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Unformatted text preview: 27. (a) To calculate the electric ﬁeld at a point very close to the center of a large, uniformly charged
conducting plate, we may replace the ﬁnite plate with an inﬁnite plate with the same area charge
density and take the magnitude of the ﬁeld to be E = σ/ε0 , where σ is the area charge density for
the surface just under the point. The charge is distributed uniformly over both sides of the original
plate, with half being on the side near the ﬁeld point. Thus,
6.0 × 10−6 C
= 4.69 × 10−4 C/m .
2(0.080 m)2 The magnitude of the ﬁeld is
4.69 × 10−4 C/m
= 5.3 × 107 N/C .
8.85 × 10−12 C2 /N · m2
2 E= The ﬁeld is normal to the plate and since the charge on the plate is positive, it points away from
(b) At a point far away from the plate, the electric ﬁeld is nearly that of a point particle with charge
equal to the total charge on the plate. The magnitude of the ﬁeld is E = q/4πε0 r2 = kq/r2 , where
r is the distance from the plate. Thus,
E= 8.99 × 109 N · m2 /C2 6.0 × 10−6 C
= 60 N/C .
(30 m)2 ...
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