1/24/2009
1
Ex:
The fig (a) shows a thin, uniform charged rod
of length
L
with total charge
Q
that can be either positive
or negative. Find the electric field strength at distance
r
in the plane that bisects the rod.
Solve: follow the 5 steps of the solving strategy
(a)
•
The
ycomponents
of the two fields cancel each other. We left with the
xcomponents of the electric field at P.
•
We divided the rod into little segments, so we can treat them as a
point charges.
•
We know how to find the electric field of a point charge (here the
segment
i
).
•
Thus, the
xcomponent of
E
i
is:
∆
∆
+
=
=
+
=
i
i
i
i
Q
r
r
Q
r
y
r
r
r
and
r
y
r
where
2
2
i
2
2
1
1
cos
θ
i
i
i
i
x
i
r
Q
E
E
θ
π
θ
cos
4
1
cos
)
(
2
0
∆
∈
=
=
⇒
∑
∑
=
=
+
∆
∈
=
=
+
∈
=
+
+
∈
=
N
i
i
N
i
x
i
x
i
i
i
x
i
r
y
Q
r
E
E
et
onent we g
net xcomp
The
r
y
r
y
r
y
E
Thus
1
2
/
3
2
2
0
1
2
/
3
2
2
0
2
2
2
2
0
)
(
4
1
)
(
,
)
(
4
4
)
(
,
π
π
π
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1/24/2009
2
•
Now we need to use he concept of
linear charge density, which gives,
•
We want to let
then the length
. Now we can replace
the sum with an integral ranging from
as shown
in the fig
,
)
(
⇒
∆
=
∆
=
∆
as,
te E
we can wri
Thus
y
L
Q
y
Q
x
λ
∑
=
+
∆
∈
=
N
i
i
x
r
y
y
r
L
Q
E
1
2
/
3
2
2
0
)
(
4
π
∞
→
N
dy
y
→
∆
2
2
L
y
to
L
y
=
−
=
in the fig.
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 Fall '08
 TAWFIQ
 Physics, Charge, Electric charge, Erod

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