Phys.219_Lec.07._2012. - Lecture 7-1 Physics 219 Question 1...

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Lecture 7 Lecture 7 -1 Physics 219 – Question 1 – February 1, 2012. 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 (Hint: Capacitance depends on the dielectric constant κ and geometric factors only: 0 ) A C d κε =
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Lecture 7 Lecture 7 -2 Dielectrics = Insulators Inserting a dielectric between the plates of a capacitor increases capacitance Q = CV holds more charges at fixed V Dielectric constant κ of a dielectric is the ratio of the capacitances when filled with it to that without it: κ > 1 always (dimensionless) 0 C C κ= In dielectric materials electrons can not be liberated from atoms thus they are insulators. However they can be polarized. (Permanent or induced.) However not all insulators are dielectrics. max At fixed V, , , E remains the same Breakdown potential determined by dielectric strength ( E before breakdown) QU Δ //
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Lecture 7 Lecture 7 -3 Material κ Dielectric strength (kV/mm) Air(1atm) 1.00054 2.5 paper 3.0 16 glass 5-10 14 Water(20 ° C) 80.4 65 Strontium titanate 310 8
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Lecture 7 Lecture 7 -4 Energy stored in capacitor with a dielectric () 2 2 1 22 Q UC V C == But 0 , A CV E d d ε κ 2 0 2 0 1 2 1 2 A UE d d volume E κε ∴= = 2 0 1 2 U volume uE = ∴≡
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Lecture 7
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This note was uploaded on 02/05/2012 for the course PHYS 219 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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Phys.219_Lec.07._2012. - Lecture 7-1 Physics 219 Question 1...

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