Electrostatic Potential and Irrotational E

Derivation

Electrostatic Potential and Irrotational E

D-285 Home PU-305 Threads fields · energy Depends on Gauss's Law from Coulomb's Law

Statement

For a time-independent (electrostatic) field the curl vanishes everywhere, \( \nabla \times \mathbf{E} = \mathbf{0} \). On any simply connected region this permits the introduction of a single-valued scalar potential \( \phi \) with \( \mathbf{E} = -\nabla \phi \). Combined with Gauss's law \( \nabla \cdot \mathbf{E} = \rho/\varepsilon_0 \) this collapses the two first-order Maxwell equations of electrostatics into one second-order equation, the Poisson equation \( \nabla^2 \phi = -\rho/\varepsilon_0 \), whose solution is unique once \( \phi \) (or \( \partial\phi/\partial n \)) is prescribed on the boundary of the region.

Why it matters

The vector field \( \mathbf{E}(\mathbf{r}) \) has three components at every point; the scalar field \( \phi(\mathbf{r}) \) has one. The curl-free property is exactly the integrability condition that lets us trade three coupled functions for a single one, so every problem in electrostatics becomes a problem of finding one scalar field subject to one linear partial differential equation.

This is the structural foundation of the entire subject. Boundary-value techniques (separation of variables, images, Green's functions, relaxation), the definition of capacitance, the energy stored in a field, and the uniqueness that guarantees a computed answer is the answer all rest on \( \nabla \times \mathbf{E} = \mathbf{0} \) and the Poisson equation it produces.

Assumptions

Static sources.All charges are at rest and densities are time-independent, so \( \partial \mathbf{B}/\partial t = \mathbf{0} \); if dropped, Faraday's law restores \( \nabla\times\mathbf{E} = -\partial\mathbf{B}/\partial t \neq \mathbf{0} \) and no single-valued scalar potential exists.

Gauss's law holds.We take \( \nabla\cdot\mathbf{E} = \rho/\varepsilon_0 \) as the prior result (from Coulomb's law); if dropped there is no source equation to convert into Poisson's equation.

Simply connected domain.The region has no topological holes threaded by anything singular, so a curl-free field is globally a gradient; if dropped (e.g. an excluded line), \( \phi \) can become multivalued around a loop even though \( \nabla\times\mathbf{E}=\mathbf{0} \) locally.

Linear vacuum-like medium.\( \varepsilon_0 \) is a constant scalar; if the medium is inhomogeneous or anisotropic one must instead write \( \nabla\cdot(\varepsilon\,\mathbf{E})=\rho_{\text{free}} \) and the clean Poisson form is modified.

Derivation

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

Coulomb field of a point charge; the prior result we build on. A

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

Direct differentiation: the gradient of \( 1/r \) with respect to the field point \( \mathbf{r} \). A

\[ \nabla\times\mathbf{E} = -\frac{q}{4\pi\varepsilon_0}\,\nabla\times\nabla\!\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right) = \mathbf{0} \]

The curl of any gradient vanishes identically, \( \nabla\times\nabla f = \mathbf{0} \), holding for \( \mathbf{r}\neq\mathbf{r}' \). B

\[ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}')\,(\mathbf{r}-\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|^{3}}\,d^3r' \quad\Rightarrow\quad \nabla\times\mathbf{E} = \mathbf{0} \]

Superposition: an arbitrary static distribution is a sum of point charges, and curl is linear, so the whole field is curl-free. B

\[ \oint_{\partial S} \mathbf{E}\cdot d\boldsymbol{\ell} = \int_S (\nabla\times\mathbf{E})\cdot d\mathbf{a} = 0 \]

Stokes's theorem: curl-free is equivalent to path-independent line integrals on a simply connected domain, the integrability condition. C

\[ \phi(\mathbf{r}) \equiv -\int_{\mathbf{r}_0}^{\mathbf{r}} \mathbf{E}\cdot d\boldsymbol{\ell} \quad\Rightarrow\quad \mathbf{E} = -\nabla\phi \]

Path-independence makes this integral a well-defined single-valued function; the Poincaré lemma guarantees a potential exists. C

\[ \nabla\cdot\mathbf{E} = \frac{\rho}{\varepsilon_0} \]

Gauss's law, the assumed prior result, is the second ingredient. A

\[ \nabla\cdot(-\nabla\phi) = \frac{\rho}{\varepsilon_0} \quad\Longrightarrow\quad \nabla^2\phi = -\frac{\rho}{\varepsilon_0} \]

