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<p> Here we learn to use Gauss's Law to calculate the Electric Field for cases with specific symmetries. We'll start with looking at planar symmetry.</p><p> Textbook Links </p><ul><li><a href="/courses/course-v1:MITx+8.02.1x+1T2019/pdfbook/0/chapter/3/11">Chapter 3.2 Gauss's Law</a></li></ul>
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Charge Distributions with Enough Symmetry for Gauss&#39;s Law
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On which of the objects shown in the figure above can you apply Gauss's Law to find the Electric Field from the given charge density? (Choose all that apply.) </p>
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<text>1. Infinite line of charge with a uniform charge density, [mathjaxinline]\lambda _0[/mathjaxinline]</text>
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<text>2. Infinite cylinder of charge with charge density [mathjaxinline]\rho =\rho _0\dfrac {cos\theta }{r}[/mathjaxinline]</text>
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<text>3. A slab of charge, infinite in [mathjaxinline]y[/mathjaxinline] and [mathjaxinline]z[/mathjaxinline] with a constant charge density [mathjaxinline]\rho[/mathjaxinline]</text>
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<text>4. A ring of charge with uniform charge density, [mathjaxinline]\lambda _0[/mathjaxinline]</text>
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<text>5. A sphere with uniform charge density, [mathjaxinline]\rho _0[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L6aQ2: Superposition of Multiple Sheets of Charge</h2>
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Superposition
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Three infinite sheets of charge are shown below. </p>
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The sheet in the middle is negatively charged with charge per unit area [mathjaxinline]-2\sigma[/mathjaxinline], and the other two sheets are positively charged with charge per unit area [mathjaxinline]+\sigma[/mathjaxinline]. </p>
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Which set of arrows (and zeros) best describes the electric field? </p>
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<h2 class="hd hd-2 unit-title">L6aQ3: Charge Density of Two Parallel Sheets</h2>
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Charge density of two parallel sheets.
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In the left figure above, two very large parallel, non-conductive sheets are separated by a distance [mathjaxinline]d[/mathjaxinline]. Sheet 1 has a uniform surface charge density [mathjaxinline]\sigma _1[/mathjaxinline], and sheet 2 has a uniform surface charge density [mathjaxinline]\sigma _2[/mathjaxinline]. A coordinate system is chosen such that the sheets are in the y-z plane, with sheet 1 at [mathjaxinline]x=0[/mathjaxinline]. In the right figure above, a graph of the x-component of the electric field as a function of [mathjaxinline]x[/mathjaxinline] is shown, where [mathjaxinline]E_0[/mathjaxinline] is a positive constant of units N/C. </p>
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The ratio [mathjaxinline]\sigma _1/\sigma _2[/mathjaxinline] is given by: </p>
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<text> [mathjaxinline]\displaystyle \frac{\sigma _1}{\sigma _2} = -\frac{1}{3}[/mathjaxinline] and [mathjaxinline]\sigma _1&lt;0[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{\sigma _1}{\sigma _2} = -\frac{5}{3}[/mathjaxinline] and [mathjaxinline]\sigma _1&lt;0[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{\sigma _1}{\sigma _2} = -\frac{5}{3}[/mathjaxinline] and [mathjaxinline]\sigma _1&gt;0[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{\sigma _1}{\sigma _2} = -\frac{1}{3}[/mathjaxinline] and [mathjaxinline]\sigma _1&gt;0[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{\sigma _1}{\sigma _2} = -\frac{2}{3}[/mathjaxinline] and [mathjaxinline]\sigma _1&gt;0[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{\sigma _1}{\sigma _2} = -\frac{2}{3}[/mathjaxinline] and [mathjaxinline]\sigma _1&lt;0[/mathjaxinline]</text>
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