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net,
22
00
92
2
1
2
122


14


2
cos45
44
(8.99 10 N m C )4(4.50 10
C)
20.6 N/C.
(5.00 10 m)
x
qq
E
rr
πε
π
−
−
⎛⎞
=°
=
⎜⎟
⎝⎠
×⋅
×
==
×
(b) By symmetry, the net field points vertically downward in the
ˆ
j
−
direction, or
90
−
°
counterclockwise from the +
x
axis.
24. Studying Sample Problem 223, we see that the field evaluated at the center of
curvature due to a charged distribution on a circular arc is given by
0
sin
4
E
r
θ
λ
ε
−
=
π
G
along the symmetry axis, with
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 Summer '10
 BB
 Physics, Charge

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