Now, by geometry,
2
2
2
1
2
r
r
a
ar
=
+
−
cosq
2
2
2
2
2
r
r
a
ar
=
+
+
 cosq
(2 10)
We take r much greater than a (
r a
) and retain terms only upto
the first order in a/r
  r
r
a
r
ra
1
2
2
2
2
1
2
=
−
+




cosθ
  ≅
−

r
a
r
2
1
2
cosθ
(2 11)
 Similarly,
  r
r
a
r
2
2
2
1
2
≅
+

cosθ
(2 12)
Using the Binomial theorem and retaining terms upto the first order
in a/r ; we obtain,
1
1 1
2
1 1
1
1 2
r
r
a
r
r
ra
≅
−

≅
+


−
cos
cos
/
θ
θ
[2