question_answer

Electric charges of \[1\mu C,\,-1\mu C\] and \[2\mu C\] are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is                        [EAMCET (Engg.) 2000]

A)            0.9 N                                        

B) 1.8 N

C)            2.7 N

D)            3.6 N

Correct Answer: B

Solution :

                     FA = force on C due to charge placed at A      \[=9\times {{10}^{9}}\times \frac{{{10}^{-6}}\times 2\times {{10}^{-6}}}{{{(10\times {{10}^{-2}})}^{2}}}=1.8\,N\] FB = force on C due to charge placed at B            \[=9\times {{10}^{9}}\times \frac{{{10}^{-6}}\times 2\times {{10}^{-6}}}{{{(0.1)}^{2}}}=1.8N\] Net force on C                    \[{{F}_{net}}=\sqrt{{{({{F}_{A}})}^{2}}+{{({{F}_{B}})}^{2}}+2{{F}_{A}}{{F}_{B}}\cos {{120}^{o}}}=1.8\,N\]