question_answer

Let ABC is a right angled triangle in which \[AB=3\,cm\]and \[BC=4\,cm\] and \[\angle ABC={{90}^{o}}.\] The three charges +15, +12 and \[-20\] esu are placed on A, B and C respectively. The force acting on B will be

A)  Zero                     

B)         25 dyne

C)  30 dyne              

D)                         150 dyne

Correct Answer: B

Solution :

The given condition can be shown as Net force on \[B,{{F}_{net}}=\sqrt{F_{A}^{2}+F_{C}^{2}}\] \[\therefore \]   \[{{F}_{A}}=\frac{5\times 12}{{{(3)}^{2}}}=20\,dyne\] and        \[{{F}_{C}}=\frac{12\times 20}{{{(4)}^{2}}}=15\,\text{dyne}\] So, \[{{F}_{net}}=\sqrt{{{(20)}^{2}}+{{(15)}^{2}}}=25\,\text{dyne}\text{.}\]