Q1 = (C 2 )V and Q2 = (2C)V 8. (a) (i) 0.5 PF (ii) 6 PC (b) (i) 2 PF (ii) 12 PC (c) (i) 8 PF (ii) 24 PC (d) (i) 0.8 PF (ii) 9.6 PC, 4.8 PC 9. (a) Two 2 PF capacitors in series with 2 u 10 V. (b) 20 V 10. (a) — 20 3 PC (b) 44.4 V 2Q = (C1 + C2 ) V 11. (b) (C1 + C2 )V1 = C1 V1 + C2 V2 V1 = (C1 V1 + C2 V2 ) ——–——— (C1 + C2 ) 5 1. C: Same charge Q on all capacitors. 3RWHQWLDOGLͿHUHQFHDFURVVFDSDFLWRU&1 , V1 = Q —– C1 2. A: U = 1 — 2 CV2 , U’ = 1 — 2 (εr C)V2 = εr U 3. D 4. C 5. A 6. B 7. B 8. C 9. (a) 5 PC Q = CV (b) 1.25 u 10–4 J U = 1 ––2 CV2 (c) 2.50 u 10–4 J W = QV (d) 1.25 u 10–4 J (iii) – (ii) 10. (a) 1.25 J U = 1 ––2 CV2 (b) 0.25 J U = 1 ––2 Q2 ——–— (C1 + C2 ) Energy dissipated as heat = 1.00 J 11. (a) (C1 + C2 ) V = C1 V1 + C2 V2 V = (C1 V1 + C2 V2 ) ——–——– (C1 + C2 ) (b) 2.0 u 10–2 J 12. (a) 1.00 u10–7 C Q = CV (b) (i) 1 000 V Vc = Q —C (ii) 4.0 u 10–5 J W = 1 ––2 C V 2 – 1 ––2 CV2 ANSWERS
Physics Term 2 STPM Chapter 13 Capacitors 79 13 13. (a) C = εr ε0 A —––– d (b) C increases 14. (a) (i) 1.20 u 10–8 C Q = CV (ii) 6.00 u 10–7 J U = 1 ––2 QV (b) (i) 70 pF C1 = 10 pF (air) C2 = 60 pF (dielectric) C = C1 + C2 (ii) 1.03 u 10–6 J U = 1 ––2 Q2 —C Work done on system against electrostatic forces. 15. (a) (i) (8 + 6) V = (8)(6.0) + (6)(12.0) V = 8.6 V (ii) Energy loss = [ 1 ––2 C1 V 2 1+ 1 ––2 C2 V 2 2 ] – 1 ––2 (C1 + C2 )V 2 = 0.058 J (b) (i) (8 + 6)V = (6)(12.0) – (8)(6.0) V = 1.7 V (ii) Energy loss = 0.56 J 6 1. D 2. B 3. C 4. C 5. D 6. B 7. C 8. C 9. D 10. A 11. (a) Before ball strikes the bat: V constant = 6.0 V When ball strikes the bat: V decreases exponentially. When ball leaves the bat: V constant. (b) (i) 0.292 V0 = 6.0 V, V = 1.75 V (ii) 2.96 u 10–3 s, V = V0 e t – —– CR 12. (a) 6.0 V (b) V decreases exponentially: Capacitor discharges through R. (c) C = 1 PF, R Nۙ τ = CR = 5.0 u 10–3 s is in the same order as time taken for bullet to travel 10 cm. (——0.10 40 = 2.5 u 10–3 s) (d) V = voltmeter reading after Y breaks = (6.0)e -t CR. Calculate t. (e) Discharge of capacitor through voltmeter negligible. STPM PRACTICE 13 1. A: Charge is conserved, Q = constant 2. D: As V across the capacitor increases, V, across R decreases to zero. 3. C: 1 ––R = 1 ––– 3.0 + 1 ––– 6.0 , R = 2.0 k: V = V0 e –t RC, ln V = ln V0 – t ––– RC Gradient = – 1 ––– RC= – 0.8 ––– 40 = – 0.02 C = 1 –––––––––––––– (2.0 u 103 )(0.02) F = 25.0 mF 4. C: U0 = 1 ––2 CV2 , U1 = 1 ––2 (4.0C)V2 = 4U0 5. D: V= V0 e –t RC, ln V = ln V0 – t ––– RC Gradient = – 1 ––– RC = – 0.4 ––– 20 , time constant = RC = 50 s 6. $ffl(ͿHFWLYHFDSDFLWDQFHfflCA = 1 ––4 C, CB = 0.4 C, CC = C, CD = C 7. C: V = VR + VC = IR + Q––C 8. D: C = εr ε0 A —––– d 9. B: Charge is conserved. (4.0 u 12.0 + 6.0 u 6.0) = (4.0 + 6.0)V V = 8.4 V 10. C: With dielectric, capacitance is increased. 11. B: C1 = Hr C0 , U = Q2 ––– 2C 12. B: Charge is conserved, new V = Q –––—— C1 + C2 , some energy dissipated as heat. 13. A 14. A 15. B 16. C: Induced charges on opposite faces of dielectric VHWVXSDUHYHUVHÀHOGWKDWORZHUVWKHHOHFWULFÀHOG in the dielectric. 