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JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    An electric dipole of length 1 cm is placed with axis making an angle of \[30{}^\circ \] 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)  \[-0.245\] J                   

    B)  \[-2.45\times {{10}^{-4}}\] J

    C)  \[-0.0245\] J                 

    D)  \[-24.5\times {{10}^{-4}}\] J

    Correct Answer: D

    Solution :

    Torque \[\tau \,=pE\,\sin \theta \,\Rightarrow \,P=\frac{\tau }{E\sin \theta }\]Potential energy \[U=-pE\cos \theta \,=\frac{-\tau }{E\,\sin \,\theta }.\,E\cos \theta \] \[=\frac{-\tau }{E\sin \theta }\,=\frac{-10\sqrt{2}}{{{10}^{4}}\,\tan {{30}^{0}}}\,=-24.5\times {{10}^{-4}}J\]


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