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PU-102 / L04 · lecture

The Divergence Theorem and Local Gauss's Law

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BEFORE YOU START

You should be comfortable with the integral form of Gauss's law and the concept of electric flux. Familiarity with basic vector calculus operations, specifically divergence and surface integrals, is assumed.

Why this matters

Moving from global flux to local source density reveals the true nature of electric fields.

While the integral form of Gauss's law is powerful for symmetric charge distributions, it treats the electric field as a bulk property over a closed surface. The differential form, derived via the Divergence Theorem, provides a point-by-point description of how charge density generates the electric field. This local perspective is essential for solving complex electrostatic problems where symmetry is absent and for understanding the fundamental relationship between sources and fields in classical electrodynamics.

From Global to Local: The Intuitive Leap

Imagine a river flowing through a landscape. The integral form of Gauss's law is like measuring the total amount of water flowing out of a large, sealed container. It tells you the net flow across the boundary, which depends on the total amount of water source inside. However, it does not tell you where the water is coming from within the container. The differential form is like placing a tiny sensor at every point in the river to measure the local rate at which water is diverging from that specific spot. If the sensor reads zero, water is flowing through that point but not originating there. If it reads a positive value, water is being created at that point. In electrostatics, the "water" is the electric field, and the "source" is the electric charge. The Divergence Theorem is the mathematical bridge that allows us to translate the total flow measurement into these local source measurements.

The Divergence Theorem: Mathematical Bridge

The Divergence Theorem (also known as Gauss's Theorem) states that the volume integral of the divergence of a vector field over a region \( V \) is equal to the surface integral of the vector field over the boundary surface \( \partial V \). Mathematically, for a continuously differentiable vector field \( \mathbf{F} \):

\[ \int_V (\nabla \cdot \mathbf{F}) \, dV = \oint_{\partial V} \mathbf{F} \cdot d\mathbf{A} \]

This theorem is the key to converting global statements into local ones. It asserts that the sum of all the local "sources" (divergence) inside a volume equals the total "flow" (flux) crossing the boundary. In the context of electrostatics, we apply this to the electric field \( \mathbf{E} \). The left-hand side represents the total divergence of the electric field within the volume, while the right-hand side is the electric flux through the surface enclosing that volume.

Deriving the Local Gauss's Law

We begin with the integral form of Gauss's law, which states that the net electric flux through any closed surface is proportional to the enclosed charge \( Q_{\text{enc}} \):

\[ \oint_{\partial V} \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]

To express this locally, we must rewrite both sides in terms of volume integrals. First, we apply the Divergence Theorem to the left-hand side. Assuming \( \mathbf{E} \) is sufficiently smooth, we have:

\[ \oint_{\partial V} \mathbf{E} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{E}) \, dV \]

Next, we express the enclosed charge \( Q_{\text{enc}} \) on the right-hand side. The total charge is the integral of the volume charge density \( \rho \) over the volume \( V \):

\[ Q_{\text{enc}} = \int_V \rho \, dV \]

Substituting these volume integrals back into Gauss's law gives:

\[ \int_V (\nabla \cdot \mathbf{E}) \, dV = \frac{1}{\varepsilon_0} \int_V \rho \, dV \]

Since this equation holds for any arbitrary volume \( V \), the integrands must be equal at every point within the volume. Therefore, we arrive at the differential form of Gauss's law:

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

This equation states that the divergence of the electric field at any point is directly proportional to the charge density at that same point. It is a local law, valid pointwise wherever the fields are differentiable.

Interpreting Divergence: Sources and Sinks

The physical interpretation of \( \nabla \cdot \mathbf{E} \) is crucial. A positive divergence indicates a net outward flux from an infinitesimal volume, implying the presence of positive charge (a source). A negative divergence indicates a net inward flux, implying the presence of negative charge (a sink). If the divergence is zero, the net flux through an infinitesimal volume is zero; field lines pass through the volume but do not originate or terminate there. This is consistent with the fact that electric field lines begin on positive charges and end on negative charges. In regions where \( \rho = 0 \), \( \nabla \cdot \mathbf{E} = 0 \), meaning the electric field is solenoidal (divergence-free) locally, even though the field itself may be non-zero. This distinction between the field value and its divergence is a common source of confusion for students transitioning from integral to differential forms.

