Irrotationality of the Electrostatic Field

Derivation

Irrotationality of the Electrostatic Field

D-042 Home PU-102 Threads fields · energy · symmetry Depends on Electric Field from Coulomb's Law, stokes-theorem

Statement

For any static charge distribution, the electrostatic field \(\vec{E}(\vec{r})\) satisfies \(\nabla \times \vec{E} = \vec{0}\) everywhere off the source points. Equivalently, the field is a gradient, \(\vec{E} = -\nabla \varphi\), and its line integral around any closed loop vanishes, \(\oint_{C} \vec{E}\cdot d\vec{\ell} = 0\). The electrostatic field is therefore conservative.

Why it matters

Irrotationality is the structural fact that lets us replace the three components of \(\vec{E}\) with a single scalar potential \(\varphi\). Every problem in electrostatics — capacitance, boundary-value problems, stored energy — rests on the existence of \(\varphi\), and \(\varphi\) exists precisely because \(\nabla \times \vec{E} = \vec{0}\).

It also draws the sharp line between electrostatics and electrodynamics. The moment magnetic flux changes in time, Faraday's law gives \(\nabla \times \vec{E} = -\partial_t \vec{B} \neq \vec{0}\), the loop integral no longer vanishes, and no single-valued potential exists. Knowing why the static curl is zero tells you exactly what breaks when fields become dynamic.

Assumptions

Static sources.If charges move so that \(\partial_t \vec{B} \neq \vec{0}\), Faraday's law makes \(\nabla \times \vec{E}\) nonzero and no scalar potential exists.

Coulomb's law in its central, inverse-square form.The proof uses that each point-charge field points radially and depends only on \(|\vec{r}-\vec{r}'|\); a non-central force law would generally not be a pure gradient.

Superposition (linearity of Maxwell's equations).If fields did not add linearly, curl-freeness of each point-charge field would not transfer to the total field.

Field evaluated off the sources / simply connected domain.The potential \(\varphi\) is single-valued only where \(\vec{E}\) is smooth; at a charge location \(\vec{E}\) diverges and the curl identity is understood distributionally away from that point.

Derivation

\[ \vec{E}_1(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\, q\, \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}} \]

Start from the Coulomb field of a single point charge \(q\) at \(\vec{r}'\), an assumed prior result. A

\[ \nabla\left(\frac{1}{|\vec{r}-\vec{r}'|}\right) = -\,\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}} \]

Compute the gradient of the inverse distance. Writing \(s=|\vec{r}-\vec{r}'|\), one has \(\partial s/\partial x_i = (x_i-x_i')/s\), so \(\nabla(1/s) = -(1/s^2)\nabla s = -(\vec{r}-\vec{r}')/s^3\). B

\[ \vec{E}_1(\vec{r}) = -\,\nabla\!\left(\frac{q}{4\pi\varepsilon_0}\,\frac{1}{|\vec{r}-\vec{r}'|}\right) \equiv -\nabla \varphi_1 \]

Substitute Step 2 into Step 1: the point-charge field is exactly minus the gradient of the scalar \(\varphi_1 = q/(4\pi\varepsilon_0 |\vec{r}-\vec{r}'|)\). B

\[ \nabla \times \vec{E}_1 = -\,\nabla \times (\nabla \varphi_1) = \vec{0} \]

Apply the identity \(\nabla\times\nabla f = \vec{0}\) for any twice-differentiable scalar \(f\); it follows from equality of mixed partials \(\partial_i\partial_j f = \partial_j\partial_i f\). Valid wherever \(\varphi_1\) is \(C^2\), i.e. away from \(\vec{r}=\vec{r}'\). C

\[ \vec{E}(\vec{r}) = \sum_{k} \vec{E}_k(\vec{r}) = -\nabla\!\sum_k \varphi_k \equiv -\nabla\varphi \]

Superpose over all source charges (or integrate over a continuous \(\rho\)). The gradient is linear, so a sum of gradients is the gradient of the sum; define the total potential \(\varphi=\sum_k\varphi_k\). B

\[ \nabla \times \vec{E} = -\,\nabla\times(\nabla\varphi) = \vec{0} \]

Curl is linear, so the curl of the superposed field is the sum of the individual curls, each zero by Step 4. Hence the total electrostatic field is irrotational. C

\[ \oint_{C} \vec{E}\cdot d\vec{\ell} \;=\; \int_{S} (\nabla\times\vec{E})\cdot d\vec{A} \;=\; 0 \]

