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Physics 2113 Lecture: 16 MON 23 FEB

Physics 2113 Jonathan Dowling. Physics 2113 Lecture: 16 MON 23 FEB. Capacitance I. 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 2113 Lecture: 16 MON 23 FEB

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  1. Physics 2113 Jonathan Dowling Physics 2113 Lecture: 16 MON 23 FEB Capacitance I

  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 (μF), pico-Farads (pF) — 10–12 F New technology: compact 1 F capacitors Q = CV where C = capacitance For fixed Q the larger the capacitance the larger the stored voltage. 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. Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors. It does not depend on the voltage across it or the charge on it at all. • Same (b) Same

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

  6. Parallel Plate Capacitor We want capacitance: C = Q/V  E field between conducting plates: Area of each plate = A Separation = d charge/area = σ = Q/A Relate E to potential difference V: –Q +Q σ d What is the capacitance C ?

  7. Capacitance and Your iPhone!

  8. 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 = ε0A/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!

  9. –Q +Q Isolated Parallel Plate Capacitor: ICPP • A parallel plate capacitor of capacitance C is charged using a battery. • Charge = Q, potential voltage difference = V. • Battery is then disconnected. If the plate separation is INCREASED, does the capacitance C: (a) Increase? (b) Remain the same? (c) Decrease? If the plate separation is INCREASED, does the capacitance C: (a) Increase? (b) Remain the same? (c) Decrease? • Q is fixed! • d increases! • C decreases (= ε0A/d) • V=Q/C; V increases.

  10. –Q +Q Parallel Plate Capacitor & Battery: ICPP • 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. • V is fixed constant by battery! • C decreases (=ε0A/d) • Q=CV; Q decreases • E = σ/ε0 = Q/ε0A decreases Does the Electric Field Inside: (a) Increase? (b) Remain the Same? (c) Decrease? Battery does work on capacitor to maintain constant V!

  11. 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:

  12. 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,

  13. 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

  14. Summary • Any two charged conductors form a capacitor. • Capacitance : C= Q/V • Simple Capacitors:Parallel plates: C = ε0 A/dSpherical: C = 4πε0 ab/(b-a)Cylindrical: C = 2πε0 L/ln(b/a)]

  15. Parallel plates: C = ε0 A/dSpherical: • Cylindrical: C = 2πε0 L/ln(b/a)] Hooked to battery V is constant. Q=VC (a) d increases -> C decreases -> Q decreases (b) a inc -> separation d=b-a dec. -> C inc. -> Q increases (c) b increases -> separation d=b-a inc.-> C dec. -> Q decreases

  16. V = VAB = VA –VB C1 Q1 VA VB A B C2 Q2 VC VD D C V = VCD = VC –VD Ceq Qtotal ΔV=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) • PARALLEL: • V is same for all capacitors • Total charge = sum of Q

  17. 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

  18. PARALLEL: • V is same for all capacitors • Total charge = sum of Q • SERIES: • Q is same for all capacitors • Total potential difference = • sum of V • Parallel: Voltage is same V on each but charge is q/2 on each since • q/2+q/2=q. (b) Series: Charge is same q on each but voltage is V/2 on eachsince V/2+V/2=V.

  19. C1 Q1 C2 Q2 Qeq Q1 Q2 Ceq C1 C2 “Compiling” 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

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

  21. Series: Isolated Islands (Constant Q) Example: Parallel or Series C2=20μF C3=30μF C1=10μF 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 μF (solve above equation) • Q = CeqV = (5.46 μF)(120V) = 655 μC • V1= Q/C1 = (655 μC)/(10 μF) = 65.5 V • V2= Q/C2 = (655 μC)/(20 μF) = 32.75 V • V3= Q/C3 = (655 μC)/(30 μF) = 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)

  22. 10 μF 10μF 10μF 10V 5μF 5μF 10V Example: Series or Parallel? Neither: Circuit Compilation Needed! In the circuit shown, what is the charge on the 10μF capacitor? • The two 5μF capacitors are in parallel • Replace by 10μF • Then, we have two 10μF capacitors in series • So, there is 5V across the 10 μF capacitor of interest • Hence, Q = (10μF )(5V) = 50μC

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