Edits to “Statement”, “Solution”, “Answer 1”

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−### Statement
+### 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 $\pm \sigma$?
+\\\\\\\$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\\\\\\\$?
+![ For problem \\\\\\\$6.2.10\\\\\\\$ |525x215, 42%](../../img/6.2.10/statement.png)
+
### Solution
−The electric field intensity for a single infinite plane with surface charge density $\sigma$ is:
+Consider the following figure...
−$$E = \frac{\sigma}{2\epsilon_0}$$
+![ Field Analysis |601x583, 51%](../../img/6.2.10/figure.png)
−#### Region 1
−In the $y$-direction:
−$$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\epsilon_0}(1-\cos{\alpha})$$
+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}\\\\\\\$.
−In the $x$-direction:
−$$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$
+For region 1 (see part \\\\\\\$a\\\\\\\$ of above figure):
−Using $E_{R1} = \sqrt{E_{R1x}^2 + E_{R1y}^2}$ and the identity $\sin{\frac{\alpha}{2}} = \sqrt{\frac{1-\cos{\alpha}}{2}}$:
+\\\\\\\$\\\\\\\$
+E_{R1y} = E - E \\\\\\\\cos{\\\\\\\\alpha} = \\\\\\\\frac{\\\\\\\\sigma}{2\\\\\\\\varepsilon_0}(1-\\\\\\\\cos{\\\\\\\\alpha})
+\\\\\\\$\\\\\\\$
−#### Answer 1:
−$$E_{R1} = \frac{\sigma}{\epsilon_0}\sin{\frac{\alpha}{2}}$$
+and for \\\\\\\$x\\\\\\\$-direction,
+\\\\\\\$\\\\\\\$
+E_{R1x} = E \\\\\\\\sin{\\\\\\\\alpha} = \\\\\\\\frac{\\\\\\\\sigma}{2\\\\\\\\varepsilon_0}\\\\\\\\sin{\\\\\\\\alpha}.
+\\\\\\\$\\\\\\\$
−#### Region 2
−In the $y$-direction:
−$$E_{R2y} = E(1+\cos{\alpha}) = \frac{\sigma}{2\epsilon_0}(1+\cos{\alpha})$$
+So, as
−In the $x$-direction:
−$$E_{R2x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$
+\\\\\\\$\\\\\\\$
+E_{R1} = \\\\\\\\sqrt{E_{R1x}^2+E_{R1y}^2},
+\\\\\\\$\\\\\\\$
−Using $E_{R2} = \sqrt{E_{R2x}^2 + E_{R2y}^2}$ and the identity $\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$:
+and taking in account that
−#### Answer 2:
−$$E_{R2} = \frac{\sigma}{\epsilon_0}\cos{\frac{\alpha}{2}}$$
+\\\\\\\$\\\\\\\$
+\\\\\\\\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}}
+\\\\\\\$\\\\\\\$
+
+
+
+
+
+