Take any closed loop \(C\) bounding a surface \(S\) in the source-free region and apply Stokes' theorem (assumed prior result). Since the integrand vanishes pointwise, the circulation is zero: the field does no net work around a loop. C

Result

\[ \nabla \times \vec{E} = \vec{0} \quad\Longleftrightarrow\quad \vec{E} = -\nabla\varphi \quad\Longleftrightarrow\quad \oint_C \vec{E}\cdot d\vec{\ell}=0 \]

Reading. The three statements are equivalent ways of saying the electrostatic field is conservative. Because each Coulomb contribution is the gradient of an inverse-distance potential, and the curl annihilates any gradient, the total field has no circulation. The work done moving a test charge between two points is path-independent and defines a potential difference.

Units check. \(\nabla\times\vec{E}\) has units \(\mathrm{(V\,m^{-1})/m = V\,m^{-2}}\); it equals \(\vec{0}\), dimensionally consistent with any target. The potential \(\varphi\) carries volts \(\mathrm{V}\), so \(-\nabla\varphi\) has \(\mathrm{V\,m^{-1}}\), matching \(\vec{E}\). The circulation \(\oint\vec{E}\cdot d\vec\ell\) has \(\mathrm{V\,m^{-1}\cdot m = V}\), consistent with an EMF that here equals zero.

Limiting cases

Single point charge: \(\vec E\) is purely radial and \(\varphi\propto 1/r\); \(\nabla\times\vec E=\vec0\) trivially by spherical symmetry.
Uniform field (parallel-plate interior): \(\vec E=\text{const}\Rightarrow\varphi=-\vec E\cdot\vec r\), a linear potential, curl zero.
Quasi-static limit: if charges vary slowly enough that \(\partial_t\vec B\) is negligible, \(\nabla\times\vec E\approx\vec0\) still holds to leading order.
Far field of any neutral cluster: dominated by the dipole term \(\varphi\propto 1/r^2\), still a gradient, still curl-free.

Breaks when

Magnetic flux varies in time. Faraday's law gives \(\nabla\times\vec E=-\partial_t\vec B\neq\vec0\); the loop integral equals the induced EMF and no single-valued potential exists. This is the entire basis of transformers and induction.
At the source points themselves. \(\vec E\) diverges as \(1/s^2\) at each charge, so \(\varphi\) is not \(C^2\) there; the identity \(\nabla\times\nabla\varphi=\vec0\) must be read distributionally and the naive pointwise proof fails at \(\vec r=\vec r'\).
Multiply connected domains with excluded flux. If a region carrying changing flux is bored out of the domain, a loop encircling it can have nonzero circulation even where \(\vec E\) is locally curl-free, and the potential becomes multivalued.

Failure modes

Confusing \(\nabla\times\vec E=\vec0\) with \(\nabla\cdot\vec E=0\). The divergence is \(\rho/\varepsilon_0\), generally nonzero; only the curl vanishes. Students routinely swap the two Maxwell statements.
Assuming curl-free means field-free. A uniform field has zero curl but is far from zero. Vanishing circulation says nothing about the magnitude of \(\vec E\).
Sign error in \(\vec E=-\nabla\varphi\). Dropping the minus sign points the field from low to high potential, reversing forces on charges.
Claiming irrotationality holds in electrodynamics. It is specific to the static case; invoking a scalar potential when \(\partial_t\vec B\neq\vec0\) is a serious error.
Mistaking path-independence for zero work. Work is path-independent, but it still depends on the endpoints; "conservative" is misread as "no work done."

Discussion

The result is best seen as a statement about the structure of the field rather than about any particular charge arrangement. Any central, superposable force whose potential depends only on separation produces an irrotational field; the specific \(1/r^2\) law is not required for curl-freeness, only for the specific form of \(\varphi\). What the derivation exposes is that "conservative" (path-independent work), "curl-free" (\(\nabla\times\vec E=\vec0\)), and "gradient field" (\(\vec E=-\nabla\varphi\)) are three faces of one property, linked by Stokes' theorem on one side and the identity \(\nabla\times\nabla=\vec0\) on the other.