Substitute \( \mathbf{E}=-\nabla\phi \) into Gauss's law and use \( \nabla\cdot\nabla=\nabla^2 \). This is Poisson's equation. B

\[ \phi = \phi_1 - \phi_2,\quad \nabla^2\phi = 0 \ \text{in } V,\quad \phi = 0 \text{ or } \tfrac{\partial\phi}{\partial n}=0 \text{ on } \partial V \]

Assume two solutions with identical boundary data; their difference solves Laplace's equation with zero data. C

\[ \int_V |\nabla\phi|^2\,d^3r = \oint_{\partial V}\phi\,\nabla\phi\cdot d\mathbf{a} - \int_V \phi\,\nabla^2\phi\,d^3r = 0 \]

Green's first identity: both right-hand terms vanish under the boundary conditions, forcing \( \nabla\phi=\mathbf{0} \), hence \( \phi_1=\phi_2 \) (up to an additive constant). Uniqueness. C

Result

\[ \nabla\times\mathbf{E}=\mathbf{0}\ \Longleftrightarrow\ \mathbf{E}=-\nabla\phi,\qquad \nabla^2\phi=-\frac{\rho}{\varepsilon_0} \]

Reading. A static electric field is irrotational, so it is the gradient of a single scalar potential. Feeding that potential into Gauss's law yields Poisson's equation, a single linear second-order PDE for \( \phi \). Given \( \phi \) or its normal derivative on the boundary, the solution is unique: electrostatics is completely determined by charge in the volume plus data on the surface.

Units check. \( \phi \) is in volts (\( \mathrm{V} = \mathrm{J\,C^{-1}} \)), so \( \nabla\phi \) is \( \mathrm{V\,m^{-1}} \), matching \( \mathbf{E} \). In Poisson's equation \( \nabla^2\phi \) has units \( \mathrm{V\,m^{-2}} \); the right side is \( \rho/\varepsilon_0 = (\mathrm{C\,m^{-3}})/(\mathrm{C^2\,N^{-1}\,m^{-2}}) = \mathrm{N\,C^{-1}\,m^{-1}} = \mathrm{V\,m^{-2}} \). Consistent.

Limiting cases

Charge-free region: \( \rho=0 \) gives Laplace's equation \( \nabla^2\phi=0 \); \( \phi \) has no interior extrema (mean-value property), so field maxima sit on boundaries.
Point charge: \( \nabla^2\phi = -q\,\delta^3(\mathbf{r})/\varepsilon_0 \) is solved by \( \phi = q/(4\pi\varepsilon_0 r) \), recovering the Coulomb potential.
Uniform field: \( \phi = -\mathbf{E}_0\cdot\mathbf{r} \) satisfies \( \nabla^2\phi=0 \), giving constant \( \mathbf{E}=\mathbf{E}_0 \).
Slowly varying fields (quasistatics): as long as \( \partial\mathbf{B}/\partial t \) is negligible compared with the terms retained, the curl-free approximation and \( \phi \) remain usable.

Breaks when

Time-varying magnetic flux. With \( \partial\mathbf{B}/\partial t\neq\mathbf{0} \), Faraday's law gives \( \nabla\times\mathbf{E}=-\partial\mathbf{B}/\partial t\neq\mathbf{0} \). No single-valued \( \phi \) exists; one needs \( \mathbf{E}=-\nabla\phi-\partial\mathbf{A}/\partial t \).
Non-simply-connected domains. Around a hole (e.g. a region excluding a wire carrying changing flux) \( \nabla\times\mathbf{E}=\mathbf{0} \) can hold locally yet \( \oint\mathbf{E}\cdot d\boldsymbol{\ell}\neq 0 \), so \( \phi \) becomes multivalued and the potential description fails globally.
Radiation and full electrodynamics. For accelerating charges the transverse (radiation) field is not curl-free; the scalar potential alone cannot represent \( \mathbf{E} \).

Failure modes

Sign slip. Writing \( \mathbf{E}=+\nabla\phi \). The minus sign encodes that \( \mathbf{E} \) points from high to low potential; dropping it reverses every field.
Confusing Poisson and Laplace. Using \( \nabla^2\phi=0 \) inside a charged region. Laplace applies only where \( \rho=0 \).
Sign in Poisson's equation. Writing \( \nabla^2\phi=+\rho/\varepsilon_0 \). The correct sign follows from \( \nabla\cdot(-\nabla\phi) \).
Assuming a potential always exists. Invoking \( \phi \) for induced (curly) fields, or across a multiply connected loop, where it is not single-valued.
Forgetting the gauge constant. Treating the absolute value of \( \phi \) as physical; only potential differences and \( \nabla\phi \) are measurable.
Over-specifying boundary data. Imposing both \( \phi \) and \( \partial\phi/\partial n \) on the whole boundary, generally over-determining the problem and admitting no solution.

