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61. (a) At
x
= 0.040 m, the net field has a rightward (+
x
) contribution (computed using Eq.
2313) from the charge lying between
x
= –0.050 m and
x
= 0.040 m, and a leftward (–
x
)
contribution (again computed using Eq. 2313) from the charge in the region from
0.040 m
x
=
to
x
= 0.050 m. Thus, since
σ
=
q
/
A
=
ρ
V
/
A
=
∆
x
in this situation, we have
93
12
2
2
00
(0.090m)
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 Spring '08
 Any
 Physics, Charge

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