Electric Potential Energy - How a charge can be brought from infinity to a point without accelerating it? Question:
My Book says:

ELECTRIC POTENTIAL ENERGY: Electric potential energy of a charge($q_o$) at a point(A) in the electric field due to any charge is given by the work done by an external force to displace $q_o$ without acceleration from infinity to that point(A).

$\color{red}{\text{How it is possible to displace a charge from infinity to a point without accelerating it}?}$
Suppose a point test charge q is located at:
(1) Infinity

Then the moment an external force $F_{external}$, the charge gets accelerated.

(2)A point in Q's electric field

Then the charge Q would exert an electrostatic force($F_Q$) and point charge q would accelerate due to this force. My book says now an external force $F_{external}$ is exerted to move it without any acceleration. Book does not specifies which force is larger.

Now three cases arrive:
$$(1) F_{external}>F_Q$$$$(2)F_{external}=F_Q$$ $$(3) F_{external}<F_Q$$
In (1) and (2) cases there would be some net force, so there would be acceleration.
In (3) case the net force would be zero so the charge would be at rest.


 A: 
How is it possible to displace a charge from infinity to a point
without accelerating it.

The statement in the book is somewhat misleading. Assuming the charge begins at rest at infinity, the requirement is to bring the charge to rest at the point so that the change in kinetic energy between the two points is zero.
To start the movement requires an acceleration over time (to achieve a velocity) and to stop the movement requires an equal deceleration over time (to bring the object to rest). In so doing the net work done on the object equals zero and the change in kinetic energy equals zero per the work energy theorem. All the work done by the external force winds up as stored electrical potential energy of the charge.
In terms of your three cases, the sequence is:
$$(1) F_{external}>F_Q$$
$$(2) F_{external}=F_Q$$
$$(3)F_{external}<F_Q$$
Where
$F_{external}$ of case (1) gives the charge a velocity $v$
$F_{external}$ of case (2) maintains the charge at velocity $v$
$F_{external}$ of case (3) brings the charge to rest $v=0$.
No work is done in case (2). The positive work done on the charge in case (1) equals the negative work done on the charge in case (3) for a net work of zero and an overall change in kinetic energy of zero.
Finally, note that as the difference between $F_{external}$ and $F_Q$ approaches zero the acceleration approaches zero, and it takes an infinite amount of time to bring the charge from infinity to the point.
The gravity analogy is taking a mass at rest and elevating the object to rest at height $h$ above its starting point.

How the work done can be zero? $_{∞}=_{}−_{∞} = _{}−0=0$
This shows that electric potential energy would always be 0.

I didn't say the work done by the external force is zero. I said the net work done by the combination of the external force and the force of the electric field is zero. The distinction is important. The net work has to be zero in order for the charge to begin and end at rest. The work in your equation is only the work done by the external agent. That does equal the change in electric potential energy as long as the electric field does an equal amount of negative work, otherwise the work done by the external force will equal the sum of the change in potential energy plus any change in kinetic energy equal to the net work done.
Hope this helps.
A: That wording is poor.  By definition, you can't take any object at rest and move it without imparting acceleration.
Further, since the electric field is a conservative field, the work needed to move the particle is both path- and acceleration- independent -- other, of course for the final velocity if you don't apply a "braking force" at the end of the journey.
A: Only because $q_0$ is very far away, it doesn't mean that it must be at rest with respect to the point $A$. I can suppose that no matter how far I imagine $q_0$, it has a radial velocity $v$ in direction to $A$.
Suppose the charges have opposed signals. Some control device attached to $q_0$ makes sure that the increasing force due to decreasing distance is balanced by an equal force opposing to it. So, the net force on $q_0$ is always zero, and the work done can be thought be made by the eletrostatic force or the opposing force, because they are always equal in modulus.
A: You can assume that in the beginning you have a very, very small acceleration  resulting in v, from then on you have no force so your last equation ist true, at the end you still have m/2v^2 left for your energy balance.
