Sign conundrum while deriving electrostatic potential

Question

Consider a fixed, positive Point charge $q1$, kept at the origin. Another (positive) charge, $q2$, is being brought from $\infty$ to the point $(r,0)$, by an external agent slowly. We wish to calculate the work done by the external agent (and thus derive the "potential" of the point charge $q1$, being defined as $w_{ext}/q2$ (or as $-w_{electric}/q2$)). Suppose we consider a position $(x,0)$:

• The magnitude of force is going to be $kq1q2/x^2$. We will thus have, $\vec{F}=\dfrac{-kq1q2}{x^2}\hat{i}$.
• When we displace it from a position $(x,0)$ to $(x-dx,0)$, the displacement vector$(\vec{ds})$ will be $(x-dx)\hat{i}-x\hat(i)=-dx\hat{i}$.
• Using $dw$=$\vec{F}.\vec{ds}$, we will get $dw=\dfrac{kq1q2}{x^2}dx$. Upon integrating from $\infty$ (initial position) to $r$ (final position), we get : $$w_{ext}=-\dfrac{kq1q2}{r}$$ and thus $$V(r)= w_{ext}/q2 =-\dfrac{kq1}{r}$$ which is completely absurd. I tried to be as rigorous as possible with the definitions, vectors etc and yet a -ve sign has crept in somewhere.

The only issue seems to be with the treatment of $dx$. Although , I took $dx$ to be the magnitude of displacement, and accounted for the direction by using $-\hat{i}$.A possible argument seems to be "$x$ decreases, so $dx$ is a negative quantity. So the "magnitude" should be $-dx$. My two concerns:

• What is then, the issue with displacement=$\vec{r_{final}} - \vec{r_{initial}}$ that simply yields $-dx\hat{i}$?
• Simply "putting" a - sign before $dx$ after claiming "$dx$ is negative" seems to be arbitrary. There should be an argument (like I presented in the previous bullet point) that will produce the - sign for the magnitude, and thus making the vector $(-dx)(-\hat{i})$.

The main essence of this problem seems to be rigorously defining what $dx$ actually represents, for a quantity $x$.

I believe the entire thing can be summarized by one question:

What is wrong in writing displacement=$\vec{r_{final}} - \vec{r_{initial}}$ that simply yields $-dx\hat{i}$? If I had $(x+dx)$ instead of $(x-dx)$,then the derivation would be correct. But why is this the Case?

Answer 1

I suspect that there is a confusion in the use and/or interpretation of $d\vec s$

• as an infinitesimal element of a directed path

and

• as a specific displacement in the problem.
  
  

You want to write $$ \begin{align} W_{ext} &=\int_P^Q \vec F_{ext}\cdot d\vec s\\ &=\int_{x_P}^{x_Q} (F_{ext,x}\ \hat\imath)\cdot (dx\ \hat\imath)\\ &=\int_{x_P}^{x_Q} F_{ext,x}\ dx\\ \end{align} $$ by expressing the symbolic-integrand in coordinates, say, your x-coordinate running to the right.
The element of path $d\vec s$ (which is not the specific displacement of your problem) is being expressed in terms of the x-coordinate.
I have not made any statement about $P$ or $Q$.
The specific values of $P$ and $Q$ do not determine how the element of path is expressed in terms of the x-coordinate.

For simplicity, if $F_{ext,x}$ were constant, then this evaluates to $$\begin{align} W_{ext} = F_{ext,x} (x_Q-x_P) \end{align}$$ This work-by-the-external-force is positive when

• $F_{ext,x}>0$ and $(x_Q-x_P)>0$
  (the force pushes to the right, and the particle is displaced to the right... so $x_Q > x_P$),
  

or when

• $F_{ext,x}<0$ and $(x_Q-x_P)<0$
  (the force pushes to the left, and the particle is displaced to the left... so $x_Q < x_P$)

In your specific problem essentially maintaining equilibrium of $q_2$ during transport, $$\vec F_{ext\ \rm [on\ q_2]} \stackrel{Newton\\ 2^{nd}\ Law}{=} -\vec F_{elec,\ \rm on\ q_2} = -\left(\frac{ k q_1 q_2}{x^2} \hat\imath \right),$$ you have $$ \begin{align} W_{ext} &=\int_P^Q F_{ext,x}\ dx\\ &=\int_{x_P}^{x_Q} \left(-\frac{k q_1 q_2}{x^2}\right) \ dx\\ &=kq_1q_2\left(\frac{1}{x}\right|_{x_P}^{x_Q}\\ &=kq_1q_2\left(\frac{1}{x_Q}-\frac{1}{x_P}\right) \end{align} $$ Now is the time to specify your initial and final values, $x_P$ and $x_Q$.
With $x_P=\infty$, $$ \begin{align} W_{ext} &=kq_1q_2\frac{1}{x_Q}>0 \end{align} $$

(for a related answer, see
Positive work along path
and
https://www.physicsforums.com/threads/dot-product-in-the-gravitational-potential-energy-formula.815230/ )