Derive an expression for the electric potential in a electric field of positive point charge at distance r.

Explanation

Consider a point charge Q at the origin as shown in figure.

The work done in bringing a unit positive test charge from infinity to the point P. For Q which is positive the work done against the repulsive force on the test charge is positive. Since the work done is independent of the path we choose a convenient path along the radial direction from infinity to the point P .
At some intermediate point P. on the path, the electrostatic force on a unit positive charge is,
`F=(kQxx1)/((r)^(2))hatr`
Where f is the unit vector along OP. Work done against this force from r. to r +` Deltar` is
`DeltaW= -(kQ)/((r)^(2)). Deltar` [ From W = F r cos`theta]`
Here `Deltar lt 0 ` so `DeltaW gt 0` in this formula .
Total work done by the external force is obtained by integrating above equation from
`r = prop ` to r = r .
`:. -int_(oo)^(r)(kQ)/((r)^(2)).dr [underset(Deltar rarr0)lim=dr]`
`:. W =- kQ [-(1)/(r)]_(oo)^(r) = (kQ)/(r)`
`:. W= (Q)/(4pi in_(0)r)`
From definition of electric potential
`V(r) = (W)/(q)=(W)/(I) [ because q=1C]`
`:. V(r) = W= (kQ)/(r)`