Q. $ABC$ is a right angled triangle in which $AB = 3cm$ and $BC = 4cm.$ And$\ \angle ABC = \pi/2.$ The three charges $+ 15, + 12$ and $- 20\ e.s.u$ are placed respectively on $A,B$ and $C.$ The force acing on $B$ is

A) $125\ dynes$
B) $35\ dynes$
C) $25\ dynes$
D) Zero
Explanation

(c) Net force on $B\ F_{net} = \sqrt{F_{A}^{2} + F_{C}^{2}}$
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$F_{A} = \frac{15\ \times \ 12}{(3)^{2}} = 20\ dyne,\ F_{C} = \frac{12\ \times \ 20\ }{(4)^{2}} = 15dyne$
$\Rightarrow F_{net} = \sqrt{F_{A}^{2} + F_{C}^{2}} = \sqrt{(20)^{2} \times (15)^{2}} = 25dyne$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.