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Консультация № 182612
24.03.2011, 20:07
55.00 руб.
0
1
1
Здравствуйте, уважаемые эксперты! Прошу вас помочь решить задачку:
По тонкому кольцу радиусом R=5,0 см равномерно распределен заряд Q=0,0229 мкКл. На оси, перпендикулярной плоскости кольца и проходящей через его центр, расположен тонкий стержень длины L=10,0 см. Стержень равномерно заряжен с линейной плотностью t=2,0мкКл/м. Центр стержня удален от центра кольца на расстояние Xo=12 см.
1) Введите координатную ось X, совпадающую со стержнем. Начало координат поместите в центр кольца.
2) Изобразите качественно направление вектора силы, действующей на стержень со стороны кольца.
3) Получите аналитическое выражение F(X) и найдите значение силы.
По тонкому кольцу радиусом R=5,0 см равномерно распределен заряд Q=0,0229 мкКл. На оси, перпендикулярной плоскости кольца и проходящей через его центр, расположен тонкий стержень длины L=10,0 см. Стержень равномерно заряжен с линейной плотностью t=2,0мкКл/м. Центр стержня удален от центра кольца на расстояние Xo=12 см.
1) Введите координатную ось X, совпадающую со стержнем. Начало координат поместите в центр кольца.
2) Изобразите качественно направление вектора силы, действующей на стержень со стороны кольца.
3) Получите аналитическое выражение F(X) и найдите значение силы.
Обсуждение
25.03.2011, 16:18
общий
это ответ
Здравствуйте, Pagon!
Элемент кольца dl будет иметь заряд dq и создавать в точке, лежащей от него на расстоянии r поле напряженностью
причем вектор dE будет направлен от кольца вдоль линии, соединяющей элемент dl с этой точкой. Этот вектор можно разложить на две составляющие - dE[sub]x[/sub] и dE[sub]y[/sub], причем составляющие dE[sub]y[/sub] для любых двух противоположных элементов dl взаимно уничтожаются. Поэтому для точки, лежащей на оси x имеем
Элемент стержня dx будет иметь заряд dq = [$964$] dx, при этом на него будет действовать сила dF = E dq = E [$964$] dx. На весь стержень будет действовать сила
В данном случае Q = 2.29·10[sup]-8[/sup] Кл, [$964$] = 2·10[sup]-6[/sup] Кл/м, R = 0.05 м, X[sub]0[/sub]-L/2 = 0.07 м, X[sub]0[/sub]+L/2 = 0.17 м. Тогда
Итак, стержень будет отталкиваться от кольца с силой F [$8776$] 2.5 мН.
Элемент кольца dl будет иметь заряд dq и создавать в точке, лежащей от него на расстоянии r поле напряженностью
причем вектор dE будет направлен от кольца вдоль линии, соединяющей элемент dl с этой точкой. Этот вектор можно разложить на две составляющие - dE[sub]x[/sub] и dE[sub]y[/sub], причем составляющие dE[sub]y[/sub] для любых двух противоположных элементов dl взаимно уничтожаются. Поэтому для точки, лежащей на оси x имеем
Элемент стержня dx будет иметь заряд dq = [$964$] dx, при этом на него будет действовать сила dF = E dq = E [$964$] dx. На весь стержень будет действовать сила
В данном случае Q = 2.29·10[sup]-8[/sup] Кл, [$964$] = 2·10[sup]-6[/sup] Кл/м, R = 0.05 м, X[sub]0[/sub]-L/2 = 0.07 м, X[sub]0[/sub]+L/2 = 0.17 м. Тогда
Итак, стержень будет отталкиваться от кольца с силой F [$8776$] 2.5 мН.
Форма ответа
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