Derivation of the Relation Between Electric Field and Electric Potential Due to a Point Charge

We will derive the relation between electric field (E\mathbf{E}) and electric potential (VV) for a point charge using basic electrostatic principles.

Step 1: Expression for Electric Potential (V) Due to a Point Charge

Consider a point charge qq placed at the origin. The electric potential VV at a distance rr from the charge is given by:

V=14πε0⋅qr

where:

• qq = Charge creating the field (Coulombs)
• rr = Distance from the charge (meters)
• ε0\varepsilon_0 = Permittivity of free space (8.85×10−12 F/m

Step 2: Electric Field as the Negative Gradient of Potential

The electric field is defined as the negative rate of change of potential with respect to distance:

E=−dVdr

Substituting V=14πε0⋅qr

dVdr=ddr(14πε0⋅qr)

Differentiating:

dVdr=−14πε0⋅qr2

Since E=−dVdr we get:

E=14πε0⋅qr2

Step 3: Interpretation

• This result is the well-known Coulomb’s law formula for the electric field due to a point charge.
• The electric field is radial and directed outward for positive charges and inward for negative charges.
• The magnitude of the electric field is directly proportional to the charge qq and inversely proportional to the square of the distance rr.

Final Relation Between E and V:

E=−dVdr

For a point charge:

E=Vr

This shows that the electric field is the negative gradient (rate of change) of electric potential in the radial direction.

This relation is crucial in understanding electrostatics, equipotential surfaces, and the behavior of charges in an electric field.