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      <text>29.3.2.2 Gauss’ theorem engineering examples</text>
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      <text><![CDATA[<h4 class="HELM_example">Engineering Example</h4>
<p><h5>Gauss’ law</h5></p>
<p><strong>Introduction</strong></p>
<p>From Gauss’ theorem, it is possible to derive a result which can be used to gain insight into situations arising in Electrical Engineering. Knowing the electric field on a closed surface, it is possible to find the electric charge within this surface. Alternatively, in a sufficiently symmetrical situation, it is possible to find the electric field produced by a given charge distribution. <br> Gauss’ theorem states </p>

<p>\(\qquad\displaystyle \int \int_S \underline{F} \cdot \underline{dS} = \int\!\!\int\!\!\int_V \underline{\nabla}\cdot \underline{F} \; dV\) </p>
<p>If \(\underline{F} = \underline{E}\), the electric field, it can be shown that, </p>
<p>\(\qquad\displaystyle \underline{\nabla}\cdot \underline{F} = \underline{\nabla}\cdot \underline{E} = \frac{q}{\varepsilon_0}\) </p>
<p>where \(q\) is the amount of charge per unit volume, or charge density, and \(\varepsilon_0\) is the permittivity of free space: \(\varepsilon_0= 10^{-9}/36\pi\) F m\(^{-1} \approx\) 8.84\(\times 10^{-12}\) F m\(^{-1}\). Gauss’ theorem becomes in this case </p>
<p>\(\qquad\displaystyle \int \int_S \underline{E} \cdot \underline{dS}=\int\!\!\int\!\!\int_V\underline\nabla \cdot \underline E\;dV= \int\!\!\int\!\!\int_V \frac{q}{\varepsilon_0} \; dV = \frac{1}{\varepsilon_0} \int\!\!\int\!\!\int_V q \; dV = \frac{Q}{\varepsilon_0}\) </p>
<p> i.e. </p>
<p> \(\qquad\displaystyle \int \int_S \underline{E} \cdot \underline{dS} = \frac{Q}{\varepsilon_0}\) </p>
<p> which is known as Gauss’ law. Here \(Q\) is the total charge inside the surface \(S\). </p>
<p> Note: this is one of the important Maxwell’s Laws.</p>

<p><strong>Problem in words</strong></p>
A point charge lies at the centre of a cube. Given the electric field, find the magnitude of the charge, using Gauss’ law .
</p>
<p><strong>Mathematical statement of problem</strong></p>
<p>Consider the cube \(-\frac{1}{2} \leq x \leq \frac{1}{2}\), \(-\frac{1}{2} \leq y \leq \frac{1}{2}\), \(-\frac{1}{2} \leq z \leq \frac{1}{2}\) where the dimensions are in metres. A point charge \(Q\) lies at the centre of the cube. If the electric field on the top face \((z=\frac{1}{2})\) is given by ,</p>
<p> \(\qquad\displaystyle \underline{E} =10 \frac{x\underline{i} + y\underline{j} + z\underline{k}}{\left(x^2 + y^2 + z^2\right)^\frac{3}{2}}\) </p>
<p> find the charge \(Q\) from Gauss’ law . </p>
<p> \({\displaystyle \left[ \mbox{Hint : } \int_{x=-\frac{1}{2}}^{\frac{1}{2}} \int_{y=-\frac{1}{2}}^{\frac{1}{2}} \left( x^2 + y^2 + \frac{1}{4} \right)^{-\frac{3}{2}} \; \mathrm{d}y \; \mathrm{d}x = \frac{4\pi}{3}\right]}\)
</p>

<p><strong>Mathematical analysis</strong></p>

<p>From Gauss’ law </p>
<p>\(\qquad \displaystyle \int \int_S \underline{E} \cdot \underline{dS} = \frac{Q}{\varepsilon_0}\) </p>
<p>so </p>
<p>\(\qquad\displaystyle Q = \varepsilon_0 \int \int_S \underline{E} \cdot \underline{dS} = 6 \varepsilon_0 \int \int_{S(top)} \underline{E} \cdot \underline{dS}\) </p>
<p>since, using the symmetry of the six faces of the cube, it is possible to integrate over just one of them (here the top face is chosen) and multiply by 6. On the top face </p>
<p>\(\qquad\displaystyle \underline{E} =10 \frac{x\underline{i} + y\underline{j} + \frac{1}{2}\underline{k}}{\left(x^2 + y^2 + \frac{1}{4}\right)^\frac{3}{2}}\) </p>
<p>and </p>
<p>\(\begin{aligned} \qquad \underline{dS} &amp;= \mbox{(element of surface area)} \times \mbox{(unit normal)}\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\ &amp;= \mathrm{d}x \; \mathrm{d}y \; \underline{k}\end{aligned}\)</p>

