6.2.10 | revision #17710 | Savchenko Solutions

Statement

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

$\sin{\frac{\alpha}{2}} = \sqrt{\frac{1-\cos{\alpha}}{2}},$

it is obtained

Answer 1

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

Region 2

In the $$y$$ -direction:
$E_{R2y} = E(1+\cos{\alpha}) = \frac{\sigma}{2\epsilon_0}(1+\cos{\alpha})$

In the $$x$$ -direction:
$E_{R2x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$

Using $E_{R2} = \sqrt{E_{R2x}^2 + E_{R2y}^2}$ and the identity $\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$ :

Answer 2

$E_{R2} = \frac{\sigma}{\epsilon_0}\cos{\frac{\alpha}{2}}$
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