Lecture_7

Lecture_7 - Lecture 7-1 Physics 219 Question 1 February 3...

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Lecture 7-1 Physics 219 – Question 1 – February 3, 2010 We have an air-filled parallel plate capacitor of capacitance C=1 µ F that is made of two square plates. We could double the capacitance by a) doubling the separation d b) inserting a dielectric of dielectric constant 1.4 c) halving potential difference between the plates V d) increasing the edge length a of the plates by a factor of 1.4 e) doubling the charges placed on the plates d a A a Q Q
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Lecture 7-2 Dielectrics = Insulator Inserting a dielectric between the plates of a capacitor increases capacitance Q=C V holds more charges at fixed V Dielectric constant κ of a dielectric is the ratio of the capacitance when filled with it to that without it: κ > 1 always (dimensionless) 0 C C κ= Material κ air (1 atm) 1.00054 paper 3.0 glass 5 - 10 water (20 ° C) 80.4 strontium titanate 310 At fixed V , , QU  , E remains the same Breakdown potential determined by dielectric strength (E max before breakdown)
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Lecture 7-3 Energy stored in capacitor with a dielectric ( ) 2 2 1 22 Q U CV C = = But 0 , A C V Ed d ε κ = ( ) ( ) 2 0 2 0 1 2 1 2 A U Ed d volume E κε ∴= = 2 0 1 2 U volume uE = ∴≡
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Lecture 7-4 Resistors in Parallel 11 2 2 3 3 iR ε= = ( ) 123 eq i i ε = ++ 1 111 eq eq
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This note was uploaded on 04/23/2011 for the course PHYS 219 taught by Professor Staff during the Spring '08 term at Purdue.

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Lecture_7 - Lecture 7-1 Physics 219 Question 1 February 3...

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