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on an element
dx
of the rod contains charge
dq
=
λ
dx
. By symmetry, we conclude that all
horizontal field 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
=
R
/
r
where
22
rx
R
=+
.
(a) Using Eq. 223 (with the 2 and sin
factors just discussed) the magnitude is
()
2
00
2
2
32
2
0
0
2
2
2
2s
i
n
44
21
4
2
LL
L
L
dq
dx
y
E
R
xR
qLR
Rd
x
x
Rx R
qL
q
LR
R
LR
πε
§·
λ
==
¨¸
+
©¹
+
ªº
λ
«»
+
+
¬¼
+
+
³³
³
G
where the integral may be evaluated by elementary means or looked up in Appendix E
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 Fall '10
 Rebello,NobelS
 Charge

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