Elliptik silindrning elektr maydoni
Elliptik silindrning elektr maydoni Cheksiz uzun silindr sirti boʻylab zaryadlar bir jinsli taqsimlangan boʻlsin. Zaryadlarning sirt zichligi ρ {\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: φ ( x , z ) = ρ ∫ − a a d x ′ ∫ − b 1 − x ′ 2 a 2 b 1 − x ′ 2 a 2 d z ′ ln ⁡ 1 ( x − x ′ ) 2 + ( z − z ′ ) 2 {\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: x 2 a 2 + z 2 b 2 = 1 + δ           ( 0 < δ ≪ 1 ) {\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {z^{2}}{b^{2}}}=1+\delta \ \ \ \ \ (0<\delta \ll 1)} Uni quyidagicha ifodalashimiz mumkin: E 1 ( x , z ) ≡ − ∂ φ ∂ z = ρ ∫ − a a d x ′ ∫ − b 1 − x ′ 2 a 2 b 1 − x ′ 2 a 2 d z ′ ∂ ∂ z ′ ln ⁡ 1 ( x − x ′ ) 2 + ( z − z ′ ) 2 = ρ ∫ − a a d x ′ ln ⁡ ( x − x ′ ) 2 + ( z + b 1 − x ′ 2 a 2 ) 2 ( x − x ′ ) 2 + ( z − b 1 − x ′ 2 a 2 ) 2 {\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: x = a cos ⁡ ψ , x ′ = a cos ⁡ ψ ′ , b / a = ε , δ + sin 2 ⁡ ψ = sin 2 ⁡ φ {\displaystyle x=a\cos \psi ,x'=a\cos \psi ',b/a=\varepsilon ,\delta +\sin ^{2}\psi =\sin ^{2}\varphi } U holda 1 − x ′ 2 a 2 = sin ⁡ ψ ′ ,   z = b sin ⁡ φ {\displaystyle {\sqrt {1-{\frac {x'^{2}}{a^{2}}}}}=\sin \psi ',\ z=b\sin \varphi } boʻladi hamda E z = a ρ ∫ 0 π sin ⁡ ψ ′ d ψ ′ ln ⁡ ( cos ⁡ ψ ′ − cos ⁡ ψ ) 2 + ε 2 ( sin ⁡ ψ ′ + sin ⁡ φ ) 2 ( cos ⁡ ψ ′ − cos ⁡ ψ ) 2 + ε 2 ( sin ⁡ ψ ′ − sin ⁡ φ ) 2 = a ρ ∫ 0 π sin ⁡ ψ ′ d ψ ′ ln ⁡ sin 2 ⁡ ψ ′ + ψ 2 sin 2 ⁡ ψ ′ − ψ 2 + ε 2 sin 2 ⁡ ψ ′ + φ 2 cos 2 ⁡ ψ ′ − φ 2 sin 2 ⁡ ψ ′ + ψ 2 sin 2 ⁡ ψ ′ − ψ 2 + ε 2 sin 2 ⁡ ψ ′ + φ 2 sin 2 ⁡ ψ ′ − φ 2 {\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}}}}} δ → 0 {\displaystyle \delta \rightarrow 0} boʻlganda, φ → ψ {\displaystyle \varphi \rightarrow \psi } . Bu oʻtishning uzluksizligidan foydalanib integral ostidagi ifodani quyidagicha yozib olishimiz mumkin: E z = a ρ ∫ 0 π sin ⁡ ψ ′ d ψ ′ ln ⁡ sin 2 ⁡ ψ + ψ ′ 2 [ sin 2 ⁡ ψ − ψ ′ 2 + ε 2 cos 2 ⁡ ψ − ψ ′ 2 ] sin 2 ⁡ ψ − ψ ′ 2 [ sin 2 ⁡ ψ + ψ ′ 2 + ε 2 cos 2 ⁡ ψ + ψ ′ 2 ] = a ρ ∫ 0 π sin ⁡ φ d φ [ 1 − cos ⁡ ( φ + ψ ) 1 − cos ⁡ ( φ − ψ ) 1 + ε 2 − 1 ε 2 + 1 cos ⁡ ( φ − ψ ) 1 + ε 2 − 1 ε 2 + 1 cos ⁡ ( φ + ψ ) ] = a ρ ∫ π + φ π − φ sin ⁡ ( φ + ψ ) d φ ln ⁡ 1 + ε 2 − 1 ε 2 + 1 cos ⁡ φ 1 − cos ⁡ ψ + a ρ ∫ ψ π + ψ sin ⁡ ( φ − ψ ) d φ ln ⁡ 1 − cos ⁡ ψ 1 + ε 2 − 1 ε 2 + 1 cos ⁡ φ {\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: φ ′ = π − φ {\displaystyle \varphi \prime =\pi -\varphi } . Natijada E z = a ρ ∫ ψ π + ψ sin ⁡ ( φ − ψ ) d φ [ ln ⁡ 1 + ε 2 − 1 ε 2 + 1 cos ⁡ φ 1 − ε 2 − 1 ε 2 + 1 cos ⁡ φ + ln ⁡ 1 − cos ⁡ φ 1 + cos ⁡ φ ] = a ρ [ Φ ( ψ , ε 2 − 1 ε 2 + 1 ) + Φ ( ψ , 1 ) ] {\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 Φ ( ψ , α ) = ∫ ψ π + ψ sin ⁡ ( φ − ψ ) d φ ln ⁡ 1 − α cos ⁡ φ 1 + α cos ⁡ φ {\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: Φ ( ψ , α ) = − cos ⁡ ( φ − ψ ) ln ⁡ 1 − α cos ⁡ φ 1 + α cos ⁡ φ | ψ π + ψ + 2 π ∫ ψ π + ψ cos ⁡ ( φ + ψ ) sin ⁡ φ 1 − α 2 cos 2 ⁡ φ + 2 α sin ⁡ ψ ∫ ψ π + ψ sin 2 ⁡ φ d φ 1 − α 2 cos 2 ⁡ φ {\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 ( 0 , ψ ) {\displaystyle (0,\psi )} va ( π , π + ψ ) {\displaystyle (\pi ,\pi +\psi )} intervallardagi qiymati bir-biriga mos keladi. Natijada, Φ ( ψ , α ) {\displaystyle \Phi (\psi ,\alpha )} ni quyidagi koʻrinishda yozishimiz mumkin: Φ ( ψ , α ) = 2 α sin ⁡ ψ ∫ 0 π sin 2 ⁡ φ d φ 1 − α 2 cos 2 ⁡ φ {\displaystyle \Phi (\psi ,\alpha )=2\alpha \sin \psi \int \limits _{0}^{\pi }{\frac {\sin ^{2}\varphi d\varphi }{1-\alpha ^{2}\cos ^{2}\varphi }}} α = 1 {\displaystyle \alpha =1} hol uchun integral oson hisoblanadi va 2 π sin ⁡ ψ {\displaystyle 2\pi \sin \psi } ga teng. α ≠ 1 {\displaystyle \alpha \neq 1} uchun esa Φ ( ψ , α ) = 2 π sin ⁡ ψ α [ 1 − ( 1 − α 2 ) ∫ 0 π d φ 1 − α 2 cos 2 ⁡ φ ] {\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 Φ ( ψ , α ) = 2 π sin ⁡ ψ α ( 1 − 1 − α 2 ) {\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: E x ( x , z ) = 4 π ρ a b + a z E z ( x , z ) = 4 π ρ b b + a x {\displaystyle E_{x}(x,z)=4\pi \rho {\dfrac {a}{b+a}}zE_{z}(x,z)=4\pi \rho {\dfrac {b}{b+a}}x}