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uniform, and we neglect fringing effect. Symmetry can be used to show that the electric
field is radial, both between the cylinder and the shell and outside the shell. It is zero, of
course, inside the cylinder and inside the shell.
(a) We take the Gaussian surface to be a cylinder of length
L
, coaxial with the given
cylinders and of larger radius
r
than either of them. The flux through this surface is
2,
rLE
Φ= π
where
E
is the magnitude of the field at the Gaussian surface. We may
ignore any flux through the ends. Now, the charge enclosed by the Gaussian surface is
q
enc
=
Q
1 +
Q
2
= –
Q
1
= –3.40
×
10
−
12
C. Consequently, Gauss’ law yields
0e
n
c
rL
E q
πε
=
or
12
enc
12
2
2
3
0
3.40 10
C
0.214 N/C,
2
2 (8.85 10
C / N m )(11.0 m)(20.0 1.30 10 m)
q
E
Lr
επ
−
−−
−×
==
=
−
π×
⋅
×
×
or 
 0.214 N/C.
E
=
(b) The negative sign in
E
indicates that the field points inward.
(c) Next, for
r
= 5.00
R
1
, the charge enclosed by the Gaussian surface is
q
enc
Q
1
3.40
×
10
−
12
C. Consequently, Gauss’ law yields
n
c
=
or
12
enc
12
2
2
3
0
3.40 10
C
0.855 N/C.
2
2 (8.85 10
C / N m )(11.0 m)(5.00 1.30 10 m)
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 Spring '10
 Rebello,NobelS
 Charge

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