P25_054 - 54. (a) The magnitude of the electric field is...

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Unformatted text preview: 54. (a) The magnitude of the electric field is (3.0 × 10−8 C) 8.99 × 109 q σ = = E= ε0 4πε0 R2 (0.15 m)2 N·m2 C2 = 1.2 × 104 N/C . (b) V = RE = (0.15 m)(1.2 × 104 N/C) = 1.8 × 103 V. (c) Let the distance be x. Then ∆V = V (x) − V = which gives x= q 4πε0 1 1 − R+x R = −500 V , R ∆V (0.15 m)(−500 V) = = 5.8 × 10−2 m . −V − ∆V −1800 V + 500 V ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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