Hence ohtaln the work done in bringing charge of `2 xx 10^(-9)` C . From infinity to the point P. Does the answer depend on the path along which the charge is brought ?

Explanation

Let V(P) = and `V(oo) =0`
`:. V(P)-V(oo) = DeltaV=V=4xx10^(4)V`
Now the work done in bringing a charge of `2xx10^(9)` C from infinity to the point P,
W= q `DeltaV=qV`
`= 2xx10^(-9) xx4xx10^(4)`
`:. W= 8 xx10^(-5)J`
No work done will be path independent.
Any arbitrary infinitesimal path can be resolved into two perpendicular displacements.
(1) Along parallel to displacement `vecr` and
(2) Along perpendicular to displacement `vecr`.
But work done corresponding to perpendicular component will be zero.