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PU-102 / L06 · lecture

Curl of E and the Conservative Field

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the definition of the curl operator \(\nabla \times \vec{F}\) and the fundamental theorem of line integrals. Familiarity with vector calculus identities, specifically that the curl of a gradient is zero, is essential for following the logical flow.

Why this matters

This property allows us to replace the complex vector field \(\vec{E}\) with a simpler scalar potential \(\varphi\).

In the previous lecture, we used Gauss's Law to determine the electric field for symmetric charge distributions. However, solving for \(\vec{E}\) directly involves vector components and can be cumbersome. By proving that the electrostatic field is conservative, we unlock a more powerful mathematical tool: the scalar potential. This transition from vector to scalar calculus simplifies the solution of boundary value problems and lays the groundwork for the potential formulation of electrostatics, which is central to the rest of electrostatics.

The Conservative Nature of Electrostatics

Imagine moving a test charge through an electric field. If the work done by the field depends only on the starting and ending positions, and not on the path taken, the field is called conservative. In electrostatics, the forces arise from stationary charges. Since the source configuration does not change with time, the energy landscape is static. This implies that you cannot extract net work from the field by moving a charge around a closed loop. If you could, you would have created a perpetual motion machine, violating the conservation of energy. The mathematical statement of this physical intuition is that the line integral of the electric field around any closed path is zero. This is the defining characteristic of a conservative field, and it is the key result that allows us to define a potential energy function per unit charge, known as the electric potential.

Mathematical Formulation: Zero Curl

The condition that the line integral around any closed loop vanishes is locally equivalent to the curl of the field being zero. The curl, \(\nabla \times \vec{E}\), measures the local "circulation" or rotation of the field. If the field has no local rotation, it is irrotational. For any static charge distribution, the electrostatic field satisfies this condition everywhere except at the source points where the field is singular. This result is derived from Coulomb's Law and the properties of the gradient operator. Specifically, since the electric field from a point charge is proportional to the gradient of \(1/r\), and the curl of a gradient is identically zero, the superposition of such fields also has zero curl. This establishes the equivalence between the integral form (zero circulation) and the differential form (zero curl).

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

This equivalence is a direct consequence of Stokes' Theorem, which relates the line integral of a vector field around a closed curve to the surface integral of its curl over any surface bounded by that curve. If the curl is zero everywhere on the surface, the surface integral vanishes, and thus the line integral must also vanish.

Introduction of the Scalar Potential

Because the electrostatic field is conservative, it can be expressed as the negative gradient of a scalar function \(\varphi(\vec{r})\). This scalar function is the electric potential. The negative sign is a convention chosen so that the potential decreases in the direction of the electric field, consistent with the idea that positive charges naturally move from high potential to low potential. The relationship is given by:

\[ \vec{E} = -\nabla\varphi \]

This equation is fundamental. It reduces the problem of finding three vector components of \(\vec{E}\) to finding a single scalar function \(\varphi\). The potential \(\varphi\) is defined up to an arbitrary additive constant, which corresponds to the choice of the zero level of potential energy. In practice, we often set \(\varphi = 0\) at infinity for localized charge distributions. The gradient operator \(\nabla\) points in the direction of the steepest increase of the scalar function, so the negative gradient points in the direction of the steepest decrease, which is the direction of the force on a positive test charge.

Path Independence and Potential Difference

The most practical consequence of the conservative nature of \(\vec{E}\) is the path independence of the work done. The potential difference between two points \(A\) and \(B\) is defined as the negative of the line integral of the electric field from \(A\) to \(B\):

\[ V_B - V_A = -\int_A^B \vec{E}\cdot d\vec{\ell} \]

Since the field is conservative, this integral depends only on the endpoints \(A\) and \(B\), not on the specific path chosen. This allows us to calculate potential differences using any convenient path, even if the field is not uniform or if the path is complex. For example, in a region with a uniform electric field \(\vec{E}\), the potential difference between two points separated by a displacement \(\Delta \vec{r}\) is simply \(\Delta V = -\vec{E} \cdot \Delta \vec{r}\). This linear relationship is a direct result of the constant field and the conservative property. It also implies that equipotential surfaces, where \(\varphi\) is constant, are always perpendicular to the electric field lines, as the gradient is normal to level surfaces.

