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<p> In this Lesson we introduce the idea of Gauss's Law and start to explore its implications.</p><p> Textbook Links </p><ul><li><a href="/courses/course-v1:MITx+8.02.1x+1T2019/pdfbook/0/chapter/3/3">Chapter 3.1-3.2 Gauss's Law</a></li></ul>
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<h2 class="hd hd-2 unit-title">L5Q2: Ranking Electric Flux</h2>
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Ranking Electric Flux.
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Consider a rectangular open surface of area [mathjaxinline]A[/mathjaxinline] which sits in an electric field which is uniform in magnitude and direction. We measure the electric flux through this open surface in the four cases indicated above where only the side view of the surface is shown. The uniform electric field always points in the [mathjaxinline]+\hat{i}[/mathjaxinline] direction and its magnitude is either [mathjaxinline]E_0/2[/mathjaxinline], [mathjaxinline]E_0[/mathjaxinline] or [mathjaxinline]2E_0[/mathjaxinline], as indicated. The surface has different orientations with the direction of [mathjaxinline]\hat{n}[/mathjaxinline], the normal to the surface, as shown in the figures. Rank [mathjaxinline]\Phi[/mathjaxinline], the electric flux through the surface from smallest (most negative) to largest (most positive). </p>
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<h2 class="hd hd-2 unit-title">L5v4: Electric Flux Through a Closed Surface</h2>
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<h2 class="hd hd-2 unit-title">L5Q3: Flux Through a Cylinder</h2>
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Non-Uniform E Field in Cylinders
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<p>In the diagram the electric field is increasing in magnitude from left to right. At which face(s) of the cylinders
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Charge in a Box.
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<p>A charge [mathjaxinline]Q = +5\, \mu C[/mathjaxinline] is located somewhere inside a closed hollow cube with sides [mathjaxinline]a = 30 \, cm[/mathjaxinline] long. What is the total electric flux through the closed surface of this cube?</p>
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<text> [mathjaxinline]9.0\times 10^{-6}\, N\cdot m^2/C[/mathjaxinline]</text>
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<text> [mathjaxinline]8.1\times 10^4\, N\cdot m^2/C[/mathjaxinline]</text>
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<text> [mathjaxinline]4.5\times 10^4\, N\cdot m^2/C[/mathjaxinline]</text>
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<text> [mathjaxinline]5.6\times 10^5\, N\cdot m^2/C[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">Gauss's Law Summary</h2>
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<h2>W2D2 Summary</h2><h2 class="heading">Electric Flux</h2><p>The surface integral of a vector field over an area is called "flux". The word "flux" was first used in studies of velocity fields where there was an actual flow of matter across a surface. In electrostatics, nothing is flowing, but electric flux is still the name used for this surface integral. Instead, we can think of electric flux as describing the number of electric field lines penetrating the surface. Flux is typically denoted by [mathjaxinline]\Phi[/mathjaxinline] (capital Phi), and electric flux by [mathjaxinline]\Phi_{E}[/mathjaxinline]. Mathematically, it is defined as</p>
[mathjax]\Phi_{E} = \iint_S \vec{E} \cdot d\vec{A}.[/mathjax]
<p class="imagecaption"><img src="/assets/courseware/v1/7dfa9f8f2cf9d0e9668bedf50ed802ac/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_W2D3_preClass-fig003.jpg" width="330"/><br/><b>Figure 3.1.1</b> Electric field lines passing through a surface of area <i>A</i>.