17. D 18. A 19. C 20. B 21. C 22. C 23. (a) 1 –––Ce = 1 ––– 3.0 + 1 ––– 4.0 , C= 1.71 PF (b) Same charge on both capacitors, Q = CV = (1.71 PF)(12.0 V) = 20.5 PC (c) With dielectric between the plates, capacitance increases. Equivalent capacitance increases. Charge stored in both capacitors increases. 24. (a) (i) 50 PF and 100 PF capacitors are in series. Charge of 50 PF = charge on 100 PF capacitor = 700 PC (ii) Equivalent capacitance of 200 PF capacitor and 150 PF capacitor (in parallel) is 350 PF Charge on 350 PF capacitor = (350 PF)V = 700 PC Hence V = 2.0 V Charge on 200 PF capacitor = (200 PF)(2.0 V) = 400 PC (b) —1 C = VXY = Q —C = (700 PC)(—–1 50 + —–– 1 350 + —–– 1 100 ) = 23.0 V
Physics Term 2 STPM Chapter 13 Capacitors 80 13 25. (a) (i) C increases (ii) Q unchanged (iii) V decreases. (b) (i) 1 ––C = —–1 20 + 1 7.0(20) C ۚ) (ii) U = 1 ––2 Q2 —– C , Q ۚ& C1 ۚ)U1 = 1 ––2 (17.5 u 10–6 u 12.0)2 —–—––––––––– (20 u 10–6) J = 1.10 mJ C2 = (20 uۚ) U2 = 0.158 mJ (c) (i) Time constant = CR = 3.0 s (ii) V = V0 e-t/CR 4.5 = 6.0e-t/3.0 t = 0.86 s (iii) Smaller R, shorter time constant. &DSDFLWRUGLVFKDUJHVIDVWHU%ULJKWHUÁDVK 26. (a) (i) Capacitance: C = Q —V , Ratio of the charge RQWKHFDSDFLWRUWRWKHSRWHQWLDOGLͿHUHQFH across the capacitor. (ii) Factors: area of each plate and the separation between the plates. Charged stored is increased because the capacitance increases. (b) (i) —1 C = —–1 20 + —–1 30 , C = 12 PF Charge from battery, Q = CV = (12 PF)(6.0 V) = 72 PC (ii) V1 = 1———– C2 C1 + C2 2V = 3.6 V (c) (i) Time constant = CR = (300 P) ۙ = 0.15 s (ii) t = 0.50 s, V = (6.0)e-0.20/0.15 V = 1.58 V Energy dissipated from resistor = —1 2 CV2 0 – —1 2 CV 2 = —1 2 (300 u 10-6)(6.02 – 1.582 ) J = 5.03 mJ 27. 6.38 V V = (0.9)6 V0 28. (a) Refer to page 50. (b) (i) — 1 C = — 1 C1 + —1 C2 gives C flۚ) Same charge on both capacitors, Q = CV ffiۚ& (ii) U = —1 2 CV2 = 0.576 mJ (iii) Energy supplied = QV = 1.15 mJ (1.15 – 0.576) mJ dissipated as heat. (c) (i) I0 = 12.0 5.0 u103 A = 2.4 mA (ii) Energy dissipated = Q(average V) ffiۚ& —–– 12.0 2 V) J = 0.576 mJ 29. (a) Q = 100 ۚ&ZKHQV = 1 V (b) 2 mC Q = CV (c) 2 u 10–2 J U = 1 ––2 CV2 (d) Insulation of dielectric would not break down. 30. (a) (i) 2 F 2 F + – 100 V A B (ii) 400 V A B (iii) 300 V A B (b) (i) 5.0 u 103 J (ii) 8.0 u 104 J (iii) 9.0 u 104 J 31. (a) Refer page 66 (b) (i) Draw graph of ln I against t. I = I0 e t – —– CR ln I = ln I 0 – (—–1 CR )t Gradient = – —–1 CR τ = 55 s (ii) C = 5.5 u 10–4 F (c) (i) 20 PC Q = CV (ii) 2.71 PC τ = CR = 4.0 s 32. (a) Refer (i) page 45, (ii) page 47 (b) (i) R = ρd —–A (c) (i) 0.368 Q = Q0 e t – —– CR (ii) 0.0472
14 81 CHAPTER ELECTRIC CURRENT 14 81 Concept Map (OHFWULF&XUUHQW &RQGXFWLRQRI(OHFWULFLW\ 'ULIW9HORFLW\ (OHFWULFDO&RQGXFWLYLW\ DQG5HVLVWLYLW\ &XUUHQW'HQVLW\ Bilingual Keywords *VUK\J[HUJL!2VUK\R[HUZ *\YYLU[KLUZP[`!2L[\TWH[HUHY\Z +YPM[]LSVJP[`!/HSHQ\OHU`\[ ,SLJ[YPJJ\YYLU[!(Y\ZLSLR[YPR ,SLJ[YPJHSJVUK\J[PVU!2VUK\RZPLSLR[YPR ,SLJ[YPJHSJVUK\J[P]P[`!2LRVUK\RZPHULSLR[YPR 6OT»ZSH^!/\R\T6OT 9LZPZ[HUJL!9PU[HUNHU 9LZPZ[P]P[`!2LYPU[HUNHU In this chapter you will learn the microscopic aspect of electric current in metals and semiconductors. You will be introduced to development in the fi eld of superconductivity. INTRODUCTION
