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25. Studying Sample Problem 224, we see that the field evaluated at the center of
curvature due to a charged distribution on a circular arc is given by
G
E
r
=
−
λ
π
4
0
2
2
ε
θ
sin
/
/
along the symmetry axis
where
λ
=
q
/
r
with
in radians. In this problem, each charged quartercircle produces a
field of magnitude
[]
/4
2
00

1
1 22
s
i
n
.
/24
4
qq
E
rr
r
επ
π
−
==
G
π
π
ππ
That produced by the positive quartercircle points at – 45°, and that of the negative
quartercircle points at +45°.
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Fall '10 term at Kansas State University.
 Fall '10
 Rebello,NobelS
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