Equipotential lines are always at right angles with the electric field (most clearly shown in the centermost equipotential line). This implies that if a charge were to move along an equipotential line then throughout the entire journey $F_{electric} \perp dr $ and hence, $F_{electric} \cdot dr = 0$.
To move a charge along an equipotential line, you'd need to supply two forces: one to cancel out the net force from the two charges, and the other to move it along the equipotential. So: $F = -F_{electric} + F_{tangential}$. Calculating the total work done:
$$W = \int{F \cdot dr} = \int{(-F_{electric} \cdot dr)} + \int{F_{tangential} \cdot dr}$$
As we previously argued, $F_{electric} \perp dr$ so
$$W = \int{F_{tangential} \cdot dr} $$
which is nonzero.
EDIT: So just to clarify, the electric field's contribution to the total work is zero, but the electric field on its own will never be able to pull the particle down from infinity.
The short answer to your question is potential energy. Absolutely everything is always trying to get to the lowest potential energy state possible.
Let's start with the fundamental difference between electric potential and electric potential energy: electric potential energy is always in reference to a specific charge (the energy of the charge,) while electric potential is in reference to a region of space (the voltage across a battery.)
A bit of a problem here is that your allowing the amount of charge to change when connecting the two charges with a wire. In that scenario, to truly talk about the potential energy, you would have to add up the potential energy of each individual electron/proton. We can kind of do this in the case without charge A, it explains why all the charges get evenly spread: the charges want to be as far as possible from each other, so they all end up compromising and being equidistant from one another. However in the case of having charge A, talking about the electric potential energy of each individual charge is not practical. This is where electric potential comes in.
The electric potential describes what will happen to charges once they are put into that region of space. In this case, the potential from charge A will decrease inverse of the distance from charge A. Positive charge wants to get to lower electric potential (because that allows it to get to a lower potential energy.) This means the charge will want to get as far away as possible and you'll end up with 7.5 C of charge concentrated at each end of the wire, evenly spread by the same argument as above.
Conversely, negative charge wants to get to the highest electric potential possible, because that gets it to a lower potential energy. In this case by lower potential energy I mean a more negative potential energy. If charge B is -10 C, then you'll have partial cancelation with charge C and end up with -5 C that will be concentrated at the point closest to charge A on the wire.
This reveals why it is often better to talk about electric potential rather than electric potential energy in circuits. You don't want to worry about what each individual electron is doing, you just want an idea of what the charge will do overall.
In the given figure, a test charge $+Q$ is brought from infinity to a point $P$ in the vicinity of a positively charged body. If $W$ joule of work is done in bringing the test charge $Q$ coulomb from infinity to the point $P$ in the vicinity of a positively charged body then the electrical potential at that point is given as $V=\frac WQ$.
Best Answer
Yes, TO ACTUALLY MAKE IT MOVE a net force must be greater, however let's say I provide a force infinitesimal greater, it will have some velocity. And then moving along that distance I will have done X amount of work against gravity.
The way you should think of it, is that IF an object moves through a distance, the Electric field is going to either be doing positive or negative work per unit charge.
Meaning if lets say in moving through a distance A to B the field does -W work per unit charge. The total amount of energy I need to provide to overcome that negative work, will be "W" aka the negative of the work done by the field. As W+(-W) = 0 so if I give W work over a distance, then the field will equally "Steal" that work from me, causing no net gain of KE by the object
Its all about measuring how much work is either done or lost on an object by the field IF it were to move along a certain path. The negative of this is the total amount of energy that "i" put in, to overcome this
Thought experiment:
Given an object moves vertically with an initial velocity V, as the object moves gravity will do negative work on it, such that at a height "H" the total amount of negative work done by gravity is equal to the objects kinetic energy, aka the object STOPS at height H
now imagine that now, the object is at the ground with the same initial velocity. But this time, I am applying an EQUAL BUT OPPOSITE force to gravity. The net force is zero, so the object will continue to move upwards at a constant velocity. When the object reaches a certain distance upwards "A", I will then STOP that applied force, clearly now the object is at a height "A" and will then continue to move upwards by the same amount as the first scenario and then stop.
Meaning that I have now applied a force -F over a distance A but this time the object has now has a maximum height A+h
Aka, A meters higher. Clearly the work that I have put in, has caused the object to move A meters heigher, than the previous scenario.
Aka, that is the total amount of work I have to do in order to overcome the force of gravity through a distance A.
More mathematically.
Consider the equation
$1/2 m v^2 + \int_{A}^{B} F \cdot dr = 0$
Aka, an object with an initial Ke, moves through the distance a to b, in the presence of a force field, such that when it reaches B, it stops. Well how much energy do I need to give the object such that in the presence of F, it stops when it moves through a to b?
Clearly
$1/2 m v^2 = - \int_{A}^{B} F \cdot dr $
this is exactly the same as if I apply the work to the object over the distance, instead of at the very start in the form of Ke