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Show curl of E is zero

Short answer

For a point charge, E is radial and depends only on r, so its curl in spherical coordinates is zero. By superposition, any static charge distribution yields ∇×E=0. Equivalently, ∮E·dl=0 around any closed path, and Stokes' theorem gives ∇×E=0 everywhere.

Show that ∇×E = 0 for the static electric field.

Statement. For a static electric field the curl is zero, ∇×E = 0, so the field is conservative.

Proof using the point charge field

A static field is the superposition of fields of point charges, so it is enough to show it for one charge Q at the origin:

E = Q/(4πε₀r²) a_r      (E_r depends on r only; E_θ = E_φ = 0)

In spherical coordinates:

∇×E = (1/(r sinθ))[∂(sinθ E_φ)/∂θ − ∂E_θ/∂φ] a_r
    + (1/r)[(1/sinθ) ∂E_r/∂φ − ∂(rE_φ)/∂r] a_θ
    + (1/r)[∂(rE_θ)/∂r − ∂E_r/∂θ] a_φ

E_r has no θ or φ dependence and E_θ = E_φ = 0, so every term vanishes: ∇×E = 0. By superposition it holds for any static charge distribution.

Integral form (Stokes' theorem)

The work done in moving a charge from A to B is −Q∫E·dl = Q(V_B − V_A), which depends only on the end points. Around a closed path A back to A, the work is zero:

∮ E·dl = 0

By Stokes' theorem, ∮E·dl = ∫(∇×E)·dS = 0 for every surface, so ∇×E = 0. This is Maxwell's equation for electrostatics, and it allows E = −∇V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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