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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)
+![ For problem $6.2.10$ |525x215, 42%](../../img/6.2.10/statement.png)
### Solution
Consider the following figure...
![ Field Analysis |601x583, 51%](../../img/6.2.10/figure.png)
−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}\\\\\\\$.
+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):
+For region 1 (see part $a$ of above figure):
−\\\\\\\$\\\\\\\$
−E_{R1y} = E - E \\\\\\\\cos{\\\\\\\\alpha} = \\\\\\\\frac{\\\\\\\\sigma}{2\\\\\\\\varepsilon_0}(1-\\\\\\\\cos{\\\\\\\\alpha})
−\\\\\\\$\\\\\\\$
+$$
+E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})
+$$
−and for \\\\\\\$x\\\\\\\$-direction,
+and for $x$-direction,
−\\\\\\\$\\\\\\\$
−E_{R1x} = E \\\\\\\\sin{\\\\\\\\alpha} = \\\\\\\\frac{\\\\\\\\sigma}{2\\\\\\\\varepsilon_0}\\\\\\\\sin{\\\\\\\\alpha}.
−\\\\\\\$\\\\\\\$
+$$
+E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}.
+$$
So, as
−\\\\\\\$\\\\\\\$
−E_{R1} = \\\\\\\\sqrt{E_{R1x}^2+E_{R1y}^2},
−\\\\\\\$\\\\\\\$
+$$
+E_{R1} = \sqrt{E_{R1x}^2+E_{R1y}^2},
+$$
and taking in account that
−\\\\\\\$\\\\\\\$
−\\\\\\\\sin{\\\\\\\\alpha} = \\\\\\\\sqrt{\\\\\\\\frac{1-\\\\\\\\cos{\\\\\\\\alpha}}{2}},
−\\\\\\\$\\\\\\\$
+$$
+\sin{\frac{\alpha}{2}} = \sqrt{\frac{1-\cos{\alpha}}{2}},
+$$
it is obtained
#### Answer 1
−\\\\\\\$\\\\\\\$
−E_{R1} = \\\\\\\\frac{\\\\\\\\sigma}{\\\\\\\\varepsilon_0}\\\\\\\\sin{\\\\\\\\frac{\\\\\\\\alpha}{2}}
−\\\\\\\$\\\\\\\$
+$$
+E_{R1} = \frac{\sigma}{\varepsilon_0}\sin{\frac{\alpha}{2}}
+$$
−For region 2, for \\\\\\\$y\\\\\\\$-direction:
+For region 2, for $y$-direction:
−\\\\\\\$\\\\\\\$
−E_{R2y} = E(1+\\\\\\\\cos{\\\\\\\\alpha})=\\\\\\\\frac{\\\\\\\\sigma}{2\\\\\\\\varepsilon_0}(1+\\\\\\\\cos{\\\\\\\\alpha})
−\\\\\\\$\\\\\\\$
+$$
+E_{R2y} = E(1+\cos{\alpha})=\frac{\sigma}{2\varepsilon_0}(1+\cos{\alpha})
+$$
−and for \\\\\\\$x\\\\\\\$-axis,
+and for $x$-axis,
−\\\\\\\$\\\\\\\$
−E_{R2x} = E\\\\\\\\sin{\\\\\\\\alpha} = \\\\\\\\frac{\\\\\\\\sigma}{2\\\\\\\\varepsilon_0}\\\\\\\\sin{\\\\\\\\alpha}.
−\\\\\\\$\\\\\\\$
+$$
+E_{R2x} = E\sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}.
+$$
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}}
−\\\\\\\$\\\\\\\$
−
−
−
−
−
−
+$$