Physics Term 2 STPM Chapter 14 Electric Current 14 82 14.1 Conduction of Electricity Students should be able to: t EFmOFFMFDUSJDDVSSFOU BOEVTFUIFFRVBUJPOI = dQ dt t FYQMBJOUIFNFDIBOJTNPGDPOEVDUJPOPGFMFDUSJDJUZJONFUBMT Learning Outcomes 1. In Chapter 12 you learned about electrostatics – electric charges at rest. When electric charge fl ows, a current is produced. 2. In metals, electric current is produced by the fl ow of free electrons. 3. Electric current I is the rate of fl ow of electric charge. Current, I = dQ dt When the electric current is produced by the fl ow of free electrons, then Current I = (rate of fl ow of electrons) u (charge of the electron) = Ne 4. Electric current is a base quantity in SI, and its unit is the ampere (A). Ampere = coulomb second A = C s–1 C = A s Mechanism of Conduction of Electricity in Metals 1. In metals such as copper, there are many free electrons, 1029 free electrons in a metre cube of copper. 2. These free electrons are in constant random motion, colliding with each other and with the copper ions in the crystal lattice. 3. At any instant the number of free electrons having a defi nite velocity in a certain direction is equal to that moving with the same speed but in the opposite direction. Hence the mean velocity of the free electrons is zero, and there is no net transfer of charge in the copper piece. 4. On the other hand, the mean speed of the free electrons in copper is of the order 105 m s–1. Figure 14.1 Random thermal motion of free electrons – – – – 5. The mean free path of free electrons in a metal is the mean distance between successive collisions of the free electrons with the metal ions. VIDEO 0U[YVK\J[PVU[V ,SLJ[YPJP[`
Physics Term 2 STPM Chapter 14 Electric Current 14 83 6. Figure 14.2 shows what happens to the free electrons in a piece of copper of uniform cross section A and length l when a potential difference is applied across the conductor. s !N ELECTRIC l ELD E is set up in the conductor. s 4HE FREE ELECTRONS EXPERIENCE A NET FORCE IN THE DIRECTION OPPOSITE TO THE ELECTRIC l ELD E. s 4HE FREE ELECTRONS drift in the direction opposite to E. s #HARGE IS TRANSFERRED AND A CURRENT I fl ows in the direction of E. E A v v v + – Free electron l I Figure 14.2 Example 1 The relative atomic mass of copper is 63.5 and its density is 8.93 u 103 kg m–3. Each m3 of copper contains 8.34 u 1028 free electrons. Calculate to 3 signifi cant fi gures, the number of free electrons contributed by each copper atom. Solution: From Avogadro number, NA = 6.02 u 1023 mol–1, Number of copper atoms in 1 mole = 6.02 u 1023 Mass of 1 mole of copper = 63.5 u 10–3 kg Mass of 1 m3 of copper = 8.93 u 103 kg Hence, number of copper atoms in 1 m3 (or 8.93 u 103 kg) = ( 8.93 u 103 ————— 63.5 u 10–3 )(6.02 u 1023) = 8.466 u 1028 8.466 u 1028 copper atoms contribute 8.34 u 1028 free electrons. Hence 1 copper atom contributes 8.34 u 1028 —————— 8.466 u 1028 free electrons = 1 free electron Quick Check 1 1. The current in a wire is 0.20 A. (a) Calculate the charge that fl ows through a cross section of the wire in 5.0 minutes. (b) The charge of an electron is 1.60 u 10–19 C. What is the number of electrons per second that fl ow through a cross section of the wire?
