q A =-2µC
1
C q A =- 2µC q B =+3µC q>0 E B (C ) F E B (C)=k.q B /BC 2 E B (C)=3.10 7 N/C (E=F/q) 6 cm ↔ 3.10 7 N/C E B (C ) 2 cm ↔ 10 7 N/C E A (C ) E C (C)=7,2.10 6 N/C 1,44 cm ↔ 7,2.10 6 N/C E tot = E A + E B E tot 5,2 cm ↔ 2,6.10 7 N/C q C =- 4µC F=q C . E q C <0 donc F opposé à E F=│q│.E F= 4.10 -6 .2,6.10 7 F= 104 N Échelle pour les forces : 100 N↔ 10 cm 104 N ↔ 10,4 cm F A/C
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E B (C). E tot. E B (C)= k.q B /BC 2 E B (C)=3.10 7 N/C (E=F/q) 6 cm ↔ 3.10 7 N/C. E B (C). q C 0. F opposé à E. 2 cm ↔ 10 7 N/C. E A (C). - PowerPoint PPT Presentation
Transcript of q A =-2µC
C
qA=-2µC qB=+3µC
q>0
EB(C)
F
EB(C)=k.qB/BC2
EB(C)=3.107N/C (E=F/q)
6 cm ↔ 3.107N/C
EB(C)
2 cm ↔ 107N/C EA(C)
EC(C)=7,2.106N/C
1,44 cm ↔ 7,2.106N/C
Etot= EA + EB
Etot
5,2 cm ↔ 2,6.107N/C
qC=-4µC
F=qC.EqC<0 donc F opposé à E
F=│q│.EF= 4.10-6.2,6.107
F= 104 N
Échelle pour les forces : 100 N↔ 10 cm104 N ↔ 10,4 cm
FA/C
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