1. AL Capacitor P.34

5. A capacitor is an electrical device for storing electric charge and energy. - consists of two parallel metal plates with an insulator (dielectric) between the plates Dielectric – air, paper, wax, ceramic, mica Capacitor

6. P.34 Air capacitorDisc ceramic capacitor Film capacitor Dipped mica capacitor Metallized paper capacitor Electrolytic capacitor Types of capacitors

7. P.34 fixed capacitor variable capacitor Types of capacitors

8. P.34 Charging ++++++ ------ VxVx QσE=σ/εV=Ed

9. P.35 Discharging ++++++ ------ VxVx QσE=σ/εV=Ed

10. P.35 ε/R -ε/R

11. P.35 Capacitance C QσE=σ/εV=Ed Unit : Farad (F) For a point charge, The capacitance of a capacitor is one farad (1 F), if the capacitor stores one coulomb (1 C) of charge when there is a potential difference of one volt (1 V) applied across it.

12. P.36 Experiment ++++++ - - - R max I= V/R Set R to maxI is min. Reduce RKeep I constant VcVc VRVR ε = V C + V R

13. P.37 If Q is constant, ++++++++++++++++++++ -------------------- A +Q -Q B E is constant, If d is reduced,V is reduced C is increased If A is increased,E is reduced C is increased V is reduced

14. P.37 Increase C by: (1) increasing overlapping area (A) (2) placing two plates closer (d) (3) replacing dielectric with higher permittivity (  )

15. P.38 Dielectric material External E field E field by dipole Resultant E field For all insulators,  r > 1.0 dielectric increases capacitance (charge-storing ability) - molecules of dielectric are polarized - one end of each molecule has excess +ve charge - other end has excess –ve charge - charges appear on surface - potential difference is induced - By C = Q/V, C increases

16. P.38 Dielectric material V+V+ V-V- V C = V + - V - More and more charges are needed to increase V C to ε Q

17. P.38 Material Relative permittivity (  r ) Vacuum1.0000 Air1.0005 Oil2 – 5 Paper2 – 6 Glass8 Ceramic80 – 1 200 Capacitance of capacitors with different materials between plates

18. Conducting material External E field E field by induced charges Zero resultant E field P.38

19. P.39

21. a. b. c. d.

22. P.39 ++++++ ++++++ A +?+? B ++++++++ ++++ +900μ+100μ Q=400μ

23. P.39

24. P.40 Q is constant Overlapping area is decreased C is decreased V is increased

25. P.40 Q is constant d is decreasedC is increased V is decreased E remains unchanged

26. P.40 V C is constant d is increased C is decreased Q is decreased E is decreased V P is decreased W is decreased

27. P.41 Use of reed switch for measuring capacitance Charging +-+- Discharging Vibrating switch - vibrates between A and B at frequency f - at A, capacitor is charged Q = CV - at B, capacitor discharges A B

28. P.41 Use of reed switch for measuring capacitance Charging Discharging VCVC Protective resistor ->reduce I Mean V C Mean I C

29. P.42 If C = 2  F, V = 6 V and the average current I = 0.5 mA, what should be the frequency f ?

30. Capacitance exists in (1) an insulated conductor and metal framework (2) two conductors in an electric cable (3) the turns of a coil of wire electric field lines leak to conductors - gives rise to stray capacitance C=C o +C s Determination of capacitance by reed switch (Stray capacitance) P.42

31. P.43

33. P.44 CRO measures V C + V R = E during charging Reading of CRO is constant during charging CRO measures V C + V R = 0 during discharging Reading of CRO is zero during discharging

34. P.44 Parallel Connection V is the same V +QC - Q If C 1 < C 2 then Q 1 < Q 2

35. P.44 Parallel Connection V is the same V +Q- Q C

36. P.44 Series Connection Q is the same V -Q+ Q C

37. P.44 Series Connection Q is the same V -Q+ Q C

38. P.45

39. What is the connection ? V is the sameParallel connection

40. P.45 V is the same

41. P.45 What is the type of connection ? Parallel connection AB + - Q A Q B Q P

42. P.45 A. Yes. Charges cannot be destroyed or created B. Yes. Voltage will be the same. By the equation. Q = C V, comparing the capacitance, C 2 > C 1, then Q 2 > Q 1 C. Yes. C 1 shares charges with C 2, so V 1 decreases as Q 1 decreases. D. Yes. Both of them are fully charged. No p.d. between them, so V 1 = V 2 E. No. There is charge flowing through R. P=I 2 R, there is energy loss.

43. P.46 Q 22 and Q 8 are the same

44. P.46 Q=I t, x-axis relates to Q

45. P.46 Connected to X1 and 2 will be charging up 1 2 3 4 Connected in series Charge in C 12 Connected to Y3 and 4 will be charging up1 will be discharging By conservation of charges

46. P.46 12 3

47. P.47 Connected in series Switch K is on 5μF5μF 10μF +5μ +10μ

48. P.47 V C/2(1), V 2C(2) and E are the same when steady state is obtained. Whole circuit is independent of R when steady state is obtained.

