Electrostatics. Flux and Gauss's law

We define the flux of a field through some surface as:

$$\Phi ={\int }_S\text{F}\cdot d\text{a}$$

The rigorous way to compute is by parametrizing the surface, and taking the dot product with the field as described in Chapter 7.3 - Parametrized Surfaces & Chapter 7.6 - Vector Field Integral over a Surface.

It is not hard to show that the flux of a single charge through a sphere centered at it is $\Phi =q/{ϵ}_0$. However, we conjecture even stronger arguments - the result is independent of a surface and its size. One way to prove it is to show that $\nabla \cdot \text{E}=0$. 1) Intuitively recall what is divergence and how we can use this result to “modify” a sphere.

Gauss’s law

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Now, let’s consider bunch of charges. Their field is the sum of individual fields, or superposition principle (Electrostatics. Charges, Energy and Fields). Hence,

$$\Phi ={\int }_S\text{E}\cdot d\text{a}={\int }_S({\text{E}}_1+{\text{E}}_2+{\text{E}}_3+...)\cdot d\text{a}=\sum _i^N\frac{q_i}{{ϵ}_0}$$

Note that the proof (the last equality) hinges on the fact that the Electric field is inverse square law. So, it holds for every inverse square laws (gravity). As a reminder recall the fields of a flat sheet, thin wire, and uniform sphere.

Energy stored in a field

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Energy stored in a field is given by:

$$U=\frac{{ϵ}_0}{2}\int E^{2}dv$$

Problem

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It would be a fine exercise to calculate the energy of a uniformly charged sphere ($\rho$) from scratch. Problems solved about Electrostatics (Purcell Ch. 1)

Answers to questions

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1. Zero divergence means that the net flux through a given surface is zero. So, we may attach any weirdly shaped object to a sphere, and this would not change the Flux.