Решебник по геометрии 11 класс. Атанасян ФГОС 627

\[\boxed{\mathbf{627.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\ \]

\[\text{ABCD}A_{1}B_{1}C_{1}D_{1} -\]

\[параллелепипед;\ \ \]

\[AB = AD = a;\ \]

\[\text{\ A}A_{1} = 2a;\]

\[в\ вершинах\ B_{1}\ и\ D_{1}\ помещены\ \]

\[заряды\ q;\ \ \]

\[в\ вершине\ A - заряд\ 2q.\]

\[Найти:\ \ \]

\[результирующую\ \]

\[напряженность\ поля\]

\[\textbf{а)}\ в\ точке\ A_{1};\ \ \]

\[\textbf{б)}\ в\ точке\ C;\ \ \]

\[\textbf{в)}\ в\ центре\ грани\ A_{1}B_{1}C_{1}D_{1};\]

\[\textbf{г)}\ в\ центре\ грани\ ABCD.\]

\[Решение.\]

\[\textbf{а)}\ Результирующая\ \]

\[напряженность\ поля\ в\ \]

\[точке\ A_{1}:\]

\[Пусть\ вектор\ \frac{1}{4}\overrightarrow{AA_{1}} = \overrightarrow{d};\ \ \ \]

\[\left| \overrightarrow{d} \right| = \frac{1}{4} \bullet 2a = \frac{a}{2}:\]

\[\overrightarrow{B_{1}A_{1}} + \overrightarrow{D_{1}A_{1}} = - \overrightarrow{A_{1}B_{1}} - \overrightarrow{A_{1}D_{1}} =\]

\[= - \left( \overrightarrow{A_{1}B_{1}} + \overrightarrow{A_{1}D_{1}} \right) = - \overrightarrow{A_{1}C_{1}} =\]

\[= \overrightarrow{C_{1}A_{1}};\]

\[\left| \overrightarrow{C_{1}A_{1}} \right| = \sqrt{a^{2} + a^{2}} = a\sqrt{2}.\]

\[Пусть\ вектор\ \overrightarrow{C_{1}A_{1}} = \overrightarrow{e};\ \ \ \]

\[\overrightarrow{e} + \overrightarrow{d} = \overrightarrow{f};\ \text{\ \ }\overrightarrow{d}\bot\overrightarrow{e}:\]

\[\left| \overrightarrow{f} \right| = \sqrt{\left| \overrightarrow{e} \right|^{2} + \left| \overrightarrow{d} \right|^{2}} =\]

\[= \sqrt{2a^{2} + \frac{a^{2}}{4}} = \sqrt{\frac{9a^{2}}{4}} = \frac{3a}{2}.\]

\[Таким\ образом:\ \]

\[\overrightarrow{E} = \frac{\text{kq}}{a^{3}} \bullet \frac{3a}{2} = \frac{3kq}{2a^{2}}.\]

\[\textbf{б)}\ Результирующая\ \]

\[напряженность\ поля\ в\ \]

\[точке\ C:\]

\[= \frac{\text{kq}}{a^{3}}\left( \frac{\overrightarrow{\text{AC}}}{\sqrt{2}} + \frac{\overrightarrow{B_{1}C} + \overrightarrow{D_{1}C}}{5\sqrt{5}} \right) =\]

\[= \frac{\text{kq}}{a^{3}}\left( \frac{\overrightarrow{A_{1}C_{1}}}{\sqrt{2}} + \frac{\overrightarrow{A_{1}D} + \overrightarrow{A_{1}B}}{5\sqrt{5}} \right).\]

\[A_{1}D = A_{1}B = \sqrt{2a^{2} + a^{2}} = a\sqrt{5}:\]

\[A_{1}BND - ромб;\]

\[\overrightarrow{A_{1}B} + \overrightarrow{A_{1}D} = \overrightarrow{A_{1}N}.\]

\[По\ теореме\ синусов\ в\ \mathrm{\Delta}BA_{1}D:\]

\[2a^{2} = 10a^{2} - 10a^{2} \bullet \cos a\]

\[10\cos a = 8\]

\[\cos a = \frac{4}{5}.\]

\[По\ теореме\ косинусов\ в\ \mathrm{\Delta}A_{1}BN:\]

\[= 10a^{2} + 10a^{2} \bullet \cos a =\]

\[= 10a^{2} + 10a^{2} \bullet \frac{4}{5} = 18a^{2}\ \]

\[\left| \overrightarrow{A_{1}N} \right| = 3\sqrt{2}\text{a.}\]

\[Пусть\ \overrightarrow{e} = \frac{\overrightarrow{A_{1}D} + \overrightarrow{A_{1}B}}{5\sqrt{5}} = \frac{\overrightarrow{A_{1}N}}{5\sqrt{5}} =\]

\[= \frac{3\sqrt{2}a}{5\sqrt{5}};\]

\[\overrightarrow{d} = \frac{\overrightarrow{A_{1}C_{1}}}{\sqrt{2}} = \frac{a\sqrt{2}}{\sqrt{2}} = a;\]

\[\angle NA_{1}C_{1} = \beta:\]

\[\cos\beta = \frac{a}{\sqrt{2}} \bullet \frac{1}{OA_{1}} =\]

\[= \sqrt{\left( \frac{a}{\sqrt{2}} \right)^{2} + 4a^{2}} = \sqrt{\frac{a^{2}}{2} + 4a^{2}} =\]

\[= \frac{3a}{\sqrt{2}}\]

