Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2006

  • question_answer
    An electric dipole is placed along the x-axis at the origin 0. A point P is at a distance of 20 cm from this origin such that OP makes an angle \[\text{ }\!\!\pi\!\!\text{ /3}\]with the x-axis. If the electric field at P makes an angle \[\text{ }\!\!\theta\!\!\text{ }\] with the x-axis, the value of \[\text{ }\!\!\theta\!\!\text{ }\] would be

    A) \[\frac{\pi }{3}\]                                              

    B) \[\frac{\pi }{3}+{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]

    C) \[\frac{2\pi }{3}\]                                            

    D) \[{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]

    Correct Answer: B

    Solution :

    Component of electric field at point P parallel to X-axis, \[{{E}_{X}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2(p\,\cos \,\pi /3)}{{{r}^{3}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{p}{{{r}^{3}}}\] Component of electric field of point P perpendicular to -X-axis,                 \[{{E}_{Y}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{p\,\sin \,\pi /3}{{{r}^{3}}}\]                 \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sqrt{3}p}{2{{r}^{3}}}\] \[\therefore \]  \[\tan \theta =\frac{{{E}_{Y}}}{{{E}_{X}}}=\frac{\sqrt{3}}{2}\] \[\therefore \]  \[\theta ={{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]


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