JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    The dimension of (1/2) \[{{\varepsilon }_{0}}{{E}^{2}}({{\varepsilon }_{0}}\]: permittivity of free space; \[E\]: electric field) is    [IIT-JEE (Screening) 2000; KCET 2000]

    A)            \[ML{{T}^{^{-1}}}\]           

    B)            \[M{{L}^{2}}{{T}^{-2}}\]

    C)            \[M{{L}^{-1}}{{T}^{-2}}\]

    D)                                      \[M{{L}^{2}}{{T}^{-1}}\]

    Correct Answer: C

    Solution :

                 Energy density \[=\frac{\text{Energy}}{\text{Volume}}\] so it?s dimensions are \[\frac{M{{L}^{2}}{{T}^{-2}}}{{{L}^{3}}}=[M{{L}^{-1}}{{T}^{-2}}]\]


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