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| <h3 id="back-link"><a href="../../#6.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#6.2">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $6.2.11.$ 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. | | $6.2.11.$ 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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| loading="lazy" width="350" /> | | loading="lazy" width="350" /> |
| <figcaption> | | <figcaption> |
| For problem $6.2.11$ | | For problem $6.2.11$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| The absolute value of the total charge of each plate is $\rho S h$, where $S$ is the surface of the plate's biggest faces. So, applying Gauss Law, the absolute value of electric field intensity generated by this plate is: | | The absolute value of the total charge of each plate is $\rho S h$, where $S$ is the surface of the plate's biggest faces. So, applying Gauss Law, the absolute value of electric field intensity generated by this plate is: |
| </p> | | </p> |
| <p> | | <p> |
| $$E = \frac{\rho h}{\varepsilon_0}$$ | | $$E = \frac{\rho h}{\varepsilon_0}$$ |
| </p> | | </p> |
| <p> | | <p> |
| Then, the field generated by one of these biggest faces of the plate is half of $E$. Let's consider now a point inside of one of these infinite plates (for example, in the upper one). The resultant field over this point is the vectorial sum of the fields generated by a lower negative plate, a plate with thickness $x$ (where $x$ is the distance from the boundary between plates and the point), and a upper plate with thickness $h-x$: | | Then, the field generated by one of these biggest faces of the plate is half of $E$. Let's consider now a point inside of one of these infinite plates (for example, in the upper one). The resultant field over this point is the vectorial sum of the fields generated by a lower negative plate, a plate with thickness $x$ (where $x$ is the distance from the boundary between plates and the point), and a upper plate with thickness $h-x$: |
| </p> | | </p> |
| <p> | | <p> |
| $$E_R(x) = \frac{\rho h}{2\varepsilon_0} + \frac{\rho (h-x)}{2\varepsilon_0} - \frac{\rho x}{2\varepsilon_0}$$ | | $$E_R(x) = \frac{\rho h}{2\varepsilon_0} + \frac{\rho (h-x)}{2\varepsilon_0} - \frac{\rho x}{2\varepsilon_0}$$ |
| </p> | | </p> |
| <p> | | <p> |
| $$E_R(x) = \frac{\rho}{\varepsilon_0}(h-x)$$ | | $$E_R(x) = \frac{\rho}{\varepsilon_0}(h-x)$$ |
| </p> | | </p> |
| <p> | | <p> |
| Hence, maximum field intensity is achieved when $x=0$, i.e., the point is in the boundary between plates. | | Hence, maximum field intensity is achieved when $x=0$, i.e., the point is in the boundary between plates. |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$E_{max} = \frac{\rho h}{\varepsilon_0}$$ | | $$E_{max} = \frac{\rho h}{\varepsilon_0}$$ |
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| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
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