L05 - Chapter 25 Capacitance Real Parallel Plates Chapter...

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Unformatted text preview: Chapter 25 Capacitance Real Parallel Plates Chapter 25 CAPACITANCE Depends on geometry and materials Units Farad = Coulomb/Volt [F = C/V] (F = 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 Q C= V Circuits and Capacitors Parallel Plate Capacitor Coaxial Capacitor cylindrical symmetry !!! Calculating Capacitance Use Gauss' LAW r r qencl = E dA = 0 Use relationship btwn V B and r E r r V f - Vi = - E ds A Parallel Plates Gauss' Law r r qencl = E dA = 0 E= = A o o Q r r V f - Vi = - E ds B A V = Ed Q A o = V d Q V= Assume A1/2 >>d, then E is constant o Ad A o C= d Coaxial Conductors Determine the potential difference b V = - ln 2 o l a Solve for the ratio of Q to V Consider a Gaussian Surface between the inner and outer conductor then Q Q Q 2 o l = b V ln a l E= = 2 o r 2 o r C 2 o = l ln b a 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 Parallel The charge on each capacitor must add up to the total charge on the equivalent capacitance Q = Q1 + Q 2 +...+ Q n C eq V = C1 V + C 2 V +...+ C n V The potential difference C eq each capacitor must match the battery potential difference = C1 + C 2 +...+C n Equivalent Capacitance Series The charge on each capacitor is the same as the charge on the equivalent capacitance Q Q Q Q = + + ... + C eq C1 C 2 Cn Potential differences must add to the total of the battery V = V1 + V2 + ... + Vn 1 1 1 1 = + + ... + C eq C1 C 2 Cn Equivalent Capacitance Equivalent Capacitance What is the equivalent capacitance between a and b? 15 F 3 F 20 F b Example C1 and 6 F in parallel a 6 F C 2 = 2 . 5 F + 6 F C 2 = 8 .5 F C2 and 20 F in series 15 F and 3 f in series 1 1 1 = + C1 15F 3F 1 6 = C1 15 F 1 1 1 = + Ceq 8.5F 20 F C 1 = 2 .5 F Ceq = 5.97 F Example 25-2 Energy Stored by a Capacitor The work done by an outside agent to move a small amount of charge from one plate of a capacitor to the other plate is d W = Vdq To charge the entire capacitor q W = dq C 0 Q W = Vdq 0 Q Q W= 2C 2 2 2 Q CV QV U= = = 2C 2 2 Energy Stored by a Parallel Plate Capacitor A oV U = 2d 2 Ad is the volume of the capacitor A o E 2d 2 U = 2d CV U = 2 2 Ad o E U= 2 u= oE 2 2 2 u = energy/volume or energy density 2 A o C= d oE U = Ad 2 Problems 1. What is the energy stored in a 1.0 mF capacitor charged to 500 V? 2. What is the energy density if the area of the capacitor is 0.1 m2 and the plates are separated by 0.02 mm? 3. What is the electric field strength between the plates of the capacitor? Dielectrics (Atomic View) Dielectrics (Macroscopic View) r r Eo Vo V= E= Start with a charge Q which won't change Eo E' The dielectric ( > 1) increases the capacitance to a larger value than the original value since C > Co r r r E = Eo - E Q = A = Q Q C = = = C0 V V0 / Example 25-7 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 according to its marked polarity What is the dielectric strength? ...
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This note was uploaded on 03/17/2008 for the course PHYS 212 taught by Professor Mahlon,gregoryda during the Fall '07 term at Pennsylvania State University, University Park.

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