\[\cos\beta = \frac{a}{\sqrt{2}} \bullet \frac{\sqrt{2}}{3a} = \frac{1}{3}.\]

\[\overrightarrow{A_{1}H} = \overrightarrow{d} + \overrightarrow{e};\]

\[= a^{2} + a^{2} \bullet \frac{18}{125} + \frac{2 \bullet 3\sqrt{2} \bullet a^{2}}{5\sqrt{5}} \bullet \frac{1}{3} =\]

\[= \frac{143a^{2}}{125} + \frac{2\sqrt{2}a^{2}}{5\sqrt{5}} =\]

\[= \frac{143a^{2}}{125} + \frac{2\sqrt{2} \bullet 5\sqrt{5}}{\left( 5\sqrt{5} \right)^{2}}a^{2} =\]

\[= \frac{143 + 10\sqrt{10}}{125}a^{2}.\]

\[Таким\ образом:\]

\[\overrightarrow{E} = \frac{\text{kq}}{a^{3}} \bullet \frac{a}{5\sqrt{5}} \bullet \sqrt{143 + 10\sqrt{10}} =\]

\[= \frac{\sqrt{143 + 10\sqrt{10}}}{5\sqrt{5}a^{2}}\text{kq.}\]

\[\textbf{в)}\ Результирующая\ \]

\[напряженность\ поля\ в\ \]

\[точке\ O_{1}:\]

\[\overrightarrow{D_{1}O_{1}} = - \overrightarrow{B_{1}O_{1}}:\]

\[\overrightarrow{E} = \frac{k2q}{27a^{3}} \bullet 2\sqrt{2} \bullet \overrightarrow{AO_{1}};\]

\[\overrightarrow{E} = \frac{4\sqrt{2}}{27a^{3}} \bullet \frac{3a}{\sqrt{2}} = \frac{4}{9a^{3}} \bullet kq.\]

\[\textbf{г)}\ Результирующая\ \]

\[напряженность\ поля\ в\ \]

\[точке\ O:\]

\[AO = \frac{1}{2}AC = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}};\ \ \ \]

\[D_{1}O = B_{1}O = \frac{3a}{\sqrt{2}};\]

\[\overrightarrow{B_{1}O} + \overrightarrow{D_{1}O} = \overrightarrow{D_{1}O} + \overrightarrow{D_{1}E} = \overrightarrow{D_{1}N}.\]

\[По\ теореме\ косинусов\ в\ \mathrm{\Delta}OD_{1}E:\]

\[2a^{2} = 2 \bullet \frac{9a^{2}}{2} - 2 \bullet \frac{9a^{2}}{2} \bullet \cos a\ \]

\[9\cos a = 7\]

\[\cos a = \frac{7}{9}.\]

\[По\ теореме\ косинусов\ в\ \mathrm{\Delta}D_{1}EN:\]

\[\left| \overrightarrow{D_{1}N} \right|^{2} =\]

\[= 2 \bullet \frac{9a^{2}}{2} - 2 \bullet \frac{9a^{2}}{2} \bullet \cos(180{^\circ} - a) =\]

\[= 9a^{2} + 9a^{2} \bullet \frac{7}{9} = 9a^{2} + 7a^{2} =\]

\[= 16a^{2}\]

\[\left| \overrightarrow{D_{1}N} \right| = 4a.\]

\[Пусть\ \overrightarrow{d} = \frac{2\sqrt{2}}{27}\left( \overrightarrow{D_{1}O} + \overrightarrow{B_{1}O} \right) =\]

\[= \frac{2\sqrt{2}}{27}\overrightarrow{D_{1}N}:\]

\[\left| \overrightarrow{d} \right| = \frac{2\sqrt{2}}{27} \bullet 4a = \frac{8\sqrt{2}a}{27};\]

\[Пусть\ \overrightarrow{\text{AO}} \bullet 4\sqrt{2} = \overrightarrow{e}:\ \]

\[\left| \overrightarrow{e} \right| = 4\sqrt{2} \bullet \frac{a}{\sqrt{2}} = 4a;\]

\[\left| \overrightarrow{e} + \overrightarrow{d} \right|^{2} = \left| \overrightarrow{e} \right|^{2} + \left| \overrightarrow{d} \right|^{2} =\]

\[= (4a)^{2} + \left( \frac{8\sqrt{2}a}{27} \right)^{2} =\]

\[= 16a^{2} + \frac{128a^{2}}{729} =\]

\[= \frac{11664 + 128}{729}a^{2} =\]

\[= \frac{11792}{729}a^{2} = \frac{16 \bullet 737}{729}a^{2}\]

\[\left| \overrightarrow{e} + \overrightarrow{d} \right| = \sqrt{\frac{16 \bullet 737}{729}a^{2}} =\]

\[= \frac{4\sqrt{737}a}{27}.\]

\[Таким\ образом\]

\[\overrightarrow{E} = \frac{\text{kq}}{a^{3}} \bullet \frac{4a\sqrt{737}}{729} = \frac{4\sqrt{737}}{27a^{2}}\text{kq.}\]

\[Ответ:\ \ а)\frac{3}{2a^{2}}kq;\ \ \]

\[\textbf{б)}\ \frac{\sqrt{143 + 10\sqrt{10}}}{5\sqrt{5}a^{2}}kq;\ \ \]

\[\textbf{в)}\frac{4}{9a^{2}}kq;\ \ г)\frac{4\sqrt{737}}{27a^{2}}\text{kq.}\]