Elliptik silindrning elektr maydoni

Andoza:Manba Elliptik silindrning elektr maydoni

Cheksiz uzun silindr sirti boʻylab zaryadlar bir jinsli taqsimlangan boʻlsin. Zaryadlarning sirt zichligi $\rho$. 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:

$\phi (x,z)=\rho \underset{-a}{\overset{a}{\int }}d{x}^{\prime }\underset{-b\sqrt{1-\frac{x{\text{'}}^2}{a^2}}}{\overset{b\sqrt{1-\frac{x{\text{'}}^2}{a^2}}}{\int }}d{z}^{\prime }\ln \frac{1}{(x-x^{\prime }{)}^2+(z-z^{\prime }{)}^2}$

Endi esa, zaryadlar silindrdan tashqarida hosil qilgan maydonning z oʻqi boʻyicha tashkil etuvchisini koʻrib chiqamiz: $\frac{x^2}{a^2}+\frac{z^2}{b^2}=1+\delta (0<\delta \ll 1)$ Uni quyidagicha ifodalashimiz mumkin: $E_1(x,z)\equiv -\frac{\partial \phi }{\partial z}=\rho \underset{-a}{\overset{a}{\int }}d{x}^{\prime }\underset{-b\sqrt{1-\frac{x{\text{'}}^2}{a^2}}}{\overset{b\sqrt{1-\frac{x{\text{'}}^2}{a^2}}}{\int }}d{z}^{\prime }\frac{\partial }{\partial z^{\prime }}\ln \frac{1}{(x-x^{\prime }{)}^2+(z-z^{\prime }{)}^2}=\rho \underset{-a}{\overset{a}{\int }}dx\prime \ln \frac{(x-x^{\prime }{)}^2+{(z+b\sqrt{1-\frac{x{\text{'}}^2}{a^2}})}^2}{(x-x^{\prime }{)}^2+{(z-b\sqrt{1-\frac{x{\text{'}}^2}{a^2}})}^2}$ Bu integralni hisoblash uchun yangi oʻzgaruvchilar kiritamiz: $x=a\cos \psi ,x^{\prime }=a\cos {\psi }^{\prime },b/a=\epsilon ,\delta +{\sin }^2\psi ={\sin }^2\phi$

U holda $\sqrt{1-\frac{x{\text{'}}^2}{a^2}}=\sin {\psi }^{\prime },z=b\sin \phi$ boʻladi hamda $E_z=a\rho \underset{0}{\overset{\pi }{\int }}\sin {\psi }^{\prime }d{\psi }^{\prime }\ln \frac{(\cos {\psi }^{\prime }-\cos \psi {)}^2+{\epsilon }^2(\sin {\psi }^{\prime }+\sin \phi {)}^2}{(\cos {\psi }^{\prime }-\cos \psi {)}^2+{\epsilon }^2(\sin {\psi }^{\prime }-\sin \phi {)}^2}=a\rho \underset{0}{\overset{\pi }{\int }}\sin {\psi }^{\prime }d{\psi }^{\prime }\ln \frac{{\sin }^2\frac{{\psi }^{\prime }+\psi }{2}{\sin }^2\frac{{\psi }^{\prime }-\psi }{2}+{\epsilon }^2{\sin }^2\frac{{\psi }^{\prime }+\phi }{2}{\cos }^2\frac{{\psi }^{\prime }-\phi }{2}}{{\sin }^2\frac{{\psi }^{\prime }+\psi }{2}{\sin }^2\frac{{\psi }^{\prime }-\psi }{2}+{\epsilon }^2{\sin }^2\frac{{\psi }^{\prime }+\phi }{2}{\sin }^2\frac{{\psi }^{\prime }-\phi }{2}}$

$\delta \to 0$ boʻlganda, $\phi \to \psi$. Bu oʻtishning uzluksizligidan foydalanib integral ostidagi ifodani quyidagicha yozib olishimiz mumkin:

