Consider a point charge Q at the origin (Fig.). For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P.
Fig. Work done in bringing a unit positive test charge from infinity to the point P, against the repulsive force of charge Q (Q > 0), is the potential at P due to the charge Q.
For Q > 0, the work done against the repulsive force on the test charge is positive.
Since 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 = Q (1) / 4 π ε0 r' ^2 r'ˆ
where r'ˆ is the unit vector along OP'.
Work done against this force from r' to r' + Δ r' is
ΔW = - Q / 4 π ε0 r' ^2 Δ r'ˆ
The negative sign appears because for Δr' < 0, ΔW is positive . Total work done (W) by the external force is obtained by integrating from r' =∞ to r' = r,
This, by definition is the potential at P due to the charge Q
V(r) = Q / 4 π ε0 r
This equation is true for any sign of the charge Q, though we considered Q > 0 in its derivation.
For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive.
[This is as it should be,since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are in the same direction.]
Finally, we note that equation V(r) = Q / 4 π ε0 r is consistent with the choice that potential at infinity be zero.
Above Figure shows how the electrostatic potential (α1/r) and the electrostatic field (α1/r 2 ) varies with r.
