Derive an expression for electrostatic potential due to a point charge .

Explanation

Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q. This is shown in Figure.

Electrostatic potential at a point P
(ii) The electric potential at the point P is
`V=underset(oo)overset(r)int(-vec(E)).dvecr=-underset(oo)overset(r)intvec(E).dvecr" "......(1)`
Electric field due to positive point charge q is
`vec(E)=(1)/(4piepsi_(0))(q)/(r^(2))hatr`
`V=(-1)/(4piepsi_(0))underset(oo)overset(r)int(q)/(r^(2))hatr.dvecr`
The infinitesimal displacement vector `dvecr=drhatr` and using `hatr.hatr=1`, we have
`V=(1)/(4piepsi_(0))underset(oo)overset(r)int(q)/(r^(2))hatr.drhatr=(1)/(4piepsi_(0))underset(oo)overset(r)int(q)/(r^(2))dr`
After the intergration,
`V=(1)/(4piepsi_(0))q{-(1)/(r)}_(oo)^(r)=(1)/(4piepsi_(0))(q)/(r)`
Henece the electric potential due to a point charge q at a distance r is
`V=(1)/(4piepsi_(0))(q)/(r)" ".......(2)`
Important points :
If the source charge q is positive, `Vgt0`. If q is negative, then V is negative and equal to `v=(1)/(4piepsi_(0))(q)/(r)`
(ii) It is clear that the potential due to positive charge decreases as the distance increases, but for a negative charge the potential increases as the distance is increased. At infinity, r= electrostatic potential is zero (V=0).
(iii) A positive charge moves from a point of higher electrostatic potential to higher electrostatic potential.
(iv) The electric potential at a point P due to a collection of charges `q_(1),q_(2),q_(3).......q_(n)` is equal to sum of the electric potentials due to individual charges.