Derive a relation between electric field and potential

Explanation

Figure with directions and finding workdone.
Or
The work done to move a unit positive charge from one point to another point against the field `vecE` through a displacement `vec(dl)` is `dW=vecE.vec(dl)=-Edl`.
This is equal to the potential difference dV, therefore
dV=dW
dV=-E dl
Or `E=-(dV)/(dl)`
Consider a point charge .+d.. Let .P. be a point initially at infinity. Let .A be a point inside the field region. Let .+1 C. be a unit positive charge moved from the point at infinity to point A. Let A. be at a distance of .y from the given point charge. Let .B. be another point at distance of .dr from .a. towards the point charge +q.

The work done in moving positive test charge from A to B, against the force of repulsion is dW =-F dr. The-ve sign indicates that the work is done against the direction of the force and dr is the displacement of +1 C of test charge in the direction opposite to the electric field.
But electric potential `dv=(dW)/(+q_0)` and electric field `E=F/(+q_0)`
When `q_0`=+1C then dW=dV,E=F
Hence dV=-Edr or `E=-(dV)/(dr)`
i.e., Electric field intensity at a point is the negative potential gradient at that point and electric field intensity is in the direction of decreasing electrostatic potential.