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Unformatted text preview: 48. (a) We consider the radial ﬁeld produced at points within a uniform cylindrical distribution of charge.
The volume enclosed by a Gaussian surface in this case is Lπr2 . Thus, Gauss’ law leads to
E= ρ Lπr2
qenc 
ρr
=
.
=
ε0 Acylinder
ε0 (2πrL)
2 ε0 (b) We note from the above expression that the magnitude of the radial ﬁeld grows with r.
(c) Since the charged powder is negative, the ﬁeld points radially inward.
(d) The largest value of r which encloses charged material is rmax = R.
0.0011 C/m3 and R = 0.050 m, we obtain
Emax = Therefore, with ρ = ρR
= 3.1 × 106 N/C .
2 ε0 (e) According to condition 1 mentioned in the problem, the ﬁeld is high enough to produce an electrical
discharge (at r = R). ...
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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, Charge

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