question_answer

Three identical charges of magnitude 2 \[\mu \]C are placed at the corner of a right angled triangle ABC whose base BC and height BA are respectively 4 cm and 3 cm. Forces on the charge at the right angled corner B due to the charges at A and C are respectively F1 and F2. The angle between their resultant force and F2 is:

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 \[\tan \theta =\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4\times {{10}^{-12}}}{{{(0.03)}^{2}}}N}{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4\times {{10}^{-12}}}{{{(0.04)}^{2}}}N}\] \[={{\left( \frac{0.04}{0.03} \right)}^{2}}=\frac{16}{9}\] \[\theta ={{\tan }^{-1}}\left( \frac{16}{9} \right)\]