Elliptik silindrning elektr maydoni

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:

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

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

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

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

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

{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: $\varphi\prime=\pi-\varphi$.

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

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

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

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

$\alpha=1$ hol uchun integral oson hisoblanadi va $2\pi\sin\psi$ ga teng. $\alpha\ne1$ uchun esa

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

yoki

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

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

${\displaystyle E_{x}(x,z)=4\pi\rho\dfrac{a}{b+a}z$

E_{z}(x,z)=4\pi\rho\dfrac{b}{b+a}x}

Yana qarang

• Zaryad elektr maydonining asosiy qonuni

Adabiyotlar

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