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<h3> Statement </h3>
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$6.2.11$ Two infinite plates of thickness h are charged uniformly in volume and stacked together. The bulk charge density of the first plate is $\rho$, and the second plate -$\rho$. Find the maximum electric field intensity.
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For problem 6.2.11
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<h3>Solution</h3>
<p>
Absolute value of total charge of each plate is $\rho S h$, where $S$ is the surface of plate's biggest faces. So, applying Gauss Law, the absolute value of electric field intensity generated by this plate is:
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$$E = \frac{\rho h}{\varepsilon_0}$$
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Then, field generated by one of these biggest faces is the half of $E$. Let's consider now a point inside of one of these infinite plates. The resultant field over this point is the vectorial sum of the fields generated by lower negative plate, a plate with thickness $x$ (where $x$ is the distance from boundary between plates and the point) and a upper plate with thickness $h-x$:
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$$E_R(x) = \frac{\rho h}{2\varepsilon_0} + \frac{\rho (h-x)}{2\varepsilon_0} - \frac{\rho x}{2\varepsilon_0}$$
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$$E_R(x) = \frac{\rho}{\varepsilon_0}(h-x)$$
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Hence, maximum field intensity is achieved when $x=0$, i.e., point is in boundary between plates.
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<h4>Answer</h4>
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$$E_{max} = \frac{\rho h}{\varepsilon_0}$$
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<p style="text-align: right; font-style: italic; font-size: 14;">
BSc. Luis Daniel Fernández Quintana<br>
Physics Department (FCNE)<br>
Universidad de Oriente, Cuba<br>
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