JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    An electric field of 1000 V/m is applied to an electric dipole at angle of \[45{}^\circ \]. The value of electric dipole moment is \[{{10}^{-29}}Cm.\]What is the potential energy of the electric dipole?       [JEE Main 11-Jan-2019 Evening]

    A) \[-10\times {{10}^{-29}}J\]

    B) \[-7\times {{10}^{-27}}J\]

    C) \[-20\times {{10}^{-18}}J\]

    D) \[-9\times {{10}^{-20}}J\]

    Correct Answer: B

    Solution :

    [b] \[E=1000V/m,p={{10}^{-29}}cm,\theta ={{45}^{o}}\]
    Potential energy stored in the dipole,
    \[U=-\vec{p}.\vec{E}=-pE\cos \theta \]
    \[=-{{10}^{-29}}\times 1000\times \cos {{45}^{o}}=-\frac{1}{\sqrt{2}}\times {{10}^{-26}}\]
    \[=-0.707\times {{10}^{-26}}J\approx -7\times {{10}^{-27}}J\]


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