6.2.10 | before revision #10493 | Savchenko Solutions

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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    <h3> Statement </h3>
    <p>
        $6.2.10.$ Two infinite planes intersecting at an angle $\alpha$ divide space into four regions. What is the electric field strength in regions 1 and 2 if the surface charge density of the planes is $\pm \sigma$?

Solution

Consider the following figure...

It's known that the electric field intensity for one of the faces of a infinite plane with surface charge density is $E = \frac{\sigma}{2\varepsilon_0}$ .

For region 1 (see part $$a$$ of above figure):

$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})$

and for $$x$$ -direction, $E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}.$ So, as $E_{R1} = \sqrt{E_{R1x}^2+E_{R1y}^2},$ and taking in account that $\sin{\alpha} = \sqrt{\frac{1-\cos{\alpha}}{2}},$ it is obtained

Answer 1

$E_{R1} = \frac{\sigma}{\varepsilon_0}\sin{\frac{\alpha}{2}}$

For region 2, for $$y$$ -direction: $E_{R2y} = E(1+\cos{\alpha})=\frac{\sigma}{2\varepsilon_0}(1+\cos{\alpha})$ and for $$x$$ -axis, $E_{R2x} = E\sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}.$

Again, $E_{R2} = \sqrt{E_{R2x}^2+E_{R2y}^2}$ and taking in account that $\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$

    <h4>Answer 2</h4>
    <p>
        $$E_{R2} = \frac{\sigma}{\varepsilon_0}\cos{\frac{\alpha}{2}}$$
    </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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