The
potential energy  expression then reads
( )
( )
(
)
2
1
2
[
]
4
2
q
U
q V
V
a
θ
ε0
=
−
−
′
π
×
r
r
(2 33)
Here, r1 and r2 denote the position vectors of +q and –q Now, the
potential difference between positions r1 and r2 equals the work done
in bringing a unit positive charge against field from r2 to r1 The
displacement parallel to the force is 2a cosq