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{\\\\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}}
\\$\\$
Answer 2
\\$\\$
E_{R2} = \\\\frac{\\\\sigma}{\\\\varepsilon_0}\\\\cos{\\\\frac{\\\\alpha}{2}}
\\$\\$