Помогите разобраться с задачей..
На расстоянии
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\[a = 10\]$ $ % MathType!MTEF!2!1!+-
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\[a = 10\]$](https://dxdy-02.korotkov.co.uk/f/1/7/8/178d7271ff2cfe39e40fed657122cbcc82.png)
см от бесконечной, равномерно заряженной плоскости находится центр равномерно заряженного кольца радиуса
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\[R = 5\]$ $% MathType!MTEF!2!1!+-
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\[R = 5\]$](https://dxdy-03.korotkov.co.uk/f/a/b/7/ab72ace30752ff85b5f1d186097e0d7082.png)
. Ось кольца параллельна плоскости. Определить напряженность результирующего поля в точке, находящейся на оси кольца на расстоянии в=R от центра кольца. Поверхностная плотность заряда на плоскости
![$% MathType!MTEF!2!1!+-
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\[\sigma \]=2 мкКл/м^2$ $% MathType!MTEF!2!1!+-
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\[\sigma \]=2 мкКл/м^2$](https://dxdy-02.korotkov.co.uk/f/9/f/2/9f29db677c4c7366008fda246bc38f9782.png)
, линейная плотность заряда на кольце
![$% MathType!MTEF!2!1!+-
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\[\tau \]=1 мкКл/м
$ $% MathType!MTEF!2!1!+-
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\[\tau \]=1 мкКл/м
$](https://dxdy-02.korotkov.co.uk/f/5/0/6/506e39b39fb68b74bda0081c9ca4cccf82.png)
Дано
![$ % MathType!MTEF!2!1!+-
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\[a = 10\]$ $ % MathType!MTEF!2!1!+-
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\[a = 10\]$](https://dxdy-02.korotkov.co.uk/f/1/7/8/178d7271ff2cfe39e40fed657122cbcc82.png)
см = 2R
![$% MathType!MTEF!2!1!+-
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\[R = 5\]$ $% MathType!MTEF!2!1!+-
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\[R = 5\]$](https://dxdy-03.korotkov.co.uk/f/a/b/7/ab72ace30752ff85b5f1d186097e0d7082.png)
см
![$% MathType!MTEF!2!1!+-
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% da9iaadkfacqGH9aqpcaaI1aaaaa!3A51!
\[ b= R = 5\]$ $% MathType!MTEF!2!1!+-
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\[ b= R = 5\]$](https://dxdy-04.korotkov.co.uk/f/f/4/f/f4f6891e544739b8634e85f5ad37507782.png)
см
![$% MathType!MTEF!2!1!+-
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\[\tau \]=1 мкКл/м $% MathType!MTEF!2!1!+-
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% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiXdqhaaa!37B9!
\[\tau \]=1 мкКл/м](https://dxdy-03.korotkov.co.uk/f/a/4/5/a4557fbeb01784ef72f78865bec5273d82.png)
![$% MathType!MTEF!2!1!+-
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\[\varepsilon = 1\]$ $% MathType!MTEF!2!1!+-
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\[\varepsilon = 1\]$](https://dxdy-03.korotkov.co.uk/f/a/3/3/a33951ec881d9c992412c5a120119d7f82.png)
- Диэлектрическая проницаемость
![$% MathType!MTEF!2!1!+-
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\[\mathop \varepsilon \nolimits_0 = 8.85*{10^{ - 12}}\]$ $% MathType!MTEF!2!1!+-
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\[\mathop \varepsilon \nolimits_0 = 8.85*{10^{ - 12}}\]$](https://dxdy-03.korotkov.co.uk/f/2/4/6/2467254762d4d24480d9ca3d4491ed4282.png)
ф/м - Электрическая проницаемость.
Найти
![$% MathType!MTEF!2!1!+-
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\[E\]$ $% MathType!MTEF!2!1!+-
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\[E\]$](https://dxdy-03.korotkov.co.uk/f/e/1/5/e15fc2039f7b2255bf29e2e614c91b7a82.png)
- ?
Решение
Посчитав, у меня получилось напряженность поля на оси кольца
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\[\mathop E\nolimits_1 = \mathop E\nolimits_1 = \int {dE*\cos \alpha = \int {dE\frac{a}{{\sqrt {{a^2} + {R^2}} }}} } = \int {\frac{{aq}}{{4\pi \varepsilon \mathop \varepsilon \nolimits_0 (\sqrt {{a^2} + {R^2}{)^2}} }}} = \frac{a}{{4\pi \varepsilon \mathop \varepsilon \nolimits_0 (\sqrt {{a^2} + {R^2}{)^3}} }}\int {dq} \]$ $% MathType!MTEF!2!1!+-
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\[\mathop E\nolimits_1 = \mathop E\nolimits_1 = \int {dE*\cos \alpha = \int {dE\frac{a}{{\sqrt {{a^2} + {R^2}} }}} } = \int {\frac{{aq}}{{4\pi \varepsilon \mathop \varepsilon \nolimits_0 (\sqrt {{a^2} + {R^2}{)^2}} }}} = \frac{a}{{4\pi \varepsilon \mathop \varepsilon \nolimits_0 (\sqrt {{a^2} + {R^2}{)^3}} }}\int {dq} \]$](https://dxdy-02.korotkov.co.uk/f/9/a/2/9a2c73017e6ff3a5bdd3d884bf5fb02d82.png)
Продолжение
Напряженность поля на бесконечной плоскости
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\[\mathop E\nolimits_2 = \frac{\sigma }{{2\mathop \varepsilon \nolimits_0 }} = \frac{{2*{{10}^{ - 6}}}}{{2\mathop \varepsilon \nolimits_0 }} = 0.112\]$ $% MathType!MTEF!2!1!+-
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\[\mathop E\nolimits_2 = \frac{\sigma }{{2\mathop \varepsilon \nolimits_0 }} = \frac{{2*{{10}^{ - 6}}}}{{2\mathop \varepsilon \nolimits_0 }} = 0.112\]$](https://dxdy-04.korotkov.co.uk/f/7/7/b/77b24804136ed741039b32a6ccd8a1f782.png)
МВ
....
Подскажите, может ли так много получаться?
И еще, как применить принцип суперпозиции в данной задаче...