Pre-Quantum Electrodynamics
Gauss's Law: the divergence of \({\bf E}\)ems.es.ef.Gl
- PM 1.10, 2.9
- Gr 2.2
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,
- PM (1.27)
\[ \oint {\bf E} \cdot d{\bf a} = \frac{1}{4\pi\varepsilon_0} \int_{\cal S} \frac{q}{r^2} \hat{\bf r} \cdot \hat{\bf r} ~r^2 \sin \theta d\theta d\varphi = \frac{q}{\varepsilon_0} \label{Gr(2.12)} \]
so by superposition, we obtain
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
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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Created: 2024-02-27 Tue 10:31