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Unformatted text preview: rge Distribution –
another use of the Gauss law
• Select a sphere as the
Gaussian surface
• For r> a qin
Φ E = ∫ E ⋅ dA = ∫ EdA =
εo
Q
Q
E=
= ke 2
2
4 πεor
r Outside the sphere Example 224: Solid sphere of
charge.
An electric charge Q is distributed
uniformly throughout a
nonconducting sphere of radius r0.
Determine the electric field (a)
outside the sphere
(r > r0) and (b) inside the sphere (r
< r0). • Select a sphere as
the Gaussian surface,
r<a
• qin < Q
• qin = ρ (4/3πr3)
ρ π qin
ΦE = ∫ E ⋅ d A = ∫ E dA =
εo
E= qin
4πεo r 2 = keQ
a 3 r Uniform Charge Distribution
• Inside the sphere, E
varies linearly with r
– E → 0 as r → 0
• The field outside the
sphere is equivalent to
that of a point charge
located at the center of
the sphere λ. Field Due to a
Line of Charge, cont
• The end view of the
field: perpendicular to
the curved surface
• The field through the
ends of the cylinder is
0 since the field is
parallel to these
surfaces Field Due to a
Line of Charge, final
• Use Gauss Law to find the field
qin
ΦE = ∫ E ⋅ d A = ∫ E dA =
εo λ
E ( 2π r ) =
εo
λ
λ
E=
= 2ke
2πεo r
r λ is the linear charge
density on the line σσ
dQ/dA E must be perpendicular to the plane and must have the same
magnitude at all points
equidistance from the plane
Choose a sma...
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This document was uploaded on 02/21/2014 for the course CHEM 110A 110a at UCLA.
 Winter '13

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