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20. We denote the radius of the thin cylinder as
R
=0
.
015 m. Using Eq. 2412, the net electric Feld for
r>R
is given by
E
net
=
E
wire
+
E
cylinder
=
−
λ
2
πε
0
r
+
λ
0
2
πε
0
r
where
−
λ
=
−
3
.
6nC
/
m is the linear charge density of the wire and
λ
0
is the linear charge density of the
thin cylinder. We note that the surface and linear charge densities of the thin cylinder are related by
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Unformatted text preview: q cylinder = λ L = σ (2 πRL ) = ⇒ λ = σ (2 πR ) . Now, E net outside the cylinder will equal zero, provided that 2 πRσ = λ , or σ = λ 2 πR = 3 . 6 × 10 − 9 C / m (2 π )(0 . 015 m) = 3 . 8 × 10 − 8 C / m 2 ....
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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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