problem23_60

University Physics with Modern Physics with Mastering Physics (11th Edition)

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Unformatted text preview: 23.60: (a) Use ∆Vab to get λ : ∆V = ∫a E ⋅ dl = ∫a λ= E= 2πε0 ∆V ln b a b b λ λ dr = ln 2πε0 2πε0 b a 2πε 0 ∆V ln b a λ ∆V = = 2πε 0 r 2πε0 r r ln b a 0.127 mm at outer surface of the wire, r = a = 2 850 V E= = 2.65 × 10 6 V m 00 ( 0.000127 m ) ln ( 1..0127cm ) 2 0 2 cm (b) at the inner surface of the cylinder, r = 1.00 cm , which gives E = 1.68 × 10 4 V m ...
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