Relation between electric field and potential: Consider the electric field due to charge +q located at the origin O. Let A and B be two adjacent points separated by distance dr. the two points are so close that electric field E between them remains constant. Let V and V + dV be the potentials at the two points.

The external force required to move the test charge q₀ against the electric field E is given by

The work done to move the test charge from A to B is

W = F .dr = -- q₀E.dr

Also the work done in moving the test charge from A to B is

W = charge * potential difference

q₀(V_B - V_A) = q₀ dV

Equating the two work done, we get

-q₀E.dr = q₀ dV

Or

The quantity dV/dr  in the rate of change potential with distance and is called potential gradient and negative sing shows that the direction of electric field in the direction of decreasing potential.

Electric potential from electric field: The relation between electric field and potential is E=-dV / dr

Integrating the above equation between points r₁ and r₂ we get

Where V₁ and V₂ are the potential at r₁ and r₂ respectively. If we take r₁, at infinity

Then V₁ = 0

And put r₂ = r. then

Hence by knowing the electric field at any point, we can calculate the electric potential at that point.

SI unit of electric field:

Electric field at any point is equal to the negative of the potential gradient. It suggest that the SI unit of electric field is volt per meter, but electric field is also defined as the force experienced by a unit positive charge so SI unit of electric field is N/Coulomb. So