Edits to “Answer 1”, “Region 2”, “Answer 2”

Anonymous edited Mar 18, 2026, 4:08 AM

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			### 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 $\pm \sigma$?

			![ 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 $E = \frac{\sigma}{2\varepsilon_0}$.

			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

			$$
		@@ -44,18 +44,15 @@Answer 1
			E_{R1} = \frac{\sigma}{\varepsilon_0}\sin{\frac{\alpha}{2}}
			$$

		−	~~For region 2, for~~ $y$-direction:
		+	#### Region 2
		+	In the $y$-direction:
		+	$$E_{R2y} = E(1+\cos{\alpha}) = \frac{\sigma}{2\epsilon_0}(1+\cos{\alpha})$$

		−	$$
		−	E_{R2y} = E~~(1+\cos~~{\alpha})=\frac{\sigma}{2~~\varepsilon~~_0}~~(1+\cos~~{\alpha})
		−	$$
		+	In the $x$-direction:
		+	$$E_{R2x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$

		−	and for $x$-axis,
		+	Using $E_{R2} = \sqrt{E_{R2x}^2 + E_{R2y}^2}$ and the identity $\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$:

		−	$$
		−	E_{R2x} = ~~E\sin{\alpha} = \frac~~{\sigma}{~~2\varepsilon~~_0}~~\sin{\alpha}.~~
		−	$$
		−
		−	Again,
		−
		+	#### Answer 2:
		+	$$E_{R2} = \frac{\sigma}{\epsilon_0}\cos{\frac{\alpha}{2}}$$
			$$

	### 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		### Solution

	Consider the following figure...		Consider the following figure...

	![ Field Analysis \|601x583, 51%](../../img/6.2.10/figure.png)		![ 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		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		and taking in account that

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

	it is obtained		it is obtained

	#### Answer 1		#### Answer 1

	$$		$$
	@@ -44,18 +44,15 @@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:		#### Region 2
			In the $y$-direction:
			$$E_{R2y} = E(1+\cos{\alpha}) = \frac{\sigma}{2\epsilon_0}(1+\cos{\alpha})$$

	$$		In the $x$-direction:
	E_{R2y} = E~~(1+\cos~~{\alpha})=\frac{\sigma}{2~~\varepsilon~~_0}~~(1+\cos~~{\alpha})		$$E_{R2x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$
	$$

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

	$$		#### Answer 2:
	E_{R2x} = ~~E\sin{\alpha} = \frac~~{\sigma}{~~2\varepsilon~~_0}~~\sin{\alpha}.~~		$$E_{R2} = \frac{\sigma}{\epsilon_0}\cos{\frac{\alpha}{2}}$$
	$$

	Again,

	$$		$$