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

Explanation

Consider a point charge .+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= Fdr. 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=E/(+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.