**Elliptik silindrning elektr maydoni**

Cheksiz uzun silindr sirti boʻylab zaryadlar bir jinsli taqsimlangan boʻlsin. Zaryadlarning sirt zichligi {\displaystyle \rho }. Shu bir jinsli cheksiz uzun silindr sirtidagi elektr maydonini hisoblaymiz.

 Ellipsning x va z oʻqlari boʻyicha yarim oʻqlari a va b ga teng deylik. U holda, zaryadlar hosil qiladigan potensial quyidagi koʻrinishda boʻladi:

{\displaystyle \varphi (x,z)=\rho \int \limits \_{-a}^{a}dx'\int \limits \_{-b{\sqrt {1-{\frac {x'^{2}}{a^{2}}}}}}^{b{\sqrt {1-{\frac {x'^{2}}{a^{2}}}}}}dz'\ln {\frac {1}{(x-x')^{2}+(z-z')^{2}}}}

Endi esa, zaryadlar silindrdan tashqarida hosil qilgan maydonning z oʻqi boʻyicha tashkil etuvchisini koʻrib chiqamiz:
 {\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {z^{2}}{b^{2}}}=1+\delta \ \ \ \ \ (0<\delta \ll 1)}
 Uni quyidagicha ifodalashimiz mumkin:
 {\displaystyle E\_{1}(x,z)\equiv -{\dfrac {\partial \varphi }{\partial z}}=\rho \int \limits \_{-a}^{a}dx'\int \limits \_{-b{\sqrt {1-{\frac {x'^{2}}{a^{2}}}}}}^{b{\sqrt {1-{\frac {x'^{2}}{a^{2}}}}}}dz'{\dfrac {\partial }{\partial z'}}\ln {\frac {1}{(x-x')^{2}+(z-z')^{2}}}=\rho \int \limits \_{-a}^{a}dx\prime \ln {\frac {(x-x')^{2}+\left(z+b{\sqrt {1-{\frac {x'^{2}}{a^{2}}}}}\right)^{2}}{(x-x')^{2}+\left(z-b{\sqrt {1-{\frac {x'^{2}}{a^{2}}}}}\right)^{2}}}}
 Bu integralni hisoblash uchun yangi oʻzgaruvchilar kiritamiz:
 {\displaystyle x=a\cos \psi ,x'=a\cos \psi ',b/a=\varepsilon ,\delta +\sin ^{2}\psi =\sin ^{2}\varphi }

 U holda {\displaystyle {\sqrt {1-{\frac {x'^{2}}{a^{2}}}}}=\sin \psi ',\ z=b\sin \varphi } boʻladi hamda
 {\displaystyle E\_{z}=a\rho \int \limits \_{0}^{\pi }\sin \psi 'd\psi '\ln {\frac {(\cos \psi '-\cos \psi )^{2}+\varepsilon ^{2}(\sin \psi '+\sin \varphi )^{2}}{(\cos \psi '-\cos \psi )^{2}+\varepsilon ^{2}(\sin \psi '-\sin \varphi )^{2}}}=a\rho \int \limits \_{0}^{\pi }\sin \psi 'd\psi '\ln {\frac {\sin ^{2}{\frac {\psi '+\psi }{2}}\sin ^{2}{\frac {\psi '-\psi }{2}}+\varepsilon ^{2}\sin ^{2}{\frac {\psi '+\varphi }{2}}\cos ^{2}{\frac {\psi '-\varphi }{2}}}{\sin ^{2}{\frac {\psi '+\psi }{2}}\sin ^{2}{\frac {\psi '-\psi }{2}}+\varepsilon ^{2}\sin ^{2}{\frac {\psi '+\varphi }{2}}\sin ^{2}{\frac {\psi '-\varphi }{2}}}}}

{\displaystyle \delta \rightarrow 0} boʻlganda, {\displaystyle \varphi \rightarrow \psi }. Bu oʻtishning uzluksizligidan foydalanib integral ostidagi ifodani quyidagicha yozib olishimiz mumkin:

