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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.
 Spring '11
 Barwick

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