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Научный форум dxdyМатематика, Физика, Computer Science, Machine Learning, LaTeX, Механика и Техника, Химия,Биология и Медицина, Экономика и Финансовая Математика, Гуманитарные науки |
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| Страница 2 из 3 |
[ Сообщений: 34 ] | На страницу Пред. 1, 2, 3 След. |
03.12.2014, 14:54 
![$\[{{z^2}}\]$](https://dxdy-04.korotkov.co.uk/f/f/5/1/f5178210cb6317f21fc34cfd3319cf6b82.png "$\[{{z^2}}\]$")
)
, по-моему, вышло бы проще).![$\[\begin{array}{l}
\varphi (z) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\frac{{\ln (2\sqrt 1 \sqrt {{z^2} - 2{z_0}z + {R^2} + z_0^2} + 2z - 2{z_0})}}{{\sqrt 1 }}\mathop {\left. {} \right|}\limits_{ - h}^h = \\
= \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{2\sqrt {{h^2} - 2{z_0}h + {R^2} + z_0^2} + 2h - 2{z_0}}}{{2\sqrt {{h^2} + 2{z_0}h + {R^2} + z_0^2} - 2h - 2{z_0}}} = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - {z_0})}^2} + {R^2}} + h - {z_0}}}{{\sqrt {{{(h + {z_0})}^2} + {R^2}} + h - {z_0}}}
\end{array}\]$](https://dxdy-03.korotkov.co.uk/f/2/5/3/25325586b56adc0f023d8312e33ce0f782.png "$\[\begin{array}{l}
\varphi (z) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\frac{{\ln (2\sqrt 1 \sqrt {{z^2} - 2{z_0}z + {R^2} + z_0^2} + 2z - 2{z_0})}}{{\sqrt 1 }}\mathop {\left. {} \right|}\limits_{ - h}^h = \\
= \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{2\sqrt {{h^2} - 2{z_0}h + {R^2} + z_0^2} + 2h - 2{z_0}}}{{2\sqrt {{h^2} + 2{z_0}h + {R^2} + z_0^2} - 2h - 2{z_0}}} = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - {z_0})}^2} + {R^2}} + h - {z_0}}}{{\sqrt {{{(h + {z_0})}^2} + {R^2}} + h - {z_0}}}
\end{array}\]$")
![$\[\varphi (z) = \frac{\sigma }{{2{\varepsilon _0}}}(\sqrt {{{(z + h)}^2} + {R^2}} - \left| {z + h} \right|)\]$](https://dxdy-04.korotkov.co.uk/f/b/3/b/b3b825b1a8127f4174824c03efa1f17c82.png "$\[\varphi (z) = \frac{\sigma }{{2{\varepsilon _0}}}(\sqrt {{{(z + h)}^2} + {R^2}} - \left| {z + h} \right|)\]$")
- потенциал диска в точке ![$\[z = -h\]$](https://dxdy-04.korotkov.co.uk/f/3/9/1/391b5a4af29e5910a2b136948298c1d782.png "$\[z = -h\]$")
в формуле?![$\[{z_0}\]$](https://dxdy-01.korotkov.co.uk/f/8/6/1/861575a6adc3d1fca56a37d6428322b482.png "$\[{z_0}\]$")
- это точка центра кольца/диска.![$\[{z_0} = - h\]$](https://dxdy-01.korotkov.co.uk/f/4/e/7/4e7e9b6faaefc395f20fd3b95fa0c6df82.png "$\[{z_0} = - h\]$")
и 
, но никаких ![$\[e\]$](https://dxdy-03.korotkov.co.uk/f/e/4/f/e4f49ad878d2295008a8ffbfbcf52f6882.png "$\[e\]$")
вдоль оси ![$\[Oz\]$](https://dxdy-01.korotkov.co.uk/f/8/8/4/884a580ff95b7370603cfc8dc92d8d6982.png "$\[Oz\]$")
из точки с координатой ![$\[z = 0\]$](https://dxdy-01.korotkov.co.uk/f/8/2/0/820a12c6faf5e8d6ae910aca4de01fe882.png "$\[z = 0\]$")
в точку ![$\[z = 2h\]$](https://dxdy-04.korotkov.co.uk/f/7/f/2/7f2cdf06264c80c95d4062502ee92c4a82.png "$\[z = 2h\]$")
? 
, и их центр совмещен с началом координат. Перепишите эти формулы для объекта, лежащего в плоскости 
, с центром в произвольной точке оси 
.![$\[d\phi = \frac{{{q_1}\frac{{dz}}{{2h}}}}{{4\pi {\varepsilon _0}{{({R^2} + {{(z - {z_0})}^2})}^{1/2}}}} = \frac{{{q_1}dz}}{{8h\pi {\varepsilon _0}{{({R^2} + {{(z - {z_0})}^2})}^{1/2}}}}\]$](https://dxdy-01.korotkov.co.uk/f/4/3/c/43cc9a0e29a36f03f54b6a9c3fabeedf82.png "$\[d\phi = \frac{{{q_1}\frac{{dz}}{{2h}}}}{{4\pi {\varepsilon _0}{{({R^2} + {{(z - {z_0})}^2})}^{1/2}}}} = \frac{{{q_1}dz}}{{8h\pi {\varepsilon _0}{{({R^2} + {{(z - {z_0})}^2})}^{1/2}}}}\]$")
.
