# From conductor 0 d q0 q y e bb t y e0 d e0 t v

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Unformatted text preview: f distance V ( y ) = − ∫ E ⋅ dy from the bottom conductor. from conductor 0 d +Q0 +Q y E BB t y -E0 d E=0 t V - Q0 21 What is ∆V = V(d)? A) ∆V = E0d B) ∆V = E0(d – t) C) ∆V = E0(d + t) y The integral = area under the curve Physics 212 Lecture 7, Slide 16 16 Back to Preflight 8 BB A) Q1 < Qo B) Q1 = Qo “The distance for Q1 is smaller therefore the charge The must decrease to compensate for the change in distance “ C) Q1 > Qo 50 40 “Since the potential remains the same, there is no change Since in the charge of Q. “ 30 “the field through the conductor is zero, so it has the constant potential. Because of this it must have greater charge so the total V is that same. “ 10 20 0 Physics 212 Lecture 7, Slide 17 17 Preflight 10 BB 50 40 30 20 A) C1 > Co B) C1 = Co C) C1 < Co 0 We can determine C from either case same V (preflight) same Q (lecture) C depends only on geometry !! Same Q: 10 E0 = Q0/ε0A V0 = E0d C0 = Q0/E0d V1 = E0(d – t) (d C1 = Q0/(E0(d – t)) (d C0 = ε0A/d C1 = ε0A/(d – t) A/(d Physics 212 Lecture 7, Slide 18 Physics Energy in Capacitors BANG 31 Physics 212 Lecture 7, Slide 19 Physics cross-section cross a4 a3 Calculation A capacitor is constructed from two co...
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## This note was uploaded on 02/09/2012 for the course PHYSICS 212 taught by Professor Mestre during the Spring '11 term at University of Illinois at Urbana–Champaign.

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