The derivation

◆ DERIVATION EMBED · D-041 · not re-derived here

Differential Gauss's Law via the Divergence Theorem

From the integral form of Gauss's law, \(\oint_{\partial V}\mathbf{E}\cdot d\mathbf{A}=\dfrac{Q_{\text{enc}}}{\varepsilon_0}\), valid for every closed surface, the divergence theorem yields the local (differential) law \(\nabla\cdot\mathbf{E}=\dfrac{\rho}{\varepsilon_0}\), holding pointwise wherever \(\mathbf{E}\) is continuously differentiable and \(\rho\) is the volume charge density.

\[ \boxed{\ \nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_0}\ } \]
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Tier 3 ยท watch out

The differential form \( \nabla \cdot \mathbf{E} = \rho/\varepsilon_0 \) is strictly valid only where the electric field is continuously differentiable. At locations with point charges or surface charge densities, the field is singular or discontinuous, and the divergence is not defined in the classical sense. In such cases, one must use the integral form or introduce the Dirac delta function to represent the charge density. Applying the differential form directly at a point charge location is mathematically invalid and leads to undefined results. C

Worked example

Problem. A uniform volume charge density \( \rho = 5.0 \times 10^{-9} \ \mathrm{C/m^3} \) is distributed throughout a spherical region of radius \( R = 0.10 \ \mathrm{m} \). Calculate the divergence of the electric field at the center of the sphere and at a point just outside the surface. Use \( \varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m} \).

  1. Identify the charge density at the center. Since the charge is uniformly distributed, \( \rho_{\text{center}} = 5.0 \times 10^{-9} \ \mathrm{C/m^3} \).
  2. Apply the differential Gauss's law at the center: \( \nabla \cdot \mathbf{E} = \rho / \varepsilon_0 \).
  3. Calculate the value: \( \nabla \cdot \mathbf{E} = \frac{5.0 \times 10^{-9}}{8.85 \times 10^{-12}} \approx 565\ \mathrm{V/m^2} \).
  4. Identify the charge density just outside the surface. Outside the charged sphere, the volume charge density is zero, so \( \rho_{\text{outside}} = 0 \).
  5. Apply the differential Gauss's law outside: \( \nabla \cdot \mathbf{E} = 0 / \varepsilon_0 = 0 \).
\[ \nabla \cdot \mathbf{E}_{\text{center}} = 5.65 \times 10^2 \ \mathrm{V/m^2}, \quad \nabla \cdot \mathbf{E}_{\text{outside}} = 0 \]

Units check. The unit of \( \rho \) is \( \mathrm{C/m^3} \). The unit of \( \varepsilon_0 \) is \( \mathrm{C/(V \cdot m)} \). Thus, \( \rho/\varepsilon_0 \) has units \( \frac{\mathrm{C/m^3}}{\mathrm{C/(V \cdot m)}} = \frac{\mathrm{V \cdot m}}{\mathrm{m^3}} = \mathrm{V/m^2} \). Since \( \mathbf{E} \) is in \( \mathrm{V/m} \), its divergence \( \nabla \cdot \mathbf{E} \) is in \( \mathrm{V/m^2} \). The units are consistent.

CHECK YOURSELF

Three quick tests

  1. What is the physical meaning of a positive divergence of the electric field?
    AnswerIt indicates the presence of positive charge density, acting as a source of electric field lines.
  2. If \( \nabla \cdot \mathbf{E} = 0 \) in a region, does this imply \( \mathbf{E} = 0 \)?
    AnswerNo. It implies that there is no net charge density in that region, but field lines can still pass through it.
  3. Why can we equate the integrands in the derivation of differential Gauss's law?
    AnswerBecause the integral equation holds for any arbitrary volume \( V \), the integrands must be equal at every point within the volume.
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-041 · Threads: force, energy, fields