Übungsaufgaben zum elektrischen Feld - Lösungen F N kN = 1, 4 ⋅103 = 1, 4 q C C 1 E= 2 F = q ⋅ E = 6,3 ⋅10−2 N = 63mN Fel = FG 3 4 q⋅ E = m⋅ g Fel qMü ⋅ E = = 3, 2 ⋅1012 FG mMü ⋅ g qMü = 5 6 m⋅ g = 6, 4 ⋅10 −19 C E ⇒q= mMü ⋅ g = 1, 6 ⋅10−4 C E Fel s ≈ FG l q⋅E s ≈ m⋅ g l ⇒E≈ siehe 5) ⇒s= m⋅ g ⋅s N = 630 l ⋅q C q ⋅ E ⋅l = 2,1cm m⋅ g Gleichmäßig beschleunigte Bewegung: 1 Aus v = a ⋅ t und s = ⋅ a ⋅ t 2 folgt v = 2 ⋅ a ⋅ s . 2 7 Aus der wirkenden Feldkraft mP ⋅ a = qP ⋅ E ergibt sich die Beschleunigung a = Also ist v = 8 2 ⋅ qP ⋅ E ⋅ s km = 6, 2 mP s Q µC = 22 2 A m Q kV = 24 E= A ⋅ε0 cm U = E ⋅ d = 1, 2kV E= U kV = 7,5 d cm 9 Q µC = ε 0 ⋅ E = 6, 64 2 A m Q = ε 0 ⋅ E ⋅ A = 319nC 10 ε0 = Q⋅d C = 8,86 ⋅10−12 A ⋅U Vm F= 1 Q1 ⋅ Q2 Q2 = 4π ⋅ ε 0 r 2 4π ⋅ ε 0 ⋅ r 2 11 a) F = 7, 6 ⋅10 2 N = 760 N b) F = 9 ⋅109 N = 9GN qP ⋅ E . mP Übungsaufgaben zum elektrischen Feld - Lösungen a) Fel = 12 Q2 = 2,3 ⋅10 −8 N 2 4π ⋅ ε 0 ⋅ r b) FG = γ ⋅ mP ⋅ mE = 1, 0 ⋅10−47 N 2 r Fel Q2 = = 2, 3 ⋅1039 , das Verhältnis ist vom Abstand der Teilchen FG 4π ⋅ ε 0 ⋅ γ ⋅ mP ⋅ mE unabhängig. c) 1 Q Q Q Q ( 21 − 22 ) = 0 folgt 21 = 22 ; mit r1 + r2 = 1m ergibt sich 4π ⋅ ε 0 r1 r2 r1 r2 2 r2 = m . 3 Aus E = E1 + E2 = 13 1 r1 = m; 3
© Copyright 2024 ExpyDoc