$E_z=a\rho \underset{0}{\overset{\pi }{\int }}\sin {\psi }^{\prime }d{\psi }^{\prime }\ln \frac{{\sin }^2\frac{\psi +{\psi }^{\prime }}{2}[{\sin }^2\frac{\psi -{\psi }^{\prime }}{2}+{\epsilon }^2{\cos }^2\frac{\psi -{\psi }^{\prime }}{2}]}{{\sin }^2\frac{\psi -{\psi }^{\prime }}{2}[{\sin }^2\frac{\psi +{\psi }^{\prime }}{2}+{\epsilon }^2{\cos }^2\frac{\psi +{\psi }^{\prime }}{2}]}=a\rho \underset{0}{\overset{\pi }{\int }}\sin \phi d\phi \left[\frac{1-\cos (\phi +\psi )}{1-\cos (\phi -\psi )}\frac{1+\frac{{\epsilon }^2-1}{{\epsilon }^2+1}\cos (\phi -\psi )}{1+\frac{{\epsilon }^2-1}{{\epsilon }^2+1}\cos (\phi +\psi )}\right]=a\rho \underset{\pi +\phi }{\overset{\pi -\phi }{\int }}\sin (\phi +\psi )d\phi \ln \frac{1+\frac{{\epsilon }^2-1}{{\epsilon }^2+1}\cos \phi }{1-\cos \psi }+a\rho \underset{\psi }{\overset{\pi +\psi }{\int }}\sin (\phi -\psi )d\phi \ln \frac{1-\cos \psi }{1+\frac{{\epsilon }^2-1}{{\epsilon }^2+1}\cos \phi }$

Birinchi integralda integrallash oʻzgaruvchisini almashtiramiz: $\phi \prime =\pi -\phi$.

Natijada $E_z=a\rho \underset{\psi }{\overset{\pi +\psi }{\int }}\sin (\phi -\psi )d\phi [\ln \frac{1+\frac{{\epsilon }^2-1}{{\epsilon }^2+1}\cos \phi }{1-\frac{{\epsilon }^2-1}{{\epsilon }^2+1}\cos \phi }+\ln \frac{1-\cos \phi }{1+\cos \phi }]=a\rho [\mathrm{\Phi }(\psi ,\frac{{\epsilon }^2-1}{{\epsilon }^2+1})+\mathrm{\Phi }(\psi ,1)]$

bu yerda $\mathrm{\Phi }(\psi ,\alpha )=\underset{\psi }{\overset{\pi +\psi }{\int }}\sin (\phi -\psi )d\phi \ln \frac{1-\alpha \cos \phi }{1+\alpha \cos \phi }$ Yuqoridagi integralni boʻlaklab integrallash mumkin: $\mathrm{\Phi }(\psi ,\alpha )=-\cos (\phi -\psi )\ln \frac{1-\alpha \cos \phi }{1+\alpha \cos \phi }{|}_{\psi }^{\pi +\psi }+2\pi \underset{\psi }{\overset{\pi +\psi }{\int }}\frac{\cos (\phi +\psi )\sin \phi }{1-{\alpha }^2{\cos }^2\phi }+2\alpha \sin \psi \underset{\psi }{\overset{\pi +\psi }{\int }}\frac{{\sin }^2\phi d\phi }{1-{\alpha }^2{\cos }^2\phi }$

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

$\mathrm{\Phi }(\psi ,\alpha )=2\alpha \sin \psi \underset{0}{\overset{\pi }{\int }}\frac{{\sin }^2\phi d\phi }{1-{\alpha }^2{\cos }^2\phi }$

$\alpha =1$ hol uchun integral oson hisoblanadi va $2\pi \sin \psi$ ga teng. $\alpha \ne 1$ uchun esa

$\mathrm{\Phi }(\psi ,\alpha )=\frac{2\pi \sin \psi }{\alpha }[1-(1-{\alpha }^2)\underset{0}{\overset{\pi }{\int }}\frac{d\phi }{1-{\alpha }^2{\cos }^2\phi }]$

yoki

$\mathrm{\Phi }(\psi ,\alpha )=\frac{2\pi \sin \psi }{\alpha }(1-\sqrt{1-{\alpha }^2})$

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

$E_x(x,z)=4\pi \rho \frac{a}{b+a}z{E}_z(x,z)=4\pi \rho \frac{b}{b+a}x$

Yana qarang

• Zaryad elektr maydonining asosiy qonuni

Adabiyotlar

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