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Given, (need figure)
Three equal charge (q1=q2=q3=4∗10−7C)
Distance (AB)=8cm=0.08m
Distance (BC)=6cm=0.06m
Distance (AC)=10cm=0.1m
Angle ፈABC=90∘
Force (F)=?
FAB=14πϵ0q1∗q2R2=9∗109∗(0.4∗10−6)2(0.08)2=0.225N
FBC=14πϵ0q1∗q2R2=9∗109∗(0.4∗10−6)2(0.06)2=0.40N
Force exerted on the charge located at 90∘ is,
F=√(FAB)2+(FBC)2+2FABFBCCosθ
F=√(0.225)2+(0.4)2+2∗0.225∗0.4∗Cos90
⇒ F=0.46N
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