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 4, 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, \[{{F}_{net}}=\sqrt{F_{A}^{2}+F_{C}^{2}}\] \[\therefore \]  \[{{F}_{A}}=\frac{5\times 12}{{{(3)}^{2}}}=15\] dyne and        \[{{F}_{C}}=\frac{12\times 20}{{{(4)}^{2}}}=15\] dyne So, \[{{F}_{net}}=\sqrt{{{(20)}^{2}}+{{(15)}^{2}}}=25\] dyne.