The derivation

◆ DERIVATION EMBED · D-042 · not re-derived here

Irrotationality of the Electrostatic Field

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.

\[ \nabla \times \vec{E} = \vec{0} \quad\Longleftrightarrow\quad \vec{E} = -\nabla\varphi \quad\Longleftrightarrow\quad \oint_C \vec{E}\cdot d\vec{\ell}=0 \]
Open full derivation →
Tier 3 ยท watch out

The condition \(\nabla \times \vec{E} = \vec{0}\) holds strictly only for electrostatic fields. If the magnetic field changes with time, Faraday's Law of Induction states that \(\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}\). In such cases, the electric field is no longer conservative, the line integral around a closed loop is non-zero, and the concept of a single-valued scalar potential \(\varphi\) alone is insufficient to describe the field. One must introduce a vector potential \(\vec{A}\) as well. Do not apply the conservative field results to time-varying magnetic fields without accounting for the induced electric field component. C

Worked example

Problem. A uniform electric field \(\vec{E} = 50\ \mathrm{V/m}\) points in the \(+x\)-direction. Calculate the potential difference \(V_B - V_A\) between point \(A\) at \((0, 0, 0)\) and point \(B\) at \((2\ \mathrm{m}, 3\ \mathrm{m}, 0)\). Verify that the result is independent of the path by calculating the integral along a path that first moves in the \(y\)-direction and then in the \(x\)-direction.

  1. Identify the electric field vector: \(\vec{E} = 50\ \hat{i}\ \mathrm{V/m}\).
  2. Calculate the displacement vector \(\Delta \vec{r} = \vec{r}_B - \vec{r}_A = (2\ \mathrm{m})\hat{i} + (3\ \mathrm{m})\hat{j}\).
  3. Compute the dot product \(\vec{E} \cdot \Delta \vec{r} = (50\ \mathrm{V/m})(2\ \mathrm{m}) + (0)(3\ \mathrm{m}) = 100\ \mathrm{V}\).
  4. Apply the potential difference formula: \(V_B - V_A = -\vec{E} \cdot \Delta \vec{r} = -100\ \mathrm{V}\).
  5. Verify path independence: Path 1 (along \(y\) then \(x\)). Segment 1: \(d\vec{\ell} = dy\ \hat{j}\), \(\vec{E} \cdot d\vec{\ell} = 0\). Integral is 0. Segment 2: \(d\vec{\ell} = dx\ \hat{i}\), \(\vec{E} \cdot d\vec{\ell} = 50\ dx\). Integral is \(\int_0^2 50\ dx = 100\ \mathrm{V}\). Total integral is \(100\ \mathrm{V}\). Potential difference is \(-100\ \mathrm{V}\). The results match.
\[ V_B - V_A = -100\ \mathrm{V} \]

Units check. The electric field is in \(\mathrm{V/m}\) and distance is in \(\mathrm{m}\). The product \(\mathrm{V/m} \times \mathrm{m}\) yields \(\mathrm{V}\), which is the correct unit for potential difference.

CHECK YOURSELF

Three quick tests

  1. If \(\nabla \times \vec{E} = \vec{0}\), is the line integral of \(\vec{E}\) around a closed loop zero?
    AnswerYes, by Stokes' Theorem, the line integral equals the surface integral of the curl, which is zero.
  2. Can the electric potential be uniquely determined from the electric field?
    AnswerNo, it is determined up to an arbitrary additive constant. Only potential differences are physically meaningful.
  3. Does the conservative nature of \(\vec{E}\) hold for a time-varying magnetic field?
    AnswerNo, Faraday's Law introduces a non-zero curl, making the field non-conservative.
PROBLEMS

Problem set for PU-102 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-042 · Threads: force, energy, fields