This is precisely the second of the two static Maxwell equations. Together with Gauss's law \(\nabla\cdot\vec E=\rho/\varepsilon_0\) it determines \(\vec E\) completely given boundary conditions, because the Helmholtz theorem fixes a field from its divergence and its curl. Electrostatics is thus the study of a field with prescribed divergence and identically zero curl — which is why it collapses to a single scalar Poisson equation \(\nabla^2\varphi=-\rho/\varepsilon_0\).

Energetically, irrotationality is what makes electrostatic potential energy well-defined: the work \(W=q\int\vec E\cdot d\vec\ell\) depends only on the endpoints, so \(U=q\varphi\) is a genuine state function. A field with circulation could pump a charge around a loop and extract energy indefinitely, which is impossible for a static conservative system.

At the deepest level the property is geometric. The field \(\vec E\) is dual to a 1-form \(E = E_i\,dx^i\), and \(\nabla\times\vec E=\vec0\) is the closedness condition \(dE=0\). On a simply connected (contractible) domain the Poincaré lemma guarantees that every closed form is exact, \(E=-d\varphi\), which is the coordinate-free origin of the potential. The multiply-connected caveat above is exactly the failure of the Poincaré lemma when the domain carries nontrivial first de Rham cohomology — the same mathematics underlying the Aharonov–Bohm phase.

Common misconceptions. "Curl-free" is not "sourceless": charges are still present and \(\nabla\cdot\vec E\neq0\); it is the rotation, not the divergence, that vanishes. And irrotationality is a property of the electrostatic field specifically, not a universal law of electric fields — the induced field around a changing magnetic flux is genuinely rotational.

Worked examples

\[ \vec E=\frac{1}{4\pi\varepsilon_0}\frac{q\,\hat r}{r^2},\qquad \oint_C\vec E\cdot d\vec\ell \stackrel{?}{=}0 \]

Take a point charge \(q=+5.0\ \mathrm{nC}\) at the origin and verify zero circulation on a loop made of two radial legs (\(r=0.10\,\mathrm m\) to \(r=0.30\,\mathrm m\)) joined by two circular arcs. A

\[ \int_{\text{arc}}\vec E\cdot d\vec\ell = 0 \]

On each arc \(d\vec\ell\) is azimuthal while \(\vec E\) is radial, so \(\vec E\cdot d\vec\ell=0\). The arcs contribute nothing, whatever their radius. B

\[ \int_{0.10}^{0.30}\!E\,dr + \int_{0.30}^{0.10}\!E\,dr = 0 \]

The two radial legs cover the same \(r\)-range in opposite directions with the same integrand \(E(r)\), so they cancel. Each has magnitude \(\frac{q}{4\pi\varepsilon_0}\!\left(\frac{1}{0.10}-\frac{1}{0.30}\right)=(44.9)(10-3.33)=299\,\mathrm V\), opposite in sign. B

\[ \oint_C \vec E\cdot d\vec\ell = +299\ \mathrm V - 299\ \mathrm V = 0 \]

Reading. The circulation vanishes leg-by-leg, confirming \(\nabla\times\vec E=\vec0\) for the Coulomb field. The units are volts, as expected for a line integral of \(\vec E\).

\[ \vec E=(a y,\;a x,\;0),\qquad a=200\ \mathrm{V\,m^{-2}} \]

Test whether a proposed field is a legitimate electrostatic field by computing its curl. First a candidate that should pass. A

\[ (\nabla\times\vec E)_z=\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}=a-a=0 \]

Only the \(z\)-component can be nonzero here, and it evaluates to \(a-a=0\). The field is curl-free, with \(\vec E=-\nabla\varphi\) for \(\varphi=-a\,xy\). B

\[ \vec E'=(-b y,\;b x,\;0),\qquad (\nabla\times\vec E')_z=b-(-b)=2b \]

Now a superficially similar field. Its curl is \(2b\neq0\), so it is rotational and cannot be an electrostatic field — no scalar potential exists for it. B

\[ \nabla\times\vec E=\vec0\ \text{(valid)},\qquad \nabla\times\vec E'=2b\,\hat z\neq\vec0\ \text{(invalid)} \]

Reading. The curl test cleanly separates admissible electrostatic fields from inadmissible ones. With \(b=200\ \mathrm{V\,m^{-2}}\), \(\vec E'\) has curl \(400\ \mathrm{V\,m^{-2}}\); such a field would require a changing magnetic flux \(\partial_t B_z=-400\ \mathrm{V\,m^{-2}}\) and is not static.

