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Консультация № 183724
27.06.2011, 17:17
62.86 руб.
0
7
1
Здравствуйте! Прошу помощи в следующем вопросе:
Внутри шар полый [$958$]1=1, [$958$]2=2,[$961$]-объемная плотность.
Построить зависимость E и [$966$] от х.
Внутри шар полый [$958$]1=1, [$958$]2=2,[$961$]-объемная плотность.
Построить зависимость E и [$966$] от х.
Обсуждение
27.06.2011, 21:25
общий
это ответ
Здравствуйте, Голицина Дарья!
Применив теорему Остроградского-Гаусса с учётом сферической симметрии объекта выводим следующее выражение напряжённости
E(x)=q(x)/(4[$183$][$960$][$183$][$949$]0[$183$][$949$][$183$]x2), где
x - расстояние от центра симметрии
q(x) - заряд, заключённый внутри сферы с радиусом x, центр которой совпадает с центром симметрии
При x<R1 имеем q(x)=0 и E(x)=0
При R1<x<R2 заряд равен q(x)=[$961$][$183$](4/3)[$183$][$960$][$183$](x3-R13)
E(x)=q(x)/(4[$183$][$960$][$183$][$949$]0[$183$][$949$]2[$183$]x2)=[$961$][$183$](x-R13/x2)/(3[$183$][$949$]0[$183$][$949$]2)
При x>R2 внутри сферы радиусом х оказывается весь заряд объекта
q(x)=[$961$][$183$](4/3)[$183$][$960$][$183$](R23-R13)
E(x)=q(x)/(4[$183$][$960$][$183$][$949$]0[$183$][$949$]1[$183$]x2)=[$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]x2)
Потенциал рассчитывается по формуле
[$966$]=x[$8734$][$8747$]E(r)dr
при x[$8805$]R2
[$966$]=x[$8734$][$8747$][$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]r2)dr=
=[$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]x)
при R1[$8804$]x[$8804$]R2
[$966$]=R2[$8734$][$8747$][$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]r2)dr+xR2[$8747$][$961$][$183$](r-R13/r2)/(3[$183$][$949$]0[$183$][$949$]2)dr=
=[$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]R2)+[$961$][$183$](R22/2+R13/R2-x2/2-R13/x)/(3[$183$][$949$]0[$183$][$949$]2)
при x<R1
[$966$](x)=[$966$](R1)=[$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]R2)+[$961$][$183$](R22/2+R13/R2-1,5R12)/(3[$183$][$949$]0[$183$][$949$]2)
Пример решения аналогичной задачи с численными значениями величин и графиками напряжённости, смещения и потенциала смотрите здесь.
Применив теорему Остроградского-Гаусса с учётом сферической симметрии объекта выводим следующее выражение напряжённости
E(x)=q(x)/(4[$183$][$960$][$183$][$949$]0[$183$][$949$][$183$]x2), где
x - расстояние от центра симметрии
q(x) - заряд, заключённый внутри сферы с радиусом x, центр которой совпадает с центром симметрии
При x<R1 имеем q(x)=0 и E(x)=0
При R1<x<R2 заряд равен q(x)=[$961$][$183$](4/3)[$183$][$960$][$183$](x3-R13)
E(x)=q(x)/(4[$183$][$960$][$183$][$949$]0[$183$][$949$]2[$183$]x2)=[$961$][$183$](x-R13/x2)/(3[$183$][$949$]0[$183$][$949$]2)
При x>R2 внутри сферы радиусом х оказывается весь заряд объекта
q(x)=[$961$][$183$](4/3)[$183$][$960$][$183$](R23-R13)
E(x)=q(x)/(4[$183$][$960$][$183$][$949$]0[$183$][$949$]1[$183$]x2)=[$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]x2)
Потенциал рассчитывается по формуле
[$966$]=x[$8734$][$8747$]E(r)dr
при x[$8805$]R2
[$966$]=x[$8734$][$8747$][$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]r2)dr=
=[$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]x)
при R1[$8804$]x[$8804$]R2
[$966$]=R2[$8734$][$8747$][$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]r2)dr+xR2[$8747$][$961$][$183$](r-R13/r2)/(3[$183$][$949$]0[$183$][$949$]2)dr=
=[$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]R2)+[$961$][$183$](R22/2+R13/R2-x2/2-R13/x)/(3[$183$][$949$]0[$183$][$949$]2)
при x<R1
[$966$](x)=[$966$](R1)=[$961$][$183$](R23-R13)/(3[$183$][$949$]0[$183$][$949$]1[$183$]R2)+[$961$][$183$](R22/2+R13/R2-1,5R12)/(3[$183$][$949$]0[$183$][$949$]2)
Пример решения аналогичной задачи с численными значениями величин и графиками напряжённости, смещения и потенциала смотрите здесь.
Неизвестный
28.06.2011, 00:08
общий
28.06.2011, 00:08
Адресаты:
1 курс
[$958$] - диэлектрическая проницаемость
[$958$] - диэлектрическая проницаемость
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