Physics Term 2 STPM Chapter 14 Electric Current 14 84 14.2 Drift Velocity Students should be able to: t FYQMBJOUIFDPODFQUPGESJGUWFMPDJUZ t EFSJWFBOEVTFUIFFRVBUJPOI = Anev Learning Outcomes 1. The drift velocity, v of free electrons in a metal is the mean velocity of the free electron when a potential difference is applied across the metal. 2. Figure 14.3 shows free electrons moving with a drift velocity v in a metal of cross section A when a potential difference is applied across the metal. If n = number of free electrons per unit volume of the metal, then the number of free electrons in a length l of the metal = n u (volume) = n u (Al) Each free electron carries a charge e, hence the charge carried by (nAl) electrons is (nAl)e. If v = drift velocity, then the time taken for an electron to move through a distance l is t = — l v The current I = rate of fl ow of charge = charge ——— time = (nAle) ——— — l v I = nAve Example 2 (a) Calculate the drift velocity of free electrons in a copper wire of cross section 1.0 mm2 when the current in the wire is 2.0 A. (Number of free electrons in copper = 1 u 1029 m–3) (b) Calculate the time taken for a free electron to move through a distance of 0.50 m in the wire. Solution: (a) Using I = nAve, drift velocity, v = —–– I nAe = ——————————————— 2.0 (1.0 u 1029)(1.0 u 10–6)(1.6 u 10–19) = 1.25 u 10–4 m s–1 (b) Time taken, t = ———— distance v = ————— 0.50 1.25 u 10–4 = 4.00 u 103 s A e e e e e e e v P Q + – Free electron l I e Figure 14.3 2014/P2/Q5, 2015/P2/Q5, 2016/P2/Q4, 2018/P2/Q16
Physics Term 2 STPM Chapter 14 Electric Current 14 85 Quick Check 2 1. When a certain current fl ows in a copper wire, the drift velocity of free electrons in the wire is v. If the same current fl ows in another copper wire of twice the radius, the drift velocity is A —v 4 B — v 2 C 2v D 4v 2. When a current of 5 A fl ows in a copper wire, the drift velocity of free electron is of the order of 10-4 m s-1. Which statement is the CORRECT EXPLANATION FOR THE DIFFERENCE IN THE orders of the values of the current and the drift velocity? A There is constant collisions between the free electrons with the lattice atoms. B Free electrons are dispersed by vibration of the lattice atoms. C The number density of free electrons in copper is large. D The random velocity of the free electrons is of the order of 105 m s-1. 3. A constant current I fl ows in a piece of metal of uniform thickness and has the shape shown in the fi gure. LMN l l I If the drift velocity of the free electrons at the cross sectional L, M, and N are vL , vM, and vN respectively, then A vL = vM = vN C vL = vN < vM B vL = vN > vM D vL > vM > vN 4. The drift velocity of free electrons in copper wire is of the order of 10–4 m s–1, but a bulb is lit with the fl ick of a switch. Which is the CORRECT EXPLANATIONfi A The large number of free electrons in copper B The random speed of the free electrons is of the order 106 m s–1 C High rate of collision between free electrons D All free electrons drift in the same direction simultaneously 14.3 Current Density Students should be able to: t EFmOFFMFDUSJDDVSSFOUEFOTJUZBOEDPOEVDUJWJUZ t VTFEUIFSFMBUJPOTIJQJ = σE Learning Outcomes 1. The current density in a conductor is the current per unit cross-sectional area. Current density, J = —I A where A = cross-sectional area. 2. From the equation, current, I = nAve Current density, J = —I A = nAve ——v J = nve 2012/P1/Q27, 2016/P2/Q16, 2017/P2/Q16 Exam Tips In the formula 0 = U(]L, the area ( must be perpendicular to the current 0.