Units check. \(\partial E/\partial x\) has \(\mathrm{(V\,m^{-1})/m}=\mathrm{V\,m^{-2}}\), the correct units for a curl of \(\vec E\).

Problems

Show directly, using \(\vec E=\dfrac{q}{4\pi\varepsilon_0}\dfrac{\vec r}{r^3}\) in Cartesian components, that \((\nabla\times\vec E)_z=0\).
Solution
With \(E_x=\frac{q}{4\pi\varepsilon_0}\frac{x}{r^3}\), \(E_y=\frac{q}{4\pi\varepsilon_0}\frac{y}{r^3}\) and \(r=(x^2+y^2+z^2)^{1/2}\): \(\partial E_y/\partial x = \frac{q}{4\pi\varepsilon_0}\,y\cdot(-3)r^{-4}\frac{x}{r}= -\frac{3q}{4\pi\varepsilon_0}\frac{xy}{r^5}\). By the same computation \(\partial E_x/\partial y = -\frac{3q}{4\pi\varepsilon_0}\frac{xy}{r^5}\), identical. Hence \((\nabla\times\vec E)_z=\partial E_y/\partial x-\partial E_x/\partial y=0\). The \(x\)- and \(y\)-components vanish by the same argument.
A field is given by \(\vec E=(c\,x^2,\;0,\;0)\) with \(c=50\ \mathrm{V\,m^{-3}}\). Is it a valid electrostatic field? If so, find \(\varphi\).
Solution
The only candidate curl components involve \(\partial E_x/\partial y=0\) and \(\partial E_x/\partial z=0\), with \(E_y=E_z=0\), so \(\nabla\times\vec E=\vec0\): valid. Then \(\varphi=-\int E_x\,dx=-\frac{c}{3}x^3+\text{const}\). Check: \(-\partial\varphi/\partial x=c x^2=E_x\). Taking \(\varphi(0)=0\), at \(x=0.20\,\mathrm m\), \(\varphi=-\frac{50}{3}(0.20)^3=-0.133\ \mathrm V\).
Compute \(\oint_C\vec E\cdot d\vec\ell\) for the uniform field \(\vec E=E_0\hat x\), \(E_0=1000\ \mathrm{V\,m^{-1}}\), around a square in the \(xy\)-plane of side \(0.50\,\mathrm m\).
Solution
Bottom leg (\(+x\), length \(0.50\)): \(+E_0(0.50)=+500\ \mathrm V\). Top leg (\(-x\)): \(-500\ \mathrm V\). The two legs along \(\hat y\) give \(\vec E\cdot d\vec\ell=0\). Total \(=500-500+0+0=0\), consistent with \(\nabla\times\vec E=\vec0\) for a constant field.
For \(\vec E=k(y\hat x+x\hat y)\), find the potential and evaluate the work per unit charge moving from \((0,0)\) to \((1,2)\,\mathrm m\), with \(k=30\ \mathrm{V\,m^{-2}}\).
Solution
Curl check: \((\nabla\times\vec E)_z=\partial E_y/\partial x-\partial E_x/\partial y=k-k=0\), curl-free. Potential: \(\varphi=-kxy+\text{const}\), since \(-\nabla(-kxy)=k(y,x,0)=\vec E\). Work per charge \(=\varphi(0,0)-\varphi(1,2)=0-(-k\cdot1\cdot2)=2k=60\ \mathrm V\). Path-independent by conservativeness.
A field \(\vec E=(\alpha y,\;\beta x,\;0)\) is claimed to be electrostatic. Find the relation between \(\alpha\) and \(\beta\) this requires, and state what physical situation arises if it is violated with \(\alpha=0\), \(\beta=150\ \mathrm{V\,m^{-2}}\).
Solution
\((\nabla\times\vec E)_z=\partial E_y/\partial x-\partial E_x/\partial y=\beta-\alpha\). Electrostatics requires \(\beta=\alpha\). With \(\alpha=0,\ \beta=150\) the curl is \(150\ \mathrm{V\,m^{-2}}\,\hat z\neq\vec0\), so no scalar potential exists. By Faraday's law this can occur only if \(\partial B_z/\partial t=-150\ \mathrm{V\,m^{-2}}\): a magnetic field decreasing in time drives an induced, rotational \(\vec E\) — an electrodynamic, not electrostatic, situation.