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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 $r_2$.

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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.

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Your equations and reasoning are correct.

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  • $\begingroup$ Thank you so much! You have no idea how confused I was because of this. :) $\endgroup$
    – Rajdeep Sindhu
    Commented Jun 23, 2020 at 5:04
  • $\begingroup$ A little question though, how is potential difference in an electric field due to a point charge difference from potential difference in a circuit? Is it different at all? $\endgroup$
    – Rajdeep Sindhu
    Commented Jun 23, 2020 at 5:12
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    $\begingroup$ It's not different (as long as the currents/fields are not changing very fast). The potential at a given point is still found by integrating the electric field, or equivalently, by adding up the potential contributions of all charges with Coulomb's law. In conductive materials (resistors), current flows from high potential to low potential, since E-field (and hence the force on positive charges) points from high potential to low potential. $\endgroup$
    – Puk
    Commented Jun 23, 2020 at 5:19

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