AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

The position of charges is shown below.The electrostatic force FAB is calculated as,
FAB=

kQ AQB

rAB2

FAB=

k(15×10−6)(12×10−6)

9×10−4

FAB=k (20×10−8)N
Similarly, the electrostatic force FBC is calculated as,
FBC=

k(20×10−6)(12×10−6)

16×10−4

FBC=k(15×10−8 )N
The resultant electrostatic force at B is,
FB=√ FAB2+FBC2 ‌‌‌(As,θ=90°)
FB=2250N