BCECE Medical BCECE Medical Solved Papers-2010

  • question_answer
    Find the dimensions of electric permittivity.

    A)  \[[{{A}^{2}}{{M}^{-1}}{{L}^{-3}}{{T}^{4}}]\] 

    B)  \[[{{A}^{2}}{{M}^{-1}}{{L}^{-3}}{{T}^{0}}]\]

    C)  \[[A{{M}^{-1}}{{L}^{-3}}{{T}^{4}}]\]     

    D)  \[[{{A}^{2}}{{M}^{0}}{{L}^{-3}}{{T}^{4}}]\]

    Correct Answer: A

    Solution :

    From Coulombs law, The force of attraction or repulsion between two point charges q, q separated by distance r is                 \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}}\] \[\Rightarrow \] \[{{\varepsilon }_{0}}=\frac{1}{4\pi }\frac{{{q}^{2}}}{F{{r}^{2}}}\] where \[{{\varepsilon }_{0}}\] is electric permittivity. Dimensions of \[{{\varepsilon }_{0}}=\frac{{{[AT]}^{2}}}{[ML{{T}^{-2}}]\,[{{L}^{2}}]}\]


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