Задача по теме "Электростатика"

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Помогите разобраться с задачей..

На расстоянии $ % MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2 % da9iaaigdacaaIWaaaaa!3955! \[a = 10\]$ см от бесконечной, равномерно заряженной плоскости находится центр равномерно заряженного кольца радиуса $% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9iaaiwdaaaa!3890! \[R = 5\]$. 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!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeq4Wdmhaaa!37B7! \[\sigma \]=2 мкКл/м^2$, линейная плотность заряда на кольце $% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiXdqhaaa!37B9! \[\tau \]=1 мкКл/м $

Дано

$ % MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2 % da9iaaigdacaaIWaaaaa!3955! \[a = 10\]$ см = 2R

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9iaaiwdaaaa!3890! \[R = 5\]$ см

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOmeiabg2 % da9iaadkfacqGH9aqpcaaI1aaaaa!3A51! \[ b= R = 5\]$ см

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeq4Wdmhaaa!37B7! \[\sigma \]=2 мкКл/м^2$

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiXdqhaaa!37B9! \[\tau \]=1 мкКл/м

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyTduMaey % ypa0JaaGymaaaa!395C! \[\varepsilon = 1\]$ - Диэлектрическая проницаемость

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaubeaeqale % aacaaIWaaabeqdbaGaeqyTdugaaOGaeyypa0JaaGioaiaac6cacaaI % 4aGaaGynaiaacQcacaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaG % ymaiaaikdaaaaaaa!4154! \[\mathop \varepsilon \nolimits_0 = 8.85*{10^{ - 12}}\]$ ф/м - Электрическая проницаемость.



Найти

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyraaaa!36BE! \[E\]$ - ?



Решение


Посчитав, у меня получилось напряженность поля на оси кольца

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaubeaeqale % aacaaIXaaabeqdbaGaamyraaaakiabg2da9maavababeWcbaGaaGym % aaqab0qaaiaadweaaaGccqGH9aqpdaWdbaqaaiaadsgacaWGfbGaai % OkaiGacogacaGGVbGaai4Caiabeg7aHjabg2da9maapeaabaGaamiz % aiaadweadaWcaaqaaiaadggaaeaadaGcaaqaaiaadggadaahaaWcbe % qaaiaaikdaaaGccqGHRaWkcaWGsbWaaWbaaSqabeaacaaIYaaaaaqa % baaaaaqabeqaniabgUIiYdaaleqabeqdcqGHRiI8aOGaeyypa0Zaa8 % qaaeaadaWcaaqaaiaadggacaWGXbaabaGaaGinaiabec8aWjabew7a % LnaavababeWcbaGaaGimaaqab0qaaiabew7aLbaakiaacIcadaGcaa % qaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGsbWaaWba % aSqabeaacaaIYaaaaOGaaiykamaaCaaaleqabaGaaGOmaaaaaeqaaa % aaaeqabeqdcqGHRiI8aOGaeyypa0ZaaSaaaeaacaWGHbaabaGaaGin % aiabec8aWjabew7aLnaavababeWcbaGaaGimaaqab0qaaiabew7aLb % aakiaacIcadaGcaaqaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGH % RaWkcaWGsbWaaWbaaSqabeaacaaIYaaaaOGaaiykamaaCaaaleqaba % GaaG4maaaaaeqaaaaakmaapeaabaGaamizaiaadghaaSqabeqaniab % gUIiYdaaaa!74CD! \[\mathop E\nolimits_1 = \mathop E\nolimits_1 = \int {dE*\cos \alpha = \int {dE\frac{a}{{\sqrt {{a^2} + {R^2}} }}} } = \int {\frac{{aq}}{{4\pi \varepsilon \mathop \varepsilon \nolimits_0 (\sqrt {{a^2} + {R^2}{)^2}} }}} = \frac{a}{{4\pi \varepsilon \mathop \varepsilon \nolimits_0 (\sqrt {{a^2} + {R^2}{)^3}} }}\int {dq} \]$

