Elliptik silindrning elektr maydoni sahifasining manbasini koʻrish

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'''Elliptik silindrning elektr maydoni'''

Cheksiz uzun silindr sirti boʻylab zaryadlar bir jinsli taqsimlangan boʻlsin. Zaryadlarning sirt zichligi <math>\rho</math>. 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:

	<math>\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}}</math>

Endi esa, zaryadlar silindrdan tashqarida hosil qilgan maydonning z oʻqi boʻyicha tashkil etuvchisini koʻrib chiqamiz:
	<math>
	\frac{x^{2}}{a^{2}}+\frac{z^{2}}{b^{2}}=1+\delta \ \ \ \ \ (0<\delta\ll 1)
	</math>
	Uni quyidagicha ifodalashimiz mumkin:
	<math>
	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}}}
	</math>
	Bu integralni hisoblash uchun yangi oʻzgaruvchilar kiritamiz:
	<math>
	x=a\cos\psi,  x'=a\cos\psi',  b/a=\varepsilon, \delta+\sin^{2}\psi=\sin^{2}\varphi
	</math>

	U holda <math>\sqrt{1-\frac{x'^{2}}{a^{2}}}=\sin\psi',\ z=b\sin\varphi</math> boʻladi hamda
	<math>
	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}}
	</math>

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

	<math>
	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}
	</math>

Birinchi integralda integrallash oʻzgaruvchisini almashtiramiz: <math>\varphi\prime=\pi-\varphi</math>.

Natijada
<math>
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]
</math>

	bu yerda
	<math>
	\Phi(\psi,\alpha)=
	\int\limits_{\psi}^{\pi+\psi}
	\sin(\varphi-\psi)d\varphi\ln
	\frac{1-\alpha\cos\varphi}
		 {1+\alpha\cos\varphi}
	</math>
Yuqoridagi integralni boʻlaklab integrallash mumkin:
	<math>
	\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}
	</math>

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

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

	<math>\alpha=1</math> hol uchun integral oson hisoblanadi va <math>2\pi\sin\psi</math> ga teng. <math>\alpha\ne1</math> uchun esa

	<math>
	\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]
	</math>

	yoki

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

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

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

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


== Yana qarang ==
* [[Zaryad elektr maydonining asosiy qonuni]]

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