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q A =-2µC

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 donc. F. 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. q C =-4µC. C. q>0. F opposé à E. 2 cm ↔ 10 7 N/C. E A (C).

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q A =-2µC

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  1. EB(C) Etot EB(C)=k.qB/BC2 EB(C)=3.107N/C (E=F/q) 6 cm ↔ 3.107N/C EB(C) qC<0 donc F F=│q│.E F= 4.10-6.2,6.107 F= 104 N Échelle pour les forces : 100 N↔ 10 cm 104 N ↔ 10,4 cm qC=-4µC C q>0 F opposé à E 2 cm ↔ 107N/C EA(C) EC(C)=7,2.106N/C 1,44 cm ↔ 7,2.106N/C qA=-2µC qB=+3µC Etot= EA + EB F=qC.E FA/C 5,2 cm ↔ 2,6.107N/C

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