<p> So </p>
<p>\(\begin{aligned} \qquad \underline{E} \cdot \underline{dS} &amp;= 10 \frac{\frac{1}{2}}{\left(x^2 + y^2 + \frac{1}{4}\right)^{\frac{3}{2}}} \; \mathrm{d}y \;\mathrm{d}x \\ &amp;= 5\left(x^2 + y^2 + \frac{1}{4}\right)^{-\frac{3}{2}} \; \mathrm{d}y \; \mathrm{d}x\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{aligned}\)</p>

<p> Now </p>
<p>\(\begin{aligned} \qquad \int \int_{S(top)} \underline{E} \cdot \underline{dS} &amp;= \int_{x=-\frac{1}{2}}^{\frac{1}{2}} \int_{y=-\frac{1}{2}}^{\frac{1}{2}} 5\left( x^2 + y^2 + \frac{1}{4} \right)^{-\frac{3}{2}} \; \mathrm{d}y \; \mathrm{d}x \qquad\qquad\qquad\qquad\qquad \\ &amp;= 5 \times \frac{4\pi}{3} \;\; \mbox{ (using the hint)} \\ &amp;= \frac{20\pi}{3}\end{aligned}\)</p>
<p> So, from Gauss’ law, </p>
<p>\(\qquad\displaystyle Q = 6 \varepsilon_0 \times \frac{20\pi}{3} = 40\pi \varepsilon_0 \approx 10^{-9}\mbox{C}\)</p>

<p><strong>Interpretation</strong></p>
<p>Gauss’ law can be used to find a charge from its effects elsewhere.</p>
<p>The form of \(\displaystyle \underline{E} =10 \frac{x\underline{i} + y\underline{j} + \frac{1}{2}\underline{k}}{\left(x^2 + y^2 + \frac{1}{4}\right)^\frac{3}{2}}\) comes from the fact that \(\underline E\) is radial and equals \(10\dfrac{\underline{r}}{r^3}=10\dfrac{\hat{\underline r}}{r^2}\)</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Verify Gauss’ theorem for the vector field \(\underline{F} = y^2 \underline{j} - x z \underline{k}\) and the triangular prism with vertices at \((0,0,0)\), \((2,0,0)\), \((0,0,1)\), \((0,4,0)\), \((2,4,0)\) and \((0,4,1)\) (see figure below). </p><img src="@@PLUGINFILE@@/plane%20in%203%20dimensions.svg" alt="plane in 3 dimensions" width="255" height="113" class="img-fluid atto_image_button_text-bottom"><p></p>
<h4 class="HELM_solution">Solution</h4>
<p>As \(\underline{F} = y^2 \underline{j} - x z \underline{k}\), \({\underline \nabla} \cdot \underline{F} = 0 + 2y - x = 2y - x\).
Thus \begin{eqnarray*}   \displaystyle\int\!\! \displaystyle\int\!\! \displaystyle\int_V {\underline \nabla} \cdot 
\underline{F}   dV &amp;=&amp;  \displaystyle\int_{x=0}^2  \displaystyle\int_{y=0}^4 
 \displaystyle\int_{z=0}^{1-x/2} (2y - x)  \mathrm{d}z  \mathrm{d}y  \mathrm{d}x \\ 
&amp;=&amp;  \displaystyle\int_{x=0}^2  \displaystyle\int_{y=0}^4 \left[\phantom{\dfrac{1}{1}}\!\!\! 2y z - x z 
\right]_{z=0}^{1-x/2}   \mathrm{d}y  \mathrm{d}x =  \displaystyle\int_{x=0}^2  \displaystyle\int_{y=0}^4 ( 2y 
- x y - x + \frac{1}{2}x^2 )  \mathrm{d}y  \mathrm{d}x \\ 
&amp;=&amp;  \displaystyle\int_{x=0}^2  \left[ y^2 - \frac{1}{2}x y^2 - x y + 
\frac{1}{2}x^2 y \right]_{y=0}^4   \mathrm{d}x  =  \displaystyle\int_{x=0}^2  ( 16 - 12 x 
 + 2x^2 )   \mathrm{d}x \\ 
 &amp;=&amp; \left[ 16 x - 6 x^2 + \frac{2}{3}x^3 \right]_0^2 = 
 \frac{40}{3} 
\end{eqnarray*} </p>

