Potential gradient gives electric field

Idea. The potential difference between two nearby points gives the work per unit charge. Since the electric field is the force per unit charge, E is the (negative) space rate of change of V.

Derivation

The potential difference V_AB = −∫E·dL (from B to A). For a small displacement dL, the change in potential is

dV = −E·dL                                   (1)

In Cartesian coordinates, V = V(x, y, z), so by the chain rule

dV = (∂V/∂x)dx + (∂V/∂y)dy + (∂V/∂z)dz       (2)

and with dL = dx a_x + dy a_y + dz a_z,

E·dL = E_x dx + E_y dy + E_z dz              (3)

Comparing (1), (2), (3) for arbitrary dx, dy, dz:

E_x = −∂V/∂x ,  E_y = −∂V/∂y ,  E_z = −∂V/∂z

E = −(∂V/∂x a_x + ∂V/∂y a_y + ∂V/∂z a_z) = −∇V

Interpretation

• The gradient ∇V is the vector pointing in the direction of the maximum increase of V, with magnitude equal to the maximum rate of change dV/dn. E is the negative of it, so E points in the direction of the fastest decrease of potential, from high to low potential.
• E is perpendicular to the equipotential surfaces (since along them dV = 0, so E·dL = 0).
• In other coordinates: cylindrical E = −(∂V/∂ρ a_ρ + (1/ρ)∂V/∂φ a_φ + ∂V/∂z a_z); spherical E = −(∂V/∂r a_r + (1/r)∂V/∂θ a_θ + (1/(r sinθ))∂V/∂φ a_φ).

Example

For a point charge, V = Q/(4πε₀r), so E = −∂V/∂r a_r = Q/(4πε₀r²) a_r, which is Coulomb's law.