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80. Let
q
1
denote the charge at
y
=
d
and
q
2
denote the charge at
y
= –
d
. The individual
magnitudes
G
E
1
and
G
E
2
are figured from Eq. 223, where the absolute value signs for
q
are unnecessary since these charges are both positive. The distance from
q
1
to a point on
the
x
axis is the same as the distance from
q
2
to a point on the
x
axis:
rx
d
=+
22
.
By
symmetry, the
y
component of the net field along the
x
axis is zero. The
x
component of
the net field, evaluated at points on the positive
x
axis, is
E
q
xd
x
x
=
F
H
G
I
K
J
+
F
H
G
I
K
J
+
F
H
G
I
K
J
2
1
4
0
π
ε
where the last factor is cos
θ
x
/
r
with
being the angle for each individual field as
measured from the
x
axis.
(a) If we simplify the above expression, and plug in
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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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