A) \[{{\tan }^{-1}}\left( \frac{9}{16} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{16}{9} \right)\]
C) \[si{{n}^{-1}}\left( \frac{16}{9} \right)\]
D) \[{{\cos }^{-1}}\left( \frac{16}{9} \right)\]
Correct Answer: B
Solution :
Three equal charges each \[2\mu C\]are placed the comers of a right angled triangle ABC. The force on charge at B due to charge at A is \[{{\vec{F}}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(2\mu C)(2\mu C)}{{{(AB)}^{2}}}\] (along BA) \[{{\vec{F}}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4\times {{10}^{-12}}}{{{(0.03)}^{2}}}N\] (along BA) The force on charge at B due to charge at C is \[{{\vec{F}}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(2\mu C)(2\mu C)}{{{(BC)}^{2}}}\] (along BC) \[{{\vec{F}}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4\times {{10}^{-12}}}{{{(0.04)}^{2}}}\] (along BC) If resultant F of force \[{{F}_{1}}\]and \[{{F}_{2}}\]makes an angle \[\theta \] with the direction of \[{{F}_{2}},\]then \[\tan \theta =\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4\times {{10}^{-12}}}{{{(0.03)}^{2}}}N}{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4\times {{10}^{-12}}}{{{(0.04)}^{2}}}N}\] \[={{\left( \frac{0.04}{0.03} \right)}^{2}}=\frac{16}{9}\] \[\theta ={{\tan }^{-1}}\left( \frac{16}{9} \right)\]You need to login to perform this action.
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