Продолжение

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaubeaeqale % aacaaIXaaabeqdbaGaamyraaaakiabg2da9maalaaabaGaamyyaiaa % dghaaeaacaaI0aGaeqiWdaNaeqyTdu2aaubeaeqaleaacaaIWaaabe % qdbaGaeqyTdugaaOGaaiikamaakaaabaGaamyyamaaCaaaleqabaGa % aGOmaaaakiabgUcaRiaadkfadaahaaWcbeqaaiaaikdaaaGccaGGPa % WaaWbaaSqabeaacaaIZaaaaaqabaaaaOGaeyypa0ZaaSaaaeaacaaI % YaGaamOuaiaadghaaeaacaaI0aGaeqiWdaNaeqyTduMaaiikamaaka % aabaGaaGinaiaadkfadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG % sbWaaWbaaSqabeaacaaIYaaaaOGaaiykamaaCaaaleqabaGaaG4maa % aaaeqaaaaakiabg2da9maalaaabaGaaGOmaiaadkfacqaHepaDcaaI % YaGaeqiWdaNaamOuaaqaaiaaisdacqaHapaCdaqfqaqabSqaaiaaic % daaeqaneaacqaH1oqzaaGccaWGsbWaaWbaaSqabeaacaaIZaaaaOGa % aiOkaiaaicdacaGGUaGaaGimaiaaiwdacaGGQaGaaGimaiaac6caca % aI1aaaaiabg2da9maalaaabaGaeqiXdqhabaWaaubeaeqaleaacaaI % WaaabeqdbaGaeqyTdugaaOGaamOuaiaacQcacaaIWaGaaiOlaiaaic % dacaaI1aGaaiOkaiaaicdacaGGUaGaaGynaaaacqGH9aqpdaWcaaqa % aiaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI2aaaaaGcbaGaaG % ioaiaac6cacaaI4aGaaGynaiaacQcacaaIXaGaaGimamaaCaaaleqa % baGaeyOeI0IaaGymaiaaikdaaaGccaGGQaGaaGimaiaac6cacaaIWa % GaaGynaiaacQcacaaIWaGaaiOlaiaaiwdaaaGaeyypa0JaaGinaiaa % c6cacaaI1aGaaGOmaaaa!9031! \[\mathop E\nolimits_1 = \frac{{aq}}{{4\pi \varepsilon \mathop \varepsilon \nolimits_0 (\sqrt {{a^2} + {R^2}{)^3}} }} = \frac{{2Rq}}{{4\pi \varepsilon (\sqrt {4{R^2} + {R^2}{)^3}} }} = \frac{{2R\tau 2\pi R}}{{4\pi \mathop \varepsilon \nolimits_0 {R^3}*0.05*0.5}} = \frac{\tau }{{\mathop \varepsilon \nolimits_0 R*0.05*0.5}} = \frac{{{{10}^{ - 6}}}}{{8.85*{{10}^{ - 12}}*0.05*0.5}} = 4.52\] МВ$


Напряженность поля на бесконечной плоскости

$% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaubeaeqale % aacaaIYaaabeqdbaGaamyraaaakiabg2da9maalaaabaGaeq4Wdmha % baGaaGOmamaavababeWcbaGaaGimaaqab0qaaiabew7aLbaaaaGccq % GH9aqpdaWcaaqaaiaaikdacaGGQaGaaGymaiaaicdadaahaaWcbeqa % aiabgkHiTiaaiAdaaaaakeaacaaIYaWaaubeaeqaleaacaaIWaaabe % qdbaGaeqyTdugaaaaakiabg2da9iaaicdacaGGUaGaaGymaiaaigda % caaIYaaaaa!4BFA! \[\mathop E\nolimits_2 = \frac{\sigma }{{2\mathop \varepsilon \nolimits_0 }} = \frac{{2*{{10}^{ - 6}}}}{{2\mathop \varepsilon \nolimits_0 }} = 0.112\]$ МВ



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Подскажите, может ли так много получаться?



И еще, как применить принцип суперпозиции в данной задаче...

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Оформляйте читабельно. Переехали.