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Physics 2102 Lecture: 08 THU 11 FEB

Physics 2102 Jonathan Dowling. Physics 2102 Lecture: 08 THU 11 FEB. Capacitance I. 25.1–4. Capacitors and Capacitance. Capacitor: any two conductors, one with charge +Q , other with charge –Q. –Q. Potential DIFFERENCE between conductors = V. +Q.

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Physics 2102 Lecture: 08 THU 11 FEB

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  1. Physics 2102 Jonathan Dowling Physics 2102 Lecture: 08 THU 11 FEB Capacitance I 25.1–4

  2. Capacitors and Capacitance Capacitor: any two conductors, one with charge +Q, other with charge –Q –Q Potential DIFFERENCE between conductors = V +Q Uses: storing and releasing electric charge/energy. Most electronic capacitors: micro-Farads (F), pico-Farads (pF) — 10–12 F New technology: compact 1 F capacitors Q = CV where C = capacitance Units of capacitance: Farad (F) = Coulomb/Volt

  3. +Q –Q Capacitance • Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors • e.g. Area of conductors, separation, whether the space in between is filled with air, plastic, etc. (We first focus on capacitors where gap is filled by AIR!)

  4. Electrolytic (1940-70) Electrolytic (new) Paper (1940-70) Capacitors Variable air, mica Ceramic (1930 on) Tantalum (1980 on) Mica (1930-50)

  5. Parallel Plate Capacitor We want capacitance: C = Q/V  E field between the plates: (Gauss’ Law) Area of each plate = A Separation = d charge/area =  = Q/A Relate E to potential difference V: +Q -Q What is the capacitance C ?

  6. Capacitance and Your iPhone!

  7. Parallel Plate Capacitor — Example • A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mm • What is the capacitance? C = 0A/d = (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m) = 2.21 x 10–9 F (Very Small!!) Lesson: difficult to get large values of capacitance without special tricks!

  8. –Q +Q Isolated Parallel Plate Capacitor • A parallel plate capacitor of capacitance C is charged using a battery. • Charge = Q, potential difference = V. • Battery is then disconnected. • If the plate separation is INCREASED, does Potential DifferenceV: (a) Increase? (b) Remain the same? (c) Decrease? • Q is fixed! • C decreases (=0A/d) • V=Q/C; V increases.

  9. –Q +Q Parallel Plate Capacitor & Battery • A parallel plate capacitor of capacitance C is charged using a battery. • Charge = Q, potential difference = V. • Plate separation is INCREASED while battery remains connected. Does the Electric Field Inside: (a) Increase? (b) Remain the Same? (c) Decrease? • V is fixed by battery! • C decreases (=0A/d) • Q=CV; Q decreases • E = Q/0A decreases

  10. Spherical Capacitor What is the electric field inside the capacitor? (Gauss’ Law) Radius of outer plate = b Radius of inner plate = a Concentric spherical shells: Charge +Q on inner shell, –Q on outer shell Relate E to potential difference between the plates:

  11. Spherical Capacitor What is the capacitance? C = Q/V = Radius of outer plate = b Radius of inner plate = a Concentric spherical shells: Charge +Q on inner shell, –Q on outer shell Isolated sphere: let b >> a,

  12. Cylindrical Capacitor What is the electric field in between the plates? Gauss’ Law! Radius of outer plate = b Radius of inner plate = a Length of capacitor = L +Q on inner rod, –Q on outer shell Relate E to potential difference between the plates: cylindrical Gaussian surface of radius r

  13. Summary • Any two charged conductors form a capacitor. • Capacitance : C= Q/V • Simple Capacitors:Parallel plates: C = 0 A/dSpherical: C = 4e0 ab/(b-a)Cylindrical: C = 20 L/ln(b/a)]

  14. V = VAB = VA –VB C1 Q1 VA VB A B C2 Q2 VC VD D C V = VCD = VC –VD Ceq Qtotal V=V Capacitors in Parallel: V=Constant • An ISOLATED wire is an equipotential surface: V=Constant • Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge! • VAB = VCD = V • Qtotal = Q1 + Q2 • CeqV = C1V + C2V • Ceq = C1 + C2 • Equivalent parallel capacitance = sum of capacitances PAR-V (Parallel V the Same)

  15. Q1 Q2 B C A C1 C2 Capacitors in Series: Q=Constant Isolated Wire: Q=Q1=Q2=Constant • Q1 = Q2 = Q = Constant • VAC = VAB + VBC SERI-Q (Series Q the Same) Q = Q1 = Q2 • SERIES: • Q is same for all capacitors • Total potential difference = sum of V Ceq

  16. C1 Q1 C2 Q2 Qeq Q1 Q2 Ceq C1 C2 Capacitors in parallel and in series • In parallel : • Cpar = C1 + C2 • Vpar= V1 = V2 • Qpar= Q1 + Q2 • In series : 1/Cser = 1/C1 + 1/C2 Vser= V1  + V2 Qser= Q1 = Q2

  17. Parallel: Circuit Splits Cleanly in Two (Constant V) C1=10 F C2=20 F C3=30 F 120V Example: Parallel or Series? • Qi = CiV • V = 120V = Constant • Q1 = (10 F)(120V) = 1200 C • Q2 = (20 F)(120V) = 2400 C • Q3 = (30 F)(120V) = 3600 C What is the charge on each capacitor? • Note that: • Total charge (7200 C) is shared between the 3 capacitors in the ratio C1:C2:C3— i.e. 1:2:3

  18. Series: Isolated Islands (Constant Q) Example: Parallel or Series C2=20mF C3=30mF C1=10mF What is the potential difference across each capacitor? • Q = CserV • Q is same for all capacitors • Combined Cser is given by: 120V • Ceq = 5.46 F (solve above equation) • Q = CeqV = (5.46 F)(120V) = 655 C • V1= Q/C1 = (655 C)/(10 F) = 65.5 V • V2= Q/C2 = (655 C)/(20 F) = 32.75 V • V3= Q/C3 = (655 C)/(30 F) = 21.8 V Note: 120V is shared in the ratio of INVERSE capacitances i.e. (1):(1/2):(1/3) (largest C gets smallest V)

  19. 10 F 10F 10F 10V 5F 5F 10V Example: Series or Parallel? Neither: Circuit Compilation Needed! In the circuit shown, what is the charge on the 10F capacitor? • The two 5F capacitors are in parallel • Replace by 10F • Then, we have two 10F capacitors in series • So, there is 5V across the 10 F capacitor of interest • Hence, Q = (10F )(5V) = 50C

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