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Консультация № 178768
31.05.2010, 01:31
0.00 руб.
0
2
2
Уважаемые эксперты помогите решить, а вернее пояснить формулу готового решения.
Находящийся в вакууме очень тонкий прямой стержень длины 2а заряжен с постоянной линейной плотностью λ . Для точек, лежащих на прямой, перпендикулярной оси стержня и проходящей через его центр, найти модуль E напряженности поля как функцию расстояния r от центра стержня.
Вот тут я прикрепил изображение этой самой формулы:
http://s14.radikal.ru/i187/1005/49/f4d95d46175d.jpg
Находящийся в вакууме очень тонкий прямой стержень длины 2а заряжен с постоянной линейной плотностью λ . Для точек, лежащих на прямой, перпендикулярной оси стержня и проходящей через его центр, найти модуль E напряженности поля как функцию расстояния r от центра стержня.
Вот тут я прикрепил изображение этой самой формулы:
http://s14.radikal.ru/i187/1005/49/f4d95d46175d.jpg
Обсуждение
31.05.2010, 09:34
общий
это ответ
Здравствуйте, Самойленко Николай Алексеевич.
Вам нетрудно будет сделать вывод нужной формулы самостоятельно, рассмотрев пару примеров. Обратитесь, пожалуйста, по следующим ссылкам:
https://rfpro.ru/upload/2544
https://rfpro.ru/upload/2545
Фактически осталось воспользоваться формулами (19.9) и (19.10), помня, что в Вашем случае (в принятых в примерах обозначениях) sin α = a/√(r2 + a2), cos α = r/√(r2 + a2), α2 = α1, E = √(Ex2 + Ey2), одна из составляющих напряженности равна нулю.
С уважением.
Вам нетрудно будет сделать вывод нужной формулы самостоятельно, рассмотрев пару примеров. Обратитесь, пожалуйста, по следующим ссылкам:
https://rfpro.ru/upload/2544
https://rfpro.ru/upload/2545
Фактически осталось воспользоваться формулами (19.9) и (19.10), помня, что в Вашем случае (в принятых в примерах обозначениях) sin α = a/√(r2 + a2), cos α = r/√(r2 + a2), α2 = α1, E = √(Ex2 + Ey2), одна из составляющих напряженности равна нулю.
С уважением.
Об авторе:
Facta loquuntur.
Facta loquuntur.
31.05.2010, 09:38
общий
это ответ
Здравствуйте, Самойленко Николай Алексеевич.
Выделим малый участок dl. На нем находится заряд dq=dl*[$955$]
Из чертежа dl=b/cosa, где b=r*da
Тогда получаем:
dq=r*da*[$955$]/cosa
Этот заряд создает напряженность:
dE=1/(4п[$958$]0)*dq/r2
Из чертежа: r=d/cosa. Разложим dE на проекции. В силу симметрии [$8721$]dEx=0
E=[$8721$]dEy=[$8721$]dE*cosa=[$8721$]cosa/(4п[$958$]0)*dq/r2=[$8721$]cosa*[$955$]*da/(4п[$958$]0d)
Интегрируем от a1 до a2:
E=a1a2[$8747$]cosa*[$955$]*da/(4п[$958$]0d)=[$955$]/(4п[$958$]0d)*(sina2-sina1)
Или окончательно имеем:
E=[$955$]/(2п[$958$]0d)*sina
Если использовать ваши обозначения мы как раз получаем искомую формулу:
E=[$955$]/(2п[$958$]0r)*cos[$966$]
Выделим малый участок dl. На нем находится заряд dq=dl*[$955$]
Из чертежа dl=b/cosa, где b=r*da
Тогда получаем:
dq=r*da*[$955$]/cosa
Этот заряд создает напряженность:
dE=1/(4п[$958$]0)*dq/r2
Из чертежа: r=d/cosa. Разложим dE на проекции. В силу симметрии [$8721$]dEx=0
E=[$8721$]dEy=[$8721$]dE*cosa=[$8721$]cosa/(4п[$958$]0)*dq/r2=[$8721$]cosa*[$955$]*da/(4п[$958$]0d)
Интегрируем от a1 до a2:
E=a1a2[$8747$]cosa*[$955$]*da/(4п[$958$]0d)=[$955$]/(4п[$958$]0d)*(sina2-sina1)
Или окончательно имеем:
E=[$955$]/(2п[$958$]0d)*sina
Если использовать ваши обозначения мы как раз получаем искомую формулу:
E=[$955$]/(2п[$958$]0r)*cos[$966$]
Форма ответа
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