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 Thus, [V(r1)–V (r2)] =
–E × 2a cosq We thus obtain,
( )
2
2
cos
4
2
4
2
θ
θ
ε
ε
0
0
= −
−
= −
−
′
π
×
π
×
p