Shouldn't the electric potential energy decrease with separation regardless of whether the charges are positive or negative?

Question

I was revising Electric Fields and it came up that if a positive charge moves in the direction of the electric field (so away from a positive charge), then the electric potential energy will decrease but if the charge is negative, it will increase.

This adds up when you consider work has to be done on the negative charge to move it against the attractive in this case whereas, for the positive charge, the potential energy is used to work.

However, looking at it from another perspective using $E=kQq/r$, the potential energy will decrease with separation regardless of the charge. Most people answer this kind of question by bringing in the sign but I want someone to conceptually answer this problem:

We know that two charges have the maximum magnitude of Electric Force when they are closest together, whether attractive or repulsive. That means they will have maximum potential to accelerate under this force by $F=ma$. Maximum potential to accelerate means maximum Potential Energy that can be converted into Kinetic Energy. Since Energy is scalar, the direction does not matter here.

Please let me know what is wrong with the above assessment so that I may understand why the topmost paragraph stands true.

Edit: Can I get an answer that debunks my explanation in bold which is contradicting the quantitative answer.

Answer 1

If $\text{potential energy}=kQq/r$ and $Qq>0$ then as $r$ increases the potential energy decreases but if $Qq<0$ then as $r$ increases the potential energy increases by becoming less negative with the increase in $r$.

Answer 2

We know that two charges have the maximum magnitude of Electric Force when they are closest together, whether attractive or repulsive.

That is true. But the direction of the force will be different depending on whether the charges have the same or different sign.

That means they will have maximum potential to accelerate under this force by $F=ma$.

This is true in the case of the two charges close together that have the same sign, in which case the potential will be maximum and the field will do work accelerating the charge converting potential to kinetic energy. But it is not the case for two charges close together with different sign, in which case the potential will be a minimum and the charges bound by the force of attraction. An external agent is needed to separate the charges to give them potential energy.

Maximum potential to accelerate means maximum Potential Energy that can be converted into Kinetic Energy.

Again, that only applies to the like charges where where the potential energy is a maximum and where, upon release of the charges, the electric field does work converting potential energy to kinetic energy. For unlike charges that are close together, the potential is a minimum and the charges are bound to each other by the force of attraction.

Since Energy is scalar, the direction does not matter here.

Potential energy is a scalar because it is not a vector quantity having direction. But it can have a sign which has nothing to do with direction, with that sign as a consequence of the equation for potential energy

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

Which has a negative sign for unlike charges and a positive sign for like charges.

Regardless of the sign of the potential energy, which is due in part to the selection of a reference (zero) potential, as a practical matter the only thing that matters is the change in potential energy between points.

For a positive test charge the potential energy decreases with separation as the the field does work converting potential energy to kinetic energy (See FIG 1 below). For a negative test charge the potential energy increases (becomes more positive or less negative) with separation as work is required by an external agent to pull the charges apart (See FIG 2 below). It should be noted, however, that the increase in potential energy decreases with increasing separation as the work required to separate the charge goes to zero.

Hope this helps.