P24_022

# P24_022 - 22. To evaluate the ﬁeld using Gauss’ law, we...

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Unformatted text preview: 22. To evaluate the ﬁeld using Gauss’ law, we employ a cylindrical surface of area 2π r L where L is very large (large enough that contributions from the ends of the cylinder become irrelevant to the calculation). The volume within this surface is V = π r2 L, or expressed more appropriate to our needs: dV = 2π r L dr . The charge enclosed is, with A = 2.5 × 10−6 C/m5 , r A r2 2π r L dr = qenc = 0 π A L r4 . 2 By Gauss’ law, we ﬁnd Φ = |E |(2πrL) = qenc /ε0 ; we thus obtain E= A r3 . 4 ε0 (a) With r = 0.030 m, we ﬁnd |E | = 1.9 N/C. (b) Once outside the cylinder, Eq. 24-12 is obeyed. To ﬁnd λ = q/L we must ﬁnd the total charge q. Therefore, 0.04 q 1 A r2 2π r L dr = 1.0 × 10−11 C/m . = L L0 And the result, for r = 0.050 m, is |E | = λ/2πε0 r = 3.6 N/C. ...
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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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