Punjab Medical Punjab - MET Solved Paper-2008

  • question_answer Three charges \[1\mu C,\,\,1\mu C\] and \[2\mu C\] are kept at the vertices \[A,\,\,B\] and \[C\] of an equilateral triangle \[ABC\] of \[10cm\] side, respectively. The resultant force on the charge at \[C\] is

    A) \[0.9N\]                              

    B) \[1.8N\]

    C) \[2.72N\]                            

    D) \[3.6N\]

    Correct Answer: D

    Solution :

    The situation is shown in figure. Force on \[C\] due to \[A\]                 \[{{F}_{AC}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{A}}{{q}_{C}}}{r_{AC}^{2}}\]                 \[=\frac{9\times {{10}^{9}}\times 1\times 2\times {{10}^{-12}}}{{{(0.1)}^{2}}}=1.8N\] Similarly,              \[{{F}_{BC}}=1.8N\] Hence, net force on                 \[C={{F}_{AC}}+{{F}_{BC}}=3.6N\]


You need to login to perform this action.
You will be redirected in 3 sec spinner