49. P.47 (1) Yes. By E = (1/2) Q V (2) Yes. By W = (1/2) F e (3) No. By W = P V

50. P.48

52. P.49

53. P.48

55. P.50 Charging mechanism of capacitors (1) electrons from –ve terminal of battery accumulate on one plate of capacitor (2) equal amount of +ve charges induced on opposite plate (3) until p.d. across the capacitor = e.m.f. of battery Charge stored in capacitor

56. P.50 Charging at constant rate +++ --- R set to maximum value RI=V/RQ=I t = C Vt = R C Keep I constant

57. P.50 Charging at varying current +++ --- R is not fixed, I will be changed

58. P.51 VRVR

59. Discharging at varying current +++ ---

60. P.51

61. P.52 Time constants τ Charging Discharging If t=τ, then If Q=0.5Q 0, then

62. P.52 e.m.f. of battery (E) = p.d. across R + p.d. across C= V R + V = IR + Q/C 1. Variation of I At t = 0: At t: Charging and Discharging

63. P.52 2.Variation of Q 3.Variation of V Charging and Discharging

64. P.52 Energy of a charged capacitor - Charge on each plate (q) = CV - Suppose a charge of +dq is moved from –ve plate to +ve plate:

65. P.52 Energy of a charged capacitor Difference in energy If capacitor is charged by a battery of e.m.f. (E) - energy stored in capacitor = ½ QE The energy escapes in the form of HEAT in the connecting wires and EM WAVEs are emitted

66. P.53 Energy changes (Constant V) V is the same V +QC - Q V = constant = E d Energy returns to battery

67. P.53 Energy changes (isolated capacitor) Q is the same V +QC - Q Q = constant Energy supplied by battery to overcome the attractive force between plates = constant V = E d

68. P.53 Energy changes (inserting dielectric material with constant V) V is the same V +QC - Q V = constant = E d Energy supplied by battery to accumulate more charges inside the capacitor

69. P.53 Energy changes (inserting dielectric material into isolated capacitor) Q is the same V +QC - Q Q = constant Energy returns to battery because +ve work has been done by electric force to attract the dielectric material into the capacitor. V = E d

70. P.54 C should be constant for an isolating conducting sphere

71. P.54 Q A and Q B are the samer A < r B V A > V B E A > E B C A < C B When they are connecting by wire, there is p.d.. The charges flow from high potential (A) to low potential (B) Q A < Q B

72. P.54 Energy loss on joining capacitors

73. P.55 Energy loss on joining capacitors No movement of charges, so no energy loss

74. P.55 Types of capacitors Series or parallel connection ? Parallel connection Greater total capacitance

75. P.56 Electrolytic capacitors Greater capacitance Polarity of capacitor is fixed, for d.c. only

76. P.56 Variable capacitors Varying capacitance by changing the overlapping area

77. P.57 Spooning charge Extra high internal resistance

78. P.57 As voltmeter No other connection If the meter is full scale deflection (f.s.d.), thenV=1 (V)

79. P.57 As ammeter R= 10 10 FV= 1V If the meter is full scale deflection (f.s.d.), thenI = V/R = 10 -10 A If V=0.5 V, then I = V/R = 0.5 x 10 -10 A

80. P.57 As meter to measure charge C= 10 -8 FV= 1V If the meter is full scale deflection (f.s.d.), then Q = CV = 10 -8 C If V=0.5 V, then Q = CV = 0.5 x 10 -8 C

81. P.59 Spooning charge For each spoon, a certain amount of +ve charge will be transfer to electrometer, the corresponding reading will be shown on the voltmeter Different size of spoon will have different amount of +ve charge carried. Different size of p.d. of EHT will have different amount of +ve charge carried.

82. P.59 P.d. across voltmeter equals to p.d. across capacitorCharges are isolated

83. P.60 Before K is closed Capacitors are fully charged Capacitors are connected in series Q 1 : Q 2 = 80μ : 80μ C 1 : C 2 = 10μ : 20 μ V 1 : V 2 = 2 : 1 = 8 : 4 V R1 : V R2 = 4 : 8 PT After K is closed V P = 8, V T = 4 V P = 8 = V T Q 2f = (20 μ)(8) = 160 μ Q 1f = (10 μ)(4) = 40 μ 80 μ ->40 μ 80 μ ->160 μ 80μ 40μ 120μ Charge flowing through K is 120 μC

84. P.60 Initially, C is fully charged When P breaks, C starts to discharge When Q breaks, discharging stops V C = 12 (V) V C drops V C = 3.2(V)

85. P.60 When P breaks, C starts to discharge

86. P.60 Half of W is dissipated

87. P.60

88. Charging to 6V by 10V battery Charging to 4V by 10V battery

89. P.61 Charging for 0.5s by 5V battery Charging for t s to 5V by 10V battery Charging for t s to 1.106V by 10V battery

90. P.61 Slope = V AC / t = Q / (Ct) = I / C Slope = const => I = const => V R = const Charging with const. I Fully charged, I=0 Discharging with const. I in opposite direction

91. P.61 (1) C = Q / V C = (1 C) / (1 V) = 1(F) Yes (2) E = (1/2) Q 2 /C = (1/2) (J) No (3) Q = I t = (1 A) (1 s) = 1 (C) No

92. P.62

93. (1) Q = C V = (100000μ) (20) = 2(C) Yes (2) Time const RC = (10) (100000 μ) = 1(s) No (3) E = (1/2) CV 2 = (1/2) (0.1)(20) 2 = 20 (J) No mean I= Q / t = 2 / 2000 = 1 (mA) After 1s, only 63% of initial charge has discharged

94. P.62 Time constant t = RC 2 x 10 -3 = (1 x 10 3 ) C C = (2 x 10 -6 ) F C = 2 μ F

95. P.62 I 1 = V /R No I 2 = V /(2R) = (1/2) I 1 V 1 = V 2 E dissipated in R independent of R E dissipated in R = E in Cap = (1/2) CV 2 Total charge Q stored depends on V and C only Yes

96. P.62

97. P.63

102. P.64

105. P.65