Правка разделов «Statement», «Region 1», «Answer 1»
en/6.2.10.md
+3 −6
| @@ -1,9 +1,7 @@ | |||
| ### 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 | |||
| The electric field intensity for a single infinite plane with surface charge density $\sigma$ is: | |||
| $$E = \frac{\sigma}{2\epsilon_0}$$ | |||
| #### Region 1 | |||
| In the $y$-direction: | |||
| $$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\epsilon_0}(1-\cos{\alpha})$$ | |||
| In the $x$-direction: | |||
| $$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$ | |||
| @@ -19,10 +17,9 @@Region 1 | |||
| Using $E_{R1} = \sqrt{E_{R1x}^2 + E_{R1y}^2}$ and the identity $\sin{\frac{\alpha}{2}} = \sqrt{\frac{1-\cos{\alpha}}{2}}$: | |||
| − | |||
| + | #### Answer 1: | ||
| $$E_{R1} = \frac{\sigma}{\epsilon_0}\sin{\frac{\alpha}{2}}$$ | |||
| − | --- | ||
| #### Region 2 | |||
| In the $y$-direction: | |||
| $$E_{R2y} = E(1+\cos{\alpha}) = \frac{\sigma}{2\epsilon_0}(1+\cos{\alpha})$$ | |||
| In the $x$-direction: | |||
| $$E_{R2x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$ | |||
| @@ -33,5 +30,5 @@Region 2 | |||
| 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: | ||
| $$E_{R2} = \frac{\sigma}{\epsilon_0}\cos{\frac{\alpha}{2}}$$ | |||
| @@ -1,9 +1,7 @@ | |||
| ### 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 $\pm \sigma$? | |||
| --- | |||
| ### Solution | ### Solution | ||
| The electric field intensity for a single infinite plane with surface charge density $\sigma$ is: | The electric field intensity for a single infinite plane with surface charge density $\sigma$ is: | ||
| $$E = \frac{\sigma}{2\epsilon_0}$$ | $$E = \frac{\sigma}{2\epsilon_0}$$ | ||
| #### Region 1 | #### Region 1 | ||
| In the $y$-direction: | In the $y$-direction: | ||
| $$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\epsilon_0}(1-\cos{\alpha})$$ | $$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\epsilon_0}(1-\cos{\alpha})$$ | ||
| In the $x$-direction: | In the $x$-direction: | ||
| $$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$ | $$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$ | ||
| @@ -19,10 +17,9 @@Region 1 | |||
| Using $E_{R1} = \sqrt{E_{R1x}^2 + E_{R1y}^2}$ and the identity $\sin{\frac{\alpha}{2}} = \sqrt{\frac{1-\cos{\alpha}}{2}}$: | Using $E_{R1} = \sqrt{E_{R1x}^2 + E_{R1y}^2}$ and the identity $\sin{\frac{\alpha}{2}} = \sqrt{\frac{1-\cos{\alpha}}{2}}$: | ||
| #### Answer 1: | |||
| $$E_{R1} = \frac{\sigma}{\epsilon_0}\sin{\frac{\alpha}{2}}$$ | $$E_{R1} = \frac{\sigma}{\epsilon_0}\sin{\frac{\alpha}{2}}$$ | ||
| --- | |||
| #### Region 2 | #### Region 2 | ||
| In the $y$-direction: | In the $y$-direction: | ||
| $$E_{R2y} = E(1+\cos{\alpha}) = \frac{\sigma}{2\epsilon_0}(1+\cos{\alpha})$$ | $$E_{R2y} = E(1+\cos{\alpha}) = \frac{\sigma}{2\epsilon_0}(1+\cos{\alpha})$$ | ||
| In the $x$-direction: | In the $x$-direction: | ||
| $$E_{R2x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$ | $$E_{R2x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$ | ||
| @@ -33,5 +30,5 @@Region 2 | |||
| Using $E_{R2} = \sqrt{E_{R2x}^2 + E_{R2y}^2}$ and the identity $\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$: | 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: | |||
| $$E_{R2} = \frac{\sigma}{\epsilon_0}\cos{\frac{\alpha}{2}}$$ | $$E_{R2} = \frac{\sigma}{\epsilon_0}\cos{\frac{\alpha}{2}}$$ | ||