![$\[ = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\int\limits_{ - h}^h {\frac{{dz}}{{\sqrt {{z^2} - 2{z_0}z + {R^2} + z_0^2} }}} = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\frac{{\ln (2\sqrt 1 \sqrt {{z^2} - 2{z_0}z + {R^2} + z_0^2} + 2z - 2{z_0})}}{{\sqrt 1 }}\mathop {\left. {} \right|}\limits_{ - h}^h \]$](https://dxdy-02.korotkov.co.uk/f/9/9/2/9929da1270a712f34336f63c18b4906182.png "$\[ = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\int\limits_{ - h}^h {\frac{{dz}}{{\sqrt {{z^2} - 2{z_0}z + {R^2} + z_0^2} }}} = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\frac{{\ln (2\sqrt 1 \sqrt {{z^2} - 2{z_0}z + {R^2} + z_0^2} + 2z - 2{z_0})}}{{\sqrt 1 }}\mathop {\left. {} \right|}\limits_{ - h}^h \]$")
. И вспомните, что ![$\[\varphi ({z_0}) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - {z_0})}^2} + {R^2}} + h - {z_0}}}{{\sqrt {{{(h + {z_0})}^2} + {R^2}} + h - {z_0}}}\]$](https://dxdy-01.korotkov.co.uk/f/c/4/3/c43ad7b2e72587be1762019a0e40e4a682.png "$\[\varphi ({z_0}) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - {z_0})}^2} + {R^2}} + h - {z_0}}}{{\sqrt {{{(h + {z_0})}^2} + {R^2}} + h - {z_0}}}\]$")
![$\[\varphi (0) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{h^2} + {R^2}} + {h_0}}}{{\sqrt {{h^2} + {R^2}} + {h_0}}} = 0\]$](https://dxdy-01.korotkov.co.uk/f/c/3/a/c3ab98029f3d3bc66937aaaaf71fbe5e82.png "$\[\varphi (0) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{h^2} + {R^2}} + {h_0}}}{{\sqrt {{h^2} + {R^2}} + {h_0}}} = 0\]$")
![$\[\varphi (2h) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - 2h)}^2} + {R^2}} + h - 2h}}{{\sqrt {{{(h + 2h)}^2} + {R^2}} + h - 2h}} = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{h^2} + {R^2}} - h}}{{\sqrt {9{h^2} + {R^2}} - h}}\]$](https://dxdy-02.korotkov.co.uk/f/1/e/5/1e536ea6d96d5361426aff83323dd8c882.png "$\[\varphi (2h) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - 2h)}^2} + {R^2}} + h - 2h}}{{\sqrt {{{(h + 2h)}^2} + {R^2}} + h - 2h}} = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{h^2} + {R^2}} - h}}{{\sqrt {9{h^2} + {R^2}} - h}}\]$")
![$\[\varphi ({z_0}) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - {z_0})}^2} + {R^2}} + h - {z_0}}}{{\sqrt {{{(h + {z_0})}^2} + {R^2}} {\bf- h} - {z_0}}}\]$](https://dxdy-02.korotkov.co.uk/f/1/1/e/11e1f0d33344a9ba063014a7d55abd9582.png "$\[\varphi ({z_0}) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - {z_0})}^2} + {R^2}} + h - {z_0}}}{{\sqrt {{{(h + {z_0})}^2} + {R^2}} {\bf- h} - {z_0}}}\]$")
![$\[\begin{array}{l}
\varphi ({z_0}) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - {z_0})}^2} + {R^2}} + h - {z_0}}}{{\sqrt {{{(h + {z_0})}^2} + {R^2}} - h - {z_0}}}\\
\varphi (0) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{h^2} + {R^2}} + {h_{}}}}{{\sqrt {{h^2} + {R^2}} - {h_{}}}} = 0\\
\varphi (2h) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - 2h)}^2} + {R^2}} + h - 2h}}{{\sqrt {{{(h + 2h)}^2} + {R^2}} - h - 2h}} = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{h^2} + {R^2}} - h}}{{\sqrt {9{h^2} + {R^2}} - 3h}}
\end{array}\]$](https://dxdy-01.korotkov.co.uk/f/8/8/7/8870cca5948b222daa3b9ddedc76132782.png "$\[\begin{array}{l}
\varphi ({z_0}) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - {z_0})}^2} + {R^2}} + h - {z_0}}}{{\sqrt {{{(h + {z_0})}^2} + {R^2}} - h - {z_0}}}\\
\varphi (0) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{h^2} + {R^2}} + {h_{}}}}{{\sqrt {{h^2} + {R^2}} - {h_{}}}} = 0\\
\varphi (2h) = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{{(h - 2h)}^2} + {R^2}} + h - 2h}}{{\sqrt {{{(h + 2h)}^2} + {R^2}} - h - 2h}} = \frac{{{q_1}}}{{8h\pi {\varepsilon _0}}}\ln \frac{{\sqrt {{h^2} + {R^2}} - h}}{{\sqrt {9{h^2} + {R^2}} - 3h}}
\end{array}\]$")
- это одно и тоже или нет (я этого не знаю!), и написать ответ.