{\displaystyle E\_{z}=a\rho \int \limits \_{0}^{\pi }\sin \psi 'd\psi '\ln {\frac {\sin ^{2}{\frac {\psi +\psi '}{2}}\left[\sin ^{2}{\frac {\psi -\psi '}{2}}+\varepsilon ^{2}\cos ^{2}{\frac {\psi -\psi '}{2}}\right]}{\sin ^{2}{\frac {\psi -\psi '}{2}}\left[\sin ^{2}{\frac {\psi +\psi '}{2}}+\varepsilon ^{2}\cos ^{2}{\frac {\psi +\psi '}{2}}\right]}}=a\rho \int \limits \_{0}^{\pi }\sin \varphi d\varphi \left[{\frac {1-\cos(\varphi +\psi )}{1-\cos(\varphi -\psi )}}{\frac {1+{\frac {\varepsilon ^{2}-1}{\varepsilon ^{2}+1}}\cos(\varphi -\psi )}{1+{\frac {\varepsilon ^{2}-1}{\varepsilon ^{2}+1}}\cos(\varphi +\psi )}}\right]=a\rho \int \limits \_{\pi +\varphi }^{\pi -\varphi }\sin(\varphi +\psi )d\varphi \ln {\frac {1+{\frac {\varepsilon ^{2}-1}{\varepsilon ^{2}+1}}\cos \varphi }{1-\cos \psi }}+a\rho \int \limits \_{\psi }^{\pi +\psi }\sin(\varphi -\psi )d\varphi \ln {\frac {1-\cos \psi }{1+{\frac {\varepsilon ^{2}-1}{\varepsilon ^{2}+1}}\cos \varphi }}}

Birinchi integralda integrallash oʻzgaruvchisini almashtiramiz: {\displaystyle \varphi \prime =\pi -\varphi }.

Natijada
{\displaystyle E\_{z}=a\rho \int \limits \_{\psi }^{\pi +\psi }\sin(\varphi -\psi )d\varphi \left[\ln {\frac {1+{\frac {\varepsilon ^{2}-1}{\varepsilon ^{2}+1}}\cos \varphi }{1-{\frac {\varepsilon ^{2}-1}{\varepsilon ^{2}+1}}\cos \varphi }}+\ln {\frac {1-\cos \varphi }{1+\cos \varphi }}\right]=a\rho \left[\Phi \left(\psi ,{\frac {\varepsilon ^{2}-1}{\varepsilon ^{2}+1}}\right)+\Phi (\psi ,1)\right]}

 bu yerda
 {\displaystyle \Phi (\psi ,\alpha )=\int \limits \_{\psi }^{\pi +\psi }\sin(\varphi -\psi )d\varphi \ln {\frac {1-\alpha \cos \varphi }{1+\alpha \cos \varphi }}}
Yuqoridagi integralni boʻlaklab integrallash mumkin:
 {\displaystyle \Phi (\psi ,\alpha )=-\cos(\varphi -\psi )\ln {\frac {1-\alpha \cos \varphi }{1+\alpha \cos \varphi }}{\Big |}\_{\psi }^{\pi +\psi }+2\pi \int \limits \_{\psi }^{\pi +\psi }{\frac {\cos(\varphi +\psi )\sin \varphi }{1-\alpha ^{2}\cos ^{2}\varphi }}+2\alpha \sin \psi \int \limits \_{\psi }^{\pi +\psi }{\frac {\sin ^{2}\varphi d\varphi }{1-\alpha ^{2}\cos ^{2}\varphi }}}

 Koʻrinib turibdiki, yuqorida berilgan integraldagi birinchi had nolga teng. Ikkinchi hadda integral ostidagi funksiyaning {\displaystyle (0,\psi )} va {\displaystyle (\pi ,\pi +\psi )}
 intervallardagi qiymati bir-biriga mos keladi. Natijada, {\displaystyle \Phi (\psi ,\alpha )} ni quyidagi koʻrinishda yozishimiz mumkin:

{\displaystyle \Phi (\psi ,\alpha )=2\alpha \sin \psi \int \limits \_{0}^{\pi }{\frac {\sin ^{2}\varphi d\varphi }{1-\alpha ^{2}\cos ^{2}\varphi }}}

{\displaystyle \alpha =1} hol uchun integral oson hisoblanadi va {\displaystyle 2\pi \sin \psi } ga teng. {\displaystyle \alpha \neq 1} uchun esa

{\displaystyle \Phi (\psi ,\alpha )={\frac {2\pi \sin \psi }{\alpha }}\left[1-(1-\alpha ^{2})\int \limits \_{0}^{\pi }{\dfrac {d\varphi }{1-\alpha ^{2}\cos ^{2}\varphi }}\right]}

 yoki

{\displaystyle \Phi (\psi ,\alpha )={\frac {2\pi \sin \psi }{\alpha }}\left(1-{\sqrt {1-\alpha ^{2}}}\right)}

 Shundan soʻng, (12) tenglamani (7)ifodagaa qoʻyamiz va (4)formulani hisobga olgan holda, elliptik silindr maydoni uchun quyidagi ifodani yozamiz:

{\displaystyle E\_{x}(x,z)=4\pi \rho {\dfrac {a}{b+a}}zE\_{z}(x,z)=4\pi \rho {\dfrac {b}{b+a}}x}

Yana qarang
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* Zaryad elektr maydonining asosiy qonuni

Adabiyotlar
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* *M.Reiser*, IEEE Trans. Nucl. Sci. 13, 171 (1966)