Physics Term 2 STPM Chapter 14 Electric Current 14 86 3. When you switch on a lamp, the lamp lights up immediately, but the free electrons drift with a velocity of the order of magnitude 10–4 m s–1. Electrical Conductivity 1. The ability of a material to conduct electric current is indicated by the value of its electrical conductivity. Good electrical conductors have high values of electrical conductivity. Electrical conductivity σ is defi ned as the ratio of the current density J to the electric fi eld strength E in the material. Electrical conductivity, σ = J E 2. The SI unit for electrical conductivity is siemens per metre (S m–1). The electrical conductivity of copper, which is a good conductor, is 5.96 u107 S m–1. Example 3 The electrical conductivity of copper is 6.0 u107 S m–1. The current in a piece of copper wire of crosssectional area 1.3 u10–6 m2 is 2.6 A. (a) What is the current density in the wire? (b) What is the electric fi eld strength E in the wire? (c) What is the potential difference across 0.5 m length of the wire? Solution: (a) Current density J = I A = 2.6 1.3 u 10–6 = 2.0 u106 A m–2 (b) Electrical conductivity, σ = J E Electrical fi eld, E = J σ = 2.0 u 106 6.0 u 107 = 3.3 u 10–2 V m–1 (c) E = V l Potential difference, V = El = (3.3 u 10–2)(0.5) V = 16.5 mV
Physics Term 2 STPM Chapter 14 Electric Current 14 87 Quick Check 3 1. (a) The density of copper is 8940 kg m-3 and mass of one mole of copper is 0.0635 kg. A current of 50 A fl ows in a copper cube of sides 1.0 cm. Assuming that each copper atom contributes one free electron, determine the number of free electrons in the cube. (b) (i) The free electrons in copper drift when a current fl ows, what causes the free electrons to drift? (ii) Find the drift velocity of free electrons in the copper cube. 2. (a) A copper wire has a cross sectional area A. The number of free electrons per unit volume in copper is n, and the electrons have a drift velocity v $ERIVE AN EXPRESSION FOR THE CURRENT IN the wire in terms of A, n, v and the electronic charge, e. (b) A current of 4.5 A fl ows in a copper wire of cross sectional area 3.2 u m2 . If n = 8.5 ufl m-3, fi nd (i) the current density, (ii) the drift velocity of the free electrons. 14.4 Electric Conductivity and Resistivity Students should be able to: t EFSJWFBOEVTFUIFFRVBUJPOσ = ne2 t m t EFmOFSFTJTUJWJUZ BOEVTFUIFGPSNVMBρ = RA l t TIPXUIFFRVJWBMFODFCFUXFFO0INTMBXBOEUIFSFMBUJPOTIJQJ = σE t FYQMBJOUIFEFQFOEFODFPGSFTJTUJWJUZPOUFNQFSBUVSFGPSNFUBMTBOETFNJDPOEVDUPSTCZVTJOHUIFFRVBUJPOσ = ne2 t m t EJTDVTTUIFFGGFDUTPGUFNQFSBUVSFDIBOHFPOUIFSFTJTUJWJUZPGDPOEVDUPST TFNJDPOEVDUPSTBOETVQFSDPOEVDUPST Learning Outcomes Resistivity 1. The resistance R of a piece of wire is s DIRECTLY PROPORTIONAL TO ITS LENGTH l s INVERSELY PROPORTIONAL TO ITS CROSS SECTIONAL AREA A, s DEPENDENT ON THE MATERIAL 2. Hence the resistance of a piece of wire R = l —A where is a constant which depends on the material of the wire, it is known as the resistivity of the material. 3. Resistivity, = —– RA l The resistivity of a material is the resistance of a unit length of the material which has a unit cross-sectional area. 4. The unit for resistivity is : m2 ——– m = : m. 5. The table below shows the resistivity of three metals. Metal Resistivity/: m Copper 1.69 u 10–8 Aluminium 3.21 u 10–8 Eureka (Constantan) 49 u 10–8 2011/P1/Q27, 2012/P1/Q26, 2013/P2/Q6,Q19, 2015/P2/ Q18(a), 2016/P2/Q6, 2017/P2/Q5,Q6
Physics Term 2 STPM Chapter 14 Electric Current 14 88 Example 4 A thin fi lm resistor in a solid-state circuit has a thickness of 1 Pm and is made of nichrome of resistivity 1 u 10–6 : m. Calculate the resistance between opposite edges of a 1 mm2 area of fi lm (a) if it is a square, (b) if it is a rectangle, 20 times as long as it is wide. Solution: (a) Using R = l —A , resistance between P and Q = (1 u 10–6)(1.0 u 10–3) —————————–— (1.0 u 10–6 u 1.0 u 10–3) = 1.0 : (b) Resistance between P and Q = (1 u 10–6)(20x) ——————— (1 u 10–6)(x) = 20 : Resistance between R and S = (1 u 10–6)(x) ——————— (1 u 10–6)(20x) = 0.05 : 1 μm x 20 x S P R Q Example 5 An electricity supply cable consists of a steel core of cross-sectional