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

Answer: \(V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}\)

Let a point charge \(+q\) be at O. We find the potential at P, at distance \(r\) from O. Potential at P is the work done in bringing a unit positive charge from infinity to P against the field.

At a point A at distance \(x\) from O the field is \(E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}\), directed away from O. The force on a unit positive charge is \(E\). To move it towards O through a small distance \(dx\) without acceleration, an external force equal and opposite to \(E\) is needed, so the work done is

(the negative sign because the displacement is towards O, i.e. \(x\) decreases). Total work from \(x = \infty\) to \(x = r\):

Potential is a scalar: positive for a positive charge, negative for a negative charge, and it varies as \(1/r\).