Obtain the relation between electric field and electric potential due to a point charge.

Explanation

Consider a point charges `.+ q ..` Let .P. be a point initially at infinity .Let .A. be a point inside the field region .Let .+ 1C . be a unit positive 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 =- 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 =(d W)/( + q_(0))` and electric field
` E= (f)/(+ q_(0 ))`
When ` q_(0) = + 1C ` then ` dW = d V, 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 .