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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. 2412 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.
 Fall '08
 SPRUNGER
 Physics

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