Discussion

The content of \( \nabla\times\mathbf{E}=\mathbf{0} \) is conservation of energy for the electrostatic force. Because \( \oint\mathbf{E}\cdot d\boldsymbol{\ell}=0 \), moving a test charge around any closed loop returns it with no net work done, so a potential energy \( U=q\phi \) can be assigned to each position. The scalar potential is not a mathematical convenience layered on top of the field; it is the statement that the electrostatic force is conservative.

The reduction to a single PDE is what makes electrostatics computationally tractable. Poisson's equation is linear, so superposition holds at the level of potentials, and the uniqueness theorem licenses the method of images and every other guess-and-verify technique: if a candidate \( \phi \) satisfies Poisson's equation in the volume and matches the boundary data, it is guaranteed to be the solution, however it was found. This is why physically absurd image charges outside the region of interest give exactly correct answers inside it.

The deeper structure is topological and appears throughout physics. Curl-free (closed) fields are locally gradients (exact) by the Poincaré lemma, but the promotion from local to global requires the domain's first de Rham cohomology to be trivial, i.e. simple connectivity. The gap between "locally a gradient" and "globally a gradient" is precisely what a nonzero loop integral measures; the same mathematics underlies the Aharonov-Bohm effect for the vector potential and the classification of conservative versus non-conservative fields on manifolds with holes.

Common misconceptions. A vanishing curl does not mean the field is zero or uniform, only that it has no rotational part. And \( \phi \) is defined only up to an additive constant, fixed by a reference point (often infinity); its absolute value carries no physical meaning while its gradient and differences do.

Worked examples

Verify that \( \phi=\dfrac{q}{4\pi\varepsilon_0 r} \) gives the Coulomb field and is curl-free.

Take the gradient in spherical symmetry, \( \nabla\phi = \frac{d\phi}{dr}\hat{r} \). A

\[ \mathbf{E} = -\nabla\phi = -\frac{q}{4\pi\varepsilon_0}\frac{d}{dr}\!\left(\frac{1}{r}\right)\hat{r} = \frac{q}{4\pi\varepsilon_0 r^2}\hat{r} \]

Since \( \mathbf{E} \) is a pure gradient, \( \nabla\times\mathbf{E}=-\nabla\times\nabla\phi=\mathbf{0} \) automatically. B

Numbers: \( q=1.0\times10^{-9}\,\mathrm{C} \), \( r=0.10\,\mathrm{m} \), \( \varepsilon_0=8.85\times10^{-12}\,\mathrm{C^2 N^{-1}m^{-2}} \).

Evaluate \( \phi \) then \( E \). A

\[ \phi = \frac{1.0\times10^{-9}}{4\pi(8.85\times10^{-12})(0.10)} = 89.9\ \mathrm{V},\qquad E = \frac{\phi}{r} = 899\ \mathrm{V\,m^{-1}} \]

For \( 1/r \), \( E=\phi/r \) exactly. A

\[ \phi \approx 90\ \mathrm{V},\qquad E \approx 9.0\times10^{2}\ \mathrm{V\,m^{-1}}\ (\hat{r}) \]

Reading. The curl-free Coulomb potential reproduces the inverse-square field and points radially outward for positive \( q \).

Units check. \( \mathrm{C}/(\mathrm{C^2N^{-1}m^{-2}\cdot m}) = \mathrm{N\,m\,C^{-1}} = \mathrm{V} \); dividing by \( r \) gives \( \mathrm{V\,m^{-1}} \).

Parallel-plate gap, \( 0\le x\le d \), no charge between plates: solve \( \nabla^2\phi=0 \) with \( \phi(0)=0 \), \( \phi(d)=V_0 \).

Charge-free region, so Laplace's equation applies; symmetry makes \( \phi=\phi(x) \). B

\[ \frac{d^2\phi}{dx^2}=0 \ \Rightarrow\ \phi(x)=Ax+B \]

General solution of the one-dimensional Laplace equation. A

\[ B=0,\quad A=\frac{V_0}{d}\ \Rightarrow\ \phi(x)=\frac{V_0}{d}\,x,\quad \mathbf{E}=-\frac{d\phi}{dx}\hat{x}=-\frac{V_0}{d}\hat{x} \]

Apply the two boundary values; uniqueness theorem guarantees this is the only solution. B

Numbers: \( V_0=12\,\mathrm{V} \), \( d=2.0\times10^{-3}\,\mathrm{m} \).

