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

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(C) In the $CGS$ system,Coulomb's law is given by $F = \frac{q_1 q_2}{r^2}$.The force on charge $B$ $(+12\,e.s.u.)$ due to charge $A$ $(+15\,e.s.u.)$

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$25$

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(C) In the $CGS$ system,Coulomb's law is given by $F = \frac{q_1 q_2}{r^2}$.The force on charge $B$ $(+12\,e.s.u.)$ due to charge $A$ $(+15\,e.s.u.)$ is repulsive and acts along $BA$ (downwards in the diagram). Its magnitude is:$F_A = \frac{15 	imes 12}{3^2} = \frac{180}{9} = 20\,dyne$.The force on charge $B$ $(+12\,e.s.u.)$ due to charge $C$ $(-20\,e.s.u.)$ is attractive and acts along $BC$ (towards $C$). Its magnitude is:$F_C = \frac{|12 	imes (-20)|}{4^2} = \frac{240}{16} = 15\,dyne$.Since the angle between $BA$ and $BC$ is $90^\circ$,the net force on $B$ is:$F_{net} = \sqrt{F_A^2 + F_C^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25\,dyne$.

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

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