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Physics 2102 Lecture: 11 FRI 06 FEB

Physics 2102 Jonathan Dowling. Physics 2102 Lecture: 11 FRI 06 FEB. Capacitance I. 25.1–3. Tutoring Lab Now Open Tuesdays. Lab Location: 102 Nicholson (across the hall from class) Lab Hours: MTWT: 12:00N–5:00PM F: 12:N–3:00PM. Capacitors and Capacitance.

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Physics 2102 Lecture: 11 FRI 06 FEB

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  1. Physics 2102 Jonathan Dowling Physics 2102 Lecture: 11 FRI 06 FEB Capacitance I 25.1–3

  2. Tutoring Lab Now Open Tuesdays • Lab Location: 102 Nicholson (across the hall from class) • Lab Hours: MTWT: 12:00N–5:00PM F: 12:N–3:00PM

  3. 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

  4. +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!)

  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 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 ?

  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 = 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!

  9. –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.

  10. –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 (=e0A/d) • Q=CV; Q decreases • E = Q/0A decreases

  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 = 0 A/dSpherical: C = 4e0 ab/(b-a)Cylindrical: C = 20 L/ln(b/a)]

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