Electrostatic potential during attraction and repulsion

Question

[This question might be similar to some other which I am unaware of. But I added this question since it was troubling me a lot. I hope that this question is not flagged for being similar if it is. I am in need of help regarding this question]

[Starting few paras are about what I have understood. If there is any misconception, please correct me]

Attraction:

If 'q' charge of the image is negative then potential at any point at a distance 'r' should be $(V=\frac{-kq}{r})$ where 'q' is the absolute value of the the charge and 'k' is the Coulomb's constant. Now this value is negative since at any point $r<<∞$, potential should be less than that of at infinity. That is because if we let a charge at any point other than infinity, it will proceed towards the charge 'q' and its kinetic energy will increase gradually. And other than its own potential energy there is no other energy to be converted into kinetic energy. So which mean potential energy decrease as we progress towards the charge 'q' and maximum potential energy is at infinity which is zero in this case

Max potential energy=0

Potential energy of the test charge=(max potential energy - work done by field to move the charge from infinity to r)=$(0-\frac{kq}{r})=\frac{-kq}{r}$

Repulsion:

If the charge 'q' is a positive charge then it will repulse the test charge and mathematically its potential is ($V=\frac{kq}{r})$. Since repulsion is occuring, a certain sort of energy is to be given to the test charge so that it can progress towards charge 'q'. Which means potential energy at infinity which is 0 is the minimum in this case. And in this case potential energy of the charge is actually $\frac{+kq}{r}$

Max potential energy= ∞

Potential energy of the test charge= $\frac{+kq}{r}$

Now it seems that potential scaling in both cases aren't the same. So what if we think of a dipole where a test charge is located at the center of the axis of the dipole. Mathematically it is found that potential at point is 0 because $V_1+V_2=\frac{kq}{r}+\frac{-kq}{r}=0$. But as we have noticed in the above paras these scalings aren't the same. Then how rational it is to incline with this mathematical proof? I mean $ |\frac{kq}{r}| $ and $ |\frac{-kq}{r}| $ shouldn't be the same in terms of physical meaning. However we apply this.

In page no:695 of fundamentals of physics by Halliday and Resnick 10th edition, a mathematical expression to calculate what is the potential of a charge at any point in a multiple charged body system is included. In a nutshell it is the summation of all the potentials

[This is to mention that I am not concerned about superposition or anything I am just confused about the scaling of electric potential. Does operation between the potentials in two cases altogether hold firm physical meaning or is both of them of different dimensions based on physical meaning to mathematical expression?]

Please anyone who has seen this post help me to figure out this problem and get rid of the confusion.

Answer 1

The potential due to dipole given by $$\phi(r)=\frac{1}{4\pi \epsilon_0}\frac{\mathbf{\hat{r}}\cdot \mathbf{p}}{r^3}$$ It's not zero. You can't put two opposite point charges on top of each other that would similar to no charge at all and thus resulting no potential. $$\phi(\mathbf{r})=\frac{q}{4\pi \epsilon_0|\mathbf{r}-\mathbf{r}_+|}-\frac{q}{4\pi \epsilon_0|\mathbf{r}-\mathbf{r}_-|}$$ More detail of the derivation can be found here. Furthus note that multipole approximations not valid for short distances.