Derive a relation between electric field and potential

Explanation

Consider region. Let '+1C' be a unit potive charge moved from the point at infinity to point 'A' Let 'A' be at a distance of 'r' 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=-Fdr. the -ve sign indicates the work is done against the direction of the force and dr is the displacement of +1C 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 the direction of decreasign electrostatic potential.