</p>
<p>Consider the surface shown in Figure 3.1.1. Let [mathjaxinline]\vec{\mathbf{A}} = A \hat{\mathbf{n}}[/mathjaxinline] be defined as the <i>area vector</i> having a magnitude of the area of the surface, [mathjaxinline]A[/mathjaxinline], and pointing in the normal direction [mathjaxinline]\hat{\mathbf{n}}[/mathjaxinline]. The flux through the surface is</p>
[mathjax]\Phi_{E} = \vec{E} \cdot \vec{A} = E A.[/mathjax]
<p>In general, a surface <i>S</i> can be curved and the electric field [mathjaxinline]\vec{\mathbf{E}}[/mathjaxinline] may vary over the surface. We shall be interested in the case where the surface is <i>closed</i>. A closed surface is a surface that completely encloses a volume. In order to compute the electric flux, we divide the surface into a large number of infinitesimal area elements [mathjaxinline]\Delta \vec{\mathbf{A}}_{i} = \Delta A_{i} \, \hat{\mathbf{n}}_{i}[/mathjaxinline], as shown in Figure 3.1.3. Note that for a closed surface the unit vector [mathjaxinline]\hat{\mathbf{n}}_{i}[/mathjaxinline] is chosen to point in the <i>outward</i> normal direction.</p>
<p class="imagecaption"><img src="/assets/courseware/v1/e2efb7d95bd0978656e21997ec118e3b/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_W2D3_preClass-fig005.jpg" width="330"/><br/><b>Figure 3.1.3</b> Electric field passing through an area element [mathjaxinline]\Delta \vec{\mathbf{A}}_{i}[/mathjaxinline], making an angle [mathjaxinline]\theta[/mathjaxinline] with the normal to the surface.
</p>
<p>The electric flux through [mathjaxinline]\Delta \vec{\mathbf{A}}_{i}[/mathjaxinline] is [mathjaxinline]\Delta \Phi_{E} = \vec{\mathbf{E}}_i \cdot \Delta \vec{\mathbf{A}}_{i} = E_i \Delta A_i \cos \theta[/mathjaxinline].</p>
<h2 class="heading">Gauss's Law</h2>
<p>Consider a positive point charge <i>q</i> located at origin of our coordinate system, as shown in Figure 3.2.1. The electric field due to the charge <i>q</i> is [mathjaxinline]\vec{\mathbf{E}} = q \hat{\mathbf{r}} /(4\pi \epsilon_{0} r^2)[/mathjaxinline]. We enclose the charge by an imaginary sphere of radius <i>r</i> called the "Gaussian surface".</p>
<p class="imagecaption"><img src="/assets/courseware/v1/b5b84e1d28b8199ced5895582084801b/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_W2D3_preClass-fig007.jpg" width="330"/><br/><b>Figure 3.2.1</b> A spherical Gaussian surface enclosing a charge [mathjaxinline]q[/mathjaxinline].
</p>
<p>The area element [mathjaxinline]d \vec{\mathbf{A}}[/mathjaxinline] is given by</p>
[mathjax]d \vec{\mathbf{A}} = r^2 \sin \theta \, d\theta \, d\phi \, \hat{\mathbf{r}}[/mathjax]
<p>as can be seen from Figure 3.2.2. Note: Physicists typically let [mathjaxinline]\theta[/mathjaxinline] (the altitude angle) run from 0 to [mathjaxinline]\pi[/mathjaxinline], and [mathjaxinline]\phi[/mathjaxinline] (the azimuthal angle) run from 0 to [mathjaxinline]2\pi[/mathjaxinline]. This is the opposite convention to what is typically used in math!</p>
<p class="imagecaption"><img src="/assets/courseware/v1/72ed24ced12e931e14840659cf1ccfdb/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_W2D3_preClass-fig008.jpg" width="330"/><br/><b>Figure 3.2.2</b> A small area element on the surface of a sphere of radius <i>r</i>.
</p>
<p>We can compute the electric flux through a small area element [mathjaxinline]d \vec{\mathbf{A}}[/mathjaxinline] as</p>
[mathjax]d\Phi_E = \vec{\mathbf{E}} \cdot d \vec{\mathbf{A}} = E r^2 \sin \theta \, d\theta \, d\phi = \frac{q}{4 \pi \epsilon_0} \sin \theta \, d\theta \, d\phi.[/mathjax]
<p>The total flux through the entire surface comes from summing up every little bit of flux. We write this as a double integral over the area.</p>
[mathjax]\Phi _{E} =\unicode{x222F}_s \vec{\mathbf{E}} \cdot d\vec{\mathbf{A}} = \frac{q}{4\pi \epsilon_0} \int_{0}^{\pi}\sin \theta \, d\theta \int_{0}^{2\pi }d\phi = \frac{q}{\epsilon_0}[/mathjax]
<p>In the above, we have chosen a sphere to be the Gaussian surface. However, it turns out that the shape of the closed surface can be arbitrarily chosen. For the surfaces shown in Figure 3.2.3, the same result ([mathjaxinline]\Phi_{E} =q/\epsilon _{0}[/mathjaxinline]) is obtained, whether the choice is [mathjaxinline]S_{1}[/mathjaxinline], [mathjaxinline]S_{2}[/mathjaxinline] or [mathjaxinline]S_{3}[/mathjaxinline].</p>
<p class="imagecaption"><img src="/assets/courseware/v1/0f82b70a7998eb962d6e451ee0f16cec/asset-v1:MITx+8.02.1x+1T2019+type@asset+block/images_W2D3_preClass-fig011.jpg" width="330"/><br/><b>Figure 3.2.3</b> Different Gaussian surfaces with the same outward electric flux.
