32. We assume
q
> 0. Using the notation
λ
=
q
/
L
we note that the (infinitesimal) charge
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
θ
factors just discussed) the magnitude is
()
()
22
22
2
22
00
00
/2
2
32
22
2
0
22
00
0
22
2
2
00
2
2s
i
n
44
22
21
22
4
2
LL
L
L
dq
dx
y
E
rx
R
x
R
qLR
Rd
x
x
Rx R
xR
qL
q
LR
R
LR
LR
θ
πε
πε
πε
πε
πε
πε
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This note was uploaded on 06/07/2010 for the course PHYS 344 taught by Professor Bb during the Spring '10 term at The Petroleum Institute.
 Spring '10
 BB
 Physics, Charge

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