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Chapter 25 . Capacitance . Real Parallel Plates. Chapter 25 CAPACITANCE. Depends on geometry and materials Units Farad = Coulomb/Volt [F = C/V] ( m F = 10 -6 F) and picofarad (pF = 10 -12 F) Examples Parallel Plates Coaxial Conductors (cylindrical symmetry)
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Chapter 25 Capacitance
Chapter 25 CAPACITANCE • Depends on geometry and materials • Units Farad = Coulomb/Volt [F = C/V] • (mF = 10-6 F) and • picofarad (pF = 10-12 F) • Examples • Parallel Plates • Coaxial Conductors (cylindrical symmetry) • Capacitance the ratio of the charge to the electric potential difference
Calculating Capacitance • Use Gauss’ LAW • Use relationship btwn and
Parallel Plates Gauss’ Law Assume A1/2 >>d, then E is constant
Coaxial Conductors Determine the potential difference Solve for the ratio of Q to V Consider a Gaussian Surface between the inner and outer conductorthen
Wiring Schemes • Series • the elements are connected end to end • Parallel • the elements are connected with left ends together and right ends together • the elements may be encountered as one follows alternate paths
Equivalent Capacitance The charge on each capacitor mustadd up to the total charge on theequivalent capacitance Parallel The potential differenceeach capacitor must matchthe battery potential difference
Equivalent Capacitance The charge on each capacitor is the same as the charge on the equivalent capacitance Series Potential differences must add to the total of the battery
15 mF 3 mF 20 mF a b 6 mF Equivalent Capacitance What is the equivalent capacitance between a and b? C1 and 6 mF in parallel Example C2 and 20 mF in series 15 mF and 3 mf in series
Energy Stored by a Capacitor The work done by an outsideagent to move a small amountof charge from one plate of a capacitor to the other plate is To charge the entire capacitor
Energy Stored by a Parallel Plate Capacitor Ad is the volume of the capacitor u = energy/volume or energy density
Problems • What is the energy stored in a 1.0 mF capacitor charged to 500 V? • What is the energy density if the area of the capacitor is 0.1 m2 and the plates are separated by 0.02 mm? • What is the electric field strength between the plates of the capacitor?
E’ Dielectrics (Macroscopic View) The dielectric (k > 1)increases the capacitance to a larger value than the original value since C > Co Eo Start with a charge Q which won’t change e = keo
Work done while inserting the dielectric There is a net force to the right, the field does the work
Schematics of Capacitor Design c is a electrolytic capacitor which must be wired accordingto its marked polarity What is the dielectric strength?