<p>To work out \( \displaystyle\int\!\! \displaystyle\int_S \underline{F} \cdot \underline{dS}\), it  is necessary to consider the contributions from the five faces separately.</p>
<p> On the front face, \(y = 0\), \(\underline{F} = -x z \underline{k}\)  and \(\underline{dS} = -\underline{j}\) thus \(\underline{F} \cdot \underline{dS} = 0\) and the contribution to the integral is zero. </p>
<p> On the back face, \(y = 4\), \(\underline{F} =  16 \underline{j} -x z \underline{k}\) and \(\underline{dS} = \underline{j}\) thus  \(\underline{F} \cdot \underline{dS} = 16\) and the contribution to the integral is </p>
<p>\(\qquad \displaystyle\int_{x=0}^2  \displaystyle\int_{z=0}^{1-x/2} 16  \mathrm{d}z \mathrm{d}x = \displaystyle\int_{x=0}^2 \left[\phantom{\dfrac{1}{1}}\!\! 16 z \right]_{z=0}^{1-x/2} \mathrm{d}x =  \displaystyle\int_{x=0}^2 16 (1-x/2)  \mathrm{d}x = \left[ \phantom{\dfrac{1}{1}}\!\!16 x - 4x^2 \right]_0^2  = 16. \)</p>

<p>On the left face, \(x=0\),  \(\underline{F} = y^2 \underline{j} \) and \(\underline{dS} = -\underline{i}\) thus \(\underline{F} \cdot \underline{dS} = 0\) and the contribution to the integral is zero. </p>

<p> On the bottom face, \(z=0\),  \(\underline{F} = y^2 \underline{j} \) and \(\underline{dS} = -\underline{k}\) thus \(\underline{F} \cdot \underline{dS} = 0\) and the contribution to the integral is zero. </p>

<p>On the top right (sloping) face, \(z = 1 - x/2\), \(\underline{F} =  y^2 \underline{j} + (\frac{1}{2}x^2 - x) \underline{k}\) and the unit normal \(\underline{\hat{n}} = \frac{1}{\sqrt{5}}\underline{i} + \frac{2}{\sqrt{5}}\underline{k}\) </p>
<p>Thus \(\underline{dS} = \left[\frac{1}{\sqrt{5}}\underline{i} + \frac{2}{\sqrt{5}}\underline{k}\right]   \mathrm{d}y  \mathrm{d}w\) where \(\mathrm{d}w\) measures the distance along the slope for a constant \(y\). </p>
<p> As \(\mathrm{d}w = \frac{\sqrt{5}}{2} \mathrm{d}x\), \(\underline{dS} = \left[\frac{1}{2}\underline{i} + \underline{k}\right]   \mathrm{d}y  \mathrm{d}x\) thus \(\underline{F} \cdot \underline{dS} = 16\) and the contribution to the integral is </p>
<p>\(\qquad \displaystyle\int_{x=0}^2  \displaystyle\int_{y=0}^{4} (\frac{1}{2}x^2 - x)  \mathrm{d}y \mathrm{d}x =  \displaystyle\int_{x=0}^2  (2x^2 - 4x)  \mathrm{d}x = \left[ \frac{2}{3} x^3 - 2 x^2 \right]_0^2 = -\frac{8}{3} \). </p>
<p> Adding the contributions, \( \displaystyle\int\!\! \displaystyle\int_S \underline{F} \cdot \underline{dS} = 0 + 16 + 0 + 0 - \frac{8}{3} = \frac{40}{3}\). </p>

<p> Thus \( \displaystyle\int\!\! \displaystyle\int_S \underline{F} \cdot \underline{dS} =  \displaystyle\int\!\! \displaystyle\int\!\! \displaystyle\int_V {\underline \nabla} \cdot \underline{F}   dV 
= \frac{40}{3}\) hence verifying Gauss' divergence theorem. </p>


<hr>
<h4 class="HELM_example">Engineering Example</h4>
<p><h5>Field strength around a charged line</h5></p>

<p><strong>Problem in words</strong></p>
<p>Find the electric field strength at a given distance from a uniformly charged line. </p>

<p><strong>Mathematical statement of problem</strong></p>
<p>Determine the electric field at a distance \(r\) from a uniformly charged line (charge per unit length \(\rho_L\)). You may assume from symmetry that the field points directly away from the line. </p><img src="@@PLUGINFILE@@/line.svg" alt="cylinder of length 1 centred around a line" width="236" height="76" class="img-fluid atto_image_button_text-bottom"><p></p>

