Unformatted text preview: 57. (a) We use Eq. 2518 to ﬁnd the potential:
R Vwall − V = − E dr
r
R 0−V − = ρ
−
R2 − r 2
4 ε0 r −V
Consequently, V = ρ
4ε 0 ρr
2 ε0 = . R2 − r 2 . (b) The value at r = 0 is
Vcenter = −1.1 × 10−3 C/m3
(0.05 m)2 − 0 = −7.8 × 104 V .
4 (8.85 × 10−12 C/V · m) ...
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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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