RAJASTHAN PMT Rajasthan - PMT Solved Paper-2009

  • 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\[\text{/3}\]with the x-axis. If the electric field at P makes an angle \[\theta \] with the x-axis, the value of \[\theta \] 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,                 \[{{E}_{X}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{2(p\cos \pi /3)}{{{r}^{3}}}\]                 \[\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{P}{{{r}^{3}}}\] Component of electric field of point \[P\]perpendicular to \[X-\]axis,                 \[{{E}_{Y}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{p\sin \pi /3}{{{r}^{3}}}\]                     \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{\sqrt{3}p}{2{{r}^{3}}}\] \[\therefore \]  \[\tan \theta =\frac{{{E}_{Y}}}{{{E}_{X}}}=\frac{\sqrt{3}}{2}\]   \[\therefore \]\[\theta ={{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]


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