<p><strong>Mathematical analysis</strong></p>
<p>Imagine a cylinder a distance \(r\) from the line and of length \(l\) (see above Figure). From Gauss’ law </p>
<p> \(\qquad\displaystyle \int \int_S \underline{E} \cdot \underline{dS} = \frac{Q}{\varepsilon_0}\) As the charge per unit length is \(\rho_L\), then the right-hand side equals \(\rho_L l / \varepsilon_0\). On the left-hand side, the integral can be expressed as the sum </p>
<p>\(\qquad\displaystyle \int \int_S \underline{E} \cdot \underline{dS} = \int \int_{S(ends)} \underline{E} \cdot \underline{dS} \; + \; \int \int_{S(curved)} \underline{E} \cdot \underline{dS}\) </p><p>Looking first at the circular ends of the cylinder, the fact that the field lines point radially away from the charged line implies that the electric field is in the plane of these circles and has no normal component. Therefore \(\underline{E} \cdot \underline{dS}\) will be zero for these ends. </p> 
<p> Next, over the curved surface of the cylinder, the electric field is normal to it, and the symmetry of the problem implies that the strength of the electric field will be constant (here denoted by \(E\)). Therefore the integral \(=\) Total curved surface area \(\times\) Field strength \(=2\pi r l E\). So, by Gauss’ law </p>
<p> \(\qquad\displaystyle \int \int_{S(ends)} \underline{E} \cdot \underline{dS} \; + \; \int \int_{S(curved)} \underline{E} \cdot \underline{dS} = \frac{Q}{\varepsilon_0}\) or \(\qquad\displaystyle 0 \; + \; 2\pi r l E = \frac{\rho_L l}{\varepsilon_0}\) </p>

<p><strong>Interpretation</strong></p>
<p>Hence, the field strength \(E\) is given by \(\displaystyle E = \frac{\rho_L}{2 \pi \varepsilon_0 r}\)
</p>

<hr>
<h4 class="HELM_example">Engineering Example</h4>
<p><b>Field strength on a cylinder</b></p>

<p><strong>Problem in words</strong></p>
<p>Given the electric field \(\underline{E}\) on the surface of a cylinder, use Gauss’ law to find the charge per unit length.</p>

<p><strong>Mathematical statement of problem</strong></p>
<p>On the surface of a long cylinder of radius \(a\) and length \(l\), the electric field is given by </p>
<p>\(\qquad\displaystyle \underline{E} = \frac{\rho_L}{2\pi\varepsilon_0} \; \frac{\left( a + b\cos\theta \right) \underline{{\hat r}} - b\sin\theta \;\underline{\hat \theta}}{(a^2 + 2 a b \cos \theta + b^2)}\) </p>
<p> (using cylindrical polar co-ordinates) due to a line of charge a distance \(b\) (\(&lt;\) \(a\)) from the centre of the cylinder. Using Gauss’ law , find the charge per unit length.</p>
<p>Hint:-\({\displaystyle \qquad \int_0^{2\pi} \frac{a + b\cos\theta}{(a^2 + 2 a b \cos\theta + b^2)} \; d\theta = \frac{2\pi}{a}}\)</p>

<strong>Mathematical analysis</strong></p>
<p>Consider a cylindrical section - as in the previous example, there are no contributions from the ends of the cylinder since the electric field has no normal component here. However, on the curved surface </p>
<p>\(\qquad\displaystyle \underline{dS} = a \; d\theta \; \mathrm{d}z \; \underline{\hat r}\) </p>
<p> so </p>
<p>\(\qquad\displaystyle \underline{E} \cdot \underline{dS} = \frac{\rho_L}{2\pi\varepsilon_0} \; \frac{a + b\cos\theta}{(a^2 + 2 a b \cos \theta + b^2)} \; a \; d\theta \; \mathrm{d}z\) </p>
<p> Integrating over the curved surface of the cylinder </p>
<p> \(\begin{aligned} \qquad \int \int_S \underline{E} \cdot \underline{dS} &amp;= \int_{z=0}^l \int_{\theta=0}^{\theta=2\pi}\frac{a\rho_L}{2\pi\varepsilon_0} \; \frac{a + b\cos\theta}{(a^2 + 2 a b \cos \theta + b^2)} \; d\theta \; \mathrm{d}z\qquad\qquad\qquad\qquad \qquad\\ &amp;= \frac{a\rho_L l}{2\pi\varepsilon_0} \int_0^{2\pi} \frac{a+ b\cos\theta}{(a^2 + 2a b \cos\theta + b^2)} \; d\theta \\ &amp;= \frac{\rho_L l}{\varepsilon_0} \;\;\;\mbox{ using the given result for the integral.}\end{aligned}\)</p>
<p> Then, if \(Q\) is the total charge inside the cylinder, from Gauss’ law </p>
<p>\(\qquad\displaystyle \frac{\rho_L l }{\varepsilon_0} = \frac{Q}{\varepsilon_0}\) so \(\displaystyle \rho_L =\frac{Q}{l}\)as one would expect.</p>

<p><strong>Interpretation</strong></p>
<p>Therefore the charge per unit length on the line of charge is given by \(\rho_L\) (i.e. the charge per unit length is constant). </p>]]></text>
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<file name="plane in 3 dimensions.svg" path="/" encoding="base64">[encoded data, 27012 characters]</file>
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