conventionselectric-fieldselectrostaticspotential energywork
Let us say a point charge Q was moved across a potential difference V, then work done would be : QV.
This work is taken negative when done external agent, Please explain when It is negative and positive, also for the general case, if we take change in potential to be V, then by the equation :
When this will give an absolute value.
I believe you are confusing the work done by the electric field with the work done on the particle.
By the electric field: The radial force of the electric field is always pointing outwards, and the displacement of the charge in this case is going inward. Thus, the integral you've specified will be negative. That is, the work done by the electric field is negative.
On the Particle: The force that must be applied to the charge (ignoring gravity, relativity, induced vector-potential, etc.) to get it from infinity to the radial distance $r$ is, at the very least, equal and opposite to the force of the electric field acting on the charge. Thus, it is always pointing inwards. Given that the charge is going inwards, the work integral will come out positive.
You can describe the electric force it terms of potential energy, because it is a conservative force. In doing so you actually replace the concept of work done by this force by the concept of potential energy. So you can not longer use both descriptions simultaneously. If you describe the electric force as doing work, then you made positive work and the electric force negative work, so that there is no net gain of kinetic energy in the object. It is a mistake to say that in this description the particle also has potential energy, because in doing so you be considering the work made by the electric field twice (both, as doing work and as gaining potential energy. The descriptions are equivalent, but it is either one or the other. If you chose the potential energy description then you no longer deal with the work of the electric force, as it is implicitly inside the concept of potential energy.
Best Answer
If a charge $Q$ is moved from rest by external forces across a potential difference $\Delta V=V_b-V_a$ from point $a$ to point $b$ and ends up at rest, then the net work done by all forces (electrical and external) will be zero by the work-energy theorem: $$W_{ext}+ Q(V_b-V_a) = \Delta K = 0.$$
That means that $W_{ext}=-Q(V_b-V_a)$. So, the work done by the field has the opposite sign of the external work.
If a positive charge is moved in the same direction as the electric field, or from higher to lower potential, the work done by the field will be positive.