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}
Yana qarang
Adabiyotlar
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