Derive an expression for the potential at a distance r from a point charge `+q`.

Explanation

Potential due to a point charge Let P be the point at a distance r from the origin O at which the electric potential due to charge `+q` is required

The electric potential at a point P is the amount of work done in carrying a unit positive charge from `oo` to P. As work done is independent of the path, we choose a convenient path along the radial direction from infinity to the point P without acceleration. Let A be an intermediate point on this path, where OA= x. The electrostatic force on unit positive charge at A is `F= (1)/(4pi epsi_(0)) (q)/(x^(2))`, (along OA)...(i)
Small work done in moving the charge through a distance dx from A to B,
`dW= F dx = Fd x cos 180^(@) = -Fdx [ because cos 180^(@)= -1] dW= - Fdx` ....(ii)
Total work done in moving unit positive charge from `oo` to the point P is
`W = int_(oo)^(r )- Fdx= int_(oo)^(r )- (1)/(4pi epsi_(0)).(q)/(x^(2))dx`
`= -(q)/(4pi epsi_(0)) int_(oo)^(r ) x^(-3)dx= -(q)/(4pi epsi_(0)) [(-1)/(x)]_(oo)^(r) [ because int x^(-2) dx= - (1)/(x)]`
`=(q)/(4pi epsi_(0)) [(1)/(r )-(1)/(oo)]`
`rArr W= (q)/(4pi epsi_(0))` ...(iii)
From the definition of electric potential, this work is equal to the potential at point P,
`V= (q)/(4pi epsi_(0)r)` ...(iv)
A positively charged particle produces a positive electric potential. A negatively charged particle produces a negative electric potential.