Evaluate the uniform field magnitude. A

\[ E=\frac{V_0}{d}=\frac{12}{2.0\times10^{-3}}=6.0\times10^{3}\ \mathrm{V\,m^{-1}} \]

Magnitude of the constant field; direction is from the high-potential plate to the low. A

\[ \phi(x)=\frac{V_0}{d}x,\qquad E=6.0\times10^{3}\ \mathrm{V\,m^{-1}} \]

Reading. A source-free gap gives a linear potential and a uniform field; the uniqueness theorem certifies this simple answer is complete.

Units check. \( V_0/d = \mathrm{V/m} \), the correct unit for \( E \).

Problems

Show explicitly that \( \nabla\times(\nabla\phi)=\mathbf{0} \) for any twice-differentiable \( \phi \).
Solution
The \( z \)-component is \( \partial_x(\partial_y\phi)-\partial_y(\partial_x\phi) \). For \( \phi\in C^2 \), mixed partials commute (Clairaut/Schwarz), so \( \partial_x\partial_y\phi=\partial_y\partial_x\phi \) and the component vanishes; the same holds for the \( x \) and \( y \) components by cyclic permutation. Hence \( \nabla\times\nabla\phi=\mathbf{0} \) identically.
A field is claimed to be \( \mathbf{E}=(y,\,-x,\,0)\ \mathrm{V\,m^{-1}} \). Can it be electrostatic?
Solution
\( \nabla\times\mathbf{E} = (\partial_x(-x)-\partial_y(y))\hat{z}+\dots = (-1-1)\hat{z} = -2\hat{z}\neq\mathbf{0} \). Since a static field must be curl-free, this cannot be an electrostatic field. (It is a rotational, induced-type field.)
Given \( \phi(x,y,z)=k(x^2-y^2)\ \mathrm{V} \) with \( k=50\ \mathrm{V\,m^{-2}} \), find \( \mathbf{E} \) and verify it satisfies Laplace's equation. Evaluate \( \mathbf{E} \) at \( (1,1,0)\,\mathrm{m} \).
Solution
\( \mathbf{E}=-\nabla\phi = (-2kx,\,2ky,\,0) \). Then \( \nabla^2\phi = 2k-2k+0 = 0 \), so it is a valid source-free potential. At \( (1,1,0) \): \( \mathbf{E}=(-2(50)(1),\,2(50)(1),\,0)=(-100,\,100,\,0)\ \mathrm{V\,m^{-1}} \), magnitude \( 100\sqrt{2}\approx141\ \mathrm{V\,m^{-1}} \).
A sphere of radius \( R=0.050\,\mathrm{m} \) carries uniform volume charge density \( \rho=1.0\times10^{-6}\,\mathrm{C\,m^{-3}} \). Using Poisson's equation, find \( \phi(r) \) inside (choose \( \phi(R) \) to match the exterior Coulomb potential).
Solution
Inside, \( \nabla^2\phi=-\rho/\varepsilon_0 \). Spherically, \( \frac{1}{r^2}\frac{d}{dr}(r^2\frac{d\phi}{dr})=-\rho/\varepsilon_0 \), giving \( \phi(r)=-\frac{\rho r^2}{6\varepsilon_0}+C \) (regular at \( r=0 \)). Total charge \( Q=\frac{4}{3}\pi R^3\rho \). Exterior potential at \( R \) is \( \phi(R)=\frac{Q}{4\pi\varepsilon_0 R}=\frac{\rho R^2}{3\varepsilon_0} \). Matching: \( C=\frac{\rho R^2}{3\varepsilon_0}+\frac{\rho R^2}{6\varepsilon_0}=\frac{\rho R^2}{2\varepsilon_0} \), so \( \phi(r)=\frac{\rho}{6\varepsilon_0}(3R^2-r^2) \). Numerically \( \frac{\rho R^2}{2\varepsilon_0}=\frac{(1.0\times10^{-6})(0.050)^2}{2(8.85\times10^{-12})}=141\ \mathrm{V} \) at the centre.
State the uniqueness theorem and use it to justify why the method of images (a point charge \( q \) above a grounded plane, replaced by an image \( -q \)) gives the correct potential in the upper half-space.
Solution
Uniqueness: a solution of Poisson's equation in a volume \( V \) is determined uniquely by \( \rho \) in \( V \) together with \( \phi \) (Dirichlet) or \( \partial\phi/\partial n \) (Neumann) on \( \partial V \). For the grounded plane, the boundary data are \( \phi\to0 \) at infinity and \( \phi=0 \) on the plane. The image configuration (\( q \) at \( +d \), \( -q \) at \( -d \)) reproduces exactly these boundary values and has the correct source \( q \) in the upper half-space (the image lies outside \( V \), so it does not alter \( \rho \) there). By uniqueness, its potential equals the true potential everywhere in the upper half-space, even though the physical lower region contains induced surface charge rather than a point image.