area 50 mm2 WITH SIX OTHER ALUMINIUM CONDUCTORS OF THE SAME CROSS SECTIONAL AREA ARRANGED around it as shown in the diagram. Find the resistance of a 120 m length of the cable. (Resistivity: steel, 9.0 u 10–8 : m; aluminium, 2.5 u 10–8 : m) Solution: Resistance of the steel cable, Rs = l —A = (9.0 u 10–8)(120) ———————— (50 u 10–6) = 0.216 : Resistance of 6 aluminium cables is equal to resistance of 1 aluminium cable of 6 times the cross-sectional, RA = (2.5 u 10–8)(120) ———————— (6 u 50 u 10–6) = 0.010 : Hence the effective resistance of the cable R is given by —1 R = — 1 RS + — 1 RA = —–—1 0.216 + —–—1 0.010 R = 9.6 u 10–3 : 1 mm 1 mm 1 μm P Q A A A A A A S
Physics Term 2 STPM Chapter 14 Electric Current 14 89 Quick Check 4 1. A potential difference is applied across a piece of semiconductor. Which of the following decreases when the temperature is increased? A Current in the semiconductor B Conductivity of the semiconductor C Drift velocity of charge carriers in the semiconductor D Number density of charge carriers in the semiconductor 2. A copper wire of diameter 1.2 mm and length l has a resistance of 1.5 :. The resistivity of copper is 1.7 ufl : m. What is the value of l? A 85 m B 100 m C 110 m D 230 m 3. The diagram shows two squares, X and Y, cut from a sheet of metal of uniform thickness t. X and Y have sides of length L and 2L respectively. L 2 L 2 L L X Y The resistance, RX and RY , are measured between the opposite faces shaded in the diagram. What is the value of R —–X RY ? A —1 4 B —1 4 C 1 D 4 4. A wire of uniform cross-sectional area has resistance R. The wire is stretched until its length increases by 10%. Assuming that the wire thins uniformly, what is the new resistance of the wire? A 0.83R C 1.10R B 0.90R D 1.21R 5. (a) Find the resistance between two opposite edge faces of a piece of square carbon fi lm of thickness 5.0 u 10–7 m, if the resistivity of carbon is 4.0 u 10–5 : m. (b) A carbon fi lm of the same thickness is formed on an insulator rod of diameter 3.0 mm. What is the length of the rod so that the carbon fi lm on its curved surface has a resistance of 100 :? 6. When a potential difference of 100 V is applied across a piece of conductor of thickness 5.0 mm and area 0.25 mm2, a current of 2.5 u 10–12 A fl ows. Calculate the resistivity of the conductor. 7. !N ELECTRIC CABLE CONSISTS OF SIX STEEL CABLES around an aluminium core of diameter 5.0 mm. The diameter of each steel cable is 2.0 mm. Calculate the current in the steel cables when the current in the aluminium core is 8.0 A. (Resistivity: aluminium, 3.2 u 10–8 : m, steel, 12.2 u 10–8 : m) 8. The cross-sectional area of a bare wire is a square of area 1.00 mm2. A 10.0 m, length of this wire is wound close together on a wooden cylinder so that neighbouring coils are completely in contact with each other. There is a total of 200 coils. The resistivity of the material of the wire is 1.0 u 10–6 : m. Estimate the effective resistance of the arrangement. 9. A current of 10.0 A fl ows in a metal rod of uniform cross-sectional area 1.50 cm2. A galvanometer which has a resistance of 40 : with a 960 : resistor in series is connected across 20 cm length of the rod. The sensitivity of the galvanometer is 4.0 cm per PA and the deflection of the galvanometer is 8.0 cm. Calculate the resistivity of the metal.
Physics Term 2 STPM Chapter 14 Electric Current 14 90 Ohm’s Law and J = σE 1. The reciprocal of resistance is known as conductance. Conductance, G = —————–– 1 resistance, R Conductivity σ of a material is the reciprocal of resistivity, U. Electrical conductivity, σ = —————–– 1 resistivity, U 2. For an ohmic conductor, a conductor that obeys ohm's law, —V I = constant = R, resistance of conductor From U = —– RA l , conductivity, σ = —1 U = —–l RA l = ——— (— V I ) A = (—l V )(—I A ) (E = —V l , J = —I A ) = —1 E u J Current density, J = σE Hence the relation J = σE is equivalent to Ohm’s law. 3. Conversely, starting with J = σE, suppose that for a conductor of length l, and cross-sectional A, when a potential difference V is applied across it, the current in the conductor is I (Figure 14.4). Current density, J = —I A Electric field strength, E = —V l Using J = σE, —I A = σ(—V l ) —V I = —–l σA Since l, σ and A are constants when the temperature is constant, — V I = constant which is Ohm’s law. Figure 14.4 l I V A
