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24. We assume
q>
0. Using the notation
λ
=
q/L
we note that the (infnitesimal) charge on an element
dx
oF the rod contains charge
dq
=
λdx
. By symmetry, we conclude that all horizontal feld components
(due to the
dq
’s) cancel and we need only “sum” (integrate) the vertical components. Symmetry also
allows us to integrate these contributions over only halF the rod (0
≤
x
≤
L/
2) and then simply double
the result. In that regard we note that sin
θ
=
y/r
where
r
=
p
x
2
+
y
2
. Using Eq. 233 (with the 2 and
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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