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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.
 Fall '08
 SPRUNGER
 Physics

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