7DS13-Week6-DiscQuiz

No edge effects a the electric potential at the

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Unformatted text preview: de of the cylinder of radius b. 1b) The change in potential we calculate from the definite integral of E- Field from a to b : q Electric field from Gauss’ Law for r>a and r<b: E * A = εo E * 2πrL = E = ΔV = − ρ(πLa2) εo ρ(a2) 2rεo ;7 a <r<b ρ(a2) ln( b ) a 2εo 1c) The capacitance follows: C = Q/V C = ρ(πa2L) ρ(a2) ln( b ) a 2εo π L C = 2n(Lε)o , d) C = 2πn(ε )K l lb a o b a 2) Comes straight from the book: E*q*d*sin(30) Solutions to Form B: 1a) There is an easy way answer to this question. For charge distributions that you know give an E- field that fades to nothing at infinity, the usual thing is to define the Electric potential to be 0 at infinity. Now since the E- field for r>b is 0 (the net charge enclosed by a Gaussian cylinder is zero for r of gaussian cylinder greater than b), you know that the electric potential cannot change at any point outside of the cylinder of radius b. 1b) Electric field from Gauss’ Law: E*A= q εo E * 2πrD = E = −η(πDa2) εo −η(a2) 2rεo ;7 a <r<b The change in potential we calculate from the definite integral of E- Field from a to b: (Negatives cancelled) ΔV = η(a2) ln( b ) a 2εo 1c) The capacitance follows: C = Q/V C = ρ(πa2D) ρ(a2) ln( b ) a 2εo D C = 2lπDεo , d) C = 2πln(ε )K n( b ) a 2) Comes straight from the book: q*r*E*sin(45) o b a...
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This note was uploaded on 09/23/2013 for the course PHYSICS 7D 7D taught by Professor Barwick during the Spring '11 term at UC Irvine.

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