##### Gauss's Law: the divergence of \({\bf E}\)
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###### Field Lines, Flux and Gauss's Law

We have seen that the curl of the electrostatic field vanishes. In view of the Helmhotz Theorem, this is but one (actually: 3, since it's a vector equation) of the differential equations we need to fix \({\bf E}\). We also need the divergence of \({\bf E}\).

Feynman's 'bullets flying out'. 'Conservation' related to \(1/r^2\) (think of volume section in space with walls radially oriented. Flux.

If there is no charge within a volume \({\cal V}\) encompassed within the closed surface \({\cal S}\),

\[ \oint {\bf E} \cdot d{\bf a} = 0. \]

However, if there is a charge inside, we don't get zero. Considering a little sphere of radius \(r\) around the charge,

\[ \oint {\bf E} \cdot d{\bf a} = \frac{1}{4\pi\varepsilon_0} \int_{\cal S} \frac{q}{r^2} \hat{\bf r} \cdot \hat{\bf r} ~r^2 \sin \theta d\theta d\varphi = \frac{q}{\varepsilon_0} \label{Gr(2.12)} \]

so by superposition, we obtain

- Gr (2.13)

Gauss' law (in integral form)

\[ \oint_{\cal S} {\bf E} \cdot d{\bf a} = \frac{Q_{\mbox{enc}}}{\varepsilon_0} \tag{Gl_i}\label{Gl_i} \]

where \(Q_{\mbox{enc}}\) is the total charge enclosed by \({\cal S}\).

By applying the divergence theorem,

\[ \oint_{\cal S} {\bf E} \cdot d{\bf a} = \int_{\cal V} {\boldsymbol \nabla} \cdot {\bf E} ~d\tau \]

and using \(Q_{\mbox{enc}} = \int_{\cal V} \rho d\tau\), and using the fact the the choice of volume is arbitrary, we get

- Gr (2.14)

Gauss' law in differential form

\[ {\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho}{\varepsilon_0}. \tag{Gl_d}\label{Gl_d} \]

We can also compute the divergence of \({\bf E}\) directly from E_vcd:

\[ {\boldsymbol \nabla} \cdot {\bf E} = \frac{1}{4\pi \varepsilon_0} \int d\tau' \rho({\bf r}') {\boldsymbol \nabla} \cdot \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3}. \]

Using divdel,

\[ {\boldsymbol \nabla} \cdot {\bf E} = \frac{1}{4\pi \varepsilon_0} \int d\tau' \rho({\bf r}') ~4\pi \delta ({\bf r} - {\bf r}') = \frac{\rho({\bf r})}{\varepsilon_0} \label{Gr(2.16)} \]

which is Gauss's law in differential form.

As a simple self-consistency exercise, check that you can recover the integral form by integrating and applying Gauss's divergence theorem.

###### Examples of applications of Gauss's law

Gauss's law in integral form is really useful when there is spherical, cylindrical or plane symmetry.

Gaussian surfaces: respectively, concentric sphere, coaxial cylinder, pillbox.

Example 2.2: Field outside a uniformly charged sphere of radius \(R\) and total charge \(q\).

Solution: consider a Gaussian surface which is a sphere of radius \(r > R\). Use spherical symmetry:

\[ \oint_{\cal S} {\bf E} \cdot d{\bf a} = \oint_{\cal S} |{\bf E}| da = |{\bf E}| 4\pi r^2 \]

since \({\bf E} = |{\bf E}| \hat{\bf r}\) and \(d{\bf a} = da \hat{\bf r}\). Therefore,

\[ {\bf E} = \frac{1}{4\pi \varepsilon_0} q \frac{\hat{\bf r}}{r^2}. \]

Same as point charge at origin!

Example 2.3: infinitely long cylinder carrying charge density \(\rho = k s\) for some constant \(k\). Find \({\bf E}\) within the cylinder.

Solution: Gaussian cylinder of length \(l\) and radius \(s\). Enclosed charge:

\[ Q_{\mbox{enc}} = \int d\tau \rho = \int_0^l dz \int_0^{2\pi} d\varphi \int_0^s ds' (k s') s' = 2\pi k l \int_0^s ds' s'^2 = \frac{2\pi}{3} kls^3. \]

Symmetry: \({\bf E}\) must be radially outward, so \({\bf E} = |{\bf E}| \hat{\bf s}\). The surface integral has no contribution from the ends of the cylinder, and

\[ \oint {\bf E} \cdot d{\bf a} = \int |{\bf E}| da = |{\bf E}| \int da = |{\bf E}| \int_0^l dz \int_0^{2\pi} d\varphi s = |{\bf E}| ~2\pi l s. \]

Therefore,

\[ {\bf E} = \frac{1}{3\varepsilon_0} ks^2 \hat{\bf s}. \]

Example 2.4: infinite plane (defined by \(z = 0\)) with uniform surface charge density \(\sigma\). Find \({\bf E}\).

Solution: Gaussian pillbox of area \(A = l_x l_y\) extending equal distances above and below plane.

Enclosed charge: \(Q_{\mbox{enc}} = \sigma A\). Top and bottom surfaces yield \(\int {\bf E} d{\bf a} = 2A |{\bf E}|\). Sides contribute nothing. Thus,

\[ {\bf E} = \frac{\sigma}{2\varepsilon_0} \hat{\bf n} = \frac{\sigma}{2\varepsilon_0} \frac{z}{|z|} \hat{\bf z} \label{Gr(2.17)} \]

where \(\hat{\bf n}\) is a unit vector extending away from the plane. Independent of distance from plane!

Example 2.5: two infinite planes (put them vertical) carrying equal but opposite uniform surface charge densities \(\pm \sigma\).

Solution: fields cancel to left and right of both planes. Between planes: field is \(\frac{\sigma}{\varepsilon_0}\) and points from \(-\) surface to \(+\) one.

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42