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Unformatted text preview: 38. We interpret the question as referring to the ﬁeld just outside the sphere (that is, at locations roughly
equal to the radius r of the sphere). Since the area of a sphere is A = 4πr2 and the surface charge
density is σ = q/A (where we assume q is positive for brevity), then
E= σ
1
=
ε0
ε0 q
4πr2 = 1q
4πε0 r2 which we recognize as the ﬁeld of a point charge (see Eq. 233). ...
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 Fall '08
 SPRUNGER
 Physics, Charge

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