Physics Term 2 STPM Chapter 14 Electric Current 14 91 Electrical Conductivity, σ 2013/P2/Q5, 2014/P2/Q18, 2015/P2/Q6, 2017/P2/Q5, 2018/P2/Q5 1. In a metal such as copper, s EACH COPPER ATOM CONTRIBUTES ONE FREE ELECTRON s THE FREE ELECTRONS ARE IN CONSTANT random motion, colliding with each other and the copper ions in the crystal lattice. s THE MEAN VELOCITY OF THE FREE ELECTRONS IS ZERO s THERE IS NO NET TRANSFER OF FREE ELECTRONS IN ANY DIRECTION BEFORE A POTENTIAL DIFFERENCE IS APPLIED across the piece of copper. 2. When a potential difference V is applied across a copper rod a length l, cross-sectional area A, and electrical conductivity σ, the electric field produced in the copper rod is E = — V l The force on each electron, F = eE acceleration, a = —F m a = —– eE m where e = charge of electron m = mass of electron 3. If τ = mean time interval between successive collisions of free electrons with metal ions. For a free electron whose random velocity immediately after a collision is u1 , its final velocity just BEFORE THE NEXT COLLISION v1 is given by v1 = u1 + at1 (using v = u + at) Similarly for the other N FREE ELECTRONS IN THE METAL THEIR RESPECTIVE VELOCITIES JUST BEFORE THE NEXT collision are v2 = u2 + at2 v3 = u3 + at3 x x x vN = uN + atN 4. The drift velocity v of the free electron is defined as the mean of v1 , v2 , v3 ..., vN Hence, drift velocity, v = mean of (v1 , v2 , v3 ..., vN) = mean of (ui + ati ) where i = 1, 2, 3,..., N = mean of (ui ) + a (mean time interval between collisions) but mean of (ui ) = 0 because the mean of the random motion velocities before the potential difference applied is zero. Let the mean time interval between successive collisions = τ. Hence, drift velocity, v = aτ = (—– eE m ) τ 5. The current density, J = nve = n(—– eE m ) τe Also, J = σE σE = n(—– eE m ) τe σ = ne2 —––τ m
Physics Term 2 STPM Chapter 14 Electric Current 14 92 6. The formula σ = ne2 —––τ m CAN BE USED TO EXPLAIN THE VARIOUS FACTORS THAT AFFECT ELECTRICAL CONDUCTION (a) Metals are good electrical conductors 4HERE ARE MANY FREE ELECTRONS IN METALS &OR EXAMPLE THE NUMBER OF FREE ELECTRONS PER M3 of copper, n = 1029. Since electrical conductivity, σ v n, metals are good conductors of electricity. Conversely, for semiconductors such as carbon and germanium, the value of n is small compared to that of copper. Hence, semiconductors are poor conductors of electricity. (b) Temperature dependence of resistivity Resistivity of a material, U = ——————— 1 conductivity, σ U v — 1 n Hence, the resistivities of metals are small, since n is large. s 4HE VALUE OF n for a metal remains constant, irrespective of the temperature. s 7HEN THE TEMPERATURE INCREASES THE IONS IN THE CRYSTAL LATTICE VIBRATES MORE VIGOROUSLY AND with bigger amplitude of oscillation. s 4HE MEAN FREE PATH OF THE FREE ELECTRONS DECREASES 3UBSEQUENTLY THE MEAN TIME INTERVAL τ between collisions decreases, and the drift velocity decreases. From the equation σ = ne2 —––τ m , the electrical conductivity V of the metal decreases and the resistivity, U = —1 σ increases. For semiconductors s 7HEN THE TEMPERATURE INCREASES THE NUMBER OF CHARGE CARRIERS PER UNIT VOLUME n increases as there are more positive holes and free electrons. s (ENCE THE ELECTRICAL CONDUCTIVITY INCREASES AND THE RESISTIVITY DECREASES Example 6 %XPLAIN HOW THE DRIFT VELOCITY OF FREE ELECTRONS IN A METAL CARRYING A CURRENT CHANGES (a) when the potential difference across the conductor is increased, (b) when the temperature of the conductor increases but the current remains constant. Solution: (a) When the potential difference V is increased, s THE ELECTRIC l ELD STRENGTH E = —V l increases. s THE FORCE ON THE FREE ELECTRON F = eE increases. Hence the drift velocity increases. (b) When the temperature increases, s METAL IONS VIBRATE MORE VIGOROUSLY AND THE AMPLITUDE OF OSCILLATION INCREASES s MEAN FREE PATH OF THE ELECTRON DECREASES (ENCE THE DRIFT VELOCITY decreases.
