In a right-angled triangle ABC,AB = 3 cm,BC | vedclass

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(C) According to Coulomb's law in $CGS$ units,the force between two charges $q_1$ and $q_2$ separated by distance $r$ is $F = \frac{q_1 q_2}{r^2}$.The

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$25$

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(C) According to Coulomb's law in $CGS$ units,the force between two charges $q_1$ and $q_2$ separated by distance $r$ is $F = \frac{q_1 q_2}{r^2}$.The force on charge at $B$ due to charge at $A$ $(F_A)$ is:$F_A = \frac{15 	imes 12}{3^2} = \frac{180}{9} = 20 \ dynes$ (directed along $BA$).The force on charge at $B$ due to charge at $C$ $(F_C)$ is:$F_C = \frac{12 	imes 20}{4^2} = \frac{240}{16} = 15 \ dynes$ (directed along $BC$ because the force is attractive).Since the angle between $BA$ and $BC$ is $90^\circ$,the net force $F_{net}$ is:$F_{net} = \sqrt{F_A^2 + F_C^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \ dynes$.

In a right-angled triangle $ABC$,$AB = 3 \ cm$,$BC = 4 \ cm$,and $\angle ABC = \frac{\pi}{2}$. Charges of $+15 \ e.s.u.$,$+12 \ e.s.u.$,and $-20 \ e.s.u.$ are placed at $A$,$B$,and $C$ respectively. What is the net force on the charge at $B$ in $dynes$?

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