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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    An electric dipole of length 1 cm is placed with the axis making an angle of\[{{30}^{o}}\]to an electric field of strength\[{{10}^{4}}N{{C}^{-1}}\]. If it experiences a torque of\[10\sqrt{2}Nm,\]the potential energy of the dipole is

    A)  \[0.245\text{ }J\]         

    B)         \[2.45\times {{10}^{-4}}J\]

    C)  \[0.0245\text{ }J\]       

    D)         \[245.0\text{ }J\]

    E)  \[24.5\times {{10}^{-4}}J\]

    Correct Answer: E

    Solution :

    Torque,                \[\tau =pE\sin \theta \] or                            \[P=\frac{\tau }{E\sin \theta }\] Potential energy, \[U=pE\cos \theta \]                                 \[=\frac{\tau }{E\sin \theta }.E\cos \theta \]                                 \[=\frac{\tau }{E\tan \theta }=\frac{10\sqrt{2}}{{{10}^{4}}\tan {{30}^{o}}}\]                 \[=24.5\times {{10}^{-4}}J\]


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