question_answer

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

A)                    \[125\ dynes\]                        

B)                    \[35\ dynes\]           

C)                    \[25\ dynes\]               

D)                    Zero

Correct Answer: C

Solution :

 Net force on B  \[{{F}_{net}}=\sqrt{F_{A}^{2}+F_{C}^{2}}\] \[{{F}_{A}}=\frac{15\times 12}{{{\left( 3 \right)}^{2}}}=20\,\,dyne\], \[{{F}_{C}}=\frac{12\times 20}{{{\left( 4 \right)}^{2}}}=15\,\,dyne\] Þ \[{{F}_{net}}=\sqrt{F_{A}^{2}+F_{C}^{2}}=\sqrt{{{(20)}^{2}}+{{(15)}^{2}}}=25\,\,dyne\]