Let's assume that we have a positive charge $Q$ and a positive test charge $q$ is moved by an external agent from a point $A$ to $B$. The distance of $Q$ from $A$ is $r_1$ and from $Q$ to $B$ is $r_2$.
Let's say that the external agent moves $q$ from $B$ to $A$. It does some work while doing so. If it had moved the charge from a distance of $r$ to $r+dr$, where $dr$ is a very tiny distance, the work $(dW)$ would have been approximately equal to $K_e\dfrac{Qq}{r^2}dr$, where $K_e$ is Coulomb's Constant.
The sum of all these little 'works' will be the total work done by the external agent when it moves $q$ from $B$ to $A$, which will be :
$$\int_{r_1}^{r_2} K_e \dfrac{Qq}{r^2}dr$$
[Note : If this is wrong, please correct it, I'm just a beginner to calculus]
The simplification of this integral, according to my Physics textbook is $K_eQq \Bigg ( \dfrac{1}{r_1}-\dfrac{1}{r_2} \Bigg )$
Now, this is the work done by the external agent when it moves $q$ from $B$ to $A$. The potential difference between two points in an electric field is defined to be the work done to move a unit charge from one point to the other. So, it would be equal to :
$$\dfrac{\text{Work done}}{q} = K_eQ \Bigg ( \dfrac{1}{r_1} – \dfrac{1}{r_2} \Bigg )$$
I think that that should give the potential difference between two points in an electric field due to a charge $Q$.
Let me know if it's correct and if it isn't, why it's incorrect.
Best Answer
Your equations and reasoning are correct.