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
	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:

	
	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)