</p>
<p>Amazingly, this result generalizes further: the statement that the net flux through any closed surface is proportional to the net charge enclosed is known as Gauss's law. Mathematically, Gauss's law is expressed as</p>
[mathjax]\Phi _{E} = \unicode{x222F}_S \vec{\mathbf{E}} \cdot d\vec{\mathbf{A}} = \frac{q_{enc}}{\epsilon_0}[/mathjax]
<p>where [mathjaxinline]q_{enc}[/mathjaxinline] is the net charge enclosed by the surface. The circle on the integral sign, [mathjaxinline]\unicode{x222F}[/mathjaxinline], means that the surface integration is over a closed surface. Sometimes, you will see the integral written as simply [mathjaxinline]\oint[/mathjaxinline].</p>
<h2 class="heading">Steps in Applying Gauss's Law</h2>
<ol><li>Identify the symmetry associated with the charge distribution.</li><li>Determine the direction of the electric field, and a "Gaussian surface" on which the magnitude of the electric field is constant over portions of the surface (draw a diagram showing your surface!).</li><li>Divide the space into different regions associated with the charge distribution. For each region, calculate [mathjaxinline]q_{enc}[/mathjaxinline], the charge enclosed by the Gaussian surface.</li><li>Calculate the electric flux [mathjaxinline]\Phi_E[/mathjaxinline] through the Gaussian surface for each region.</li><li>Equate [mathjaxinline]\Phi_E[/mathjaxinline] with [mathjaxinline]q_{enc}/\epsilon_0[/mathjaxinline], and deduce the magnitude of the electric field.</li><li>Construct the vector expression for the vector field (remember your unit vectors!).</li></ol>
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<h2 class="hd hd-2 unit-title">L5Q5: Flux Through a Sphere</h2>
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Flux Through a Sphere
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The total electric flux through the spherical surface shown in the figure below is </p>
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<h2 class="hd hd-2 unit-title">L5Q6: Three Charges and a Gaussian Sphere</h2>
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Three Charges and a Gaussian Sphere
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Consider the three charges [mathjaxinline]+Q[/mathjaxinline], [mathjaxinline]+2Q[/mathjaxinline], [mathjaxinline]-Q[/mathjaxinline], and a mathematical spherical surface (it does not physically exist) with the charge [mathjaxinline]+Q[/mathjaxinline] located inside the sphere as shown in the figure below. </p>
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What is [mathjaxinline]\Phi[/mathjaxinline], the total electric flux through the spherical surface between the charges? Write your answer using some or all of the following: [mathjaxinline]Q[/mathjaxinline], epsilon_0 for [mathjaxinline]\epsilon _{0}[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\Phi[/mathjaxinline] = </p>
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<h2 class="hd hd-2 unit-title">L5Q7: Gauss's Law with Grass Seeds Image</h2>
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Gauss&#39;s Law
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The grass seeds figure below shows the electric field of three charges with charges [mathjaxinline]+q[/mathjaxinline], [mathjaxinline]+q[/mathjaxinline], and [mathjaxinline]-q[/mathjaxinline] (with [mathjaxinline]q&gt;0[/mathjaxinline]). The grey shaded region in the figure shows a spherical Gaussian surface containing two of the charges. The electric flux through the spherical Gaussian surface is </p>
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