At some intermediate point P¢ on the path, the
electrostatic force on a unit positive charge is
2
0
1 ˆ
4
'
Q
εr
×
′
π
r
(2 5)
where ˆ′r is the unit vector along OP¢ Work done
against this force from  r¢ to r¢ + Dr¢ is
2
0
4
Q'
W
r
εr
∆
= −
∆ ′
π
(2 6)
The negative sign appears because for Dr¢ < 0, DW is positive