Gauss’s law in differential form: the divergence of the electric field at a point equals what multiple of the local volume charge density?

Introduction / Context:
Gauss’s law is one of Maxwell’s equations and comes in integral and differential forms. The differential form links the divergence of the electric field to the local charge density, providing a point-wise relationship in space that is crucial in field theory and electrostatics.

Given Data / Assumptions:

• We are in linear, homogeneous, isotropic medium with vacuum permittivity ε0 (or in free space).
• We are asked for the mathematical proportionality between ∇·E and ρv.
• Standard SI units are used.

Concept / Approach:

Gauss’s law (differential form) states: ∇·E = ρv/ε0. This is directly derived from the integral form ∮ E·dA = Q_enclosed/ε0 by applying the divergence theorem. It applies pointwise to relate local field divergence to the local charge density.

Step-by-Step Solution:

Verification / Alternative check:

In materials, D = εE and Gauss’s law may be written ∇·D = ρv (free). In free space where D = ε0 E, dividing both sides by ε0 again yields ∇·E = ρv/ε0.

Why Other Options Are Wrong:

• 1/(4π ε0): appears in Coulomb’s law, not in Gauss’s law differential form.
• ε0 times ρv: inverted proportionality.
• 'the volume charge density' without the 1/ε0 factor is incomplete.
• 'zero everywhere': only true in charge-free regions (ρv = 0), not in general.

Common Pitfalls:

• Confusing the roles of E, D, and ε0.
• Mixing constants from Coulomb’s law with those from Maxwell’s equations in SI form.

Final Answer:

1/ε0 times the volume charge density at that point