ABC is a right-angled triangle in which AB = | vedclass

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(D) According to the question,three charge particles are placed at the vertices of a right-angled triangle $ABC$ where $Q_A = +15 \ \mu C$,$Q_B = +12

Vedclass · Accepted Answer

$2250$

Vedclass · Answer

(D) According to the question,three charge particles are placed at the vertices of a right-angled triangle $ABC$ where $Q_A = +15 \ \mu C$,$Q_B = +12 \ \mu C$,and $Q_C = -20 \ \mu C$.The force on charge $B$ due to charge $A$ is given by Coulomb's law:$F_{AB} = \frac{k |Q_A Q_B|}{r_{AB}^2} = \frac{(9 	imes 10^9) 	imes (15 	imes 10^{-6}) 	imes (12 	imes 10^{-6})}{(3 	imes 10^{-2})^2} = \frac{9 	imes 10^9 	imes 180 	imes 10^{-12}}{9 	imes 10^{-4}} = 1800 \ N$.Since $Q_A$ and $Q_B$ are both positive,this force is repulsive (directed away from $A$,i.e.,along $BC$ extended).The force on charge $B$ due to charge $C$ is:$F_{BC} = \frac{k |Q_B Q_C|}{r_{BC}^2} = \frac{(9 	imes 10^9) 	imes (12 	imes 10^{-6}) 	imes (20 	imes 10^{-6})}{(4 	imes 10^{-2})^2} = \frac{9 	imes 10^9 	imes 240 	imes 10^{-12}}{16 	imes 10^{-4}} = 1350 \ N$.Since $Q_B$ is positive and $Q_C$ is negative,this force is attractive (directed towards $C$).Since the angle between $F_{AB}$ and $F_{BC}$ is $90^{\circ}$,the resultant force is:$F_B = \sqrt{F_{AB}^2 + F_{BC}^2} = \sqrt{(1800)^2 + (1350)^2} = \sqrt{3240000 + 1822500} = \sqrt{5062500} = 2250 \ N$.Thus,the correct option is $(d)$.

$ABC$ is a right-angled triangle in which $AB = 3 \ cm$,$BC = 4 \ cm$ and the right angle is at $B$. Three charges $+15 \ \mu C$,$+12 \ \mu C$ and $-20 \ \mu C$ are placed respectively at $A$,$B$ and $C$. The force acting on the charge at $B$ is (in $N$)

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