# Electrostatic potential of a point charge

From Wikipedia

The electrostatic potential energy, UE, of one point charge q at position r in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position rref[note 1] to that position r.1:§25-1[note 2] Since the potential energy is defined as negative of work done by the conservative force. I will proceed with the proof by calculation work done by the force in bringing the point charge q from $$\infty$$ to $$r$$.

I have used $$dx$$ instead of $$ds$$ as depicted in the diagram.

$$W=F_{Qq}\cdot dx$$
Since the direction of the force by Q on q ($$F_{Qq}$$) is antiparallel the dot product is resolved as
$$W=F_{Qq}\times dx \cos(180^{\circ})$$
$$W=-F_{Qq}\times dx$$

Now, subbing in the value of $$F_{Qq}= \frac{k Qq}{x^2}$$

$$W=-\int_{\infty}^{r} \frac{k Qq}{x^2}dx$$
$$W=\bigg[\frac{kQq}{x}\bigg]_{\infty}^{r}$$
$$W=\frac{kQq}{r}$$

Since this is the work calculated therefore potential energy is its negative:

$$U=-\frac{kQq}{r}$$

However the correct expression for the potential energy of two point charges is

$$U=\frac{kQq}{r}$$

I can't find anything wrong with my proof but it is wrong, why?

• Duplicate which explains that the direction of $d\vec x$ is entirely determined by the path taken or the limits of integration. – Farcher Jan 14 '19 at 18:39

## 1 Answer

You should work with $$d\vec{r}$$ instead of $$d\vec{x}$$. As they are opposite, that's your problem.