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(b) Once outside the cylinder, Eq. 2312 is obeyed. To find
λ
=
q
/
L
we must find the total
charge
q
. Therefore,
0.04
21
1
0
1
2
1.0 10
C/m.
q
Ar
rLdr
LL
π
−
==
×
∫
And the result, for
r
= 0.050 m, is
0


/2
3.6 N/C.
Er
λπε
G
32. To evaluate the field using Gauss’ law, we employ a cylindrical surface of area 2
π
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
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This note was uploaded on 09/07/2010 for the course PHYS 247 taught by Professor Zv during the Fall '10 term at The Petroleum Institute.
 Fall '10
 ZV

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