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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    The magnitude of charges in an electric dipole are \[3.2\times {{10}^{-19}}C\]and the distance between them is 2.4\[\overset{o}{\mathop{\text{A}}}\,\]. If its placed at the electric field of intensity \[4\times {{10}^{5}}\]V/m, then the electric dipole moment will be

    A)  \[9.6\times {{10}^{-5}}C-m\]

    B)  \[12.8\times {{10}^{-14}}C-m\]

    C)  \[7.68\times {{10}^{-29}}C-m\]

    D)  \[30\times {{10}^{-24}}C-m\]

    Correct Answer: D

    Solution :

     \[\tau =p\times E\] \[=q\times 2l\times E\] \[=3.2\times {{10}^{-19}}\times 2.4\times {{10}^{-10}}\times 4\times {{10}^{-5}}\] \[=3.2\times {{10}^{-29}}C-m\]


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