P24_038 - 38. We interpret the question as referring to the...

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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. 23-3). ...
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