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Unformatted text preview: 23. The electric ﬁeld is radially outward from the central wire. We want to ﬁnd its magnitude in the region
between the wire and the cylinder as a function of the distance r from the wire. Since the magnitude
of the ﬁeld at the cylinder wall is known, we take the Gaussian surface to coincide with the wall. Thus,
the Gaussian surface is a cylinder with radius R and length L, coaxial with the wire. Only the charge
on the wire is actually enclosed by the Gaussian surface; we denote it by q . The area of the Gaussian
surface is 2πRL, and the ﬂux through it is Φ = 2πRLE . We assume there is no ﬂux through the ends
of the cylinder, so this Φ is the total ﬂux. Gauss’ law yields q = 2πε0 RLE . Thus,
q = 2π 8.85 × 10−12 C2
N · m2 (0.014 m)(0.16 m) 2.9 × 104 N/C = 3.6 × 10−9 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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