Solution

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:

$$E = \frac{\rho h}{\varepsilon_0}$$

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 $h-x$:

$$E_R(x) = \frac{\rho h}{2\varepsilon_0} + \frac{\rho (h-x)}{2\varepsilon_0} - \frac{\rho x}{2\varepsilon_0}$$

$$E_R(x) = \frac{\rho}{\varepsilon_0}(h-x)$$

Hence, maximum field intensity is achieved when $x=0$, i.e., the point is in the boundary between plates.

    <h4>Answer</h4>
    <p>
        $$E_{max} = \frac{\rho h}{\varepsilon_0}$$
    </p>

    <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>
    </p>


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