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Unformatted text preview: 7. The potential diﬀerence between the wire and cylinder is given, not the linear charge density on the
wire. We use Gauss’ law to ﬁnd an expression for the electric ﬁeld a distance r from the center of the
wire, between the wire and the cylinder, in terms of the linear charge density. Then integrate with
respect to r to ﬁnd an expression for the potential diﬀerence between the wire and cylinder in terms
of the linear charge density. We use this result to obtain an expression for the linear charge density in
terms of the potential diﬀerence and substitute the result into the equation for the electric ﬁeld. This
will give the electric ﬁeld in terms of the potential diﬀerence and will allow you to compute numerical
values for the ﬁeld at the wire and at the cylinder. For the Gaussian surface use a cylinder of radius r
and length , concentric with the wire and cylinder. The electric ﬁeld is normal to the rounded portion
of the cylinder’s surface and its magnitude is uniform over that surface. This means the electric ﬂux
through the Gaussian surface is given by 2πr E , where E is the magnitude of the electric ﬁeld. The
charge enclosed by the Gaussian surface is q = λ , where λ is the linear charge density on the wire.
Gauss’ law yields 2πε0 r E = λ . Thus,
λ
E=
.
2πε0 r
Since the ﬁeld is radial, the diﬀerence in the potential Vc of the cylinder and the potential Vw of the wire
is
r
r
w
c
rc
λ
λ
∆ V = Vw − V c = −
E dr =
ln
,
dr =
2πε0 r
2πε0 rw
r
r
c
w
where rw is the radius of the wire and rc is the radius of the cylinder. This means that
λ=
and
E= 2πε0 ∆V
ln(rc /rw ) ∆V
λ
=
.
2πε0 r
r ln(rc /rw ) (a) We substitute rc for r to obtain the ﬁeld at the surface of the wire:
= ∆V
850 V
=
−6 m) ln [(1.0 × 10−2 m)/(0.65 × 10−6 m)]
rw ln(rc /rw )
(0.65 × 10 = E 1.36 × 108 V/m . (b) We substitute rc for r to ﬁnd the ﬁeld at the surface of the cylinder:
E = ∆V
850 V
=
rc ln(rc /rw )
(1.0 × 10−2 m) ln [(1.0 × 10−2 m)/(0.65 × 10−6 m)] = 8.82 × 103 V/m . ...
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 Fall '08
 SPRUNGER
 Physics, Charge

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