Physics Term 2 STPM Chapter 14 Electric Current 14 93 Electrical Conduction in Semiconductors 2009/P2/Q5, 2010/P1/Q28, 2018/P2/Q7 1. In a pure semiconductor, known as intrinsic semiconductor, the charge carriers are free electrons and positive holes which are equal in number. A free electron gives rise to a positive hole. 2. When a potential difference is applied across a piece of intrinsic semiconductor, the free electrons drift in one direction, and the positive holes are shifted in the opposite direction. Electric charges flow in the semiconductor and this constitutes a current. 3. At room temperature, the number of charge carriers per unit volume n is rather small compared TO THAT IN METALS 4HIS EXPLAINS FOR THE LOW ELECTRICAL CONDUCTIVITY σ, and high resistivity U of semiconductors. 4. The conductivity σ of semiconductors is improved by doping semiconductors with small traces of impurities. The semiconductors obtained are known as extrinsic semiconductors. 5. )F THE IMPURITY IS A GROUP 6 ELEMENT FOR EXAMPLE ANTIMONY OR ARSENIC THE NUMBER OF FREE ELECTRONS increases. The majority charge carrier is the free electrons, and the minority charge carrier, the positive holes. The semiconductor produced is n-type. 6. A p-type semiconductor is produced by doping an intrinsic semiconductor with group III elements such as indium or gallium. The majority charge carrier is the positive holes and the minority charge carrier, the free electrons. 7. When the temperature of a semiconductor increases, more hole-electron pairs are produced, the number of charge carriers increases and electrical conductivity increases, while resistivity decreases. Superconductors 1. The resistivity of a metal increases when temperature increases. Conversely, the resistivity decreases when temperature decreases. 2. This leads scientists to believe that, the resistivity of a metal would be zero when its temperature is sufficiently low. 3. %XPERIMENTS TO MEASURE THE RESISTIVITY OF METALS AT LOW TEMPERATURES INTENSIlED AFTER WHEN Wroblewski and Olzewski succeeded in liquifying air. 4. %XPERIMENTS TO MEASURE THE RESISTIVITY OF METALS WERE CARRIED OUT AT THE FOLLOWING TEMPERATURES A BOILING POINT OF OXYGEN n # OR + B TEMPERATURE OF LIQUID NITROGEN n # OR + C TEMPERATURE OF LIQUID HYDROGEN n # OR + D TEMPERATURE OF LIQUID HELIUM n # OR + 5. )N A $UTCH SCIENTIST (EIKE +AMERLING /NNES MEASURED THE RESISTIVITY OF MERCURY AT LOW TEMPERATURES 4HE RESULTS OF THE EXPERIMENT ARE AS SHOWN IN THE GRAPH BELOW &IGURE 0 4.0 4.2 4.4 Temperature / K Resistivity ρ / m Figure 14.5 Info Physics 4VQFSDPOEVDUPST BSF VTFE UP NBLF QPXFSGVM FMFDUSPNBHOFUT CFDBVTF MBSHF DVSSFOUT DBO nPX BOE WFSZ MJUUMF IFBU JT QSPEVDFE 5IFTF FMFDUSPNBHOFUT BSF VTFE GPS MFWJUBUJPO PG USBJO BCPWFUIFUSBDLT
Physics Term 2 STPM Chapter 14 Electric Current 14 94 6. 7HEN THE TEMPERATURE DECREASES THE RESISTIVITY OF MERCURY DECREASES !T A TEMPERATURE OF + the resistivity of mercury drops to zero. 7. -ERCURY AT + HAS BECOME A superconductor 4HE TEMPERATURE OF + IS KNOWN AS THE CRITICAL temperature, Tc of mercury. 8. The phenomenon of the resistivity of a material becoming zero is known as superconductivity. 9. A superconductor has zero resistance. Hence no heat is dissipated and large currents can fl ow in a superconductor. 10. A superconductor is an element, an alloy, or compound that will conduct electricity without resistance below a certain temperature known as the critical temperature, Tc . 11. 4HE HIGHEST CRITICAL TEMPERATURE ACHIEVED IS + FOR SUPERCONDUCTIVITY CUPRATES OR COPPER OXIDES Uses of Superconductors 1. Superconductors conduct electricity without resistance. This means that there is no energy loss in the form of heat when current fl ows in a superconductor. 2. Once set in motion, electrical current will fl ow forever in a closed loop of superconducting material. 3. Superconductors can be employed in the transmission of electrical power. High-temperature superconductors (HTS) can be used to transmit millions of watt of power with practically no energy loss. The amount of superconducting wires used is very much less than that of copper wires. 4. Magnetic-levitation or MAGLEV train use strong superconducting electromagnets which are smaller, and lighter than conventional electromagnet. The train fl oats and eliminates friction between the train and tracks. MAGLEV train can achieve a speed of almost 500 km per hour. 5. Electric generators made of superconducting wires are more effi cient than conventional generators that use copper wires. In fact, their effi ciency is more than 99% and their size is much smaller. 6. High-power electric motor made of superconducting material that are smaller can be used to power electric vehicles and power tools. 7. Efforts are being made to built supercomputers known as ‘petafl op’ computer that can perform thousand-trillion fl oating point operation per second. These computers would be able to run at blistering speed. Quick Check 5 1. The electrical conductivity of a metal decreases with increase in temperature because A the concentration of conduction electrons decreases. B the mean distance between metal ions increases. C the speed of the conduction electrons increases. D the scattering of the conduction electron by the metal ions increases. 2. The process of electrical conduction in a metal may be described by a model in which the free electrons behave much like the molecules of a gas, colliding with ions of the metal. According to this model, an increase in temperature causes the conductivity of the metal to decrease because A the root-mean-square speed of the free electrons increases. B the number of free electrons increases. C the mean time between collisions of electrons with the metal ions increases. D the mean free path of the free electron decreases.