Energy expended in moving point charge in E field. Having trouble understanding an excerpt from E&M textbook 
To move charge from one point to another in an electric field, the force which we must apply is equal and opposite to the force due to the field. 
  (Quoted from Engineering Electromagnetics by Hayt.)

Here is my concern:
To move a charge, shouldn't the force that we must apply be just a little higher than the force exrted on the charge by the field? I know that electric field intensity is the amount of 'kick' a test charge feels when placed in that electric field. Now if I apply the SAME amount of force against the force due to E field, the charge, according to my understanding, will just stay there and won't move. If I apply a force just a little higher than force due to E field, only then the charge should move.
Why does the text say that to MOVE charge in E field, we should apply force EQUAL to that of the force experienced by charge because of E field.
 A: Remember that, if the net force is zero, velocity is constant (not necessarily zero!). You only need to do push a tiny bit harder for a tiny bit of time to start moving the charge. This extra amount can safely be ignored. Once the charge is moving with some nonzero velocity, equal force is enough.
A: 
To move charge from one point to another in an electric field, the force which we must apply is equal and opposite to the force due to the field. 


The sentence you provided is actually confusing. I think it should have been:  
To move charge from one point (lower potential) to another (higher potential) in an electric field, the constant force which we must apply from the lower potential point (till higher potential point) must be equal and opposite to the force due to field at the higher potential point.
A: You are thinking about energy, and not about force. The charge exerts a force upon the system, or the system exerts a force upon the charge. Energy is consumed in that process, but that isn't the point of the statement made.
Look at it through Energy $E$.(the field will be $\mathbf E$)
Lets look at these formula:
\begin{align}
E&=\int_\ell \mathbf F \operatorname{d} \ell\\
F&=q\mathbf E\\
E&=q\int_\ell \mathbf E \operatorname{d} \ell\\
V&=\int_\ell \mathbf E \operatorname{d} \ell\\
\end{align}
As you can see energy is related to force in the same way as the potential difference is related to the electric field. What has that to do with the question? Not much, it should only give you the oportunity to think in the relation $\mathbf E$ and $V$. In the definition he is talkin about moving a charge wit the the force $\mathbf F$, that't actually energy! 
Think about it as energy states which change when you  move from one place to the other. To do that we need to apply a Force, to move back we need to apply the force in the opposite direction. The thing is that even if we move with the field we will still work with a force that is opposite to it. The point that you are moving from a higher to a lower energy state, or reverse does not matter. According to the definition the charge has to have a net force of zero($\sum \mathbf F=0$), because then it can freely move according to newtons first law.
If you had a higher force than the one produced by the field, your electric field would be higher (equation 1), and the charge would accelerate! I don't like how the definition is written, there would be a lot of better ways to say it.
